Schwinger-Keldysh path integral for the quantum harmonic oscillator
SSchwinger-Keldysh path integral for the quantumharmonic oscillator
Yoni BenTov
Perimeter Institute for Theoretical Physics, Waterloo, Ontario N2L 2Y5
Abstract
I review the generating function for quantum-statistical mechanics, known as theFeynman-Vernon influence functional, the decoherence functional, or the Schwinger-Keldysh path integral. I describe a probability-conserving iε prescription from a path-integral implementation of Lindblad evolution. I also explain how to generalize the for-malism to accommodate out-of-time-ordered correlators (OTOCs), leading to a Larkin-Ovchinnikov path integral. My goal is to provide step-by-step calculations of pathintegrals associated to the harmonic oscillator. Contents a r X i v : . [ h e p - t h ] F e b Harmonic oscillator 24
J J term . . . . . . . . . . . . . . . . . . . . . . 364.2.4 Influence phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.3 Other order of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.3.1 Ground state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.4 Thermal state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.4.1 Integrals over initial field configurations . . . . . . . . . . . . . . . . . 414.4.2 Integral over final field configuration . . . . . . . . . . . . . . . . . . 434.4.3 Influence phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.5 First excited state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 t = 0 . . . . . . . . . . . . . . . . . . . . . 586.1.4 Wightman expectation value for t (cid:54) = 0 . . . . . . . . . . . . . . . . . 596.1.5 Out-of-time-ordered correlator . . . . . . . . . . . . . . . . . . . . . . 626.2 Schwinger-Keldysh path integral . . . . . . . . . . . . . . . . . . . . . . . . . 626.2.1 Integrals over initial field configurations . . . . . . . . . . . . . . . . . 626.2.2 Integral over final field configuration . . . . . . . . . . . . . . . . . . 632.2.3 Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646.2.4 Energy in the field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 666.3 Larkin-Ovchinnikov path integral . . . . . . . . . . . . . . . . . . . . . . . . 68 iε prescription 73 iε prescription . . . . . . . . . . . . . . . 768.3.1 Action on classical solutions . . . . . . . . . . . . . . . . . . . . . . . 778.3.2 Sourced damped/unstable oscillator . . . . . . . . . . . . . . . . . . . 778.3.3 The difference field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 788.3.4 Simplify the action: Boundary terms . . . . . . . . . . . . . . . . . . 798.3.5 Simplify the action: Integral over q f . . . . . . . . . . . . . . . . . . . 798.3.6 Simplify the action: Boundary terms, part 2 . . . . . . . . . . . . . . 808.3.7 Simplify the action: Simplify q − ( t ) . . . . . . . . . . . . . . . . . . . 818.3.8 Simplify the action: Remaining terms in action . . . . . . . . . . . . 818.3.9 Simplify the action: Organization . . . . . . . . . . . . . . . . . . . . 828.4 Specify the density matrix and complete the calculation . . . . . . . . . . . . 848.4.1 Simplify the influence phase . . . . . . . . . . . . . . . . . . . . . . . 85 A.1 Stationary-phase method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92A.2 Classical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92A.2.1 Homogeneous and particular solutions . . . . . . . . . . . . . . . . . 92A.2.2 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 93A.2.3 Complete solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94A.3 Action on classical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94A.4 The overall factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96A.5 Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98A.6 Time-slicing method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99A.6.1 Composition law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99A.6.2 Iterated integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100A.6.3 Free-particle factor for harmonic oscillator . . . . . . . . . . . . . . . 1013
Integrating out vs. tracing out 101
B.1 Integrating out . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102B.2 Tracing out . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102B.3 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
C List of results 103
C.1 Feynman path integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103C.1.1 Free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103C.1.2 Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104C.2 Schwinger-Keldysh path integrals . . . . . . . . . . . . . . . . . . . . . . . . 104C.2.1 Quadratic actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104C.2.2 Harmonic oscillator in ground state . . . . . . . . . . . . . . . . . . . 105C.2.3 Harmonic oscillator in thermal state . . . . . . . . . . . . . . . . . . 105C.2.4 Quenched oscillator in ground state . . . . . . . . . . . . . . . . . . . 106C.2.5 Harmonic oscillator in first excited state . . . . . . . . . . . . . . . . 106C.3 Larkin-Ovchinnikov path integrals . . . . . . . . . . . . . . . . . . . . . . . . 106C.4 Lindblad-improved iε prescription . . . . . . . . . . . . . . . . . . . . . . . . 1074 Introduction
A typical review of the Schwinger-Keldysh formalism [1, 2] pitched to high-energy theoristsgoes something like this [3, 4]. Remember the Feynman path integral? It only works if thevacuum in the far future is the same as the vacuum in the far past. But when a quantumfield theory is placed on a dynamical background—like an expanding universe or an evapo-rating black hole—there is no guarantee that the vacuum in the future will be the same asthe vacuum in the past. So the Feynman path integral must be replaced by something better.But that makes the formalism sound like witchcraft, when it is just statistical mechanics[5].The Schwinger-Keldysh formalism is about constructing generating functions for expecta-tion values instead of transition amplitudes. It has nothing to do with dynamics, equilibriumor otherwise. It is also, in its path-integral incarnation, better attributed to Feynman andVernon [7], whose approach I will follow and extend. To encapsulate my presentation within a consistent, unambiguous framework, I will startall the way from the path-integral decomposition itself, presenting the Schwinger-Keldyshformalism more or less in the style of Sections 6 and 7 of the textbook by Srednicki [9]. Without-of-time-ordered correlators (OTOCs) [10] in the collective consciousness [11, 12], I willfurther construct a generating function for 4-point expectation values, which I will call theLarkin-Ovchinnikov path integral. In this entire article I will consider only a single example: The harmonic oscillator. Again,and again, and again. I will review its Feynman path integral, then calculate its Schwinger-Keldysh and Larkin-Ovchinnikov path integrals for various choices of density matrix.Building on all of that, I will review the path-integral implementation [16, 17] of Lindbladevolution for open systems [18]; I will wrap up by formulating a probability-conserving ver-sion of the iε prescription from S -matrix theory. Although I will consider only single-particlequantum mechanics, I remind you that quantum mechanics is quantum field theory in 0 + 1dimensions. Everything in this paper should be understood with that mindset.Whether your affiliation is particle physics, cosmology, open systems, biology, or finance,I invite you to Bring Your Own Motivation. This is a paper about path integrals. I wishonly to present a series of correct mathematical steps. It is true that most who study nonequilibrium dynamics are primarily concerned with expectation values,but that concern is a choice [6]. See Kamenev and Levchenko [8] for other historical references. For an operator treatment with examples, see Aleiner et al. [13]; for further physical interpretation, seeStanford [14]. For another path-integral treatment, see Haehl et al. [15] Generating functions in quantum mechanics
Let ˆ q be a quantum-mechanical coordinate operator, and let ˆ H be the time-independentHamiltonian. The Heisenberg-picture quantum field operator ˆ q ( t ) corresponding to thetime-independent operator ˆ q is defined as the solution of the Heisenberg equation: ∂ t ˆ q ( t ) = i [ ˆ H, ˆ q ( t )] = ⇒ ˆ q ( t ) = e i ˆ Ht ˆ q e − i ˆ Ht . (2.1)Let | ψ (cid:105) and | ψ (cid:48) (cid:105) be arbitrary quantum states. The purpose of this paper is to study generatingfunctions for quantities of the form (cid:104) ψ (cid:48) | e − i ˆ Ht (cid:48) ˆ q ( t n ) ... ˆ q ( t ) e i ˆ Ht | ψ (cid:105) . (2.2)When | ψ (cid:48) (cid:105) (cid:54) = | ψ (cid:105) and t (cid:48) (cid:54) = t , those quantities are called transition amplitudes; when | ψ (cid:48) (cid:105) = | ψ (cid:105) and t (cid:48) = t , they are called expectation values. In both cases I will also call them correlationfunctions or correlators.The generating functions will be expressed as path integrals. Whether those path inte-grals obviate any need to mention the operator expressions in Eqs. (2.1) and (2.2) in the firstplace is not my concern. In Eq. (2.2), no relation among the times t , ..., t n is assumed—if the operators are to beordered in some way, I will include the appropriate ordering symbol. One theme of thisreview is the path-integral description of arbitrarily-ordered products of fields, not just oftime-ordered products.For transition amplitudes, it should always be understood that t < t , ..., t n < t (cid:48) . Forexpectation values, there are two options: Either t (cid:48) = t < t , ..., t n or t , ..., t n < t (cid:48) = t . Inthis paper I will consider only the former. Eq. (2.2) is a lot to unpack, and ultimately it will be defined by the examples that follow.But as a preliminary matter, I will set n = 1 and write down a transition amplitude and anexpectation value. For a transition amplitude, write t = t , t (cid:48) = t f , | ψ (cid:105) = | ψ (cid:105) , | ψ (cid:48) (cid:105) = | ψ f (cid:105) , and insertˆ1 = (cid:82) ∞−∞ dq | q (cid:105)(cid:104) q | to the left of | ψ (cid:105) , and ˆ1 = (cid:82) ∞−∞ dq f | q f (cid:105)(cid:104) q f | to the right of (cid:104) ψ f | . The I will generalize to Lindblad evolution in Sec. 8. See p. 217 of Feynman & Hibbs [19]. I am not sure whether the standard usage of “transition amplitude”includes the case of expectation value, but I treat the two as exclusive. See the introduction to Chapter 9 of Weinberg, Vol. I [20] for comments about unitarity whose significanceI still do not fully appreciate. (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:104) ψ f | q f (cid:105)(cid:104) q | ψ (cid:105) (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) . (2.3)The primordial task in this paper will be to review the generating function for the object (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) , (2.4)which is the 1-point transition amplitude in the coordinate basis. I call that the primordialtask because it will lead to the Feynman path integral over a finite interval, out of which allof the other path integrals are built.To make sense of the time dependence in Eq. (2.4), you may prefer to rewrite it as (cid:104) q f | e − i ( t f − t ) ˆ H ˆ q e − i ( t − t ) ˆ H | q (cid:105) . (2.5)With the evolution operator ˆ U ( t ) ≡ e − it ˆ H , you could interpret Eq. (2.5) in the Schrodingerpicture as an off-diagonal matrix element of ˆ q between the time-dependent states ˆ U ( t − t ) | q (cid:105) and ˆ U ( t f − t ) † | q f (cid:105) .But I will work in the Heisenberg picture: Operators are functions of time, and states de-scribing the system are not. The source-free Feynman path integral is the evolution operatorin the coordinate basis [see Eq. (2.33)], and Eq. (2.4) is what the Feynman path integral withsources generates [see Eq. (2.31)]. The operator ˆ q ( t ) corresponds to the path-integrationvariable q ( t ), and, as far as I am concerned, the evolution operators ˆ U ( t f ) = e − it f ˆ H andˆ U ( t ) † = e it ˆ H are there because the mathematics of path integrals over finite intervals saysso. Now for an expectation value. Return to Eq. (2.2) with n = 1, set t = t (cid:48) = t , | ψ (cid:105) = | ψ (cid:48) (cid:105) = | ψ (cid:105) , and insert ˆ1 = (cid:82) ∞−∞ dq | q (cid:105)(cid:104) q | to the left of | ψ (cid:105) , and ˆ1 = (cid:82) ∞−∞ dq (cid:48) | q (cid:48) (cid:105)(cid:104) q (cid:48) | to the right of (cid:104) ψ | . The result is (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq (cid:104) q | ψ (cid:105)(cid:104) ψ | q (cid:48) (cid:105) (cid:104) q (cid:48) | e − i ˆ Ht ˆ q ( t ) e i ˆ Ht | q (cid:105) . (2.6)Next, insert ˆ1 = e i ˆ Ht f e − i ˆ Ht f = (cid:82) ∞−∞ dq f e i ˆ Ht f | q f (cid:105)(cid:104) q f | e − i ˆ Ht f to the left of ˆ q ( t ) to get (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq (cid:104) q | ψ (cid:105)(cid:104) ψ | q (cid:48) (cid:105) (cid:104) q (cid:48) | e − i ˆ Ht e i ˆ Ht f | q f (cid:105)(cid:104) q f | e − i ˆ Ht f q ( t ) e i ˆ Ht | q (cid:105) . (2.7) As Srednicki points out [9], the object | q, t (cid:105) ≡ e i ˆ Ht | q (cid:105) is an instantaneous eigenstate of the quantumfield ˆ q ( t ); that is, ˆ q ( t ) | q, t (cid:105) = q | q, t (cid:105) . In that notation, Eq. (2.4) would read (cid:104) q f , t f | ˆ q ( t ) | q , t (cid:105) . But whenattempting to weave a continuous thread between scattering theory and open systems, I found it distractingto countenance the faintest whiff of a time-dependent state. I have concluded that the notational convenienceis usually not worth it, but I dabble (see Sec. 7.1). ρ ≡ | ψ (cid:105)(cid:104) ψ | is the density matrix for the system in state | ψ (cid:105) . In terms of adensity matrix, the expectation-value version of Eq. (2.2) would readTr (cid:104) ˆ q ( t n ) ... ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:105) , t , ..., t n > t . (2.8)I will continue to call that an expectation value regardless of whether the state is pure.Once again, I will comment on the time dependence of the operators. Given Eq. (2.1), youcould, if you wish, insert a factor of ˆ1 = e i ˆ Ht e − i ˆ Ht to the left of each ˆ q ( t i ), for i = 1 , ..., n − q ( t n − t ) ... ˆ q ( t − t ) ˆ ρ ]. But the mathematics of path integralsguides against doing so: The path-integration variable q ( t ) depends only on the parameter t , not on both t and its lower bound t .Alternatively, e.g., for n = 1, you could once again define ˆ U ( t ) = e − i ˆ Ht and rewrite the1-point expectation value asTr (cid:104) ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:105) = Tr (cid:104) ˆ q ˆ U ( t − t ) ˆ ρ ˆ U ( t − t ) † (cid:105) . (2.9)That could be interpreted in the Schrodinger picture as the expectation value of ˆ q in thetime-dependent state described by a density matrix ˆ U ( t − t ) ˆ ρ ˆ U ( t − t ) † . But as I saidbefore, I will insist on the Heisenberg picture: The density matrix for the system describesthe system’s state, and the system’s state does not evolve.Now that the formulas have marinated a bit, I will derive their path-integral expressions. I will begin with the 1-point transition amplitude in Eq. (2.4), repeated below for conve-nience: (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) . (2.10) Split the time interval into an arbitrarily large number N of arbitrarily small steps ε : T ≡ t f − t ≡ N ε . (2.11) Instructors usually begin with the 0-point amplitude, then explore time ordering from higher-point func-tions. Instead I begin with the 1-point amplitude, because its path-integral decomposition is only slightlymore complicated and yet demonstrates the inherent time ordering of operator insertions. Two calculationsfor the price of one. n be the number of steps required to get from t to t : t − t ≡ nε . (2.12)Although I will keep t f and t finite, the goal is to calculate correlation functions of fields at any intermediate times, allowing in principle the limits t → −∞ and t f → + ∞ . So n and N − n should also be taken large.Insert a copy of ˆ1 = (cid:82) ∞−∞ dq | q (cid:105)(cid:104) q | between each instance of e i ˆ Hε : (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) = (cid:104) q f | e − i ( t f − t ) ˆ H ˆ q e − i ( t − t ) ˆ H | q (cid:105) = (cid:104) q f | e − iε ˆ H ... e − iε ˆ H (cid:124) (cid:123)(cid:122) (cid:125) N − n copies ˆ q e − iε ˆ H ... e − iε ˆ H (cid:124) (cid:123)(cid:122) (cid:125) n copies | q (cid:105) = (cid:90) ∞−∞ dq N − ... (cid:90) ∞−∞ dq (cid:104) q f | e − iε ˆ H | q N − (cid:105) ... (cid:104) q n +2 | e − iε ˆ H | q n +1 (cid:105) ×(cid:104) q n +1 | e − iε ˆ H ˆ q | q n (cid:105) (cid:124) (cid:123)(cid:122) (cid:125) q n | q n (cid:105) (cid:104) q n | e − iε ˆ H | q n − (cid:105) ... (cid:104) q | e − iε ˆ H | q (cid:105) = (cid:90) ∞−∞ dq N − ... (cid:90) ∞−∞ dq q n N − (cid:89) j = 0 (cid:104) q j +1 | e − iε ˆ H | q j (cid:105) , q N ≡ q f . (2.13)Now specialize to Hamiltonians of the formˆ H = ˆ p + V (ˆ q ) , [ˆ q, ˆ p ] = i ˆ1 . (2.14)Even though I will not end up doing anything with it in this paper, I want to keep trackof where I choose to evaluate the field in each infinitesimal interval. So I will express theincremental evolution operator as e − iε ˆ H = e − iαεV (ˆ q ) e − iε
12 ˆ p e − i (1 − α ) εV (ˆ q ) e O ( ε ) , α ∈ [0 , . (2.15)Let | q (cid:105) be an eigenstate of ˆ q , and let | p (cid:105) be an eigenstate of ˆ p :ˆ q | q (cid:105) = q | q (cid:105) , ˆ p | p (cid:105) = p | p (cid:105) , (cid:104) q | p (cid:105) = e ipq . (2.16) See the Appendices of Matacz [21] for an example in which that matters. A last-minute reference sweeprevealed that he too wrote about Feynman-Vernon path integrals for oscillators [22]. To go further for the special case V (ˆ q )= m ˆ q , write e − i
12 ˆ p = e − i
14 ˆ p e − i
14 ˆ p , and use the Baker-Campbell-Hausdorff formula multiple times, as follows.First note that [ˆ q , ˆ p ] = 2 i { ˆ q, ˆ p } . Then use e A e B = e A + B + 12 [ A,B ]+ ... with A = − iαεV (ˆ q ) and B = − iε ˆ p to get e − iαεV (ˆ q ) e − iε
14 ˆ p = e − iε (cid:16)
14 ˆ p + 12 αm ˆ q (cid:17) − iαε m { ˆ q, ˆ p } + O ( ε ) ≡ e X .Similarly, use e B e A (cid:48) = e B + A (cid:48) + 12 [ B,A (cid:48) ]+ ... with B = − iε ˆ p (again), and A (cid:48) = − i (1 − α ) εV (ˆ q ) to get e − iε
14 ˆ p e − i (1 − α ) εV (ˆ q ) = e − iε (cid:16)
14 ˆ p +(1 − α ) 12 m ˆ q (cid:17) + i (1 − α ) ε m { ˆ q, ˆ p } + O ( ε ) ≡ e Y .Finally, use e X e Y = e X + Y + 12 [ X,Y ]+ ... to get e − iαεV (ˆ q ) e − iε
12 ˆ p e − i (1 − α ) εV (ˆ q ) = e − iε ˆ H − i (cid:16) α − (cid:17) ε m { ˆ q, ˆ p } + O ( ε ) .For the special case α = —that is, for midpoint regularization—the decomposition is valid to O ( ε ).As a mildly interesting aside, I point out that e − iε ˆ H ≈ (cid:16) ˆ1 − iε ˆ H (cid:17) (cid:16) ˆ1 + iε ˆ H (cid:17) − is also unitary andaccurate to O ( ε ) [23]. But interpreting the resulting path integral and its discretization error in terms of aneffective action and the renormalization group seems for the birds. ε is: C j +1 ,j ≡ (cid:104) q j +1 | e − iε ˆ H | q j (cid:105) ≈ (cid:104) q j +1 | e − iαεV (ˆ q ) e − iε
12 ˆ p e − i (1 − α ) εV (ˆ q ) | q j (cid:105) = (cid:90) ∞−∞ dp j π e − iε p j (cid:104) q j +1 | e − iαεV (ˆ q ) | p j (cid:105)(cid:104) p j | e − i (1 − α ) εV (ˆ q ) | q j (cid:105) = (cid:90) ∞−∞ dp j π e − iε p j e − iαεV ( q j +1 ) (cid:104) q j +1 | p j (cid:105) e − i (1 − α ) εV ( q j ) (cid:104) p j | q j (cid:105) = 12 π e − iε [ αV ( q j +1 )+(1 − α ) V ( q j )] (cid:90) ∞−∞ dp j e − iε p j e i ( q j +1 − q j ) p j = 12 π e − iε [ αV ( q j +1 )+(1 − α ) V ( q j )] (cid:90) ∞−∞ dp j e − iε (cid:26)(cid:104) p j − (cid:16) qj +1 − qjε (cid:17)(cid:105) − (cid:16) qj +1 − qjε (cid:17) (cid:27) = 12 π e iε (cid:26) (cid:16) qj +1 − qjε (cid:17) − [ αV ( q j +1 )+(1 − α ) V ( q j )] (cid:27) (cid:90) ∞−∞ dp e − iε p . (2.17)Notice that I did not yet do the integral: Shifting by a constant in the complex plane doesnot merit review, but rotating to get the phase right does.Consider the contour integral I C ( a, n ) ≡ (cid:90) C dz e iaz n , (2.18)with a real and positive, n an integer greater than or equal to 1, and C = C + C + C thecontour depicted in Fig. 1. (The case in Eq. (2.17) is with a <
0, but it is cleaner to makethe point with a > /n th of the upper-right quadrant: 0 < θ < θ n ≡ π n . (2.19)10he integration variable is parametrized along each segment of the contour as follows: C : z = x , x from 0 to R ; C : z = R e iθ , θ from 0 to θ n ; C : z = r e iθ n , r from R to 0 . (2.20)The integral in Eq. (2.18) encloses no poles and therefore evaluates to zero: I C ( a, n ) = (cid:90) R dx e iax n + (cid:90) θ n R e iθ i dθ e ia ( Re iθ ) n + (cid:90) R dr e iθ n e ia ( re iθn ) n = 0 . (2.21)The task is to relate the complex integral to a real integral, so I want the imaginary part of e ia ( re iθn ) n = e iar n cos( nθ n ) e − ar n sin( nθ n ) (2.22)to be zero: cos( nθ n ) ≡ ⇒ θ n = π n (2 k + 1) , (2.23)with k any integer. But I must also maintain convergence of the real part:sin( nθ n ) > ⇒ < θ n < πn . (2.24)That fixes k = 0 and hence θ n = π n . Taking R → ∞ therefore produces the desired relation: (cid:90) ∞ dx e iax n = e iθ n (cid:90) ∞ dr e − ar n . (2.25)Now I can repeat the analogous steps for the negative domain of x , take n = 2, and combinethe results to recover the usual, sloppily asserted result for the complex Gaussian integral: (cid:90) ∞−∞ dx e iax = e iπ/ (cid:90) ∞−∞ dr e − ar = (cid:114) iπa . (2.26)With the analog of that for negative a , I can write down the infinitesimal matrix element: C j +1 ,j = 1 √ πiε e iε (cid:26) (cid:16) qj +1 − qjε (cid:17) − [ αV ( q j +1 )+(1 − α ) V ( q j )] (cid:27) . (2.27)I will now choose α = 0 (endpoint regularization) and proceed as usual. Defining ˙ q j ≡ ( q j +1 − q j ) /ε and L ( q ) ≡ ˙ q − V ( q ) , (2.28)I arrive at the path-integral representation of the 1-point transition amplitude: (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) = (2 πiε ) − N/ (cid:90) ∞−∞ dq N − ... (cid:90) ∞−∞ dq q n e iε (cid:80) N − j = 0 L ( q j ) ≡ (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) q ( t ) e i (cid:82) tft dt L ( q ( t )) . (2.29) I will explain why I fixated on this, other than for basic dignity. For real-valued Gaussian integrals, thestandard method is to square the integral, switch to polar coordinates, then square-root the result. For thecomplex Gaussian integral, I C ( a,
1) = 0 would then produce ia , leading to the correct answer. But naivelyconsidering I C ( a,
2) = 0 with a contour C over the whole upper half-plane would erroneously suggest anoverall phase of e iπ/ = i instead of e iπ/ = √ i . .2.2 Feynman path integral It is standard practice in probability and statistics to calculate moments of a distribution byintroducing an auxiliary variable and taking derivatives with respect to it [24]. ScrutinizingEq. (2.29) from that point of view suggests introducing an auxiliary field J ( t ) and definingthe following generating function: Z ( q , t ; q f , t f | J ( · )) ≡ (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) e i (cid:82) tft dt [ L ( q ( t )) + J ( t ) q ( t )] . (2.30)For notational convience, I will usually suppress the time arguments on the left-hand side. In terms of Eq. (2.30), the 1-point amplitude is (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) = − i δδJ ( t ) Z ( q , q f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 . (2.31)The derivation leading to Eq. (2.29) also makes clear that, for n > Z ( q , q f | J ) generatesamplitudes of only time-ordered products of fields:( − i ) n δ n δJ ( t ) ...δJ ( t n ) Z ( q , q f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 = (cid:104) q f | e − i ˆ Ht f T (ˆ q ( t ) ... ˆ q ( t n )) e i ˆ Ht | q (cid:105) , (2.32)with T ( ... ) the time-ordering symbol. I will refer to the amplitude-generating function in Eq. (2.30) as the Feynman path inte-gral, even though the history is more complicated, and even though Feynman wrote downmany path integrals.
Often lost in the cauldron of graduate school is the relation between the Feynman path in-tegral and the wavefunction. Before continuing, I just want to point out something simple.The evolution operator ˆ U ( t − t ) = e − it ˆ H e it ˆ H has coordinate-basis matrix elements (cid:104) q | ˆ U ( t − t ) | q (cid:105) = Z ( q , t ; q, t |
0) = (cid:90) q ( t ) = qq ( t ) = q D q ( · ) e i (cid:82) tt dt (cid:48) L ( q ( t (cid:48) )) . (2.33) Further comments about notation. First, I prefer to put q on the left and q f on the right, whereasmost people seem to prefer the opposite. Second, I will usually just write J instead of J ( · ) in Z ( q , q f | J ( · )).Third, as a related matter, I do not use bracket notation for functionals— Z ( q , q f | J ) is a functional of J buta function of q and q f , so I would have to write something like Z ( q , q f )[ J ( · )], which would look ridiculous.I also say “generating function” instead of “generating functional,” because it sounds better and there is zerorisk of confusion. For two operators ˆ A ( t ) and ˆ B ( t ), the time-ordering symbol is T (cid:16) ˆ A ( t ) ˆ B ( t (cid:48) ) (cid:17) ≡ Θ( t − t (cid:48) ) ˆ A ( t ) ˆ B ( t (cid:48) ) +Θ( t (cid:48) − t ) ˆ B ( t (cid:48) ) ˆ A ( t ). It is just mathematical shorthand whose general definition for t , ..., t n is too annoying towrite. evolution operator in the coordinate basis.Meanwhile, a system in a state | ψ (cid:105) has wavefunction ψ ( q ) ≡ (cid:104) q | ψ (cid:105) . That is a fixed functionfor all time (again, Heisenberg picture). I could, for fun, combine that fixed function withEq. (2.33) and define ψ ( q, t ) ≡ (cid:90) ∞−∞ dq (cid:104) q | ˆ U ( t − t ) | q (cid:105)(cid:104) q | ψ (cid:105) = (cid:90) ∞−∞ dq ψ ( q ) (cid:90) q ( t ) = qq ( t ) = q D q ( · ) e i (cid:82) tt dt (cid:48) L ( q ( t (cid:48) )) . (2.34)That is the Schrodinger-picture wavefunction for the system. So the Feynman path integralis related to the Schrodinger-picture wavefunction, but it is the evolution operator. Instead of the left-hand side of Eq. (2.31), consider Eq. (2.8) with n = 1:Tr (cid:16) ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) . (2.35)Or consider the special case ˆ ρ = | q (cid:105)(cid:104) q | , a pure state of definite coordinate value, inwhich case Eq. (2.35) would become (cid:104) q | e − i ˆ Ht ˆ q ( t ) e i ˆ Ht | q (cid:105) . That cannot be generated fromEq. (2.30), since setting t f = t there would just leave you with 1. No problem, nothingmysterious: Just find another generating function.
I have already shown in Eq. (2.7) that the expectation value in Eq. (2.35) is built fromEq. (2.4) and the conjugate of the zero-point amplitude—the new ingredient is Eq. (2.31),the Feynman path integral. Inserting that into Eq. (2.7) leads to a new generating function:Tr (cid:16) ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) (cid:104) q | ˆ ρ | q (cid:48) (cid:105) (cid:124) (cid:123)(cid:122) (cid:125) ≡ ρ ( q ,q (cid:48) ) (cid:104) q (cid:48) | e − i ˆ Ht e i ˆ Ht f | q f (cid:105) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ( − i ) δδJ ( t ) Z ( q , q f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 ρ ( q , q (cid:48) ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48) = 0 = ( − i ) δδJ ( t ) (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.36)If I define Z ( J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ , (2.37) Skipping ahead to Eqs. (2.43) and (2.67), I should point out that (cid:104) q | e − i ˆ Ht T (cid:18) e − i (cid:82) tft dt J ( t )ˆ q ( t ) (cid:19) e i ˆ Ht | q (cid:105) could, technically, generate Eq. (2.35) with ˆ ρ = | q (cid:105)(cid:104) q | . But good luck formulating effective field theory basedon that, except as the J (cid:48) = 0 limit of Eq. (2.37). (cid:16) ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = − i δδJ ( t ) Z ( J, J (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.38)The expectation-value-generating function Z ( J, J (cid:48) ) in Eq. (2.37) is the Schwinger-Keldyshpath integral. No highfalutin appeal to nonequilibrium dynamics, no spells.
Take two derivatives of the generating function Z ( q , q f | J ) in Eq. (2.30) to get a time-orderedtransition amplitude: (cid:104) q f | e − i ˆ Ht f T (ˆ q ( t )ˆ q ( t )) e i ˆ Ht | q (cid:105) = ( − i ) δ δJ ( t ) δJ ( t ) Z ( q , q f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 . (2.39)Take two derivatives of the generating function Z ( J, J (cid:48) ) in Eq. (2.37) to get a collection ofexpectation values:Tr (cid:104) T (ˆ q ( t )ˆ q ( t )) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:105) = ( − i ) δ δJ ( t ) δJ ( t ) Z ( J, J (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , Tr (cid:104) T ∗ (ˆ q ( t )ˆ q ( t )) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:105) = (+ i ) δ δJ (cid:48) ( t ) δJ (cid:48) ( t ) Z ( J, J (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , (2.40)Tr (cid:16) ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = δ δJ ( t ) δJ (cid:48) ( t ) Z ( J, J (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , Tr (cid:16) ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = δ δJ (cid:48) ( t ) δJ ( t ) Z ( J, J (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.41)In Eq. (2.40), T ∗ is the antitime-ordering symbol. The expectation values in Eq. (2.40)are called the Feynman and Dyson correlators, and the expectation values in Eq. (2.41) arecalled the Wightman correlators.
Logarithms of generating functions are given special names: The log of the Feynman pathintegral is called the effective action, and the log of the thermal partition function is calledthe free energy. The log of the Schwinger-Keldysh path integral is called the influence phase ,Φ(
J, J (cid:48) ) ≡ − i ln Z ( J, J (cid:48) ) . (2.42)Feynman and Vernon [7] called it that because they interpreted J as the coordinate of anadditional oscillator, whereas in this paper I treat J only as an auxiliary variable for probing For two operators ˆ A ( t ) and ˆ B ( t ), the antitime-ordering symbol is T ∗ (cid:16) ˆ A ( t ) ˆ B ( t ) (cid:17) ≡ Θ( t − t (cid:48) ) ˆ B ( t (cid:48) ) ˆ A ( t ) +Θ( t (cid:48) − t ) ˆ A ( t ) ˆ B ( t (cid:48) ). Compare to Footnote 13. By the way, I struggle with whether to write “anti-time-ordering”or “antitime-ordering”—does the “anti” modify “time-ordering,” or is the operator ordered in “antitime”? Dueling precedents seem to have led me to the sociologically befuddled terminology of “Schwinger-Keldysh influence phase,” but what can I do. If it were up to me I would name the whole formalism afterFeynman and Vernon, since Keldysh did not use path integrals, and Schwinger’s paper is unreadable. Z ( q , q f | J ) ≡ (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) e i (cid:82) tft [ L ( q ( t ))+ J ( t ) q ( t )] = (cid:104) q f | e − it f ˆ H T (cid:18) e i (cid:82) tft dt J ( t )ˆ q ( t ) (cid:19) e it ˆ H | q (cid:105) , (2.43)into the definition of the expectation-value-generating function in Eq. (2.37). If the densitymatrix is normalized, then setting J (cid:48) = J will reduce the generating function to 1: Z ( J, J ) = Tr ˆ ρ = 1 . (2.44)So any influence phase will satisfy Φ( J, J ) = 0 . (2.45)To derive the second property, return to Eq. (2.37) itself, regardless of how you choose toexpress Z ( q , q f | J ), and exchange the sources: Z ( J (cid:48) , J ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J (cid:48) ) Z ( q (cid:48) , q f | J ) ∗ = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq ρ ( q (cid:48) , q ) Z ( q (cid:48) , q f | J (cid:48) ) Z ( q , q f | J ) ∗ = (cid:18)(cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq ρ ( q (cid:48) , q ) ∗ Z ( q (cid:48) , q f | J (cid:48) ) ∗ Z ( q , q f | J ) (cid:19) ∗ . (2.46)Since the density matrix is hermitian—i.e., ρ ( q (cid:48) , q ) ∗ = ρ ( q , q (cid:48) )—the quantity in parenthesesis the original generating function: Z ( J (cid:48) , J ) = Z ( J, J (cid:48) ) ∗ . (2.47)So any influence phase will also satisfyΦ( J (cid:48) , J ) = − Φ( J, J (cid:48) ) ∗ . (2.48)I will apply those properties in due time. But now I want to express the correlation functionsfrom Eqs. (2.40) and (2.41) in terms of the influence phase. Taking one derivative withrespect to J ( t ) and another with respect to J (cid:48) ( t (cid:48) ) would produce δ Z ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t (cid:48) ) = e i Φ( J,J (cid:48) ) (cid:18) i δ Φ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t (cid:48) ) − δ Φ( J, J (cid:48) ) δJ ( t ) δ Φ( J, J (cid:48) ) δJ (cid:48) ( t (cid:48) ) (cid:19) . (2.49)All influence phases I calculate in this paper will satisfy δ Φ( J, J (cid:48) ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 = δ Φ( J, J (cid:48) ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 = 0 , (2.50)15n which case δ Z ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 = i δ Φ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.51)That is the key formula, leading to the desired expressions for the 2-point correlators:Tr (cid:104) T (ˆ q ( t )ˆ q ( t )) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:105) = − i δ Φ( J, J (cid:48) ) δJ ( t ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , Tr (cid:104) T ∗ (ˆ q ( t )ˆ q ( t )) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:105) = − i δ Φ( J, J (cid:48) ) δJ (cid:48) ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , (2.52)Tr (cid:16) ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = i δ Φ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , Tr (cid:16) ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = i δ Φ( J, J (cid:48) ) δJ (cid:48) ( t ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.53) Neurosis beckons a brief rewinding to Sec. 2.3.2: What if I wanted to generate a 2-pointtransition amplitude (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) ˆ q ( t ) e i ˆ Ht | q (cid:105) ? (2.54)I would need to go forward in time with one source, then again go forward in time with adifferent source, with both sources defined over the whole range between t and t f . To dothat, it seems I must introduce an intermediate branch that goes backward: (cid:104) q f | e − i ˆ Ht f ˆ q ( t )ˆ q ( t ) e i ˆ Ht | q (cid:105) = (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht e − i ˆ Ht e i ˆ Ht f e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) = (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq (cid:48) f (cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:48) (cid:105)(cid:104) q (cid:48) | e − i ˆ Ht e i ˆ Ht f | q (cid:48) f (cid:105)(cid:104) q (cid:48) f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105) = (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq (cid:48) f (cid:20) ( − i ) δδJ (cid:48)(cid:48) ( t ) Z ( q (cid:48) , q f | J (cid:48)(cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48)(cid:48) = 0 (cid:21) × (cid:104) Z ( q (cid:48) , q (cid:48) f | J (cid:48) ) ∗ (cid:12)(cid:12) J (cid:48) = 0 (cid:105) (cid:20) ( − i ) δδJ ( t ) Z ( q , q (cid:48) f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 (cid:21) = ( − i ) δ δJ ( t ) δJ (cid:48)(cid:48) ( t ) Z ( q , q f | J, J (cid:48) , J (cid:48)(cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = J (cid:48)(cid:48) = 0 , (2.55)with Z ( q , q f | J, J (cid:48) , J (cid:48)(cid:48) ) ≡ (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq (cid:48) f Z ( q , q (cid:48) f | J ) Z ( q (cid:48) , q (cid:48) f | J (cid:48) ) ∗ Z ( q (cid:48) , q f | J (cid:48)(cid:48) ) . (2.56)That is a specimen never before witnessed by these two eyes. Is it useful? I thank Paolo Glorioso for helping me think through this. .4 Out-of-time-ordered correlators Bridging the gap between superconductors [10] and black holes [25, 26], Kitaev emphasizedarbitrarily-ordered 4-point correlation functions, known as “out-of-time-ordered correlators”or “OTOCs,” in establishing equivalences between effective field theories [12].The Schwinger-Keldysh path integral can generate arbitrarily-ordered 2-point correlators,as in Eq. (2.41). But taking more than two derivatives would time-order the correspondingoperators, so Z ( J, J (cid:48) ) could never generate a correlation function like Tr (cid:16) ˆ q ( t )ˆ q ( t )ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) . (2.57)The mentality by now should be clear: If you want a correlation function that your existinggenerating functions cannot generate, then find another generating function. To find a path integral that could generate Eq. (2.57), I will follow reasoning analogous tothat which led to Eq. (2.56):Tr (cid:16) ˆ q ( t )ˆ q ( t )ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = Tr (cid:16) e − i ˆ Ht ˆ q ( t ) e i ˆ Ht f e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht e − i ˆ Ht ˆ q ( t ) e i ˆ Ht f e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht ˆ ρ (cid:17) = (cid:90) ∞−∞ dq dq (cid:48) dq (cid:48)(cid:48) dq f dq (cid:48) f (cid:104) q (cid:48)(cid:48) | e − i ˆ Ht ˆ q ( t ) e i ˆ Ht f | q (cid:48) f (cid:105)(cid:104) q (cid:48) f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:48) (cid:105)×(cid:104) q (cid:48) | e − i ˆ Ht ˆ q ( t ) e i ˆ Ht f | q f (cid:105)(cid:104) q f | e − i ˆ Ht f ˆ q ( t ) e i ˆ Ht | q (cid:105)(cid:104) q | ˆ ρ | q (cid:48)(cid:48) (cid:105) = (cid:90) ∞−∞ dq dq (cid:48) dq (cid:48)(cid:48) dq f dq (cid:48) f (cid:20) (+ i ) δδJ (cid:48)(cid:48)(cid:48) ( t ) Z ( q (cid:48)(cid:48) , q (cid:48) f | J (cid:48)(cid:48)(cid:48) ) ∗ (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48)(cid:48)(cid:48) = 0 (cid:21) (cid:20) ( − i ) δδJ (cid:48)(cid:48) ( t ) Z ( q (cid:48) , q (cid:48) f | J (cid:48)(cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48)(cid:48) = 0 (cid:21) × (cid:20) (+ i ) δδJ (cid:48) ( t ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48) = 0 (cid:21) (cid:20) ( − i ) δδJ ( t ) Z ( q , q f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 (cid:21) ρ ( q , q (cid:48)(cid:48) )= δ δJ ( t ) δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = J (cid:48)(cid:48) = J (cid:48)(cid:48)(cid:48) = 0 , (2.58) Motivated by the work of Shenker and Stanford [11]. There is some debate about which type of arbitrarily-ordered 4-point function is important [27], rooted inpoorly chosen words by Maldacena et al. [28] It does not matter whether the partitioning of fractional powersof the density matrix is a UV or IR effect, and I am not interested in traditional definitions of chaos—I aminterested in moments of the system in the ensemble described by ρ . Who cares about the Loschmidt echo. Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ (cid:90) ∞−∞ dq dq (cid:48) dq (cid:48)(cid:48) dq f dq (cid:48) f ρ ( q , q (cid:48)(cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ Z ( q (cid:48) , q (cid:48) f | J (cid:48)(cid:48) ) Z ( q (cid:48)(cid:48) , q (cid:48) f | J (cid:48)(cid:48)(cid:48) ) ∗ . (2.59)I will call that the Larkin-Ovchinnikov path integral, because it generates Larkin-Ovchinnikovcorrelation functions [10]. Adopting the terminology of Feynman and Vernon, I will also refer toΦ(
J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ − i ln Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) (2.60)as an influence phase, modified by “generalized” or “Larkin-Ovchinnikov” depending on con-text and whimsy.I want to record for later use some expressions for Eq. (2.57) in terms of derivatives ofΦ(
J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) instead of those of Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ), analogous to Eqs. (2.52) and (2.53). First,there is the following general expression:Tr (cid:16) ˆ q ( t )ˆ q ( t )ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) = δ Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) δJ ( t ) δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = J (cid:48)(cid:48) = J (cid:48)(cid:48)(cid:48) = 0 = i δ Φ δJ ( t ) δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + i (cid:18) δ Φ δJ ( t ) δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ (cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t )+ δ Φ δJ (cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ (cid:48)(cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) (cid:19) + i (cid:18) δ Φ δJ ( t ) δJ (cid:48) ( t ) δ Φ δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t )+ δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ ( t ) δJ (cid:48)(cid:48) ( t ) δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) (cid:19) + i (cid:18) δ Φ δJ ( t ) δ Φ δJ (cid:48) ( t ) δ Φ δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t )+ δ Φ δJ (cid:48) ( t ) δ Φ δJ (cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ (cid:48)(cid:48) ( t ) δ Φ δJ (cid:48)(cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48) ( t ) + δ Φ δJ ( t ) δ Φ δJ (cid:48)(cid:48)(cid:48) ( t ) δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t )+ δ Φ δJ ( t ) δ Φ δJ (cid:48)(cid:48) ( t ) δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ (cid:48) ( t ) δ Φ δJ (cid:48)(cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48)(cid:48) ( t ) (cid:19) + i δ Φ δJ ( t ) δ Φ δJ (cid:48) ( t ) δ Φ δJ (cid:48)(cid:48) ( t ) δ Φ δJ (cid:48)(cid:48)(cid:48) ( t ) , (2.61)which just expresses the combinatorics of exponentials. (All expressions are evaluated for J = J (cid:48) = J (cid:48)(cid:48) = J (cid:48)(cid:48)(cid:48) = 0, and I have used Z (0 , , ,
0) = 1.) I am continuing the trend of naming path integrals after famous Russians who did not use path integrals.See Footnote 16. q ( t ) → − q ( t ), the odd derivatives will be zero. For an ac-tion that is quadratic in q ( t ), all derivative powers higher than two will be zero. So for allcalculations in this paper, the above formula will reduce toTr (cid:16) ˆ q ( t )ˆ q ( t )ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) = − (cid:18) δ Φ δJ ( t ) δJ (cid:48) ( t ) δ Φ δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) δ Φ δJ ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) + δ Φ δJ ( t ) δJ (cid:48)(cid:48) ( t ) δ Φ δJ (cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) (cid:19) . (2.62) On the right-hand side of Eq. (2.62) I mean the Larkin-Ovchinnikov influence phase, Φ(
J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ),not the Schwinger-Keldysh influence phase, Φ(
J, J (cid:48) ). But in the special case of quadratic ac-tions, the two are related.To establish that relation, I will recall some observations of Feynman and Vernon. Startwith the general form Φ( J, J (cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) G F ( t − t (cid:48) ) J ( t ) J ( t (cid:48) ) − G D ( t − t (cid:48) ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − G < ( t − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − G > ( t − t (cid:48) ) J (cid:48) ( t ) J ( t (cid:48) ) (cid:105) . (2.63)Imposing Φ( J, J ) = 0 [Eq. (2.45)] gives the first constraint: G F ( t ) − G D ( t ) − G < ( t ) − G > ( t ) = 0 . (2.64)Imposing Φ( J, J (cid:48) ) ∗ = − Φ( J (cid:48) , J ) [Eq. (2.48)] gives the second constraint: G D ( t ) = G F ( t ) ∗ , G < ( t ) = − G > ( t ) ∗ . (2.65)Swapping t and t (cid:48) in the integrand and recognizing that Φ( J, J (cid:48) ) must remain invariant givesthe third constraint: G F ( t ) = G F ( − t ) , G D ( t ) = G D ( − t ) , G < ( t ) = G > ( − t ) . (2.66)All right, back to Φ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ).To see what happens when sources are set equal in pairs, it is again useful to invoke theoperator form of the amplitude-generating function from Eq. (2.43). With that, I can ex- In Sec. 4.5 I will calculate the Schwinger-Keldysh path integral for a particular non-quadratic action.But I will not calculate the Larkin-Ovchinnikov path integral for that case. I assume translational invariance for now to simplify the notation. Z ( J, J (cid:48) ) = Tr (cid:20) T ∗ (cid:18) e − i (cid:82) tft dt J (cid:48) ( t ) ˆ q ( t ) (cid:19) T (cid:18) e i (cid:82) tft dt J ( t ) ˆ q ( t ) (cid:19) e it ˆ H ˆ ρ e − it ˆ H (cid:21) , and (2.67) Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = Tr (cid:20) T ∗ (cid:18) e − i (cid:82) tft dt J (cid:48)(cid:48)(cid:48) ( t ) ˆ q ( t ) (cid:19) T (cid:18) e i (cid:82) tft dt J (cid:48)(cid:48) ( t ) ˆ q ( t ) (cid:19) ×T ∗ (cid:18) e − i (cid:82) tft dt J (cid:48) ( t ) ˆ q ( t ) (cid:19) T (cid:18) e i (cid:82) tft dt J ( t ) ˆ q ( t ) (cid:19) e it ˆ H ˆ ρ e − it ˆ H (cid:21) . (2.68)There are three pairs that will reduce Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) to the Schwinger-Keldysh generatingfunction: Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48) ) = Z ( J, J (cid:48) ) , Z ( J, J, J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = Z ( J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) , Z ( J, J (cid:48) , J (cid:48) , J (cid:48)(cid:48)(cid:48) ) = Z ( J, J (cid:48)(cid:48)(cid:48) ) . (2.69)But I still need to constrain the J (cid:48) J (cid:48)(cid:48) , J J (cid:48)(cid:48) , and J (cid:48) J (cid:48)(cid:48)(cid:48) terms in the influence phase. First Iwill isolate the dependence on J and J (cid:48) . Setting J (cid:48)(cid:48)(cid:48) = J would not make the J -dependentfactors cancel; setting J = J (cid:48)(cid:48)(cid:48) = 0 instead would leave the remaining time-ordered andantitime-ordered operators in the wrong order: Z (0 , J (cid:48) , J (cid:48)(cid:48) ,
0) = Tr (cid:20) T (cid:18) e i (cid:82) tft dt J (cid:48)(cid:48) ( t ) ˆ q ( t ) (cid:19) T ∗ (cid:18) e − i (cid:82) tft dt J (cid:48) ( t ) ˆ q ( t ) (cid:19) e it ˆ H ˆ ρ e − it ˆ H (cid:21) . (2.70)I do not know how to turn that into a Schwinger-Keldysh function, but I do know that itgenerates the same two point functions: δ Z (0 , J (cid:48) , J (cid:48)(cid:48) , δJ (cid:48) ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48) = J (cid:48)(cid:48) = 0 = − Tr (cid:104) T ∗ (ˆ q ( t )ˆ q ( t )) e it ˆ H ˆ ρ e − it ˆ H (cid:105) = − δ Z ( J, J (cid:48) ) δJ (cid:48) ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 ,δ Z (0 , J (cid:48) , J (cid:48)(cid:48) , δJ (cid:48)(cid:48) ( t ) δJ (cid:48)(cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48) = J (cid:48)(cid:48) = 0 = − Tr (cid:104) T (ˆ q ( t )ˆ q ( t )) e it ˆ H ˆ ρ e − it ˆ H (cid:105) = − δ Z ( J, J (cid:48) ) δJ ( t ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 ,δ Z (0 , J (cid:48) , J (cid:48)(cid:48) , δJ (cid:48)(cid:48) ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48) = J (cid:48)(cid:48) = 0 = Tr (cid:104) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:105) = δ Z ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.71)All I need is that third one. For the J J (cid:48)(cid:48) and J (cid:48) J (cid:48)(cid:48)(cid:48) terms, I instead consider Z ( J, , J (cid:48)(cid:48) ,
0) = Tr (cid:20) T (cid:18) e i (cid:82) tft dt J (cid:48)(cid:48) ( t ) ˆ q ( t ) (cid:19) T (cid:18) e i (cid:82) tft dt J ( t ) ˆ q ( t ) (cid:19) e it ˆ H ˆ ρ e − it ˆ H (cid:21) , and Z (0 , J (cid:48) , , J (cid:48)(cid:48)(cid:48) ) = Tr (cid:20) T ∗ (cid:18) e − i (cid:82) tft dt J (cid:48)(cid:48)(cid:48) ( t ) ˆ q ( t ) (cid:19) T ∗ (cid:18) e − i (cid:82) tft dt J (cid:48) ( t ) ˆ q ( t ) (cid:19) e it ˆ H ˆ ρ e − it ˆ H (cid:21) . (2.72)The off-diagonal correlation functions are δ Z ( J, , J (cid:48)(cid:48) , δJ ( t ) δJ (cid:48)(cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48)(cid:48) = 0 = δ Z (0 , J (cid:48) , , J (cid:48)(cid:48)(cid:48) ) δJ (cid:48) ( t ) δJ (cid:48)(cid:48)(cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J (cid:48) = J (cid:48)(cid:48)(cid:48) = 0 = − Tr (cid:104) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:105) = − δ Z ( J, J (cid:48) ) δJ ( t ) δJ (cid:48) ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 . (2.73)20otice the extra minus sign. Eqs. (2.69), (2.71), and (2.73) tell me everything I need toknow. The Larkin-Ovchinnikov phase for quadratic actions isΦ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:8) + G F ( t − t (cid:48) ) [ J ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )] − G D ( t − t (cid:48) ) [ J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − G < ( t − t (cid:48) ) [ J ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) + J ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) + J (cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) − J ( t ) J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − G > ( t − t (cid:48) ) [ J (cid:48) ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48)(cid:48) ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48) ( t (cid:48) )] (cid:9) . (2.74)Inserting that into Eq. (2.62), I obtainTr (cid:16) ˆ q ( t )ˆ q ( t )ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) = − [ G < ( t − t ) G < ( t − t ) + G < ( t − t ) G < ( t − t ) + G < ( t − t ) G < ( t − t )] . (2.75)That is the form of the relation that I will end up using, but for posterity I could recall that G < ( t − t (cid:48) ) = − i Tr (cid:16) ˆ q ( t (cid:48) )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) , in which caseTr (cid:16) ˆ q ( t )ˆ q ( t )ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) = Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) + Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) + Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) . (2.76)Defining, just for the moment, the notation (cid:28) ˆ O (cid:29) ≡ Tr (cid:16) ˆ O e it ˆ H ˆ ρ e − it ˆ H (cid:17) , and relabelingthe times as ( t , t , t , t ) ≡ ( t (cid:48) , t (cid:48) , t (cid:48) , t (cid:48) ), I can express Eq. (2.76) as (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) )ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) = (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) + (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) + (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) (cid:28) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) ) (cid:29) . (2.77)Given that, I have no idea what Larkin and Ovchinnikov intended to express in Eq. (23) oftheir paper [10]. The Schwinger-Keldysh formalism is about a choice of generating function, not about nonequi-librium dynamics. To belabor the point, I will entertain the phenomenological iε prescrip-tion from S -matrix theory, which does not conserve probability. (I will explore an improved,probability-conserving, but ultimately still phenomenological version in Sec. 8.)21 .5.1 From Feynman path integral Start with Eq. (2.39) and insert a complete set of energy eigenstates | n (cid:105) , ˆ H | n (cid:105) = E n | n (cid:105) ;assume a unique ground state with energy E and a nonzero energy gap ∆ n ≡ E n − E . Thetwo-point amplitude is: (cid:104) q f | e − i ˆ Ht f T (ˆ q ( t )ˆ q ( t )) e i ˆ Ht | q (cid:105) = ∞ (cid:88) n,m = 0 e − iE n t f e + iE m t (cid:104) q f | n (cid:105)(cid:104) n |T (ˆ q ( t )ˆ q ( t )) | m (cid:105)(cid:104) m | q (cid:105) = e − iE T (cid:104) q f | (cid:105)(cid:104) q | (cid:105) ∗ (cid:104) |T (ˆ q ( t )ˆ q ( t )) | (cid:105) + (cid:88) { all beside n = m = 0 } e − i ∆ n t f e + i ∆ m t (cid:104) q f | n (cid:105)(cid:104) n |T (ˆ q ( t )ˆ q ( t )) | m (cid:105)(cid:104) m | q (cid:105) . (2.78)As explained by Srednicki [9] and countless others, the standard prescription amounts toreplacing ∆ n by (1 − iε )∆ n , with ε an arbitrarily small positive real number. Sending t f → + ∞ and t → −∞ with ε held fixed would then suppress all excited-state contributionsrelative to the ground-state contribution. As long as the ground-state wavefunction ψ ( q ) ≡(cid:104) q | (cid:105) is nonzero at q = q and q = q f , then I can divide by (cid:104) q f | (cid:105)(cid:104) q | (cid:105) ∗ when calculatingtransition amplitudes. Therefore, (cid:104) |T (ˆ q ( t )ˆ q ( t )) | (cid:105) = lim ε → ( − i ) δ δJ ( t ) δJ ( t ) Z ε ( q , q f | J ) (cid:12)(cid:12)(cid:12)(cid:12) J = 0 , (2.79)with Z ε ( q , q f | J ) ∝ (cid:90) q ( ∞ ) = q f q ( −∞ ) = q D q ( · ) e i (cid:82) ∞−∞ dt [ L ε ( q ( t ))+ J ( t ) q ( t )] , L ε ( q ) = (1+ iε ) ˙ q − (1 − iε ) m q . (2.80)Now I will adapt that reasoning to the Schwinger-Keldysh path integral.22 .5.2 From Schwinger-Keldysh path integral The arbitrarily-ordered 2-point correlator is:Tr (cid:16) ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = Tr (cid:16) e − i ˆ Ht ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ (cid:17) = ∞ (cid:88) n,n (cid:48) = 0 (cid:104) n (cid:48) | e − i ˆ Ht ˆ q ( t )ˆ q ( t ) e i ˆ Ht | n (cid:105)(cid:104) n | ˆ ρ | n (cid:48) (cid:105) = e iE t e − iE t (cid:104) | ˆ q ( t )ˆ q ( t ) | (cid:105) ρ + ∞ (cid:88) n = 1 e iE n t (cid:104) | ˆ q ( t )ˆ q ( t ) | n (cid:105) ρ n + ∞ (cid:88) n (cid:48) = 1 e − iE n (cid:48) t (cid:104) n (cid:48) | ˆ q ( t )ˆ q ( t ) | (cid:105) ρ n (cid:48) + ∞ (cid:88) n,n (cid:48) = 1 e iE n t e − iE n (cid:48) t (cid:104) n (cid:48) | ˆ q ( t )ˆ q ( t ) | n (cid:105) ρ nn (cid:48) = (cid:104) | ˆ q ( t )ˆ q ( t ) | (cid:105) ρ + e iE t ∞ (cid:88) n = 1 e i ∆ n t (cid:104) | ˆ q ( t )ˆ q ( t ) | n (cid:105) ρ n + e − iE t ∞ (cid:88) n (cid:48) = 1 e − i ∆ n (cid:48) t (cid:104) n (cid:48) | ˆ q ( t )ˆ q ( t ) | (cid:105) ρ n (cid:48) + ∞ (cid:88) n,n (cid:48) = 1 e i ∆ n t e − i ∆ n (cid:48) t (cid:104) n (cid:48) | ˆ q ( t )ˆ q ( t ) | n (cid:105) ρ nn (cid:48) . (2.81)Adapting the iε prescription to this case would amount to replacing ∆ n → (1 − iε )∆ n on theforward branch and ∆ n (cid:48) → (1+ iε )∆ n (cid:48) on the backward branch. Then t → −∞ would onceagain suppress all excited-state contributions, and as long as ρ (cid:54) = 0 I can divide by ρ .Therefore, (cid:104) | ˆ q ( t )ˆ q ( t ) | (cid:105) = δ δJ ( t ) δJ (cid:48) ( t ) Z ε ( J, J (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 , (2.82)with Z ε ( J, J (cid:48) ) ∝ (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) × (cid:90) q ( ∞ ) = q f q ( −∞ ) = q D q ( · ) (cid:90) q (cid:48) ( ∞ ) = q f q (cid:48) ( −∞ ) = q (cid:48) D q (cid:48) ( · ) e i (cid:82) ∞−∞ dt [ L ε ( q ( t )) − L − ε ( q (cid:48) ( t ))+ J ( t ) q ( t ) − J (cid:48) ( t ) q (cid:48) ( t )] . (2.83)That provides a path-integral description of Wightman correlators. I do not know whether the effective field theory of Wightman correlators is important inequilibrium, but the above would be how to formulate it. Out of equilibrium, however, Idoubt that this setup makes sense [29, 30]. For the amplitude-generating function, the usual argument goes a step further: I could change q and q f to whatever I want as long as ψ ( q ) and ψ ( q f ) remain nonzero, in which case I could once again divideby ψ ( q f ) ψ ( q ) ∗ . The boundary conditions on the field do not matter. Presumably there is an analogousargument for Z ε ( J, J (cid:48) ), where as long as the density matrix has nonzero overlap with the ground state, theinitial values of the field should not matter; the final values of the field also should not matter, providedthat they agree (i.e., q ( ∞ ) = q (cid:48) ( ∞ )). Maybe there is a kind of stationary-state argument to be made fromEq. (4.41) or Eq. (8.60). Harmonic oscillator
The formalism established, it is time for the First of Examples, Quadratic Action of the FreeField, Pinnacle of Tractability, and Mother of Hubris: V ( q ) = m q . (3.1) The first task is to calculate the Feynman path integral with sources and fixed boundaryconditions: Z ( q , q f | J ) = (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) e i (cid:82) tft dt (cid:16)
12 ˙ q − m q + Jq (cid:17) . (3.2)If you think that deriving the amplitude-generating function for the harmonic oscillator isbeneath the scope of a contemporary research article, then stop reading and calculate it.Start to finish, including the normalization. Now.There are at least two methods I know of to calculate Z ( q , q f | J ). The first is to returnto the time-sliced definition from Sec. 2.2.1 and evaluate the iterated integrals, which wouldbe analogous to calculating integrals like (cid:82) ba dx f ( x ) by pre-Riemannian brute force [31]. Itried but could not complete that calculation (hara-kiri); I review the calculation for m = 0and J = 0 in Appendix A.6.The more erudite method is to deal with continuum field theory directly, applying the methodof stationary phase. That is the approach taken by most modern books, but be warned that,in general, even (0+1)-dimensional quantum field theory needs regularization, a 1-loop de-terminant, and 2-loop counterterms [32, 33, 34, 35].I will proceed with the continuum method. Split up the integration variable into a clas-sical path satisfying the boundary conditions, plus fluctuations: q ( t ) = ¯ q ( t ) + Q ( t ) , (3.3)with ¨¯ q ( t ) + m ¯ q ( t ) = J ( t ) , ¯ q ( t ) = q , ¯ q ( t f ) = q f , (3.4)and Q ( t ) = Q ( t f ) = 0 , (3.5)with Q ( t ) otherwise unconstrained. I will review how to solve Eq. (3.4). Yes, really.24 .2.1 Green’s function
First consider the homogeneous equation for a function G ( t ) with modified initial conditions,as follows: ¨ G ( t ) + m G ( t ) = 0 , G (0) = 0 , ˙ G (0) = 1 . (3.6)Integrate that equation against a Gaussian profile in time: (cid:90) ∞−∞ dt e − σ t + iωt (cid:104) ¨ G ( t ) + m G ( t ) (cid:105) = 0 . (3.7)Integrating by parts twice produces the following equation: e − σ t + iωt (cid:104) ˙ G ( t ) − iωG ( t ) (cid:105)(cid:12)(cid:12)(cid:12) ∞ t = −∞ + 1 σ (cid:90) ∞−∞ dt e − σ t + iωt t (cid:104) ˙ G ( t ) − iωG ( t ) (cid:105) + (cid:0) − ω + m (cid:1) (cid:90) ∞−∞ dt e − σ t + iωt G ( t ) = 0 . (3.8)Since e − σ t is even in t , the boundary terms are zero. The integral in the first line ofEq. (3.8) is some number. Therefore, doing all of that and then taking σ → ∞ leaves theFourier-transformed equation: (cid:0) − ω + m (cid:1) ˜ G ( ω ) = 0 , ˜ G ( ω ) ≡ (cid:90) ∞−∞ dt e iωt G ( t ) . (3.9)Either ˜ G ( ω ) = 0 or ω = ± m , so G ( t ) = c + e imt + c − e − imt for some constants c ± . Becausethe problem is manifestly real, I will instead work with the usual trigonometric functions: G ( t ) = A cos( mt ) + B sin( mt ) . (3.10)Since G (0) = A , the requirement G (0) = 0 means A = 0 and G ( t ) = B sin( mt ). Then˙ G ( t ) = Bm cos( mt ), in which case ˙ G (0) = Bm ≡ ⇒ B = 1 /m . Therefore, the Green’sfunction is G ( t ) = 1 m sin( mt ) . (3.11)I reviewed that for two reasons. First, to avoid lazily tossing boundary terms without ex-plaining why. Second, to emphasize that, in this formulation, the Green’s function is aswritten in Eq. (3.11)—it is not the operator-inverse of ∂ t + m , and it is odd in time. This is what Haberman calls an “alternative” formulation of the Green’s function [36]. If I remembercorrectly (the book is quarantined in my office), he further imposes causality on the solution of Eq. (3.6),whereas I prefer to impose causality only on the solution of Eq. (3.4). Sure, in this case you could just start from Eq. (3.10)—I know what an ansatz is. But as a student Inever found that satisfying, and in any case I am front-loading this for Sec. 8.3.2. .2.2 Homogeneous solution The homogenous part, q h ( t ), of the classical field is defined as the general solution of¨ q h ( t ) + m q h ( t ) = 0 . (3.12)Boundary conditions are imposed on the total classical field, not just on the homogeneouspart; and they will be imposed at times t and t f , not at zero. So at this stage all I can dois repeat the form of Eq. (3.10), but shifted by t : q h ( t ) = A ˙ G ( t − t ) + B G ( t − t ) , (3.13)for some constants A and B to be determined shortly. The particular part, q p ( t ), of the classical field is defined as the general solution of¨ q p ( t ) + m q p ( t ) = J ( t ) , (3.14)modulo homogeneous solutions, and subject to causality . In terms of the Green’s function inEq. (3.11), that means q p ( t ) = (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) J ( t (cid:48) ) . (3.15)I reviewed this reasoning to make Eq. (3.15) nonnegotiable. The classical field is the sum of its homogeneous and particular parts, subject to boundaryconditions: ¯ q ( t ) = q p ( t ) + q h ( t ) , ¯ q ( t ) = q , ¯ q ( t f ) = q f . (3.16)In Eq. (3.15) the limits are such that t (cid:48) is between t and t f , in which case Θ( t − t (cid:48) ) = 0.Since G (0) = 0 and ˙ G (0) = 1, the initial condition implies¯ q ( t ) = A × B × A ≡ q . (3.17)Similarly, Θ( t f − t (cid:48) ) = 1. So the final condition will fix B in terms of q f and J ( t ):¯ q ( t f ) = q ˙ G ( t f − t ) + B G ( t f − t ) + (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) ≡ q f = ⇒ B = 1 G ( t f − t ) (cid:20) q f − q ˙ G ( t f − t ) − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:21) . (3.18)Defining T ≡ t f − t , (3.19)26nd recognizing the trigonometric identity G ( T ) ˙ G ( t − t ) − ˙ G ( T ) G ( t − t ) = G ( t f − t ) , (3.20)I arrive at the complete solution for the classical field:¯ q ( t ) = 1 G ( T ) (cid:40) q f G ( t − t ) + q G ( t f − t )+ (cid:90) t f t dt (cid:48) [ G ( T ) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J ( t (cid:48) ) (cid:41) . (3.21) Next I will need to evaluate the action S ( q ) = (cid:90) t f t dt (cid:2) ˙ q ( t ) − m q ( t ) + J ( t ) q ( t ) (cid:3) (3.22)on the classical solution in Eq. (3.21). Since ˙ q = ddt ( q ˙ q ) − q ¨ q , and since the classical solution¯ q was defined to satisfy ¨¯ q + m ¯ q = J , I find S cl ( q , q f | J ) ≡ S (¯ q ) = ¯ q ˙¯ q (cid:12)(cid:12) t f t = t + (cid:90) t f t dt J ( t )¯ q ( t ) . (3.23)Again, that is the action on classical solutions, not the classical action. Eq. (3.22) is theclassical action, which is a function of the quantum field. Confusing, I know.The derivative of Eq. (3.21) is (remember that G (0) = 0):˙¯ q ( t ) = 1 G ( T ) (cid:40) q f ˙ G ( t − t ) − q ˙ G ( t f − t )+ (cid:90) t f t dt (cid:48) (cid:104) G ( T )Θ( t − t (cid:48) ) ˙ G ( t − t (cid:48) ) − ˙ G ( t − t ) G ( t f − t (cid:48) ) (cid:105) J ( t (cid:48) ) (cid:41) . (3.24)At the initial time (remember that ˙ G (0) = 1), I find˙¯ q ( t ) = 1 G ( T ) (cid:26) q f − q ˙ G ( T ) − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:27) . (3.25)At the final time, after using ˙ G ( T ) ˙ G ( t f − t ) − ˙ G ( T ) G ( t f − t ) = G ( t − t ), I find˙¯ q ( t f ) = 1 G ( T ) (cid:26) q f ˙ G ( T ) − q + (cid:90) t f t dt (cid:48) G ( t (cid:48) − t ) J ( t (cid:48) ) (cid:27) . (3.26)27herefore,¯ q ˙¯ q | t f t = t = q f G ( T ) (cid:26) q f ˙ G ( T ) − q + (cid:90) t f t dt (cid:48) G ( t (cid:48) − t ) J ( t (cid:48) ) (cid:27) − q G ( T ) (cid:26) q f − q ˙ G ( T ) − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:27) = 1 G ( T ) (cid:26) ˙ G ( T ) (cid:0) q f + q (cid:1) − q f q + (cid:90) t f t dt [ q f G ( t (cid:48) − t ) + q G ( t f − t (cid:48) )] J ( t (cid:48) ) (cid:27) . (3.27)Meanwhile, (cid:90) t f t dt J ( t )¯ q ( t ) = 1 G ( T ) (cid:90) t f t dt [ q f G ( t − t ) + q G ( t f − t )] J ( t )+ 1 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ G ( T ) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) . (3.28)Therefore: S cl ( q , q f | J ) = 12 G ( T ) (cid:26) ˙ G ( T ) (cid:0) q f + q (cid:1) − q f q + (cid:90) t f t dt [ q f G ( t − t ) + q G ( t f − t )] J ( t )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ G ( T ) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) (cid:27) . (3.29) With the splitting of the field into a classical solution plus fluctuations [Eq. (3.3)], the actionbecomes (cid:90) t f t dt (cid:2) ˙ q ( t ) − m q ( t ) + J ( t ) q ( t ) (cid:3) = S cl ( q , q f | J ) + (cid:90) t f t (cid:16) ˙ Q − m Q (cid:17) . (3.30)The square has been completed; the fluctuations are source-free.At this stage, the generating function for the harmonic oscillator has the form Z ( q , t ; q f , t f | J ) = C ( T ) e iS cl ( q ,t ; q f ,t f | J ) , (3.31)with C ( T ) ≡ Z (0 , t ; 0 , t f |
0) = (cid:90) Q ( t f ) = 0 Q ( t ) = 0 D Q ( · ) e i (cid:82) tft dt (cid:16)
12 ˙ Q − m Q (cid:17) . (3.32)Because the time dependence will now be important, I have specified it explicitly.28irst, notice that C ( T ) does not depend on the external source J , so it will be sufficientto consider Z ( q , t ; q f , t f |
0) = C ( T ) e iS cl ( q ,t ; q f ,t f | , S cl ( q , t ; q f , t f |
0) = ˙ G ( T )( q f + q ) − q f q G ( T ) . (3.33)Second, recognize that Z ( q , t f ; q f , t f |
0) is just the transition amplitude (cid:104) q f | e − i ˆ HT | q (cid:105) , whichsatisfies a composition law: Z ( q , t ; q , t |
0) = (cid:104) q | e − it ˆ H e it ˆ H | q (cid:105) = (cid:104) q | e − it ˆ H e it ˆ H e − it ˆ H e it ˆ H | q (cid:105) = (cid:90) ∞−∞ dq (cid:104) q | e − it ˆ H e it ˆ H | q (cid:105)(cid:104) q | e − it ˆ H e it ˆ H | q (cid:105) = (cid:90) ∞−∞ dq Z ( q , t ; q , t | Z ( q , t ; q , t | . (3.34)Because that composition law is nonlinear, imposing it on Eq. (3.31) will fix C ( T ). The first thing to show is that the sum of two S cl s will collect into an S cl plus a quadraticterm to be integrated over. S cl ( q , t ; q , t |
0) + S cl ( q , t ; q , t |
0) = ˙ G ( t − t )( q + q ) − q q G ( t − t ) + ˙ G ( t − t )( q + q ) − q q G ( t − t )= G ( t − t ) (cid:104) ˙ G ( t − t )( q + q ) − q q (cid:105) + G ( t − t ) (cid:104) ˙ G ( t − t )( q + q ) − q q (cid:105) G ( t − t ) G ( t − t ) . (3.35)First observe that G ( t − t ) ˙ G ( t − t ) + G ( t − t ) ˙ G ( t − t ) = G ( t − t ) . (3.36)29hank you, trigonometry. So the numerator in Eq. (3.35) is N ≡ G ( t − t ) (cid:104) ˙ G ( t − t )( q + q ) − q q (cid:105) + G ( t − t ) (cid:104) ˙ G ( t − t )( q + q ) − q q (cid:105) = G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) q − G ( t − t ) q + G ( t − t ) q ] q = G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) (cid:26) q − (cid:20) G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:21) q (cid:27)(cid:124) (cid:123)(cid:122) (cid:125)(cid:16) q − (cid:104) G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:105)(cid:17) − (cid:104) G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:105) = G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ( q − ... ) − [ G ( t − t ) q + G ( t − t ) q ] G ( t − t ) . (3.37)The part of N that can be moved outside the integral over q is therefore: N − G ( t − t )( q − ... ) = [ G ( t − t ) ˙ G ( t − t ) − G ( t − t ) ] G ( t − t ) q + [ G ( t − t ) ˙ G ( t − t ) − G ( t − t ) ] G ( t − t ) q − G ( t − t ) G ( t − t ) q q G ( t − t ) . (3.38)With more trigonomagic, namely G ( t − t ) ˙ G ( t − t ) − G ( t − t ) = ˙ G ( t − t ) G ( t − t ) , and G ( t − t ) ˙ G ( t − t ) − G ( t − t ) = ˙ G ( t − t ) G ( t − t ) , (3.39)that will simplify further into N − G ( t − t )( q − ... ) = G ( t − t ) G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) G ( t − t ) ˙ G ( t − t ) q − G ( t − t ) G ( t − t ) q q G ( t − t )= G ( t − t ) G ( t − t ) (cid:32) ˙ G ( t − t ) ( q + q ) − q q G ( t − t ) (cid:33) = 2 G ( t − t ) G ( t − t ) S cl ( q , t ; q , t | . (3.40)30 Christmas miracle. (Or just basic quantum mechanics.) With that, I can compose twoamplitudes: (cid:90) ∞−∞ dq Z ( q , t ; q , t | Z ( q , t ; q , t | C ( t − t ) C ( t − t ) e iS cl ( q ,t ; q ,t | (cid:90) ∞−∞ dq e iG ( t − t G ( t − t G ( t − t ( q − ... )= C ( t − t ) C ( t − t ) e iS cl ( q ,t ; q ,t | (cid:115) πi G ( t − t ) G ( t − t ) G ( t − t ) ≡ Z ( q , t ; q , t |
0) = C ( t − t ) e iS cl ( q ,t ; q ,t | = ⇒ C ( t − t ) (cid:112) πi G ( t − t ) = C ( t − t ) (cid:112) πi G ( t − t ) C ( t − t ) (cid:112) πi G ( t − t ) . (3.41)The solution is C ( t ) = 1 (cid:112) πi G ( t ) . (3.42) The generating function for the harmonic oscillator is Z ( q , t ; q f , t f | J ) = 1 (cid:112) πi G ( T ) e iS cl ( q ,t ; q f ,t f | J ) , G ( t ) = 1 m sin( mt ) , T = t f − t , (3.43)with S cl ( q , t ; q f , t f | J ) = 12 G ( T ) (cid:104) ˙ G ( T ) (cid:0) q f + q (cid:1) − q f q (cid:105) + 1 G ( T ) (cid:90) t f t dt [ G ( t − t ) q f + G ( t f − t ) q ] J ( t ) − G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t )] J ( t ) J ( t (cid:48) ) . (3.44)Is that formula in Feynman and Hibbs [19]? Yes. Did any of my illustrious professors deignto make me learn it? No. A standard alternative to imposing the composition law on the overall factor for the oscilla-tor path integral is to calculate that factor up to a frequency-independent factor using moderegularization. I will review that, too.First divide by Z ≡ Z (0 , | | m = 0 , (3.45)31hich is the free-particle amplitude. Define an operator ˆ O whose matrix elements in timeare the pertinent derivative operators: (cid:104) t | ˆ O| t (cid:48) (cid:105) ≡ (cid:18) d dt + m (cid:19) δ ( t − t (cid:48) ) , (cid:104) t |O | t (cid:48) (cid:105) ≡ (cid:104) t | ˆ O| t (cid:48) (cid:105) (cid:12)(cid:12)(cid:12) m = 0 . (3.46)The ratio of generating functions is then Z (0 , | Z = (cid:115) det ˆ O det ˆ O . (3.47)The determinant of an operator is the product of its eigenvalues. The eigenfunctions f ( t ) of d dt that equal zero at t = t have the form f ( t ) = A sin [ ω ( t − t )] , (3.48)with corresponding eigenvalues ω . Demanding f ( t f ) = 0 then quantizes the frequencies: f ( t f ) = A sin( ωT ) ≡ ⇒ ω = πnT , , ± , ± , ... . (3.49)Taking only the positive-frequency solutions gives the ratio of determinants as follows:det ˆ O det ˆ O = ∞ (cid:89) n = 1 − T m π n = πz sin( πz ) (cid:12)(cid:12)(cid:12)(cid:12) z = mTπ = mT sin( mT ) = TG ( T ) . (3.50)At this point I have the unsourced path integral up to a factor: Z (0 , |
0) = Z (cid:115) TG ( T ) . (3.51)See Appendix A for two ways of calculating Z . A typical physicist would write that as Z (0 , | Z = (cid:115) det (cid:16) d dt (cid:17) det (cid:16) d dt + m (cid:17) . But just as I write D q ( · ) insteadof D q ( t ), I refuse to write expressions that should not be read literally. This is not about mathematicalpseudorigor, which I detest, but rather about writing what you mean. Aha, says the heckler who read footnote 26: In Eq. (3.51) you have a T on the right-hand side but noton the left-hand side, hypocrite! No, no. In Eq. (3.51), T is a variable on both sides of the equation; it is justsuppressed on the left-hand side because I did not feel like writing it. In contrast, the expression “det( d dt )”makes no sense in the first place, because t does not appear in it. It is an uncivilized attempt to express“det( d d · )” or what I wrote in Eqs. (3.46) and (3.47). Schwinger-Keldysh path integral
The basic formula is Eq. (2.37), which I repeat below for convenience: Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ . (4.1)Calculating that for even the oscillator in its ground state took me some time. I interruptthis broadcast for a public service announcement about the order of integration. When I first tried to verify the results from Feynman and Vernon, I looked at Eq. (4.1) andthought, well, if the state is pure—i.e., if ˆ ρ = | ψ (cid:105)(cid:104) ψ | —then the double integral over the initialfield configurations will factorize: (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ = (cid:18)(cid:90) ∞−∞ dq (cid:104) q | ψ (cid:105) Z ( q , q f | J ) (cid:19) (cid:18)(cid:90) ∞−∞ dq (cid:48) (cid:104) q (cid:48) | ψ (cid:105) Z ( q (cid:48) , q f | J (cid:48) ) (cid:19) ∗ . And if the state is ground, like beef, then | ψ (cid:105) will be Gaussian. So just do the one integral,copy it with a conjugation and prime, then collect everything for the remaining integral overthe final field configuration. Go to Loloan, have a gimlet.What a roller-coaster ride that turned out to be. I did it, and I will show you how todo it, but it is not easy.The simpler method is to forget about the density matrix, go back to Eq. (4.1), and first dothe integral over q f . Why? Revisit Eq. (3.44): Since the final field configuration is required tobe the same, the quadratic terms in S cl ( q , q f | J ) − S cl ( q (cid:48) , q f | J (cid:48) ) will drop out! The integralover q f will produce a delta function that constrains q − q (cid:48) , leaving behind only an integralover q + q (cid:48) . Only then will it matter which ρ you choose, and the resulting integral will notbe so bad.In the end, I wanted to evaluate Z ( J, J (cid:48) ) using both methods anyway (at least for the os-cillator in its ground state), both to check my work (and triple-check the reference) andto show that the order of integration does not matter. It is also important, for the sake ofmaking progress, to accept that you will never know the right path until you just try one [38].For most of the examples, I will present the more complicated method. Partly because Ido not feel like redoing everything, and partly to save you the trouble. This was also noticed by Anglin [37], undoubtedly among others. .2 Ground state Back to your regularly scheduled programming. The ground-state wavefunction of the har-monic oscillator is (cid:104) q | (cid:105) = (cid:0) mπ (cid:1) e − mq . (4.2)Preparing the oscillator in its ground state means selecting a density matrix ρ ( q, q (cid:48) ) = (cid:104) q | (cid:105)(cid:104) | q (cid:48) (cid:105) = (cid:113) mπ e − m ( q + q (cid:48) ) . (4.3) The integral over the initial condition factorizes: (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ = Ψ( q f | J ) Ψ( q f | J (cid:48) ) ∗ , (4.4)whereΨ( q f | J ) = (cid:90) ∞−∞ dq (cid:104) q | (cid:105) Z ( q , q f | J ) = (cid:16) mπ (cid:17) (cid:112) πiG ( T ) (cid:90) ∞−∞ dq e − mq + iS cl ( q ,q f | J ) . (4.5)Extract the part that does not depend on q :Ψ( q f | J ) = (cid:16) mπ (cid:17) (cid:112) πiG ( T ) e iS cl (0 ,q f | J ) Ω( q f | J ) , (4.6)withΩ( q f | J ) = (cid:90) ∞−∞ dq e − mq + i G ( T ) (cid:110) ˙ G ( T ) q +2 q (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105)(cid:111) = (cid:90) ∞−∞ dq e i G ( T ) (cid:110) [ ˙ G ( T )+ imG ( T ) ] q +2 q (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105)(cid:111) = (cid:90) ∞−∞ dq e ie imT G ( T ) (cid:110) q +2 q e − imT (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105)(cid:111) (cid:16) ˙ G ( T ) + imG ( T ) = e imT (cid:17) = e − i e − imT G ( T ) (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105) (cid:112) πiG ( T ) e − imT . (4.7)The (cid:112) πiG ( T ) from Eq. (4.7) cancels the 1 / (cid:112) πiG ( T ) from Eq. (4.6), and the √ e − imT fromEq. (4.7) cancels out of the Ω( q f | J )Ω( q f | J (cid:48) ) ∗ implied by Eq. (4.4). The generating function34s Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ = (cid:90) ∞−∞ dq f Ψ( q f | J ) Ψ( q f | J (cid:48) ) ∗ = (cid:114) mπ (cid:90) ∞−∞ dq f e i [ S cl (0 ,q f | J ) − S cl (0 ,q f | J (cid:48) ) ] Ω( q f | J ) (cid:112) πiG ( T ) Ω( q f | J (cid:48) ) ∗ (cid:112) − πiG ( T )= (cid:114) mπ (cid:90) ∞−∞ dq f e i [ S cl (0 ,q f | J ) − S cl (0 ,q f | J (cid:48) ) ] e − i e − imT G ( T ) (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105) e + i e + imT G ( T ) (cid:104)(cid:82) tft dt G ( t f − t ) J (cid:48) ( t ) − q f (cid:105) ≡ (cid:114) mπ (cid:90) ∞−∞ dq f e i G ( T ) I ( q f | J,J (cid:48) ) . (4.8) Time to organize that I ( q f | J, J (cid:48) ): I ( q f | J, J (cid:48) ) = ˙ G ( T ) q f + 2 q f (cid:90) t f t dt G ( t − t ) J ( t ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) J ( t ) J ( t (cid:48) ) − e − imT (cid:20) q f − (cid:90) t f t dt G ( t f − t ) J ( t ) (cid:21) − ˙ G ( T ) q f − q f (cid:90) t f t dt G ( t − t ) J (cid:48) ( t ) + 2 (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) )+ e + imT (cid:20) q f − (cid:90) t f t dt G ( t f − t ) J (cid:48) ( t ) (cid:21) = 2 imG ( T ) q f + 2 q f (cid:90) t f t dt (cid:8)(cid:2) G ( t − t )+ e − imT G ( t f − t ) (cid:3) J ( t ) − (cid:2) G ( t − t )+ e imT G ( t f − t ) (cid:3) J (cid:48) ( t ) (cid:9) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:110)(cid:104) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )+Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t )+ G ( t f − t ) G ( t f − t (cid:48) ) e − imT (cid:105) J ( t ) J ( t (cid:48) ) − (cid:104) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t )+ G ( t f − t ) G ( t f − t (cid:48) ) e + imT (cid:105) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:111) . (4.9)Defining α ( t ) ≡ G ( t − t ) + e imT G ( t f − t ) (4.10)and j ( J, J (cid:48) ) ≡ (cid:90) t f t dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] , (4.11)35 will need to calculate the following integral: (cid:90) ∞−∞ dq f e i G ( T ) [ imG ( T ) q f +2 j ( J,J (cid:48) ) q f ] = (cid:114) πm e − mG ( T )2 j ( J,J (cid:48) ) . (4.12)The overall factor cancels, and the influence phase Φ( J, J (cid:48) ) ≡ − i ln Z ( J, J (cid:48) ) isΦ(
J, J (cid:48) ) = 12 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:26) i mG ( T ) [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] [ α ( t (cid:48) ) ∗ J ( t (cid:48) ) − α ( t (cid:48) ) J (cid:48) ( t (cid:48) )] − (cid:2) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t ) + G ( t f − t ) G ( t f − t (cid:48) ) e − imT (cid:3) J ( t ) J ( t (cid:48) )+ (cid:2) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t ) + G ( t f − t ) G ( t f − t (cid:48) ) e + imT (cid:3) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:27) . (4.13)Now for simplification. I will simplify the J J term by hand then let you program the restinto Mathematica.
J J term
For t > t (cid:48) , the coefficient of J ( t ) J ( t (cid:48) ) in the integrand of Eq. (4.13) is K ( t, t (cid:48) ) ≡ − (cid:2) G ( t f − t ) G ( t (cid:48) − t ) + G ( t f − t ) G ( t f − t (cid:48) ) e − imT (cid:3) + i α ( t ) ∗ α ( t (cid:48) ) ∗ mG ( T )= i mG ( T ) (cid:2) G ( t − t ) + e imT G ( t f − t ) (cid:3) (cid:2) G ( t (cid:48) − t ) + e − imT G ( t f − t (cid:48) ) (cid:3) = i mG ( T ) ( R + iS ) , (4.14)with R ≡ G ( t − t ) G ( t (cid:48) − t ) + G ( t f − t ) G ( t f − t (cid:48) ) + cos( mT ) [ G ( t f − t ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) G ( t − t )] ,S ≡ sin( mT ) [ G ( t f − t ) G ( t (cid:48) − t ) − G ( t f − t (cid:48) ) G ( t − t )] . (4.15)By writing G ( t ) = m sin( mt ) = im ( e imt − e − imt ), I find that S reduces to S = − G ( T ) mG ( t − t (cid:48) ) . (4.16)The expression R requires more work. First, using sin( a − b ) = sin( a ) sin( b ) − cos( a ) sin( b )with a = m ( t f − t ) and b = m ( t f − t ), I recognize that G ( t − t ) + cos( mT ) G ( t f − t ) = G ( T ) cos[ m ( t f − t )], and G ( t f − t ) + cos( mT ) G ( t − t ) = G ( T ) cos[ m ( t − t )]. So R can berewritten as R = G ( T ) m { cos[ m ( t f − t )] sin[ m ( t (cid:48) − t )] + cos[ m ( t − t )] sin[ m ( t f − t (cid:48) )] } . (4.17)Now I can use a trigonometric cyclic identity that I do not remember from high school:cos( a − b ) sin( c − d ) + cos( b − d ) sin( a − c ) + cos( b − c ) sin( d − a ) = 0 . (4.18)36ith a = mt f , b = mt , c = mt (cid:48) , and d = mt , I therefore get R = G ( T ) ˙ G ( t − t (cid:48) ) . (4.19)So the coefficient of J ( t ) J ( t (cid:48) ) for t > t (cid:48) is K ( t, t (cid:48) ) = i G ( T )2 m (cid:104) ˙ G ( t − t (cid:48) ) − imG ( t − t (cid:48) ) (cid:105) = G ( T ) i m e − im ( t − t (cid:48) ) . (4.20)For t < t (cid:48) , the coefficient of J ( t ) J ( t (cid:48) ) is K (cid:48) ( t, t (cid:48) ) ≡ − (cid:2) G ( t f − t (cid:48) ) G ( t − t ) + G ( t f − t ) G ( t f − t (cid:48) ) e − imT (cid:3) + i α ( t ) ∗ α ( t (cid:48) ) ∗ mG ( T )= K ( t (cid:48) , t ) = G ( T ) i m e + im ( t − t (cid:48) ) . (4.21)So the coefficient of J ( t ) J ( t (cid:48) ) isΘ( t − t (cid:48) ) K ( t, t (cid:48) ) + Θ( t (cid:48) − t ) K (cid:48) ( t, t (cid:48) ) = G ( T ) i m (cid:104) Θ( t − t (cid:48) ) e − im ( t − t (cid:48) ) + Θ( t (cid:48) − t ) e im ( t − t (cid:48) ) (cid:105) = G ( T ) i m e − im | t − t (cid:48) | . (4.22)Having derived that explicitly, I will state the results. First, some notation and terminology. The positive-frequency Wightman function, or “greaterGreen’s function,” is G > ( t ) ≡ i m e − imt , (4.23)and the negative-frequency Wightman function, or “lesser Green’s function,” is G < ( t ) ≡ i m e + imt = − G > ( t ) ∗ . (4.24)The Feynman function is G F ( t ) ≡ Θ( t ) G > ( t ) + Θ( − t ) G < ( t ) = i m e − im | t | , (4.25)and the Dyson function is G D ( t ) ≡ − [Θ( t ) G < ( t ) + Θ( − t ) G > ( t )] = − i m e + im | t | = G F ( t ) ∗ . (4.26)In terms of those, the influence phase isΦ( J, J (cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) G F ( t − t (cid:48) ) J ( t ) J ( t (cid:48) ) − G D ( t − t (cid:48) ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − G < ( t − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − G > ( t − t (cid:48) ) J (cid:48) ( t ) J ( t (cid:48) ) (cid:105) . (4.27) My conventions for Wightman functions differ by factors of i from those of Birrell and Davies [39]. G F ( t ), the Feynman expectation value isTr (cid:104) T (ˆ q ( t )ˆ q ( t )) e it ˆ H ˆ ρ e − it ˆ H (cid:105) = − δ Z ( J, J (cid:48) ) δJ ( t ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 = − i δ Φ( J, J (cid:48) ) δJ ( t ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 = − iG F ( t − t ) . (4.28)More generally, because all of my intuition about effective field theory comes from pathintegrals, I define G F , G D , G < , and G > as the coefficients of J J , J (cid:48) J (cid:48) , J J (cid:48) , and J (cid:48) J in theinfluence phase. As I mentioned in Sec. 4.1, despite the way I wrote Eq. (4.1) I could have instead performedthe “final” integration first: Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Y ( q , q (cid:48) | J, J (cid:48) ) , (4.29)with Y ( q , q (cid:48) | J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ . (4.30)Since the q f term in Eq. (3.44) does not depend on q or J , the quadratic terms will dropout of the product of amplitude-generating functions in Eq. (4.30): Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ = 12 πG ( T ) e i [ S cl ( q ,q f | J ) − S cl ( q (cid:48) ,q f | J (cid:48) ) ] . (4.31)Reorganizing the action as S cl ( q , q f | J ) = 12 G ( T ) (cid:26) ˙ G ( T ) q f + 2 (cid:20)(cid:90) t f t dt G ( t − t ) J ( t ) − q (cid:21) q f + F ( q | J ) (cid:27) , (4.32)with F ( q | J ) = ˙ G ( T ) q + 2 q (cid:90) t f t dt G ( t f − t ) J ( t ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) J ( t ) J ( t (cid:48) ) , (4.33)I find for the difference in actions: S cl ( q , q f | J ) − S cl ( q (cid:48) , q f | J ) = 1 G ( T ) (cid:20)(cid:90) t f t dt G ( t − t ) ( J ( t ) − J (cid:48) ( t )) − q + q (cid:48) (cid:21) q f + 12 G ( T ) [ F ( q | J ) − F ( q (cid:48) | J (cid:48) )] . (4.34)38ecalling that (cid:82) ∞−∞ dq e ikq/a = 2 πa δ ( k ) (with a assumed positive), I learn that the integral inEq. (4.30) will constrain the difference in initial field configurations: Y ( q , q (cid:48) | J, J (cid:48) ) = 12 πG ( T ) e i G ( T ) [ F ( q | J ) − F ( q (cid:48) | J (cid:48) ) ] (cid:90) ∞−∞ dq f e iG ( T ) (cid:104)(cid:82) tft dt G ( t − t )( J ( t ) − J (cid:48) ( t )) − q + q (cid:48) (cid:105) q f = e i G ( T ) [ F ( q | J ) − F ( q (cid:48) | J (cid:48) ) ] δ (cid:20)(cid:90) t f t dt G ( t − t ) ( J ( t ) − J (cid:48) ( t )) − q + q (cid:48) (cid:21) . (4.35)The generating function becomes Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq (cid:48) ρ (cid:18) q (cid:48) + (cid:90) t f t dt G ( t − t ) ( J ( t ) − J (cid:48) ( t )) , q (cid:48) (cid:19) × e i G ( T ) (cid:104) F ( q (cid:48) + (cid:82) tft dt G ( t − t )( J ( t ) − J (cid:48) ( t )) | J ) − F ( q (cid:48) | J (cid:48) ) (cid:105) . (4.36)Shift the integration variable to symmetrize the integrand: q (cid:48) ≡ q − k , = ⇒ q (cid:48) + k = q + k ; with k ≡ (cid:90) t f t dt G ( t − t ) ( J ( t ) − J (cid:48) ( t )) . (4.37)The generating function is now Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq ρ (cid:18) q + k , q − k (cid:19) e i G ( T ) [ F ( q + k | J ) − F ( q − k | J (cid:48) ) ] . (4.38)From Eq. (4.33) and ( q + k ) − ( q − k ) = 2 qk , I find F ( q + k | J ) − F ( q − k | J (cid:48) ) = 2 q (cid:90) t f t dt (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) [ J ( t ) − J (cid:48) ( t )] + B ( J, J (cid:48) ) , (4.39)with the q -independent part being B ( J, J (cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t (cid:48) − t ) [ J ( t )+ J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] − (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) [ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] . (4.40)The generating function is now Z ( J, J (cid:48) ) = e i G ( T ) B ( J,J (cid:48) ) × (cid:90) ∞−∞ dq ρ (cid:18) q + (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] , q − (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] (cid:19) × e iG ( T ) q (cid:82) tft dt [ ˙ G ( T ) G ( t − t )+ G ( t f − t ) ] [ J ( t ) − J (cid:48) ( t )] . (4.41) Experts in the Schwinger-Keldysh formalism are probably fuming that I am not yet working in the sumand difference basis. Cool your jets and wait till Sec. 8. .3.1 Ground state Now it is time to pick ρ . I will complete this calculation only for ˆ ρ = | (cid:105)(cid:104) | , which is whythis whole business is not in its own section devoted to the alternative order of integration.With (cid:104) q | (cid:105) = ( m/π ) / e − mq as before, and with( q + k + ( q − k = 2 q + k , (4.42)I find ρ ( q + k , q − k (cid:114) mπ e − mk e − mq . (4.43)So the generating function is Z ( J, J (cid:48) ) = e i G ( T ) B ( J,J (cid:48) ) (cid:114) mπ e − mk I ( J, J (cid:48) ) , (4.44)with I ( J, J (cid:48) ) = (cid:90) ∞−∞ dq e − mq + iG ( T ) q(cid:96) , (cid:96) = (cid:90) t f t dt (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) [ J ( t ) − J (cid:48) ( t )] . (4.45)Completing the square and performing the Gaussian integral givesΦ( J, J (cid:48) ) = − i ln Z ( J, J (cid:48) ) = i G ( T ) (cid:20) B ( J, J (cid:48) ) + imG ( T ) k + imG ( T ) (cid:96) (cid:21) , (4.46)with k = (cid:82) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] from Eq. (4.37).On to simplification. First, B ( J, J (cid:48) ). I trust that after symmetrizing in t and t (cid:48) , and bytreating the t > t (cid:48) and t < t (cid:48) cases separately, you will obtain the form B ( J, J (cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ G ( t f − t (cid:48) ) G ( t − t ) − G ( t f − t ) G ( t (cid:48) − t )] × { Θ( t − t (cid:48) ) [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) )+ J (cid:48) ( t (cid:48) )] − Θ( t (cid:48) − t ) [ J ( t )+ J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] } . (4.47)I trust you further to combine that with (cid:96) + m G ( T ) k = (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:26) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) (cid:104) ˙ G ( T ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) (cid:105) + m G ( T ) G ( t − t ) G ( t (cid:48) − t ) (cid:27) [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] , (4.48)by which I mean take the resulting coefficients of J ( t ) J ( t (cid:48) ), J ( t ) J (cid:48) ( t (cid:48) ), J (cid:48) ( t ) J ( t (cid:48) ), and J (cid:48) ( t ) J (cid:48) ( t (cid:48) ),and insert them into Mathematica. The essential simplification is the following: (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) (cid:104) ˙ G ( T ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) (cid:105) + m G ( T ) G ( t − t ) G ( t (cid:48) − t )+ imG ( T ) [ G ( t f − t (cid:48) ) G ( t − t ) − G ( t f − t ) G ( t (cid:48) − t )] = G ( T ) e im ( t − t (cid:48) ) . (4.49)With that, you will arrive at Eq. (4.27) with the definitions given in that subsection.40 .4 Thermal state Ludwig E. Boltzmann, I choose you:ˆ ρ = N e − β ˆ H , N = 1 / Tr( e − β ˆ H ) . (4.50)As Feynman and Vernon pointed out, the position-space representation of e − β ˆ H is an imaginary-time transition amplitude: (cid:104) q | e − β ˆ H | q (cid:105) = Z ( q , q | | t − t →− iβ = 1[2 πiG ( − iβ )] / e i G ( − iβ ) [ ˙ G ( − iβ )( q + q ) − q q ] . (4.51)Fix N by demanding Tr( ˆ ρ ) = 1: (cid:90) ∞−∞ dq (cid:104) q | ˆ ρ | q (cid:105) = N [2 πiG ( − iβ )] / (cid:90) ∞−∞ dq e iG ( − iβ ) [ ˙ G ( − iβ ) − ] q = N (cid:110) (cid:104) ˙ G ( − iβ ) − (cid:105)(cid:111) / ≡ ⇒ N = (cid:110) (cid:104) ˙ G ( − iβ ) − (cid:105)(cid:111) / . (4.52)So the thermal density matrix in the field basis is ρ ( q , q ) = (cid:16) ˙ G ( − iβ ) − iπG ( − iβ ) (cid:17) / e i G ( − iβ ) [ ˙ G ( − iβ )( q + q ) − q q ] . (4.53) Once again, the basic formula is Eq. (2.37): Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ . (4.54)This time, however, the double integral over the initial field configurations does not factorize,so it is worth passing to a 2-by-2 matrix representation: (cid:126)Q ≡ (cid:18) q q (cid:48) (cid:19) = ⇒ ρ ( q , q (cid:48) ) = (cid:16) ˙ G ( − iβ ) − iπG ( − iβ ) (cid:17) / e i G ( − iβ ) (cid:126)Q T ˙ G ( − iβ ) − − G ( − iβ ) (cid:126)Q . (4.55)The generating function is Z ( J, J (cid:48) ) = (cid:16) ˙ G ( − iβ ) − iπG ( − iβ ) (cid:17) / πG ( T ) e i G ( T ) ( − (cid:82) tft dt (cid:82) tft dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] × (cid:90) ∞−∞ dq f e i G ( T ) (+2) (cid:82) tft dt q f G ( t − t )[ J ( t ) − J (cid:48) ( t )] I ( q f | J, J (cid:48) ) , (4.56)where I ( q f | J, J (cid:48) ) = (cid:90) R d Q e iG ( T ) (cid:16) (cid:126)Q T M (cid:126)Q + (cid:126)j T (cid:126)Q (cid:17) , (4.57) Simplify the prefactor if you want to, but that is the most convenient way to write it. M = (cid:32) ˙ G ( T ) + ˙ G ( − iβ ) G ( − iβ ) G ( T ) − G ( T ) G ( − iβ ) − G ( T ) G ( − iβ ) − ˙ G ( T ) + ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:33) , (cid:126)j = (cid:32) (cid:82) t f t dt G ( t f − t ) J ( t ) − q f − (cid:104)(cid:82) t f t dt G ( t f − t ) J (cid:48) ( t ) − q f (cid:105)(cid:33) . (4.58)The Gaussian integral produces I ( q f | J, J (cid:48) ) = 2 πG ( T ) e − i G ( T ) (cid:126)j T M − (cid:126)j , (4.59)with M − = ˙ G ( T ) − ˙ G ( − iβ ) G ( − iβ ) G ( T ) − G ( T ) G ( − iβ ) − G ( T ) G ( − iβ ) − (cid:104) ˙ G ( T ) + ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:105) . (4.60)Therefore: (cid:126)j T M − (cid:126)j = (cid:34) ˙ G ( T ) − ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:35) (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) J ( t ) J ( t (cid:48) ) − (cid:34) ˙ G ( T ) + ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:35) (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) )+ 2 G ( T ) G ( − iβ ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t − t f ) G ( t f − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − G ( T ) (cid:16) ˙ G ( − iβ ) − G ( − iβ ) (cid:17) q f − q f (cid:90) t f t dt G ( t f − t ) (cid:110)(cid:104) ˙ G ( T ) − (cid:16) ˙ G ( − iβ ) − G ( − iβ ) (cid:17) G ( T ) (cid:105) J ( t ) − (cid:104) ˙ G ( T ) + (cid:16) ˙ G ( − iβ ) − G ( − iβ ) (cid:17) G ( T ) (cid:105) J (cid:48) ( t ) (cid:111) . (4.61)With that, the generating function becomes Z ( J, J (cid:48) ) = (cid:16) ˙ G ( − iβ ) − iπG ( − iβ ) (cid:17) / e i G ( T ) ( − (cid:82) tft dt (cid:82) tft dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] × e − i G ( T ) (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) (cid:110)(cid:104) ˙ G ( T ) − ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:105) J ( t ) J ( t (cid:48) ) − (cid:104) ˙ G ( T )+ ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:105) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:111) X ( J, J (cid:48) ) , (4.62)where X ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f e i (cid:16) ˙ G ( − iβ ) − G ( − iβ ) (cid:17) q f + iG ( T ) q f (cid:82) tft dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] , (4.63)with α ( t ) = G ( t − t ) + (cid:34) ˙ G ( T ) + (cid:32) ˙ G ( − iβ ) − G ( − iβ ) (cid:33) G ( T ) (cid:35) G ( t f − t ) . (4.64)Note that when β → ∞ , α ( t ) → G ( t − t ) + e imT G ( t f − t ), recovering the α ( t ) from Eq. (4.10).42 .4.2 Integral over final field configuration With j ( J, J (cid:48) ) ≡ (cid:90) t f t dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] , (4.65)I get X ( J, J (cid:48) ) = (cid:18) iπG ( − iβ )˙ G ( − iβ ) − (cid:19) / e − iG ( − iβ )[ ˙ G ( − iβ ) − G ( T )2 j ( J,J (cid:48) ) . (4.66)The overall factor cancels, and I get the influence phase Φ( J, J (cid:48) ) ≡ − i ln Z ( J, J (cid:48) ) in the formΦ(
J, J (cid:48) ) = − G ( − iβ )[ ˙ G ( − iβ ) − G ( T ) (cid:26)(cid:90) t f t dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] (cid:27) − G ( − iβ ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t )] [ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] − G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) (cid:110)(cid:104) ˙ G ( T ) − ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:105) J ( t ) J ( t (cid:48) ) − (cid:104) ˙ G ( T )+ ˙ G ( − iβ ) G ( − iβ ) G ( T ) (cid:105) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:111) . (4.67) Now I ask Mathematica to simplify Eq. (4.67), term by term, frustration by frustration. Theresult is:Φ(
J, J (cid:48) ) = i m sinh (cid:0) βm (cid:1) (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:8) cos (cid:2) m (cid:0) | t − t (cid:48) | + i β (cid:1)(cid:3) J ( t ) J ( t (cid:48) )+ cos (cid:2) m (cid:0) | t − t (cid:48) | − i β (cid:1)(cid:3) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − (cid:0) βm (cid:1) cos [ m ( t − t (cid:48) )] J ( t ) J (cid:48) ( t (cid:48) ) (cid:9) . (4.68)Denoting by Φ the ground-state expression from Eq. (4.27), and defining the function∆( t ) ≡ e βm − im cos( mt ) , (4.69)I recover the influence phase Φ( J, J (cid:48) ) = − i ln Z ( J, J (cid:48) ) as Feynman and Vernon expressed it:Φ(
J, J (cid:48) ) = Φ ( J, J (cid:48) ) + (cid:90) t f t dt (cid:90) t f t dt (cid:48) ∆( t − t (cid:48) ) [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] . (4.70)The thermal correction respects Eqs. (2.45) and (2.48).43 .5 First excited state I sat at a desk in Pasadena and thought, you know, why not try the first excited state: (cid:104) q | (cid:105) = √ m (cid:0) mπ (cid:1) / q e − mq = √ m q (cid:104) q | (cid:105) . (4.71)An oscillator in its first excited state would have a density matrix ρ ( q, q (cid:48) ) = (cid:104) q | (cid:105)(cid:104) | q (cid:48) (cid:105) = 2 m qq (cid:48) (cid:104) q | (cid:105)(cid:104) | q (cid:48) (cid:105) = 2 m (cid:113) mπ e − m ( q + q (cid:48) )+ln( qq (cid:48) ) . (4.72)That is not quadratic! It is, however, a hopping good time. Begin with the integral ψ ( q f | J ) ≡ (cid:90) ∞−∞ dq (cid:104) q | (cid:105) e iS cl ( q ,q f | J ) = √ m (cid:16) mπ (cid:17) / e i G ( T ) (cid:104) ˙ G ( T ) q f +2 q f (cid:82) tft dt G ( t − t ) J ( t ) − (cid:82) tft dt (cid:82) tft dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) J ( t ) J ( t (cid:48) ) (cid:105) χ ( q f | J ) , (4.73)where χ ( q f | J ) ≡ (cid:90) ∞−∞ dq q e − mq + i G ( T ) (cid:110) ˙ G ( T ) q +2 q (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105)(cid:111) = iG ( T ) ∂∂q f (cid:90) ∞−∞ dq e i G ( T ) (cid:110) [ ˙ G ( T )+ imG ( T ) ] q +2 q (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105)(cid:111) (cid:124) (cid:123)(cid:122) (cid:125) ≡ χ ( q f | J ) . (4.74)With ˙ G ( T ) + imG ( T ) = e imT , completing the square gives χ ( q f | J ) = (cid:112) πiG ( T ) e − imT e − ie − imT G ( T ) (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105) . (4.75)So χ ( q f | J ) = e − imT (cid:112) πiG ( T ) e − imT (cid:20) q f − (cid:90) t f t dt G ( t − t f ) J ( t ) (cid:21) e − ie − imT G ( T ) (cid:104) q f − (cid:82) tft dt G ( t f − t ) J ( t ) (cid:105) . (4.76)Inserting that into ψ ( q f | J ) and complex-conjugating the result, I get (cid:90) ∞−∞ dq f ψ ( q f | J ) ψ ( q f | J (cid:48) ) ∗ = 2 m (cid:16) mπ (cid:17) / πG ( T ) e i G ( T ) (cid:82) tft dt (cid:82) tft dt (cid:48) ( − t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] I ( J, J (cid:48) ) , (4.77)where I ( J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f (cid:20) q f − (cid:90) t f t dt G ( t f − t ) J ( t ) (cid:21) (cid:20) q f − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J (cid:48) ( t (cid:48) ) (cid:21) e i G ( T ) X ( q f | J,J (cid:48) ) , (4.78)44ith X ( q f | J, J (cid:48) ) ≡ q f (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] − e − imT (cid:20) q f − (cid:90) t f t dt G ( t f − t ) J ( t ) (cid:21) + e imT (cid:20) q f − (cid:90) t f t dt G ( t f − t ) J (cid:48) ( t ) (cid:21) = 2 imG ( T ) q f + 2 q f (cid:90) t f t dt [ α ( t ) ∗ J ( T ) − α ( t ) J (cid:48) ( t )] − (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) (cid:2) e − imT J ( t ) J ( t (cid:48) ) − e imT J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:3) , (4.79)where α ( t ) ≡ G ( t − t ) + e imT G ( t f − t ) . (4.80)Therefore, I ( J, J (cid:48) ) = e − i G ( T ) (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) [ e − imT J ( t ) J ( t (cid:48) ) − e imT J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) ] K ( J, J (cid:48) ) , (4.81)where K ( J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f (cid:20) q f − (cid:90) t f t dt G ( t f − t ) J ( t ) (cid:21) (cid:20) q f − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J (cid:48) ( t (cid:48) ) (cid:21) × e − mq f + iqfG ( T ) (cid:82) tft dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] = (cid:26) [ − iG ( T ) ∂ µ ] − (cid:90) t f t dt G ( t f − t ) [ − iG ( T ) ∂ µ ]+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) (cid:27) M ( J, J (cid:48) ; µ ) (cid:12)(cid:12)(cid:12)(cid:12) µ = 0 , (4.82)with M ( J, J (cid:48) ; µ ) ≡ (cid:90) ∞−∞ dq f e − mq f + iG ( T ) (cid:110) µ + (cid:82) tft dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] (cid:111) q f = (cid:16) πm (cid:17) / e − j mG ( T )2 , (4.83)where j ≡ µ + (cid:82) t f t dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )]. Using ∂ µ j = 1 and ∂ µ j = 0, I find ∂ µ (cid:18) e − j mG ( T )2 (cid:19) = − j mG ( T ) e − j mG ( T )2 ,∂ µ (cid:18) e − j mG ( T )2 (cid:19) = − mG ( T ) (cid:20) − j mG ( T ) (cid:21) e − j mG ( T )2 . (4.84)So K ( J, J (cid:48) ) = (cid:16) πm (cid:17) / N e − j mG ( T )2 (cid:12)(cid:12)(cid:12)(cid:12) µ = 0 , (4.85)45here N ≡ m (cid:20) − j mG ( T ) (cid:21) − ij mG ( T ) (cid:90) t f t dt G ( t f − t ) [ J ( t )+ J (cid:48) ( t )]+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) µ = 0 = 12 m (cid:26) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ A ( t, t (cid:48) ) ∗ J ( t ) J ( t (cid:48) ) + A ( t, t (cid:48) ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) + B ( t, t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) )] (cid:27) , (4.86)with the following functions: A ( t, t (cid:48) ) ≡ − α ( t ) α ( t (cid:48) )2 mG ( T ) + i α ( t ) G ( t f − t (cid:48) ) + α ( t (cid:48) ) G ( t f − t )2 G ( T )= − m cos[ m ( t − t (cid:48) )] ,B ( t, t (cid:48) ) ≡ α ( t ) ∗ α ( t (cid:48) ) mG ( T ) − iG ( T ) [ α ( t ) ∗ G ( t f − t (cid:48) ) − α ( t (cid:48) ) G ( t f − t )] + 2 mG ( t f − t ) G ( t f − t (cid:48) )= 1 m cos[ m ( t − t (cid:48) )] . (4.87)The factor N simplifies to N = 12 m (cid:26) − m (cid:90) t f t dt (cid:90) t f t dt (cid:48) cos[ m ( t − t (cid:48) )] [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] (cid:27) . (4.88)Finally I can put everything together to obtain the generating function: Z ( J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) (cid:104) q | (cid:105)(cid:104) | q (cid:48) (cid:105) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ = 12 πG ( T ) (cid:90) ∞−∞ dq f ψ ( q f | J ) ψ ( q f | J (cid:48) ) ∗ = 2 m (cid:16) mπ (cid:17) / e i G ( T ) (cid:82) tft dt (cid:82) tft dt (cid:48) ( − t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] I ( J, J (cid:48) )= 2 m (cid:16) mπ (cid:17) / e i G ( T ) (cid:82) tft dt (cid:82) tft dt (cid:48) ( − t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] × e − i G ( T ) (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t f − t ) G ( t f − t (cid:48) ) [ e − imT J ( t ) J ( t (cid:48) ) − e imT J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) ] K ( J, J (cid:48) )= 2 mN Z ( J, J (cid:48) ) , (4.89)46here Z ( J, J (cid:48) ) = e i Φ ( J,J (cid:48) ) is the generating function for the oscillator in its ground state.So the influence function Φ ( J, J (cid:48) ) ≡ − i ln Z ( J, J (cid:48) ) is Φ ( J, J (cid:48) ) = Φ ( J, J (cid:48) ) − i ln (cid:26) − m (cid:90) t f t dt (cid:90) t f t dt (cid:48) cos[ m ( t − t (cid:48) )] [ J ( t ) − J (cid:48) ( t )] [ J (cid:48) ( t ) − J (cid:48) ( t (cid:48) )] (cid:27) . (4.91)The Wightman correlator isTr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) = δ Z ( J, J (cid:48) ) δJ ( t ) δJ ( t ) (cid:12)(cid:12)(cid:12)(cid:12) J = J (cid:48) = 0 = Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H ˆ ρ e − it ˆ H (cid:17) + 1 m cos [ m ( t − t )] . (4.92)The operator calculation for general n is straightforward and leads to (cid:104) n | ˆ q ( t )ˆ q ( t ) | n (cid:105) = 12 m e − im ( t − t ) + nm cos [ m ( t − t )] , (4.93)which agrees with the above path-integral result for n = 1. As with the “other order of integration” method from Sec. 4.3 for evaluating the Schwinger-Keldysh path integral, there is much to simplify in the Larkin-Ovchinnikov path integralbefore choosing a density matrix. For convenience, I will repeat the basic formula, Eq. (2.59): Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ (cid:90) ∞−∞ dq dq (cid:48) dq (cid:48)(cid:48) dq f dq (cid:48) f ρ ( q , q (cid:48)(cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ Z ( q (cid:48) , q (cid:48) f | J (cid:48)(cid:48) ) Z ( q (cid:48)(cid:48) , q (cid:48) f | J (cid:48)(cid:48)(cid:48) ) ∗ . (5.1)First I will use Eqs. (3.43) and (3.44) to evaluate A ( q , q i | J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f Z ( q , q f | J ) Z ( q i , q f | J (cid:48) ) ∗ . (5.2) One potentially interesting exercise would be to calculate Φ n for a density matrix ˆ ρ = | n (cid:105)(cid:104) n | , then use e − β ˆ H = ∞ (cid:88) n = 0 e − βm ( n + 12 ) | n (cid:105)(cid:104) n | (4.90)to arrive at the thermal influence phase. I thought about trying that, but then I looked up the series formof the Hermite polynomials and said forget it. This is a translationally invariant problem, so t drops out. S cl ( q , q f | J ) − S cl ( q i , q f | J (cid:48) ) = 12 G ( T ) (cid:26) ˙ G ( T )( q − q i ) + 2 (cid:90) t f t dt G ( t f − t ) [ q J ( t ) − q i J (cid:48) ( t )] − (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )]+2 (cid:20) q i − q + (cid:90) t f t dt G ( t − t ) ( J ( t ) − J (cid:48) ( t )) (cid:21) q f (cid:27) . (5.3)The only dependence on q f is linear, so the integral produces a delta function. Since (cid:82) ∞−∞ dq e iG ( T ) µq = 2 πG ( T ) δ ( µ ) (assuming positive G ( T ), which all of the previous formulasimplicitly require), the integral evaluates to A ( q , q i | J, J (cid:48) ) = δ (cid:18) q i − q + (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] (cid:19) e i G ( T ) B ( q ,q i | J,J (cid:48) ) , (5.4)with B ( q , q i | J, J (cid:48) ) = ˙ G ( T ) (cid:0) q − q i (cid:1) + 2 (cid:90) t f t dt G ( t f − t ) [ q J ( t ) − q i J (cid:48) ( t )] − (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] . (5.5)Similarly, A ( q i , q (cid:48) | J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:90) ∞−∞ dq (cid:48) f Z ( q i , q (cid:48) f | J (cid:48)(cid:48) ) Z ( q (cid:48) , q (cid:48) f | J (cid:48)(cid:48)(cid:48) ) ∗ = δ (cid:18) q (cid:48) − q i + (cid:90) t f t dt G ( t − t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] (cid:19) e i G ( T ) B ( q i ,q (cid:48) | J (cid:48)(cid:48) ,J (cid:48)(cid:48)(cid:48) ) , (5.6)with B ( q i , q (cid:48) | J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = ˙ G ( T ) (cid:0) q i − q (cid:48) (cid:1) + 2 (cid:90) t f t dt G ( t f − t ) [ q i J (cid:48)(cid:48) ( t ) − q (cid:48) J (cid:48)(cid:48)(cid:48) ( t )] − (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) [ J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] . (5.7)Because of the delta functions, the integrals over q and q (cid:48) can be done immediately: Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:90) ∞−∞ dq i (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) A ( q , q i | J, J (cid:48) ) A ( q i , q (cid:48) | J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) )= (cid:90) ∞−∞ dq i ρ ( q , q (cid:48) ) e i G ( T ) [ B ( q ,q i | J,J (cid:48) )+ B ( q i ,q (cid:48) | J (cid:48)(cid:48) ,J (cid:48)(cid:48)(cid:48) ) ] (cid:12)(cid:12)(cid:12)(cid:12) q = q i + (cid:82) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] ,q (cid:48) = q i − (cid:82) t f t dt G ( t − t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] . (5.8)48ime to simplify the argument of the exponential. First, simplify q − q (cid:48) : q − q (cid:48) = 2 q i (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )+ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )]+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) { [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] − [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] [ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] } . (5.9)The q i terms have canceled—they will come solely from the density matrix. With that, Ifind B ( q , q i | J, J (cid:48) ) + B ( q i , q (cid:48) | J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = 2 q i (cid:90) t f t dt (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) [ J ( t ) − J (cid:48) ( t )+ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )]+ ˙ G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) { [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] − [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] [ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] } + C ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) , (5.10)where to anticipate simplifying further I have separated a quantity C ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡− (cid:90) t f t dt (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) [ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )+ J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )]+ 2 (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t (cid:48) − t ) [ J ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) − J ( t ) J (cid:48) ( t (cid:48) )+ J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )] . (5.11)For t > t (cid:48) , the J J and J (cid:48)(cid:48)(cid:48) J (cid:48)(cid:48)(cid:48) terms cancel, and for t < t (cid:48) the first line is absent. So anotherway to write Eq. (5.11) is C ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = 2 (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t f − t ) G ( t (cid:48) − t ) { Θ( t − t (cid:48) ) [ J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )]+Θ( t (cid:48) − t ) [ J ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − J ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) } . (5.12)The J J , J (cid:48) J (cid:48) , J (cid:48)(cid:48) J (cid:48)(cid:48) , and J (cid:48)(cid:48)(cid:48) J (cid:48)(cid:48)(cid:48) terms are symmetric in t and t (cid:48) , so I will use that to symmetrizethe coefficients of those terms in the integrand. I will keep the J J (cid:48) and J (cid:48)(cid:48)(cid:48) J (cid:48)(cid:48) terms as written.I get: C ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) { [Θ( t − t (cid:48) ) G ( t f − t (cid:48) ) G ( t − t ) + Θ( t (cid:48) − t ) G ( t f − t ) G ( t (cid:48) − t )] [ J ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )]+ [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t )] [ J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )]+2 G ( t f − t ) G ( t (cid:48) − t ) [ J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) − J ( t ) J (cid:48) ( t (cid:48) )] } . (5.13)49ombining that with the other q i -independent terms from Eq. (5.10) gives B ( q , q i | J, J (cid:48) ) + B ( q i , q (cid:48) | J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) =2 q i (cid:90) t f t dt (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) [ J ( t ) − J (cid:48) ( t )+ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] + D ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) , (5.14)with D ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:110) (cid:104) G ( t f − t ) + ˙ G ( T ) G ( t − t ) (cid:105) G ( t (cid:48) − t ) × [ J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ) − J ( t ) J (cid:48) ( t (cid:48) )]+ (cid:104) Θ( t − t (cid:48) ) G ( t f − t (cid:48) ) G ( t − t )+Θ( t (cid:48) − t ) G ( t f − t ) G ( t (cid:48) − t )+ ˙ G ( T ) G ( t − t ) G ( t (cid:48) − t ) (cid:105) × [ J ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )]+ (cid:104) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t )+Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t )+ ˙ G ( T ) G ( t − t ) G ( t (cid:48) − t ) (cid:105) × [ J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )] (cid:111) . (5.15)To proceed I will need to specify ρ . I will need to simplify the oscillator ground-state density matrix from Eq. (4.2) using theexpressions for q and q (cid:48) produced by the delta functions. The wavefunctions are: (cid:104) q | (cid:105)| q = q i + (cid:82) tft dt G ( t − t )[ J ( t ) − J (cid:48) ( t )] = (cid:16) mπ (cid:17) / e − m (cid:104) q i +2 q i (cid:82) tft dt G ( t − t )[ J ( t ) − J (cid:48) ( t )] (cid:105) × e − m (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t − t ) G ( t (cid:48) − t )[ J ( t ) − J (cid:48) ( t )][ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] , (cid:104) q (cid:48) | (cid:105)| q (cid:48) = q i − (cid:82) tft dt G ( t − t )[ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] = (cid:16) mπ (cid:17) / e − m (cid:104) q i − q i (cid:82) tft dt G ( t − t )[ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] (cid:105) × e − m (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t − t ) G ( t (cid:48) − t )[ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )][ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] . (5.16)So the density matrix is ρ ( q , q (cid:48) ) = (cid:16) mπ (cid:17) / e − m (cid:110) q i + q i (cid:82) tft dt G ( t − t )[ J ( t ) − J (cid:48) ( t ) − J (cid:48)(cid:48) ( t )+ J (cid:48)(cid:48)(cid:48) ( t )] (cid:111) × e − m (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) { [ J ( t ) − J (cid:48) ( t )][ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )]+[ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )][ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] } . (5.17)There is the q i term, and with it the requisite factor of − m to give a Gaussian integral thatwill cancel the overall factor of ( m/π ) / . The coefficient of q i from ρ and i G ( T ) ( B + B ) leads50o j ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ (cid:90) t f t dt (cid:26) − imG ( T ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) [ J ( t ) − J (cid:48) ( t )+ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )]+ G ( t − t ) [ J ( t ) − J (cid:48) ( t ) − J (cid:48)(cid:48) ( t )+ J (cid:48)(cid:48)(cid:48) ( t )] (cid:27) = (cid:90) t f t dt (cid:40) (cid:20) G ( t − t ) − imG ( T ) (cid:16) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:17)(cid:21) [ J ( t ) − J (cid:48) ( t )] − (cid:20) G ( t − t ) + imG ( T ) (cid:16) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:17)(cid:21)(cid:124) (cid:123)(cid:122) (cid:125) ≡ α ( t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] (cid:41) = (cid:90) t f t dt { α ( t ) ∗ [ J ( t ) − J (cid:48) ( t )] − α ( t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] } . (5.18)Completing the square and doing the integral over q i gives the generating function Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = e i Φ( J,J (cid:48) ,J (cid:48)(cid:48) ,J (cid:48)(cid:48)(cid:48) ) (5.19)with generalized influence phaseΦ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = 12 G ( T ) D ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) − i m j ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) + i m (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) (cid:110) [ J ( t ) − J (cid:48) ( t )][ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )]+ [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )][ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] (cid:111) . (5.20) Now I will need to systematically collect and simplify each type of term (
J J , J J (cid:48) , and soon). The square of j gives j = (cid:90) t f t dt (cid:90) t f t dt (cid:48) { α ( t ) ∗ α ( t (cid:48) ) ∗ [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )]+ α ( t ) α ( t (cid:48) ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] [ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − α ( t ) ∗ α ( t (cid:48) ) [ J ( t ) − J (cid:48) ( t )] [ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − α ( t ) α ( t (cid:48) ) ∗ [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] } . (5.21)I want to explain a few things about simplifying that.First, focus on the last two lines of Eq. (5.21) and compare them to the second line of51qs. (5.20) and to (5.15). Notice something? The last two lines of Eq. (5.21) have no coun-terparts anywhere else in the generalized influence phase. So α ( t ) ∗ α ( t (cid:48) ) had better simplifyby itself, and indeed it does: α ( t ) ∗ α ( t (cid:48) ) = 1 m e im ( t − t (cid:48) ) . (5.22)By “simplifies” here I mean that all dependence on t and t f cancels out, and the resultdepends on t and t (cid:48) only as t − t (cid:48) (i.e., it is translationally invariant). In contrast, α ( t ) α ( t (cid:48) )does not simplify in that sense: α ( t ) α ( t (cid:48) ) = − m e − im ( t + t (cid:48) − t ) . (5.23)This lack of simplification is expected, because there are other terms in the influence phasethat will combine with J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) ), etc.The second thing I want to explain is how to simplify one of the source-diagonal terms,say the J J one. I have already symmetrized Eq. (5.15) in t and t (cid:48) to anticipate a relativelystraightforward collecting of terms. For t > t (cid:48) , the symmetrized coefficient of J ( t ) J ( t (cid:48) ) is O > ( t, t (cid:48) ) ≡ G ( T ) (cid:104) ˙ G ( T ) G ( t − t ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) G ( t − t ) (cid:105) + i m G ( t − t ) G ( t (cid:48) − t ) − i m α ( t ) ∗ α ( t (cid:48) ) ∗ . (5.24)I begin by returning to α ( t ) α ( t (cid:48) ): α ( t ) α ( t (cid:48) ) = G ( t − t ) G ( t (cid:48) − t ) − m G ( T ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) (cid:104) ˙ G ( T ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) (cid:105) + imG ( T ) (cid:104) G ( T ) G ( t − t ) G ( t (cid:48) − t ) + G ( t − t ) G ( t f − t (cid:48) ) + G ( t f − t ) G ( t (cid:48) − t ) (cid:105) . (5.25)Conjugating that and combining it with the other terms in Eq. (5.24) produces the followingreal and imaginary parts:Re[ O > ( t, t (cid:48) )] = 14 G ( T ) [ G ( t f − t (cid:48) ) G ( t − t ) − G ( t f − t ) G ( t (cid:48) − t )] = 14 G ( t − t (cid:48) ) , (5.26)Im[ O > ( t, t (cid:48) )] = m (cid:40) (cid:34) G ( T ) m G ( T ) (cid:35) G ( t − t ) G ( t (cid:48) − t )+ 1 m G ( T ) (cid:104) G ( t f − t ) G ( t f − t (cid:48) ) + ˙ G ( T ) [ G ( t − t ) G ( t f − t (cid:48) ) + G ( t f − t ) G ( t (cid:48) − t )] (cid:105) (cid:41) . (5.27)Since ˙ G ( T ) + m G ( T ) = cos ( mT ) + sin ( mT ) = 1, I can factor out an overall m G ( T ) inthe denominator. Then I can use the trigonometric identitycos( a − b ) sin( a − c ) − sin( a − b ) cos( a − c ) = sin( b − c ) (5.28)52s follows: G ( t (cid:48) − t ) + ˙ G ( T ) G ( t f − t (cid:48) ) = G ( T ) ˙ G ( t f − t (cid:48) ) ,G ( t f − t (cid:48) ) + ˙ G ( T ) G ( t (cid:48) − t ) = G ( T ) ˙ G ( t (cid:48) − t ) . (5.29)With all of that, the imaginary part becomes:Im[ O > ( t, t (cid:48) )] = 14 mG ( T ) (cid:110) G ( t − t ) (cid:104) G ( t (cid:48) − t ) + ˙ G ( T ) G ( t f − t (cid:48) ) (cid:105) + G ( t f − t ) (cid:104) G ( t f − t (cid:48) ) + ˙ G ( T ) G ( t (cid:48) − t ) (cid:105)(cid:111) = 14 mG ( T ) (cid:110) G ( t − t ) ˙ G ( t f − t (cid:48) ) + G ( t f − t ) ˙ G ( t (cid:48) − t ) (cid:111) = 14 m ˙ G ( t − t (cid:48) ) . (5.30)Putting the real and imaginary parts together, I find O > ( t, t (cid:48) ) = 14 G ( t − t (cid:48) ) + i m ˙ G ( t − t (cid:48) ) = i m e − im ( t − t (cid:48) ) . (5.31)That, remember, is only the coefficient of J ( t ) J ( t (cid:48) ) for t > t (cid:48) . But now that I have derivedthat much, I am confident that you can derive the rest (or ask Mathematica).The final result for the generalized influence phase will indeed have the form in Eq. (2.74)with the Green’s functions in Eqs. (4.23)-(4.26). The thermal density matrix from Eq. (4.53) is ρ ( q , q (cid:48) ) = (cid:16) ˙ G ( − iβ ) − iπG ( − iβ ) (cid:17) / e i G ( − iβ ) [ ˙ G ( − iβ )( q + q (cid:48) ) − q q (cid:48) ] . (5.32)Inserting q = q i + (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t )] ,q (cid:48) = q i − (cid:90) t f t dt G ( t − t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] (5.33)gives˙ G ( − iβ )( q + q (cid:48) ) − q q (cid:48) =2 (cid:104) ˙ G ( − iβ ) − (cid:105) (cid:26) q i + q i (cid:90) t f t dt G ( t − t ) [ J ( t ) − J (cid:48) ( t ) − J (cid:48)(cid:48) ( t )+ J (cid:48)(cid:48)(cid:48) ( t )] (cid:27) + E ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) , (5.34)53here E ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) (cid:110) ˙ G ( − iβ ) [( J ( t ) − J (cid:48) ( t )) ( J ( t (cid:48) ) − J (cid:48) ( t (cid:48) ))+ ( J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )) ( J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) ))] + 2 [ J ( t ) − J (cid:48) ( t )] [ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] (cid:111) . (5.35)The generating function is Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:16) ˙ G ( − iβ ) − iπG ( − iβ ) (cid:17) / e i G ( T ) D e i G ( − iβ ) E I , (5.36)where the integral to be done is I = (cid:90) ∞−∞ dq i e iG ( − iβ ) [ ˙ G ( − iβ ) − (cid:110) q i + q i (cid:82) tft dt G ( t − t )[ J ( t ) − J (cid:48) ( t ) − J (cid:48)(cid:48) ( t )+ J (cid:48)(cid:48)(cid:48) ( t )] (cid:111) × e iG ( T ) q i (cid:82) tft dt [ ˙ G ( T ) G ( t − t )+ G ( t f − t )][ J ( t ) − J (cid:48) ( t )+ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] . (5.37)I will complete the square as before, defining the thermal version of α ( t ) as α ( t ) ≡ G ( t − t ) − (cid:18) G ( − iβ )˙ G ( − iβ ) − (cid:19) (cid:32) ˙ G ( T ) G ( t − t ) + G ( t f − t ) G ( T ) (cid:33) . (5.38)Note that G ( − iβ )˙ G ( − iβ ) − = − im coth( mβ ) is imaginary, so indeed I retain the structure j ≡ (cid:90) t f t dt { α ( t ) ∗ [ J ( t ) − J (cid:48) ( t )] − α ( t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] } (5.39)from the zero-temperature case. Completing the square gives I = (cid:16) iπG ( − iβ )˙ G ( − iβ ) − (cid:17) / e − i ˙ G ( − iβ ) − G ( − iβ ) j . (5.40)So the generating function has the form Z = e i Φ with generalized influence phaseΦ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = D G ( T ) + E G ( − iβ ) − (cid:32) ˙ G ( − iβ ) − G ( − iβ ) (cid:33) j . (5.41) Simplification happens largely as before. This time I have used Mathematica to simplifyeverything and will not pretend otherwise. But there is still something to say. As an example,I will focus on the
J J term, which ends up beingΦ JJ term = i m sinh (cid:0) mβ (cid:1) (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:26) Θ( t − t (cid:48) ) cos (cid:20) m (cid:18) t − t (cid:48) + i β (cid:19)(cid:21) +Θ( t (cid:48) − t ) cos (cid:20) m (cid:18) t − t (cid:48) − i β (cid:19)(cid:21)(cid:27) J ( t ) J ( t (cid:48) ) . (5.42)54onsider the zero-temperature limit:cos[ m ( t − t (cid:48) + i β )]sinh( mβ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) β →∞ = e − im ( t − t (cid:48) ) . (5.43)Subtracting that from the nonzero-temperature expression gives a Bose-Einstein factor:cos[ m ( t − t (cid:48) + i β )]sinh( mβ ) − e − im ( t − t (cid:48) ) = 2 e βm − m ( t − t (cid:48) )] . (5.44)That is real and symmetric in t and t (cid:48) . Subtracting the zero-temperature contribution fromthe J J term givesΦ JJ term − Φ JJ term | β = ∞ = 1 e βm − (cid:90) t f t dt (cid:90) t f t dt (cid:48) i m cos [ m ( t − t (cid:48) )] J ( t ) J ( t (cid:48) ) . (5.45)Repeating that process for everything will give the final result for the generalized influencephase:Φ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = Φ ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) + 1 e βm − (cid:90) t f t dt (cid:90) t f t dt (cid:48) i m cos [ m ( t − t (cid:48) )] ×{ [ J ( t ) − J (cid:48) ( t )+ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] } . (5.46) Expectation values, expectation values, expectation values. That, not dynamics, is what thisis about. Expectation values.Understand?Congratulations, you are ready for a dynamical problem. At some initial time, I changethe oscillator frequency instantaneously from one value to another. The Hamiltonian will beˆ H ( t ) = ˆ p + m ( t ) ˆ q , m ( t ) = (cid:26) m , t < t m , t > t . (6.1)Okay, I lied: Even that is not really dynamical. Written that way, the Hamiltonian changesin time. But the perspective offered by the generating function is that Eq. (6.1) merely de-scribes evolution with the t > t constant-frequency Hamiltonian—it is just that the systemis in an excited state.A genuinely dynamical version of this idea would be to take a smooth profile in m ( t ) thatapproaches constant values at t and t f ; for an example over an infinite interval, see Jordan[40]. I will content myself with Eq. (6.1). 55 .1 Operator calculations Before diving into the path-integral calculation, it is worth reviewing the operator setup.
For t > t , the Hamiltonian isˆ H ( t > t ) = ˆ p + m ˆ q ≡ ˆ H out . (6.2)In terms of ˆ a out ≡ (cid:114) m (cid:18) ˆ q + im ˆ p (cid:19) , (6.3)the Hamiltonian is ˆ H out = m (cid:16) ˆ a † out ˆ a out + ˆ1 (cid:17) . (6.4)For t < t , the Hamiltonian is ˆ H ( t < t ) = ˆ p + m ˆ q ≡ ˆ H in . (6.5)In terms of ˆ a in ≡ (cid:114) m (cid:18) ˆ q + im ˆ p (cid:19) , (6.6)the Hamiltonian is ˆ H in = m (cid:16) ˆ a † in a in + ˆ1 (cid:17) . (6.7)When m (cid:54) = m , the operators ˆ a out and ˆ a in are different. Concretely, for complex numbers A and B , the operators can be expressed in terms of each other asˆ a out ≡ A ˆ a in + B ˆ a † in . (6.8)To solve for A and B , introduce the ground states | out (cid:105) and | in (cid:105) . They satisfyˆ a out | out (cid:105) = 0 , ˆ a in | in (cid:105) = 0 , (cid:104) out | a † out = 0 , (cid:104) in | a † in = 0 . (6.9)Therefore: ˆ a out | in (cid:105) = B ˆ a † in | in (cid:105) , (cid:104) in | ˆ a out = A (cid:104) in | ˆ a in . (6.10)In terms of the 1-particle states | out (cid:105) ≡ ˆ a † out | out (cid:105) , | in (cid:105) ≡ ˆ a † in | in (cid:105) , (6.11)those imply B = (cid:104) in | ˆ a out | in (cid:105) , A = (cid:104) in | ˆ a out | in (cid:105) . (6.12)56ut those are just formal expressions, so let me derive what they mean. To begin, I willreview how to solve the condition ˆ a out | out (cid:105) = 0 for the wavefunction (cid:104) q | out (cid:105) . Acting on theequation ˆ a out | out (cid:105) = 0 from the left with (cid:104) q | gives (cid:104) q | ˆ a out | out (cid:105) = (cid:114) m (cid:104) q | ˆ q + im ˆ p | out (cid:105) = (cid:114) m (cid:18) q + 1 m ∂ q (cid:19) (cid:104) q | out (cid:105) = 0 . (6.13)The solution is (cid:104) q | out (cid:105) = C e − mq , (6.14)with C fixed by demanding (cid:104) out | out (cid:105) = 1: (cid:104) out | out (cid:105) = (cid:90) ∞−∞ dq (cid:104) out | q (cid:105)(cid:104) q | out (cid:105) = | C | (cid:90) ∞−∞ dq e − mq = | C | (cid:114) πm ≡ ⇒ | C | = (cid:16) mπ (cid:17) / . (6.15)The overall phase in Eq. (6.14) is arbitrary; I will choose C = | C | .The wavefunction for the 1-particle state is (cid:104) q | out (cid:105) = (cid:104) q | ˆ a † out | out (cid:105) = (cid:114) m (cid:18) q − m ∂ q (cid:19) (cid:104) q | out (cid:105) = √ m q Ce − mq . (6.16)What I need are the “in” wavefunctions, which are (cid:104) q | in (cid:105) = (cid:16) m π (cid:17) / e − m q , (cid:104) q | in (cid:105) = (cid:16) m π (cid:17) / √ m q e − m q . (6.17)Therefore, the coefficient A is A = (cid:104) in | ˆ a out | in (cid:105) = (cid:90) ∞−∞ dq (cid:104) in | q (cid:105) (cid:114) m (cid:18) q + 1 m ∂ q (cid:19) (cid:104) q | in (cid:105) = (cid:114) m (cid:114) m π √ m (cid:90) ∞−∞ dq e − m q (cid:18) q + 1 m ∂ q (cid:19) (cid:18) q e − m q (cid:19) = m √ πm (cid:90) ∞−∞ dq (cid:2) ( m − m ) q + 1 (cid:3) e − m q = m √ πm [ − ( m − m ) ∂ m + 1] (cid:90) ∞−∞ dq e − m q = m √ m [ − ( m − m ) ∂ m + 1] m − / = m √ m (cid:34) m − m m / + 1 m / (cid:35) = 12 √ mm [ m − m + 2 m ]= m + m √ mm . (6.18)By an analogous calculation, the coefficient B is B = m − m √ mm . (6.19)That concludes baby’s first Bogoliubov transformation. The coefficients satisfy | A | − | B | =1; for m → m , they reduce to A → B → B . Return to Eq. (6.10) andconsider the magnitude-squared of ˆ a out | in (cid:105) . Since the number operator for “out” modes isˆ N out ≡ ˆ a † out ˆ a out , the average number of quanta with frequency m in the ground state of theoscillator with frequency m is (cid:104) in | ˆ N out | in (cid:105) = | B | . (6.20) The Heisenberg-picture field operator for t > t isˆ q ( t > t ) = 1 √ m (cid:16) e − imt ˆ a out + e imt ˆ a † out (cid:17) . (6.21)Inserting a out = A ˆ a in + B ˆ a † in givesˆ q (0) | in (cid:105) = 1 √ m ( A ∗ + B ) | in (cid:105) , (cid:104) in | ˆ q ( t ) = 1 √ m (cid:0) e − imt A + e imt B ∗ (cid:1) (cid:104) in | . (6.22)With the results for A and B I find e imt A ∗ + e − imt B = 1 √ mm [ m cos( mt ) + im sin( mt )] . (6.23)In terms of my old friend G ( t ) ≡ m sin( mt ), I define ζ ( t ) ≡ ˙ G ( t ) + im G ( t ) , (6.24)in terms of which ˆ q ( t > t ) = 1 √ m (cid:104) ζ ( t ) ∗ ˆ a in + ζ ( t ) ˆ a † in (cid:105) . (6.25)Eqs. (6.25) and (6.21) form a poignant expression of the Bogoliobov transformation. t = 0As somewhat of a warmup but mainly to make a point, I will briefly set t = 0 and calculatea few 2-point correlation functions.With Eqs. (6.21) and (6.25), I can calculate the following expectation value: (cid:104) in | ˆ q ( t )ˆ q (0) | in (cid:105) = 14 mm (cid:2) e − imt ( m + m ) + e imt ( m − m ) (cid:3) . (6.26)I can also calculate the following transition amplitude: (cid:104) out | ˆ q ( t )ˆ q (0) | in (cid:105) = √ mm ) / ( m + m ) / e − imt . (6.27)58 can also calculate, for reference, the “out-out” version of Eq. (6.26): (cid:104) out | ˆ q ( t )ˆ q (0) | out (cid:105) = 12 m e − imt , (6.28)consistent with the greater Green’s function from Eq. (4.23).When m → m , all of those will reduce to the same thing. When m (cid:54) = m , I know ofno innate reason to prefer the expectation value over the transition amplitude. If the physicscalls for an expectation value (whether “in-in” or “out-out”), calculate an expectation value;otherwise, calculate a transition amplitude. The choice is yours. t (cid:54) = 0What the Schwinger-Keldysh path integral calculates isTr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H out ˆ ρ e − it ˆ H out (cid:17) , (6.29)with t , > t . The physical setup is to put the system into the pre-quench ground state,ˆ ρ = | in (cid:105)(cid:104) in | , (6.30)and evolve the fields using the post-quench Hamiltonian, ˆ H out . The mathematical setupshould be general enough to admit a limit t → −∞ , which is what is always done whenimplementing an iε prescription. I will take t negative but keep it finite.I will calculate Eq. (6.29) in two ways: Easy and Hard.The easy way is to first calculate | ψ ( t ) (cid:105) ≡ ˆ q ( t ) e it ˆ H out | in (cid:105) = e it ˆ H out e − it ˆ H out ˆ q ( t ) e it ˆ H out | in (cid:105) = e it ˆ H out ˆ q ( t − t ) | in (cid:105) , (6.31)where the last equality follows from t > t . Using Eq. (6.25), I find | ψ ( t ) (cid:105) = e it ˆ H out √ m ζ ( t − t ) | in (cid:105) . (6.32)Therefore, Eq. (6.29) becomes:Tr (cid:16) ˆ q ( t )ˆ q ( t ) e it ˆ H out | in (cid:105)(cid:104) in | e − it ˆ H out (cid:17) = (cid:104) in | e − it ˆ H out ˆ q ( t )ˆ q ( t ) e it ˆ H out | in (cid:105) = (cid:104) ψ ( t ) | ψ ( t ) (cid:105) = 12 m ζ ( t − t ) ζ ( t − t ) ∗ . (6.33)For m = m , ζ ( t ) = e imt , and I recover the translationally invariant result G < ( t, t (cid:48) ) = i m e im ( t − t (cid:48) ) = G < ( t − t (cid:48) ). Stage cleared. Not a state, just shorthand. | n out (cid:105) ≡ √ n ! (ˆ a † out ) n | out (cid:105) , (6.34)the correlation function can be written as follows:Tr (cid:0) ˆ q ( t )ˆ q ( t ) e it H out ˆ ρ e − it H out (cid:1) = ∞ (cid:88) n = 0 (cid:104) n out | ˆ q ( t )ˆ q ( t ) e it H out ˆ ρ e − it H out | n out (cid:105) = ∞ (cid:88) n = 0 e it E out n e − it E out n (cid:104) n out | ˆ qe − it ˆ H out ˆ q ( t ) e it ˆ H out ˆ ρ | n out (cid:105) = ∞ (cid:88) n,n (cid:48) ,n (cid:48)(cid:48) = 0 e it E out n e − it E out n e − it E out n (cid:48)(cid:48) e it E out n (cid:48) (cid:104) n out | ˆ q | n (cid:48)(cid:48) out (cid:105) (cid:104) n (cid:48)(cid:48) out | ˆ q ( t ) | n (cid:48) out (cid:105) (cid:124) (cid:123)(cid:122) (cid:125) e it E out n (cid:48)(cid:48) − E out n (cid:48) ) (cid:104) n (cid:48)(cid:48) out | ˆ q | n (cid:48) out (cid:105) (cid:104) n (cid:48) out | ˆ ρ | n out (cid:105) = ∞ (cid:88) n,n (cid:48) ,n (cid:48)(cid:48) = 0 e it ( E out n − E out n (cid:48)(cid:48) )+ it ( E out n (cid:48)(cid:48) − E out n (cid:48) )+ it ( E out n (cid:48) − E out n ) (cid:104) n out | ˆ q | n (cid:48)(cid:48) out (cid:105)(cid:104) n (cid:48)(cid:48) out | ˆ q | n (cid:48) out (cid:105)(cid:104) n (cid:48) out | ˆ ρ | n out (cid:105) . (6.35)Since t <
0, the ˆ q appearing above is defined in terms of ˆ a out :ˆ q = ˆ q (0) = 1 √ m (cid:16) ˆ a out + ˆ a † out (cid:17) . (6.36)So the matrix elements of the field operator are the usual sort of thing: (cid:104) ˆ n out | ˆ q | n (cid:48)(cid:48) out (cid:105) = 1 √ m (cid:16) √ n (cid:48)(cid:48) δ n,n (cid:48)(cid:48) − + √ n (cid:48)(cid:48) +1 δ n,n (cid:48)(cid:48) +1 (cid:17) , (cid:104) ˆ n (cid:48)(cid:48) out | ˆ q | n (cid:48) out (cid:105) = 1 √ m (cid:16) √ n (cid:48) δ n (cid:48)(cid:48) ,n (cid:48) − + √ n (cid:48) +1 δ n (cid:48)(cid:48) ,n (cid:48) +1 (cid:17) . (6.37)The density matrix in Eq. (6.30) has matrix elements (cid:104) n (cid:48) out | ˆ ρ | n out (cid:105) = (cid:104) n (cid:48) out | in (cid:105)(cid:104) in | n out (cid:105) . (6.38)Calculating the overlap (cid:104) in | n out (cid:105) is the hard part. In terms of the Hermite polynomials H n ( x ) ≡ ( − n e x ∂ nx ( e − x ) , (6.39)the wavefunction for the n th excited state is (cid:104) q | n out (cid:105) = 1 √ n n ! (cid:16) mπ (cid:17) / e − mq H n ( √ m q ) . (6.40)60o the overlap I need is (cid:104) in | n out (cid:105) = ( mm ) / √ π √ n n ! (cid:90) ∞−∞ dq e −
12 ( m + m ) q H n ( √ m q ) . (6.41)Because the Hermite polynomials have definite parity, that overlap is nonzero only for even n . For n = 0 , , , ,
8, Mathematica provides the ratios r n ≡ (cid:104) in | n out (cid:105)(cid:104) in | out (cid:105) = (cid:32) , √ (cid:18) m − m m + m (cid:19) , (cid:114) (cid:18) m − m m + m (cid:19) , (cid:114) (cid:18) m − m m + m (cid:19) , (cid:114) (cid:18) m − m m + m (cid:19) (cid:33) . (6.42)Apparently those are related to the central binomial coefficients, leading to the closed-formexpression r n = (cid:32) n !2 n (cid:2)(cid:0) n (cid:1) ! (cid:3) (cid:33) / (cid:18) m − m m + m (cid:19) n/ , n = 0 , , , , , ... . (6.43)In terms of that, the density-matrix elements are (cid:104) n (cid:48) out | ˆ ρ | n out (cid:105) = (cid:26) |(cid:104) in | out (cid:105)| r n (cid:48) r n for n and n (cid:48) even0 otherwise . (6.44)The overlap between ground states is (cid:104) in | out (cid:105) = (cid:114) m + m ( mm ) / . (6.45)With that, the rest is just algebra and shifting summation indices. One thing I will note is r k +2 = (cid:115) k + 12( k + 1) (cid:18) m − m m + m (cid:19) r k . (6.46)I will need the following two infinite sums: S ≡ ∞ (cid:88) k = 0 k r k = ( m − m ) ( m + m )8( mm ) / , S ≡ ∞ (cid:88) k = 0 (2 k + 1) r k = ( m + m ) mm ) / . (6.47)Putting everything together eventually givesTr (cid:0) ˆ q ( t )ˆ q ( t ) e it H out ˆ ρ e − it H out (cid:1) = 18 m m (cid:8) ( m + m ) e − im ( t − t ) + ( m − m ) e im ( t − t ) +2( m − m ) cos[ m ( t + t − t )] (cid:9) . (6.48)When t = t and t = t = 0, that reduces to Eq. (6.26). For general t , t , and t , Isimplify Eq. (6.48) using ζ ( t ) from Eq. (6.24) by working with τ , ≡ m ( t , − t ) and usingcos( τ − τ ) = cos( τ ) cos( τ ) + sin( τ ) sin( τ ) and sin( τ − τ ) = sin( τ ) cos( τ ) − cos( τ ) sin( τ ).I will once again arrive at Eq. (6.33), which this time I will express in terms of the lesserGreen’s function: G < ( t, t (cid:48) ) = i Tr (cid:16) ˆ q ( t )ˆ q ( t ) e i ˆ Ht ˆ ρ e − i ˆ Ht (cid:17) = i m ζ ( t − t ) ζ ( t (cid:48) − t ) ∗ . (6.49)Was hard mode pointless? Probably. I thank Jacob Lin for figuring that out. .1.5 Out-of-time-ordered correlator Motivated by the steps leading to Eq. (6.33), I will consider the following object: | ψ ( t , t ) (cid:105) ≡ ˆ q ( t )ˆ q ( t ) e it ˆ H out | in (cid:105) . (6.50)Inserting factors of ˆ1 = e it ˆ H out e − it ˆ H out before and after ˆ q ( t ), recalling that t , > t , andusing Eq. (6.25), I find | ψ ( t , t ) (cid:105) = e it ˆ H out ζ ( t − t )2 m (cid:104) ζ ( t − t ) ∗ | in (cid:105) + √ ζ ( t − t ) | in (cid:105) (cid:105) . (6.51)Therefore: Tr (cid:16) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) )ˆ q ( t )ˆ q ( t ) e it ˆ H out | in (cid:105)(cid:104) in | e − it ˆ H out (cid:17) = (cid:104) ψ ( t (cid:48) , t (cid:48) ) | ψ ( t , t ) (cid:105) = ζ ( t − t ) ζ ( t (cid:48) − t ) ∗ (2 m ) (cid:104) ζ ( t (cid:48) − t ) (cid:104) in | + √ ζ ( t (cid:48) − t ) ∗ (cid:104) in | (cid:105) (cid:104) ζ ( t − t ) ∗ | in (cid:105) + √ ζ ( t − t ) | in (cid:105) (cid:105) = ζ ( t − t ) ζ ( t (cid:48) − t ) ∗ m [ ζ ( t (cid:48) − t ) ζ ( t − t ) ∗ + 2 ζ ( t (cid:48) − t ) ∗ ζ ( t − t )] . (6.52)From Eq. (6.49), I know that ζ ( t − t ) ζ ( t (cid:48) − t ) = − im G < ( t, t (cid:48) ). That factor of 2 in thesecond term of Eq. (6.52) suggests that I write 2 = 1 + 1 and pair the ζ and ζ ∗ s in twodifferent ways to get two different product of G < s. So I get:Tr (cid:16) ˆ q ( t (cid:48) )ˆ q ( t (cid:48) )ˆ q ( t )ˆ q ( t ) e it ˆ H out | in (cid:105)(cid:104) in | e − it ˆ H out (cid:17) = − [ G < ( t , t ) G < ( t (cid:48) , t (cid:48) ) + G < ( t , t (cid:48) ) G < ( t , t (cid:48) ) + G < ( t , t (cid:48) ) G < ( t , t (cid:48) )] , (6.53)consistent with Eq. (2.75). After that recapitulation of the operator formalism, I am ready to evaluate the path integral.
Given the “in” wavefunction (cid:104) q | in (cid:105) = (cid:0) m π (cid:1) / e − m q from Eq. (6.17) and the density matrixˆ ρ = | in (cid:105)(cid:104) in | , I express the generating function in its factorized form: Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f Ψ( q f | J ) Ψ( q f | J (cid:48) ) ∗ , (6.54) I am using t (cid:48) = t and t (cid:48) = t this time because the first line expresses the “scattering matrix” interpre-tation of the OTOC [41, 42]. q f | J ) ≡ (cid:90) ∞−∞ dq (cid:104) q | in (cid:105) Z ( q , q f | J )= (cid:0) m π (cid:1) / √ πiG ( T ) (cid:90) ∞−∞ dq e i [ im q + S cl ( q ,q f | J ) ]= (cid:0) m π (cid:1) / √ πiG ( T ) e i G ( T ) (cid:104) ˙ G ( T ) q f +2 q f (cid:82) tft dt G ( t − t ) J ( t ) − (cid:82) tft dt (cid:82) tft dt (cid:48) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) J ( t ) J ( t (cid:48) ) (cid:105) χ ( q f | J ) , (6.55)where [again, let ζ ( t ) ≡ ˙ G ( t ) + im G ( t ), as in Eq. (6.24)]: χ ( q f | J ) ≡ (cid:90) ∞−∞ dq e i G ( T ) (cid:110) [ ˙ G ( T )+ im G ( T ) ] q +2 q (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105) (cid:111) = (cid:90) ∞−∞ dq e iζ ( T )2 G ( T ) (cid:110) q + 2 q ζ ( T ) − (cid:20)(cid:90) t f t dt G ( t f − t ) J ( t ) − q f (cid:21) (cid:111)(cid:124) (cid:123)(cid:122) (cid:125) (cid:18) q ζ ( T ) − (cid:20)(cid:82) tft dt G ( tf − t ) J ( t ) − qf (cid:21)(cid:19) − ζ ( T ) − (cid:20)(cid:82) tft dt G ( tf − t ) J ( t ) − qf (cid:21) = (cid:112) πiG ( T ) ζ ( T ) − e − iζ ( T ) − G ( T ) (cid:104)(cid:82) tft dt G ( t f − t ) J ( t ) − q f (cid:105) . (6.56) Integrating over the initial field configurations leaves me with Z ( J, J (cid:48) ) = (cid:0) m π (cid:1) / | ζ ( T ) | (cid:90) ∞−∞ dq f e i G ( T ) I ( q f | J,J (cid:48) ) , (6.57)where I ( q f | J, J (cid:48) ) = (cid:2) − ζ ( T ) − + ζ ( T ) ∗− (cid:3) q f + 2 q f (cid:90) t f t dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] − (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:8) [2Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + ζ ( T ) − G ( t f − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) − [2Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + ζ ( T ) ∗− G ( t f − t ) G ( t f − t (cid:48) )] J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:9) , (6.58)with α ( t ) = G ( t − t ) + ζ ( T ) ∗− G ( t f − t ) . (6.59)Since ζ ( T ) = ˙ G ( T ) + im G ( T ), I find − ζ ( T ) − + ζ ( T ) ∗− = 2 im G ( T ) | ζ ( T ) | , (6.60)which is good, because that is the factor required from the Gaussian integral to cancel outthe overall factor in Z ( J, J (cid:48) ). With j ( J, J (cid:48) ) ≡ (cid:82) t f t dt [ α ( t ) ∗ J ( t ) − α ( t ) J (cid:48) ( t )] as usual, I find (cid:90) ∞−∞ dq f e i G ( T ) { [ − ζ ( T ) − + ζ ( T ) ∗− ] q f +2 q f j } = (cid:113) π | ζ ( T ) | m e − | ζ ( T ) | m G ( T )2 j . (6.61)63o the influence phase Φ( J, J (cid:48) ) ≡ − i ln Z ( J, J (cid:48) ) isΦ(
J, J (cid:48) ) = 12 G ( T ) (cid:26) i | ζ ( T ) | m G ( T ) j ( J, J (cid:48) ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:2) t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + ζ ( T ) − G ( t f − t ) G ( t f − t (cid:48) ) (cid:3) J ( t ) J ( t (cid:48) )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:2) t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + ζ ( T ) ∗− G ( t f − t ) G ( t f − t (cid:48) ) (cid:3) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (cid:27) . (6.62)That should (and does) reduce to Eq. (4.13) when m = m . Mathematica did not comply when I asked it to simplify Eq. (6.62), so I will reorganize thatexpression and try, try again. Consider the
J J term:Φ JJ term = 12 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ R ( t, t (cid:48) ) + iS ( t, t (cid:48) )] J ( t ) J ( t (cid:48) ) , (6.63)with R + iS written in its symmetrized form: R ( t, t (cid:48) ) + iS ( t, t (cid:48) ) ≡ i | ζ ( T ) | m G ( T ) α ( t ) ∗ α ( t (cid:48) ) ∗ − (cid:2) Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t f − t (cid:48) ) G ( t − t ) + ζ ( T ) − G ( t f − t ) G ( t f − t (cid:48) ) (cid:3) . (6.64)After some uninteresting algebra, I obtain the following form of the imaginary part: S ( t, t (cid:48) ) = 12 m G ( T ) (cid:8) | ζ ( T ) | G ( t − t ) G ( t (cid:48) − t ) + G ( t f − t ) G ( t f − t (cid:48) )+ ˙ G ( T ) [ G ( t − t ) G ( t f − t (cid:48) ) + G ( t f − t ) G ( t (cid:48) − t )] (cid:111) . (6.65)Even in that form, Mathematica remained obstinate unless I first told it to set m = 1. Atany rate, the simplified form is S ( t, t (cid:48) ) = G ( T )4 m (cid:26)(cid:18) m m (cid:19) cos[ m ( t − t (cid:48) )] + (cid:18) − m m (cid:19) cos[ m ( t + t (cid:48) − t )] (cid:27) . (6.66)The real part turns out to be much simpler: R ( t, t (cid:48) ) = G ( T )2 [Θ( t − t (cid:48) ) G ( t − t (cid:48) ) + Θ( t (cid:48) − t ) G ( t (cid:48) − t )] = G ( T ) G ( | t − t (cid:48) | ) . (6.67) As opposed to the rest of it, right? See, I did exercise editorial discretion. R ( t, t (cid:48) ) + iS ( t, t (cid:48) ) = G ( T )2 (cid:26) G ( | t − t (cid:48) | ) + i m (cid:20)(cid:18) m m (cid:19) ˙ G ( t − t (cid:48) ) + (cid:18) − m m (cid:19) ˙ G ( t + t (cid:48) − t ) (cid:21)(cid:27) . (6.68)When m = m , that correctly reduces to R + iS = G ( T ) i m e − im | t − t (cid:48) | .Expressing the J (cid:48) J (cid:48) term asΦ J (cid:48) J (cid:48) term = 12 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ R (cid:48) ( t, t (cid:48) ) + iS (cid:48) ( t, t (cid:48) )] J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) (6.69)and simplifying along analogous lines, I find that R (cid:48) + iS (cid:48) = − R + iS = − ( R − iS ) = − ( R + iS ) ∗ , (6.70)consistent with Eqs. (2.64)-(2.66).The off-diagonal part is the easiest to simplify:Φ JJ (cid:48) term = − G ( T ) i | ζ ( T ) | m G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ α ( t ) ∗ α ( t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) + α ( t ) α ( t (cid:48) ) ∗ J (cid:48) ( t ) J ( t (cid:48) )] . (6.71)The coefficient of J ( t ) J (cid:48) ( t (cid:48) ) simplifies to α ( t ) ∗ α ( t (cid:48) ) = G ( T ) | ζ ( T ) | (cid:110) cos[ m ( t − t )] + i m m sin[ m ( t − t )] (cid:111) (cid:110) cos[ m ( t (cid:48) − t )] − i m m sin[ m ( t (cid:48) − t )] (cid:111) . (6.72)So the influence phase has the expected formΦ( J, J (cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ G F ( t, t (cid:48) ) J ( t ) J ( t (cid:48) ) − G D ( t, t (cid:48) ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − G > ( t, t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − G < ( t, t (cid:48) ) J (cid:48) ( t ) J ( t (cid:48) )] , (6.73)with G F ( t, t (cid:48) ) = 12 (cid:26) G ( | t − t (cid:48) | ) + i m (cid:20)(cid:18) m m (cid:19) ˙ G ( t − t (cid:48) ) + (cid:18) − m m (cid:19) ˙ G ( t + t (cid:48) − t ) (cid:21)(cid:27) , (6.74) G D ( t, t (cid:48) ) = G F ( t, t (cid:48) ) ∗ , and G < ( t, t (cid:48) ) = i m (cid:104) ˙ G ( t − t ) + im G ( t − t ) (cid:105) (cid:104) ˙ G ( t (cid:48) − t ) − im G ( t (cid:48) − t ) (cid:105) , (6.75)65 > ( t, t (cid:48) ) = − G < ( t, t (cid:48) ) ∗ . To verify that those G > and G F are consistent, calculate the follow-ing: F ( t, t (cid:48) ) ≡ Θ( t − t (cid:48) ) G > ( t, t (cid:48) ) + Θ( t (cid:48) − t ) G < ( t, t (cid:48) )= i m (cid:110) Θ( t − t (cid:48) ) (cid:104) ˙ G ( t − t ) − im G ( t − t ) (cid:105) (cid:104) ˙ G ( t (cid:48) − t ) + im G ( t (cid:48) − t ) (cid:105) +Θ( t (cid:48) − t ) (cid:104) ˙ G ( t − t ) + im G ( t − t ) (cid:105) (cid:104) ˙ G ( t (cid:48) − t ) − im G ( t (cid:48) − t ) (cid:105)(cid:111) = i m [Θ( t − t (cid:48) ) + Θ( t (cid:48) − t )] (cid:124) (cid:123)(cid:122) (cid:125) = 1 (cid:104) ˙ G ( t − t ) ˙ G ( t (cid:48) − t ) + m G ( t − t ) G ( t (cid:48) − t ) (cid:105) − [Θ( t − t (cid:48) ) − Θ( t (cid:48) − t )] im (cid:104) G ( t − t ) ˙ G ( t (cid:48) − t ) − ˙ G ( t − t ) G ( t (cid:48) − t ) (cid:105)(cid:124) (cid:123)(cid:122) (cid:125) = G ( t − t (cid:48) ) = i m (cid:104) ˙ G ( t − t ) ˙ G ( t (cid:48) − t ) + m G ( t − t ) G ( t (cid:48) − t ) (cid:105) + 12 G ( | t − t (cid:48) | ) . (6.76)The real part is real good. To finish simplifying the imaginary part, use cos( a ± b ) =cos( a ) cos( b ) ∓ sin( a ) sin( b ) with a = t − t and b = t (cid:48) − t to write˙ G ( t − t ) ˙ G ( t (cid:48) − t ) = 12 (cid:104) ˙ G ( t + t (cid:48) − t ) + ˙ G ( t − t (cid:48) ) (cid:105) ,G ( t − t ) G ( t (cid:48) − t ) = − m (cid:104) ˙ G ( t + t (cid:48) − t ) − G ( t − t (cid:48) ) (cid:105) . (6.77)Therefore, ˙ G ( t − t ) ˙ G ( t (cid:48) − t ) + m G ( t − t ) G ( t (cid:48) − t ) =12 (cid:20)(cid:18) m m (cid:19) ˙ G ( t − t (cid:48) ) + (cid:18) − m m (cid:19) ˙ G ( t + t (cid:48) − t ) (cid:21) , (6.78)in which case F ( t, t (cid:48) ) = G F ( t, t (cid:48) ). Win. The intuitive picture I had before calculating the Schwinger-Keldysh path integral for thequenched oscillator is that I should be able to see the field in the “wrong” vacuum absorb oremit particles as it relaxes to the right one. I still do not know how to see that directly fromEqs. (6.73) and (6.75), but I do understand how to recover an averaged version of it. I thank Beatrice Bonga and Justin Wilson for conversations about this. H ( t > t ) = [ ∂ t ˆ q ( t )] + m ˆ q ( t ) . (6.79)The average energy in the field is then: E ( t ) ≡ Tr (cid:104) ˆ H ( t ) e it ˆ H out ˆ ρ in e − it ˆ H out (cid:105) = lim t (cid:48) → t (cid:0) ∂ t ∂ t (cid:48) + m (cid:1) Tr (cid:104) ˆ q ( t (cid:48) )ˆ q ( t ) e it ˆ H out ˆ ρ in e − it ˆ H out (cid:105) = lim t (cid:48) → t (cid:0) ∂ t ∂ t (cid:48) + m (cid:1) (cid:20) − i δ δJ ( t ) δJ (cid:48) ( t (cid:48) ) Φ( J, J (cid:48) ) (cid:21) = − i lim t (cid:48) → t (cid:0) ∂ t ∂ t (cid:48) + m (cid:1) G < ( t, t (cid:48) ) , (6.80)with G < ( t, t (cid:48) ) given in Eq. (6.75). The result of taking those derivatives and setting t (cid:48) = t is E = m + m m . (6.81)The average energy in the field remains constant, which is one manifestation of the commentI made below Eq. (6.1). To interpret Eq. (6.81) further, recall the Bogoliubov coefficients A and B from Sec. 6.1.1: A = m + m √ mm , B = m − m √ mm . (6.82)The energy in Eq. (6.81) therefore has the form E = (cid:0) | A | + | B | (cid:1) m (cid:0) | B | + (cid:1) m . (6.83)That makes a lot of sense: The is the contribution of the post-quench ground state, and | B | is the average number of post-quench quanta contained in the pre-quench ground state[recall Eq. (6.81)].It is also interesting to take limits of Eq. (6.81): E → m for m → m for m → m m m for m → . (6.84)The first case shows that sending a particle into the void releases half of the ground-stateenergy; the second case is a sanity check, recovering the full ground-state energy. The thirdcase shows that trying to confine an almost-free particle requires almost infinite energy.Poetic. 67 .3 Larkin-Ovchinnikov path integral A ground-state density matrix ˆ ρ = | in (cid:105)(cid:104) in | (6.85)will give the same ρ ( q , q (cid:48) ) as for the unquenched case except for the replacement m → m : ρ ( q , q (cid:48) ) = (cid:16) m π (cid:17) / e − m (cid:110) q i + q i (cid:82) tft dt G ( t − t )[ J ( t ) − J (cid:48) ( t ) − J (cid:48)(cid:48) ( t )+ J (cid:48)(cid:48)(cid:48) ( t )] (cid:111) × e − m (cid:82) tft dt (cid:82) tft dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) { [ J ( t ) − J (cid:48) ( t )][ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )]+[ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )][ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] } . (6.86)Completing the square and so on will proceed along the same lines as for the unquenchedcase, leading to the same kind of j = (cid:82) t f t dt { α ( t ) ∗ [ J ( t ) − J (cid:48) ( t )] − α ( t ) [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] } butwith a modified α ( t ): α ( t ) ≡ G ( t − t ) + im G ( T ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) . (6.87)The preliminary result is Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = e i Φ( J,J (cid:48) ,J (cid:48)(cid:48) ,J (cid:48)(cid:48)(cid:48) ) withΦ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = 12 G ( T ) D ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) − i m j ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) + i m (cid:90) t f t dt (cid:90) t f t dt (cid:48) G ( t − t ) G ( t (cid:48) − t ) (cid:110) [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )]+ [ J (cid:48)(cid:48) ( t ) − J (cid:48)(cid:48)(cid:48) ( t )] [ J (cid:48)(cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] (cid:111) . (6.88)Now I ask Mathematica to simplify each term. First, observe that α ( t ) ∗ α ( t (cid:48) ) = 1 m m (cid:8) m cos[ m ( t − t )] + im sin[ m ( t − t )] (cid:9) × (cid:8) m cos[ m ( t (cid:48) − t )] − im sin[ m ( t (cid:48) − t )] (cid:9) , (6.90)which correctly reduces to Eq. (5.22) when m = m . The final time t f was arbitrary andappropriately dropped out. But for this case, the initial time t is physical: It is the time atwhich the oscillator’s frequency changed. The problem is not translationally invariant, andthe resulting correlation functions should depend on t .Anticipating the final form of the influence phase, I define the functions G > ( t, t (cid:48) ) ≡ i m m (cid:8) m cos[ m ( t − t )] − im sin[ m ( t − t )] (cid:9) × (cid:8) m cos[ m ( t (cid:48) − t )] + im sin[ m ( t (cid:48) − t )] (cid:9) , (6.91) Mathematica required some coaxing to put the result into that form, at least as of version 12.0 on macOS10.15.3. I had to first collect terms in m and m , then work with the ratio m/m , then simplify. Concretely,after defining functions α [ t ] and α Star[ t ] for Eq. (6.87) and its conjugate, I wrote:Collect[ α Star[t] α [tp], { m,m0 } , FullSimplify] /. { m → µ m0 } // FullSimplify /. { µ → mm0 } // FullSimplify(6.89) G < ( t, t (cid:48) ) ≡ i m m (cid:8) m cos[ m ( t − t )] + im sin[ m ( t − t )] (cid:9) × (cid:8) m cos[ m ( t (cid:48) − t )] − im sin[ m ( t (cid:48) − t )] (cid:9) = G > ( t (cid:48) , t ) = − G > ( t, t (cid:48) ) ∗ . (6.92)Given those two, I further define G F ( t, t (cid:48) ) ≡ Θ( t − t (cid:48) ) G > ( t, t (cid:48) ) + Θ( t (cid:48) − t ) G < ( t, t (cid:48) ) , (6.93)and G D ( t, t (cid:48) ) ≡ − [Θ( t − t (cid:48) ) G < ( t, t (cid:48) ) + Θ( t (cid:48) − t ) G > ( t, t (cid:48) )] = G F ( t, t (cid:48) ) ∗ . (6.94)In terms of those, the generalized influence phase will look exactly like Eq. (2.74), which Iwill repeat using the now-required translationally noninvariant notation:Φ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:8) G F ( t, t (cid:48) ) [ J ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )] − G D ( t, t (cid:48) ) [ J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − G < ( t, t (cid:48) ) [ J ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − G > ( t, t (cid:48) ) [ J (cid:48) ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )]+ G < ( t, t (cid:48) ) [ J ( t ) J (cid:48)(cid:48) ( t (cid:48) ) + J (cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) − J ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )]+ G > ( t, t (cid:48) ) [ J (cid:48)(cid:48) ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48) ( t ) J (cid:48) ( t (cid:48) )] (cid:9) . (6.95)Long live the quench. 69 Open systems
Drainage! Drainage, Eli, you boy.—Daniel Plainview,
There Will Be Blood
The most pedestrian way to model the decay of an excited state is to add a small imagi-nary part to the Hamiltonian. For example, if a single-particle wavefunction ψ ( (cid:126)x ) were aneigenstate of a nonhermitian Hamiltonianˆ H = ˆ H − i ˆΓ (7.1)with complex energy E = E − i Γ, then the wavefunction at later times would be ψ ( t, (cid:126)x ) = ψ ( (cid:126)x ) e − i E t . The probability density would be | ψ ( t, (cid:126)x ) | = | ψ ( (cid:126)x ) | e − Γ t , and the state woulddecay with a lifetime Γ − . Drainage.But unlike oil prospectors, quantum subsystems return what they take. Consider the densitymatrix in a basis of instantaneous field eigenstates: ˆ ρ ( t ) ≡ e + i ˆ Ht ˆ ρ e − i ˆ Ht . (7.2)That time-dependent operator satisfies the differential equation ∂ t ˆ ρ ( t ) = i ( ˆ H ˆ ρ ( t ) − ˆ ρ ( t ) ˆ H ) . (7.3)Including decay as above would entail replacing the right-hand side of Eq. (7.3) by ∂ t ˆ ρ ( t ) = i (cid:16) ˆ H ˆ ρ ( t ) − ˆ ρ ( t ) ˆ H † (cid:17) . (7.4)As I said, this approach does not conserve probability. So let me try to compensate the lossof probability by transforming ˆ ρ with some other operator ˆ L while preserving ˆ ρ ( t ) † = ˆ ρ ( t ): ∂ t ˆ ρ ( t ) = i (cid:16) ˆ H ˆ ρ ( t ) − ˆ ρ ( t ) ˆ H † (cid:17) − ˆ L ˆ ρ ( t ) ˆ L † . (7.5)Imposing conservation of probability in the form ∂ t Tr[ ˆ ρ ( t )] ≡ L to Im( ˆ H ):0 = i Tr (cid:16) [ ˆ H, ˆ ρ ( t )] − i { ˆΓ , ˆ ρ ( t ) } (cid:17) − Tr (cid:16) ˆ L ˆ ρ ( t ) ˆ L † (cid:17) = Tr (cid:16)(cid:16) ˆΓ − ˆ L † ˆ L (cid:17) ˆ ρ ( t ) (cid:17) . (7.6)The goal is to conserve probability for any density matrix, so the above impliesˆΓ = ˆ L † ˆ L . (7.7) Notwithstanding some earlier footnotes.
70 included only a single ˆ L to streamline the presentation. Eq. (7.5) could be generalized to ∂ t ˆ ρ ( t ) = i (cid:16) ˆ H ˆ ρ ( t ) − ˆ ρ ( t ) ˆ H † (cid:17) − N L (cid:88) I = 1 ˆ L I ˆ ρ ( t ) ˆ L † I , (7.8)in which case imposing ∂ t Tr[ ˆ ρ ( t )] = 0 would implyˆΓ = N L (cid:88) I = 1 ˆ L † I ˆ L I . (7.9)The evolution law in Eq. (7.8) with the nonhermitian Hamiltonian in Eq. (7.1) constrainedby Eq. (7.9) is called Lindblad evolution. The objective is to construct a generating function for expectation values of Heisenberg-picture operators that evolve with Lindblad dynamics instead of Hamiltonian dynamics[44, 17]. To begin I will revisit the path-integral decomposition that led to the generat-ing function Z ( J, J (cid:48) ) in Eq. (2.37).Recall the group-theoretic relation e it ˆ H ˆ q e − it ˆ H = e it [ ˆ H, · ] ˆ q , (7.10)which relates the group elements U ( t ) ≡ e it ˆ H to infinitesimal transformations in the adjointrepresentation, δ t ˆ q = it [ ˆ H, ˆ q ]. What I want is to redo the path-integral decomposition, thistime working directly with the adjoint representation.Tr (ˆ q ( t ) ˆ ρ ( t )) = Tr (cid:16) e − it f ˆ H ˆ q ( t ) ˆ ρ ( t ) e it f ˆ H (cid:17) = Tr (cid:16) e − it f ˆ H e it ˆ H ˆ qe − it ˆ H e it ˆ H ˆ ρ e − it ˆ H e it f ˆ H (cid:17) = Tr (cid:16) e − i ( t f − t ) ˆ H ˆ qe − i ( t − t ) ˆ H ˆ ρ e i ( t − t ) ˆ H e i ( t f − t ) ˆ H (cid:17) = Tr (cid:110) e − i ( t f − t )[ ˆ H, · ] (cid:16) ˆ q e − i ( t − t )[ ˆ H, · ] ˆ ρ (cid:17)(cid:111) . (7.11)In this presentation the evolution is less “right-to-left-then-back-again” and more “outward.”For now this is just an alternative way to organize the evolution, but it will be necessary toevolve this way for an open system. This is usually written in terms of the Schrodinger-picture density matrix ˆ ρ Sch ( t ) as ∂ t ˆ ρ Sch ( t ) = − i (cid:16) ˆ H ˆ ρ Sch ( t ) − ˆ ρ Sch ( t ) ˆ H (cid:17) + (cid:80) N L I = 1 ˆ L I ˆ ρ Sch ( t ) ˆ L † I . The discrepancy is reconciled by the trivial observationthat ˆ ρ Sch ( t ) = ˆ ρ ( − t ). Also, only recently did I stumble upon yet another useful paper by Strunz, whosederivation of Lindblad evolution is like mine [43]. .2.1 Path-integral decomposition Expand the density matrix in the coordinate basis:ˆ ρ = (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) | q (cid:105)(cid:104) q (cid:48) | . (7.12)Writing t − t ≡ nε as in Eq. (2.12), I want to perform a path-integral decomposition of e − i ( t − t )[ ˆ H, · ] ( | q (cid:105)(cid:104) q (cid:48) | ). I enclosed the matrix-basis element in parentheses to emphasize thatthe commutator acts on both sides , which is why I call this “double-sided” evolution. Toperform the path-integral decomposition in this presentation, I need to insert ˆ1 = (cid:82) ∞−∞ dq | q (cid:105)(cid:104) q | simultaneously on the left and right: e − iε [ ˆ H, · ] ( | q (cid:105)(cid:104) q (cid:48) | ) = ˆ1 e − iε [ ˆ H, · ] ( | q (cid:105)(cid:104) q (cid:48) | ) ˆ1= (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) | q (cid:105)(cid:104) q | (cid:104) e − iε [ ˆ H, · ] ( | q (cid:105)(cid:104) q (cid:48) | ) (cid:105) | q (cid:48) (cid:105)(cid:104) q (cid:48) | = (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) | q (cid:105)(cid:104) q | (cid:104)(cid:16) − iε [ ˆ H, · ] (cid:17) ( | q (cid:105)(cid:104) q (cid:48) | ) (cid:105) | q (cid:48) (cid:105)(cid:104) q (cid:48) | = (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) | q (cid:105) (cid:104) (cid:104) q | q (cid:105)(cid:104) q (cid:48) | q (cid:48) (cid:105) − iε (cid:16) (cid:104) q | ˆ H | q (cid:105)(cid:104) q (cid:48) | q (cid:48) (cid:105) − (cid:104) q | q (cid:105)(cid:104) q (cid:48) | ˆ H | q (cid:48) (cid:105) (cid:17)(cid:105) (cid:104) q (cid:48) | . (7.13)With that, and with ˆ H = ˆ p + V (ˆ q ), so that I can do the momentum integrals, evolutionup to t = t gives e − i ( t − t )[ ˆ H, · ] ˆ ρ = (cid:90) ∞−∞ dq dq (cid:48) ρ ( q , q (cid:48) ) × (cid:90) ∞−∞ dq dq (cid:48) πε ... (cid:90) ∞−∞ dq n dq (cid:48) n πε | q n (cid:105) e iε (cid:80) nj = 1 (cid:16)
12 ˙ q j − − V ( q j − ) −
12 ˙ q (cid:48) j − + V ( q (cid:48) j − ) (cid:17) (cid:104) q (cid:48) n | . (7.14)Now I can act with ˆ q and perform the analogous decomposition of the operator e − i ( t f − t )[ ˆ H, · ] acting on ˆ qe − i ( t − t )[ ˆ H, · ] ˆ ρ to get the same result as before.The point is really the last line of Eq. (7.13): When ˆ H acts on the left I get the matrixelement (cid:104) q | ˆ H | q (cid:105) , and when ˆ H acts on the right I get the matrix element (cid:104) q (cid:48) | ˆ H | q (cid:48) (cid:105) . Operators like e − i ( t − t )[ ˆ H, · ] that act in this way are sometimes called “superoperators” [17]. .2.2 Phase-space path integral for Lindblad evolution Continue to focus on Eq. (7.13). For Hamiltonian evolution, I had to calculate matrixelements of the form C j,j − ≡ (cid:104) q j | (cid:16) e − iε [ ˆ H, · ] (cid:0) | q j − (cid:105)(cid:104) q (cid:48) j − | (cid:1)(cid:17) | q (cid:48) j (cid:105) == (cid:104) q j | (cid:16)(cid:16) ˆ1 − iε [ ˆ H, · ] (cid:17) (cid:0) | q j − (cid:105)(cid:104) q (cid:48) j − | (cid:1)(cid:17) | q (cid:48) j (cid:105) = (cid:104) q j | q (cid:48) j (cid:105) − iε (cid:104) q j | (cid:16) ˆ H | q j − (cid:105)(cid:104) q (cid:48) j − | − | q j − (cid:105)(cid:104) q (cid:48) j − | ˆ H (cid:17) | q (cid:48) j (cid:105) = (cid:104) q j | q (cid:48) j (cid:105) − iε (cid:16) (cid:104) q j | ˆ H | q j − (cid:105)(cid:104) q (cid:48) j − | q (cid:48) j (cid:105) − (cid:104) q j | q j − (cid:105)(cid:104) q (cid:48) j − | ˆ H | q (cid:48) j (cid:105) (cid:17) . (7.15)For Lindblad evolution, I will need to calculate the analogous matrix elements but with thereplacement[ ˆ H, · ] (cid:0) | q j − (cid:105)(cid:104) q (cid:48) j − | (cid:1) → ˆ H| q j − (cid:105)(cid:104) q (cid:48) j − | − | q j − (cid:105)(cid:104) q (cid:48) j − | ˆ H † + i N L (cid:88) I = 1 ˆ L I | q j − (cid:105)(cid:104) q (cid:48) j − | ˆ L † I . (7.16)The generating function in phase-space variables is therefore Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) × (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) (cid:90) q (cid:48) ( t f ) = q f q (cid:48) ( t ) = q (cid:48) D q (cid:48) ( · ) (cid:90) D p ( · ) (cid:90) D p (cid:48) ( · ) e i I ( q,p ; q (cid:48) , p (cid:48) | J,J (cid:48) ) , (7.17)with I ( q, p ; q (cid:48) , p (cid:48) | J, J (cid:48) ) = (cid:90) t f t dt (cid:40) p ˙ q − H ( q, p | J ) − [ p (cid:48) ˙ q (cid:48) − H ∗ ( q (cid:48) , p (cid:48) | J (cid:48) )] − i N L (cid:88) I = 1 L I ( q, p ) L ∗ I ( q (cid:48) , p (cid:48) ) (cid:41) , (7.18)and H ( q, p | J ) = H ( q, p ) − J q , H ( q, p ) = H ( q, p ) − i Γ( q, p ), with Γ( q, p ) = (cid:80) N L I = 1 | L I ( q, p ) | . iε prescription Strunz [16] performed the momentum integrals with Lindblad operators ˆ L I = α I ˆ q + β I ˆ p forgeneral coefficients α I and β I . I want to focus on a particular case:ˆ L = √ mε ˆ a , (8.1)with ˆ a = (cid:112) m (cid:0) ˆ q + im ˆ p (cid:1) . This describes the loss of a single quantum to the bath [17]. Thenonhermitian Hamiltonian is for this case isˆ H = ˆ H − i m (cid:2) (1 − iε ) ˆ a † ˆ a + ˆ1 (cid:3) . (8.2) I still call this the “Hamiltonian formalism” (i.e., q s and p s) with the understanding that it describesLindblad evolution. iε prescription to the formalism in which proba-bility is conserved. Inserting Eq. (8.1) into Eq. (7.18), performing the momentum integrals,and dropping terms of O ( ε ) and higher leads to the generating function Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) (cid:90) q (cid:48) ( t f ) = q f q (cid:48) ( t ) = q (cid:48) D q (cid:48) ( · ) e i (cid:82) tft dt [ L ( q ( t ) , q (cid:48) ( t ))+ J ( t ) q ( t )+ J (cid:48) ( t ) q (cid:48) ( t )] , (8.3)with Lagrangian L ( q, q (cid:48) ) = (1+ iε ) ˙ q − (1 − iε ) m q − (cid:2) (1 − iε ) ˙ q (cid:48) − (1+ iε ) m q (cid:48) (cid:3) − iε (cid:2) ˙ q ˙ q (cid:48) + m qq (cid:48) + im ( ˙ qq (cid:48) − ˙ q (cid:48) q ) (cid:3) . (8.4)The first line is the Lagrangian that would result from the traditional iε prescription, asdescribed in Sec. 2.5; the second line is how to fix it. Note that L ( q, q ) = 0, as required. As I qualified back in Sec. 2.5, the prescription defined by Eq. (8.4) is still phenomenological.It is still just a prescription, not a derivation from a fundamental model. A derivation would entail the following: Take a closed system, such as a self-interactingfield in a box, split the field into “fast” and “slow” modes, trace out the fast modes, andleave behind an effective Lindblad action for the slow modes. The parameter ε would begiven in terms of the fast/slow cutoff and the parameters of the underlying model.This qualification continues, I think, the remarks in Sec. 10-5 of Feynman & Hibbs [19]and explains why the work of Caldeira and Leggett does not solve the problem [46]. Anunsatisfying attempt was given by Hu et al. [47] Perhaps I should be able to adapt thework of Calzetta et al. [48], Lombardo and Mazzitelli [49], or Zanella and Calzetta [50] toderive an effective ε from φ theory or φ theory. In another life. Thank you, General Buchhold. A particle-physics analogy is treating the proton mass as a free parameter instead of deriving it from theQCD coupling and the quark masses [45]. I will return to QCD for different reasons in Sec. 9. I thank A. Zee for pointing me to that section and for a discussion about this. They set up everything correctly but then lament that separating “a single field into the high andlow momentum sectors” would be “cumbersome to carry out” and instead “consider two independent self-interacting scalar fields φ ( x ) depicting the system, and ψ ( x ) depicting the bath.” Gee whiz. And in their mainexample, they “extend the range of all integrations in the Feynman diagrams to cover the whole momentumspace,” effectively assuming that “ φ and ψ are two independent fields.” Wonderful. Maybe φ theory in 5 + 1 dimensions, as advocated for pedagogical reasons by Srednicki [9]. .2 Step 1: Keldysh basis for Hamiltonian evolution The last horse crosses the finish line: I will finally switch to the Keldysh basis.Start with the combined Lagrangian for the forward and backward fields: L ( q, q (cid:48) ) = ˙ q − m q + J q − (cid:0) ˙ q (cid:48) − m q (cid:48) + J (cid:48) q (cid:48) (cid:1) . (8.5)The Keldysh basis is defined by the sum and difference of the fields: q ± ≡ q ± q (cid:48) , J ± ≡ J ± J (cid:48) . (8.6)Then q = q + + q − q (cid:48) = q + − q − , (8.7)and similarly for J and J (cid:48) . The pertinent combinations of fields are q − q (cid:48) = q + q − , J q − J (cid:48) q (cid:48) = ( J + q − + J − q + ) . (8.8)In terms of q ± , the Lagrangian in Eq. (8.5) becomes L K ( q + , q − ) ≡ L (cid:0) q + + q − , q + − q − (cid:1) = ˙ q + ˙ q − − m q + q − + ( J + q − + J − q + ) . (8.9)Now drop the K subscript, because even I have limits. The equations of motion are: δSδq + = − (¨ q − + m q − ) + J − = 0 = ⇒ ¨ q − + m q − = J − , (8.10) δSδq − = − (¨ q + + m q + ) + J + = 0 = ⇒ ¨ q + + m q + = J + . (8.11)The action evaluated on solutions of those equations is S cl ≡ (cid:90) t f t dt L (cid:12)(cid:12)(cid:12)(cid:12) those eqs. = q + ˙ q − (cid:12)(cid:12) t f t = t + (cid:90) t f t dt (cid:2) − q + (cid:0) ¨ q − + m q − (cid:1) + ( J + q − + J − q + ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) those eqs. = q + ˙ q − (cid:12)(cid:12) t f t = t + (cid:90) t f t dt J + q − . (8.12)To verify that that reproduces the S cl from the original basis, note that q (cid:48) ˙ q − q ˙ q (cid:48) | t f t = (cid:90) t f t dt ddt ( q (cid:48) ˙ q − q ˙ q (cid:48) ) = (cid:90) t f t dt ( ˙ q (cid:48) ˙ q + q (cid:48) ¨ q − ˙ q ˙ q (cid:48) − q ¨ q (cid:48) ) = (cid:90) t f t dt ( q (cid:48) ¨ q − q ¨ q (cid:48) ) . (8.13) Others prefer the notation q + and q − to denote what I call q and q (cid:48) , and they instead call the sum anddifference fields q classical and q quantum (or some variation thereof). q − J = − m q and¨ q (cid:48) − J (cid:48) = − m q (cid:48) ), I find q (cid:48) ˙ q − q ˙ q (cid:48) | t f t = t + (cid:90) t f t dt ( J (cid:48) q − J q (cid:48) ) = (cid:90) t f t dt [ ( J (cid:48) − ¨ q (cid:48) ) q + (¨ q − J ) q (cid:48) ]= (cid:90) t f t dt (cid:0) m q (cid:48) q − m qq (cid:48) (cid:1) = 0 . (8.14)So indeed inserting the definitions of q ± and J ± into the above S cl would reproduce S cl = q ˙ q (cid:12)(cid:12) t f t = t + (cid:82) t f t dt J q − (cid:16) q (cid:48) ˙ q (cid:48) (cid:12)(cid:12) t f t = t + (cid:82) t f t dt J (cid:48) q (cid:48) (cid:17) . iε prescription Inserting the Lagrangian from Eq. (2.80) into Eq. (2.83) produces what I will call the “old”Lagrangian for the iε prescription: L old ≡ (1+ iε ) ˙ q − (1 − iε ) m q − (cid:2) (1 − iε ) ˙ q (cid:48) − (1+ iε ) m q (cid:48) (cid:3) (8.15)= (cid:0) ˙ q + ˙ q − − m q + q − (cid:1) + iε (cid:2) ˙ q + ˙ q − + m (cid:0) q + q − (cid:1)(cid:3) . (8.16)The additional cross terms I get from Lindblad evolution are L cross ≡ − iε (cid:2) ˙ q ˙ q (cid:48) + m qq (cid:48) + im ( ˙ qq (cid:48) − ˙ q (cid:48) q ) (cid:3) (8.17)= − iε (cid:2) ˙ q − ˙ q − + m ( q − q − ) + 2 im ( q + ˙ q − − q − ˙ q + ) (cid:3) . (8.18)Adding that to Eq. (8.16) produces a “new” Lagrangian, completed and corrected: L new ≡ L old + L cross = (cid:0) ˙ q + ˙ q − − m q + q − (cid:1) + iε (cid:2) ˙ q − + m q − + im ( q − ˙ q + − q + ˙ q − ) (cid:3) . (8.19)Magnifique. To that I add the source term J q − J (cid:48) q (cid:48) = ( J − q + + J + q − ), as before. The newsourced equations of motion are¨ q − − εm ˙ q − + m q − = J − , (8.20)¨ q + + 2 εm ˙ q + + m q + = J + + 2 iε (cid:0) m q − − J − (cid:1) ≡ ˆ J + . (8.21)This is where the utility of the Keldysh basis becomes evident. The difference field is an unstable oscillator, and the sum field is a damped oscillator sourced by the difference field.76 .3.1 Action on classical solutions Now I will evaluate the action on classical solutions. With ˙ q + ˙ q − = ( q + ˙ q − ) · − q + ¨ q − , ˙ q − =( q − ˙ q − ) · − q − ¨ q − , and q − ˙ q + = ( q − q + ) · − ˙ q − q + , I find: S (0)cl ≡ (cid:90) t f t dt L new (cid:12)(cid:12)(cid:12)(cid:12) eqs. of motion (8.22)= ( q + ˙ q − + iεq − ˙ q − − εmq + q − ) (cid:12)(cid:12) t f t = t + (cid:90) t f t dt (cid:2) − q + (cid:0) ¨ q − + m q − − εm ˙ q − (cid:1) + iεq − ( − ¨ q − + m q − ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) eqs. of motion (8.23)= [( q + + iεq − ) ˙ q − − εm q + q − ] (cid:12)(cid:12) t f t = t + (cid:90) t f t dt (cid:2) − q + J − + iεq − (cid:0) m q − − J − + O ( ε ) (cid:1)(cid:3) (8.24)= [( q + + iεq − ) ˙ q − − εm q + q − ] (cid:12)(cid:12) t f t = t + (cid:90) t f t dt (cid:2) − ( q + + iεq − ) J − + 2 iεm q − (cid:3) + O ( ε ) . (8.25)Adding the source terms, I get: S cl = S (0)cl + (cid:90) t f t dt ( J − q + + J + q − ) (8.26)= [( q + + iεq − ) ˙ q − − εm q + q − ] (cid:12)(cid:12) t f t = t + (cid:90) t f t dt (cid:2) ( J + − iεJ − ) q − + 2 iεm q − (cid:3) . (8.27) The equation of motion for the sourced damped/unstable oscillator is¨ q ( t ) + γ ˙ q ( t ) + m q ( t ) = J ( t ) . (8.28)I will first regale you with dimensional analysis: • S = (cid:90) dt ˙ q ∼ t q t ∼ ⇒ q ∼ √ t • m ∼ ddt ∼ t • γ ddt ∼ γ t ∼ d dt ∼ t = ⇒ γ ∼ t • J ∼ d dt q ∼ t / t ∼ t − / . (8.29)Then I will remind you of Sec. 3.2.1, in which I explained my conception of the Green’sfunction: ¨ G ( t ) + γ ˙ G ( t ) + m G ( t ) = 0 , G (0) = 0 , ˙ G (0) = 1 . (8.30) I will not use this directly, but I found it invaluable for catching mistakes. Unless you are much smarterthan I am, you will make mistakes.
77y reason for invoking calligraphy will become evident in a New York minute. Fourier-transforming the defining homogeneous equation in Eq. (8.30) gives (cid:0) − ω − iγω + m (cid:1) ˜ G ( ω ) = 0 , (8.31)which implies that ˜ G ( ω ) = 0 or ω = − i γ ± (cid:113) m − (cid:0) γ (cid:1) . (8.32)In terms of cosines and sines, the solution is therefore G ( t ) = e − γt (cid:20) A cos (cid:18)(cid:113) m − γ t (cid:19) + B sin (cid:18)(cid:113) m − γ t (cid:19)(cid:21) . (8.33)Imposing G (0) = 0 and ˙ G (0) = 1 produces the standard result: G ( t ) = e − γt (cid:113) m − γ sin (cid:18)(cid:113) m − γ t (cid:19) . (8.34)The approximation I will use is G ( t ) ≈ e − γt G ( t ) , G ( t ) = 1 m sin( mt ) . (8.35)That relation is why I chose the notation G ( t ) for this Green’s function. After a colosseumof trial and error, I have determined that it is best to express everything in terms of the verysame G ( t ) from Eq. (3.11).Eq. (8.35) establishes the approximation scheme for the iε prescription and the sign con-ventions for stability of the oscillator. Use Eqs. (8.28) and (8.35) to interpret Eqs. (8.20)and (8.21). Given the Green’s function in Eq. (8.35) and the steps in Secs. 3.2.2-3.2.4, I trust you tosolve Eq. (8.20) for the difference field: q − ( t ) = 1 G ( T ) (cid:34) q − e εm ( t − t ) G ( t f − t )+ (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) (cid:16) G ( T ) G ( t − t (cid:48) )Θ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) ) (cid:17) J − ( t (cid:48) ) (cid:35) , (8.36)with G ( t ) ≡ m sin( mt ). The time-derivative is˙ q − ( t ) = 1 G ( T ) (cid:110) q − e εm ( t − t ) (cid:104) − ˙ G ( t f − t ) + εmG ( t f − t ) (cid:105) + (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) (cid:104) G ( T ) ˙ G ( t − t (cid:48) )Θ( t − t (cid:48) ) − ˙ G ( t − t ) G ( t f − t (cid:48) )+ εm ( G ( T ) G ( t − t (cid:48) )Θ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )) (cid:105) J − ( t (cid:48) ) (cid:111) . (8.37)78he values of the derivative at t and t f are:˙ q − ( t ) = − G ( T ) (cid:26) q − (cid:104) ˙ G ( T ) − εmG ( T ) (cid:105) + (cid:90) t f t dt e − εm ( t − t ) G ( t f − t ) J − ( t ) (cid:27) , ˙ q − ( t f ) = 1 G ( T ) (cid:26) − q − e εmT + (cid:90) t f t dt e εm ( t f − t ) (cid:104) G ( T ) ˙ G ( t f − t ) − ˙ G ( T ) G ( t f − t ) (cid:105) J − ( t ) (cid:27) = e εmT G ( T ) (cid:26) − q − + (cid:90) t f t dt e − εm ( t − t ) G ( t − t ) J − ( t ) (cid:27) . (8.38)A blessing from Eq. (8.27) is that the explicit solution for the sum field is not required. Given Eq. (8.38), I can start to simplify Eq. (8.27). The boundary terms are: (cid:104) ( q + + iεq − ) ˙ q − − εmq + q − (cid:105)(cid:12)(cid:12)(cid:12) t f t = t = 2 q f ˙ q − ( t f ) − [( q +0 + iεq − ) ˙ q − ( t ) − εmq +0 q − ]= 2 q f e εmT G ( T ) (cid:26) − q − + (cid:90) t f t dt e − εm ( t − t ) G ( t − t ) J − ( t ) (cid:27) + εmq +0 q − + q +0 + iεq − G ( T ) (cid:26) q − (cid:104) ˙ G ( T ) − εmG ( T ) (cid:105) + (cid:90) t f t dt e − εm ( t − t ) G ( t f − t ) J − ( t ) (cid:27) = 2 q f e εmT G ( T ) (cid:26) − q − + (cid:90) t f t dt e − εm ( t − t ) G ( t − t ) J − ( t ) (cid:27) + q +0 + iεq − G ( T ) (cid:26) q − ˙ G ( T ) + (cid:90) t f t dt e − εm ( t − t ) G ( t f − t ) J − ( t ) (cid:27) . (8.39) q f There are no terms quadratic in q f . If I do the integral over q f , the terms linear in q f willresult in a delta function that sets q − = (cid:90) t f t dt e − εm ( t − t ) G ( t − t ) J − ( t ) . (8.40)So instead of trying to insert q − ( t ) into S cl , I could first do the integral over q f and insertEq. (8.40) into q − ( t ). That integral is (cid:90) ∞−∞ dq f e iq f eεmTG ( T ) (cid:104) − q − + (cid:82) tft dt e − εm ( t − t G ( t − t ) J − ( t ) (cid:105) =2 πG ( T ) e − εmT δ (cid:18) q − − (cid:90) t f t dt e − εm ( t − t ) G ( t − t ) J − ( t ) (cid:19) . (8.41)Instead of calculating the overall factor from first principles, this time I will just fix it byrequiring Z = 1 when J (cid:48) = J [recall Eq. (2.44)]. In terms of J + and J − , that would mean79 = 1 when J − = 0. From now on I will write Z K ( J + , J − ) ≡ Z ( J + + J − , J + − J − ) and drop the K , as I already did after Eq. (8.9): Z ( J + , J − ) = N (cid:90) ∞−∞ dq +0 (cid:90) ∞−∞ dq − ρ (cid:0) q +0 + q − , q +0 − q − (cid:1) (cid:90) ∞−∞ dq f e iS cl ( q +0 ,q − ,q f | J + ,J − ) = 2 πG ( T ) e − εmT N × (cid:90) ∞−∞ dq +0 ρ (cid:0) q +0 + q − , q +0 − q − (cid:1) e iS cl ( q +0 ,q − , | J + ,J − ) (cid:12)(cid:12)(cid:12)(cid:12) q − = (cid:82) tft dt e − εm ( t − t G ( t − t ) J − ( t ) . (8.42) With Eq. (8.40), I find˙ G ( T ) q − + (cid:90) t f t dt e − εm ( t − t ) G ( t f − t ) J − ( t ) = (cid:90) t f t dt e − εm ( t − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) J − ( t ) , (8.43)an expression that looks reassuringly like something I found for ε = 0. At this point itis useful to organize terms in powers of ε (without expanding any of the exponentials, asusual—powers of ε , not powers of εmt ). The O ( ε ) part of the boundary terms is q +0 G ( T ) (cid:26) ˙ G ( T ) q − + (cid:90) t f t dt e − εm ( t − t ) G ( t f − t ) J − ( t ) (cid:27) = q +0 G ( T ) (cid:90) t f t dt e − εm ( t − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) J − ( t ) , (8.44)and the O ( ε ) part is iεq − G ( T ) (cid:26) ˙ G ( T ) q − + (cid:90) t f t dt e − εm ( t − t ) G ( t f − t ) J − ( t ) (cid:27) = iε G ( T ) (cid:26)(cid:90) t f t dt (cid:48) e − εm ( t (cid:48) − t ) G ( t (cid:48) − t ) J − ( t (cid:48) ) (cid:27) (cid:90) t f t dt e − εm ( t − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) J − ( t )= iε G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − εm ( t + t (cid:48) − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) G ( t (cid:48) − t ) J − ( t ) J − ( t (cid:48) ) . (8.45)80 .3.7 Simplify the action: Simplify q − ( t )That expression for q − will greatly simplify q − ( t ): q − ( t ) = 1 G ( T ) (cid:26)(cid:20)(cid:90) t f t dt (cid:48) e − εm ( t (cid:48) − t ) G ( t (cid:48) − t ) J − ( t (cid:48) ) (cid:21) e εm ( t − t ) G ( t f − t )+ (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) [ G ( T ) G ( t − t (cid:48) )Θ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J − ( t (cid:48) ) (cid:27) = 1 G ( T ) (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) G ( T ) G ( t − t (cid:48) )Θ( t − t (cid:48) ) + G ( t f − t ) G ( t (cid:48) − t ) − G ( t f − t (cid:48) ) G ( t − t ) (cid:124) (cid:123)(cid:122) (cid:125) = − G ( T ) G ( t − t (cid:48) ) J − ( t (cid:48) )= − (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) J − ( t (cid:48) ) . (8.46)Compare that to Eq. (8.37). With Eq. (8.46), I can simplify the nonboundary terms in the action: S nonboundarycl = (cid:90) t f t dt (cid:2) ( J + ( t ) − iεJ − ( t )) q − ( t ) + 2 iεm q − ( t ) (cid:3) . (8.47)The q − term will require special attention, since there will be an intermediate integral: (cid:90) t f t dt (cid:48)(cid:48) q − ( t (cid:48)(cid:48) ) = (cid:90) t f t dt (cid:48)(cid:48) (cid:20)(cid:90) t f t dt e εm ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t )Θ( t − t (cid:48)(cid:48) ) J − ( t ) (cid:21) (cid:20)(cid:90) t f t dt (cid:48) e εm ( t (cid:48)(cid:48) − t (cid:48) ) G ( t (cid:48)(cid:48) − t (cid:48) )Θ( t (cid:48) − t (cid:48)(cid:48) ) J − ( t (cid:48) ) (cid:21) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − εm ( t + t (cid:48) ) J − ( t ) J − ( t (cid:48) ) (cid:90) t f t dt (cid:48)(cid:48) e εmt (cid:48)(cid:48) G ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t (cid:48) ) Θ( t − t (cid:48)(cid:48) )Θ( t (cid:48) − t (cid:48)(cid:48) ) . (8.48)That product of step functions says that t (cid:48)(cid:48) must be less than t and less than t (cid:48) , so the integralover t (cid:48)(cid:48) has two cases: t > t (cid:48) and t < t (cid:48) . Considering each separately, I rewrite the integral as I ( t, t (cid:48) ) ≡ (cid:90) t f t dt (cid:48)(cid:48) e εmt (cid:48)(cid:48) G ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t (cid:48) ) Θ( t (cid:48)(cid:48) − t )Θ( t (cid:48)(cid:48) − t (cid:48) )= Θ( t − t (cid:48) ) (cid:90) t (cid:48) t dt (cid:48)(cid:48) e εmt (cid:48)(cid:48) G ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t (cid:48) ) + Θ( t (cid:48) − t ) (cid:90) tt dt (cid:48)(cid:48) e εmt (cid:48)(cid:48) G ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t (cid:48) ) . (8.49)81hat integral will have an overall factor of ε − , making its contribution to the action include O ( ε ) and O ( ε ) terms. The coefficients of Θ( t − t (cid:48) ) and Θ( t (cid:48) − t ) are (cid:90) t (cid:48) t dt (cid:48)(cid:48) e εmt (cid:48)(cid:48) G ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t (cid:48) )= 14 m ε (cid:110)(cid:16) e εmt (cid:48) − e εmt (cid:17) ˙ G ( t − t (cid:48) ) + εm (cid:104) e εmt (cid:48) G ( t − t (cid:48) ) − e εmt G ( t + t (cid:48) − t ) (cid:105)(cid:111) + O ( ε ) , (8.50) (cid:90) tt dt (cid:48)(cid:48) e εmt (cid:48)(cid:48) G ( t (cid:48)(cid:48) − t ) G ( t (cid:48)(cid:48) − t (cid:48) )= 14 m ε (cid:110)(cid:0) e εmt − e εmt (cid:1) ˙ G ( t − t (cid:48) ) − εm (cid:2) e εmt G ( t − t (cid:48) ) + e εmt G ( t + t (cid:48) − t ) (cid:3)(cid:111) + O ( ε ) . (8.51)Therefore, e − εm ( t + t (cid:48) ) I ( t, t (cid:48) )= 14 m ε (cid:110) Θ( t − t (cid:48) ) (cid:104)(cid:16) e − εm ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) (cid:17) ˙ G ( t − t (cid:48) )+ εm (cid:16) e − εm ( t − t (cid:48) ) G ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) G ( t + t (cid:48) − t ) (cid:17)(cid:105) + Θ( t (cid:48) − t ) (cid:104)(cid:16) e εm ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) (cid:17) ˙ G ( t − t (cid:48) ) − εm (cid:16) e εm ( t − t (cid:48) ) G ( t − t (cid:48) ) + e − εm ( t + t (cid:48) − t ) G ( t + t (cid:48) − t ) (cid:17)(cid:105)(cid:111) + O ( ε ) . (8.52) The way to organize all of that is to remember that there is still an integral over q +0 to bedone against the density matrix. So I will organize S cl into three terms: A part that dependson q +0 , a q +0 -independent part of O ( ε ), and a q +0 -independent part of O ( ε ). S cl = q +0 G ( T ) (cid:90) t f t dt e − εm ( t − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) J − ( t ) + A ( J + , J − ) + iεB ( J − ) , (8.53)82ith A ( J + , J − ) = − (cid:90) t f t dt (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) J + ( t ) J − ( t (cid:48) )+ i m (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) Θ( t − t (cid:48) ) (cid:16) e − εm ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) (cid:17) + Θ( t (cid:48) − t ) (cid:16) e εm ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) (cid:17) (cid:105) ˙ G ( t − t (cid:48) ) J − ( t ) J − ( t (cid:48) )= − (cid:90) t f t dt (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) J + ( t ) J − ( t (cid:48) )+ i m (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) + Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) (cid:105) ˙ G ( t − t (cid:48) ) J − ( t ) J − ( t (cid:48) ) , (8.54)and (do not forget the part from the boundary terms): B ( J − ) = 12 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − εm ( t + t (cid:48) − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) G ( t (cid:48) − t ) J − ( t ) J − ( t (cid:48) )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) J − ( t ) J − ( t (cid:48) )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) Θ( t − t (cid:48) ) (cid:16) e − εm ( t − t (cid:48) ) G ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) G ( t + t (cid:48) − t ) (cid:17) − Θ( t (cid:48) − t ) (cid:16) e εm ( t − t (cid:48) ) G ( t − t (cid:48) ) + e − εm ( t + t (cid:48) − t ) G ( t + t (cid:48) − t ) (cid:17)(cid:105) J − ( t ) J − ( t (cid:48) )= 12 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − εm ( t + t (cid:48) − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) G ( t (cid:48) − t ) J − ( t ) J − ( t (cid:48) )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) (cid:16) Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) +Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) (cid:17) G ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) G ( t + t (cid:48) − t ) (cid:105) J − ( t ) J − ( t (cid:48) ) (8.55)= 12 G ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − ε ( t + t (cid:48) − t ) (cid:104) ˙ G ( T ) G ( t − t ) G ( t (cid:48) − t ) + G ( t f − t ) G ( t (cid:48) − t ) − G ( t + t (cid:48) − t ) (cid:105) J − ( t ) J − ( t (cid:48) )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:16) Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) +Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) (cid:17) G ( t − t (cid:48) ) J − ( t ) J − ( t (cid:48) ) . (8.56)Since J − ( t ) J − ( t (cid:48) ) is symmetric in t and t (cid:48) , I should write G ( t f − t ) G ( t (cid:48) − t ) → [ G ( t f − t ) G ( t (cid:48) − t )+ G ( t f − t (cid:48) ) G ( t − t )] and behold that2 ˙ G ( T ) G ( t − t ) G ( t (cid:48) − t ) + G ( t f − t ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) G ( t − t ) − G ( t + t (cid:48) − t ) = 0 . (8.57)83o the first line is zero. Meanwhile, the expression in the second line, M ( t, t (cid:48) ) ≡ (cid:16) Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) +Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) (cid:17) G ( t − t (cid:48) ) , (8.58)is antisymmetric: M ( t (cid:48) , t ) = (cid:16) Θ( t (cid:48) − t ) e − εm ( t (cid:48) − t ) + Θ( t − t (cid:48) ) e εm ( t (cid:48) − t ) (cid:17) G ( t (cid:48) − t )= − (cid:16) Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) + Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) (cid:17) G ( t − t (cid:48) )= − M ( t, t (cid:48) ) . (8.59)So (cid:82) t f t dt (cid:82) t f t dt (cid:48) M ( t, t (cid:48) ) J − ( t ) J − ( t (cid:48) ) = 0, and I learn that B ( J − ) = 0: The O ( ε ) part is zero.At this stage the generating function is Z ( J + , J − ) = 2 πG ( T ) e − εmT N e iA ( J + ,J − ) × (cid:90) ∞−∞ dq ρ (cid:18) q + (cid:82) tft dt e − εm ( t − t G ( t − t ) J − ( t )2 , q − (cid:82) tft dt e − εm ( t − t G ( t − t ) J − ( t )2 (cid:19) × e i G ( T ) q (cid:82) tft dt e − εm ( t − t [ ˙ G ( T ) G ( t − t )+ G ( t f − t ) ] J − ( t ) . (8.60)I suspect there is physics to be gleaned from that general form. I will complete this calculation only for the ε = 0 oscillator ground state: ρ ( q, q (cid:48) ) = (cid:114) mπ e − m ( q + q (cid:48) ) . (8.61)Now to fix the normalization factor in Eq. (8.60). Let k = (cid:82) t f t dte − εm ( t − t ) G ( t − t ) J − ( t ) and j = (cid:82) t f t dte − εm ( t − t ) [ ˙ G ( T ) G ( t − t ) + G ( t f − t )] J − ( t ). The integral over q is: (cid:90) ∞−∞ dq ρ (cid:0) q + k , q − k (cid:1) e ij G ( T ) q = (cid:114) mπ e − mk (cid:90) ∞−∞ dq e − m ( q − ijmG ( T ) q )= 2 e − m k e − m j m G ( T )2 . (8.62)So the normalization factor should be N = 14 πG ( T ) e − εmT , (8.63)leaving an influence phaseΦ( J + , J − ) = A ( J + , J − ) + imk ij mG ( T ) . (8.64)84 .4.1 Simplify the influence phase Φ( J + , J − ) = − (cid:90) t f t dt (cid:90) t f t dt (cid:48) e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) J + ( t ) J − ( t (cid:48) )+ i m (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) + Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) − e − εm ( t + t (cid:48) − t ) (cid:105) ˙ G ( t − t (cid:48) ) J − ( t ) J − ( t (cid:48) )+ im (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − εm ( t + t (cid:48) − t ) G ( t − t ) G ( t (cid:48) − t ) J − ( t ) J − ( t (cid:48) )+ i mG ( T ) (cid:90) t f t dt (cid:90) t f t dt (cid:48) e − εm ( t + t (cid:48) − t ) (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) × (cid:104) ˙ G ( T ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) (cid:105) J − ( t ) J − ( t (cid:48) ) . (8.65)Everything in this problem is translationally invariant, so everything that depends on t and t f had better cancel. Guess what: (cid:104) ˙ G ( T ) G ( t − t ) + G ( t f − t ) (cid:105) (cid:104) ˙ G ( T ) G ( t (cid:48) − t ) + G ( t f − t (cid:48) ) (cid:105) + m G ( T ) G ( t − t ) G ( t (cid:48) − t ) = G ( T ) ˙ G ( t − t (cid:48) ) . (8.66)So all of the terms proportional to e − εm ( t + t (cid:48) − t ) will, in fact, cancel. Writing Θ( t − t (cid:48) ) e − εm ( t − t (cid:48) ) + Θ( t (cid:48) − t ) e εm ( t − t (cid:48) ) = e − εm | t − t (cid:48) | and expressing the J + ( t ) J − ( t (cid:48) ) term as the sum of J + ( t ) J − ( t (cid:48) )and J + ( t (cid:48) ) J − ( t ) terms, I arrive at the result:Φ( J + , J − ) = i m (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:104) e − εm | t − t (cid:48) | ˙ G ( t − t (cid:48) ) J − ( t ) J − ( t (cid:48) )+ im e εm ( t − t (cid:48) ) G ( t − t (cid:48) ) Θ( t (cid:48) − t ) J + ( t ) J − ( t (cid:48) ) + im e εm ( t (cid:48) − t ) G ( t (cid:48) − t ) Θ( t − t (cid:48) ) J − ( t ) J + ( t (cid:48) ) (cid:105) . (8.67)No J + J + terms, sublime.To garnish the omelette, I will return to the original basis. With J ± = J ± J (cid:48) , the productsof sources become: J − ( t ) J − ( t (cid:48) ) = [ J ( t ) − J (cid:48) ( t )][ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] = J ( t ) J ( t (cid:48) ) + J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J ( t (cid:48) ) ,J + ( t ) J − ( t (cid:48) ) = [ J ( t )+ J (cid:48) ( t )][ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] = J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48) ( t ) J ( t (cid:48) ) ,J − ( t ) J + ( t (cid:48) ) = [ J ( t ) − J (cid:48) ( t )][ J ( t (cid:48) )+ J (cid:48) ( t (cid:48) )] = J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) + J ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J ( t (cid:48) ) . (8.68)The J + J − terms become: e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) J + ( t ) J − ( t (cid:48) ) + e εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) ) J − ( t ) J + ( t (cid:48) )= (cid:104) e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) + e εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) ) (cid:105) [ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )]+ (cid:104) − e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) + e εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) ) (cid:105) [ J ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J ( t (cid:48) )] . (8.69)85ote that the coefficient of J J − J (cid:48) J (cid:48) is symmetric, not antisymmetric: C ( t, t (cid:48) ) ≡ e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) + e εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) ) (8.70) C ( t (cid:48) , t ) = e εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t )= e εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) + e εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) )= C ( t, t (cid:48) ) . (8.71)Therefore, I find: e − εm | t − t (cid:48) | ˙ G ( t − t (cid:48) )[ J ( t ) J ( t (cid:48) ) + J (cid:48) ( t ) J (cid:48) ( t (cid:48) )] + imC ( t, t (cid:48) )[ J ( t ) J ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48) ( t (cid:48) )]= (cid:110) e − εm ( t − t (cid:48) ) (cid:104) ˙ G ( t − t (cid:48) ) − im G ( t − t (cid:48) ) (cid:105) Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) (cid:104) ˙ G ( t − t (cid:48) )+ im G ( t − t (cid:48) ) (cid:105) Θ( t (cid:48) − t ) (cid:111) J ( t ) J ( t (cid:48) )+ (cid:110) e − εm ( t − t (cid:48) ) (cid:104) ˙ G ( t − t (cid:48) )+ im G ( t − t (cid:48) ) (cid:105) Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) (cid:104) ˙ G ( t − t (cid:48) ) − im G ( t − t (cid:48) ) (cid:105) Θ( t (cid:48) − t ) (cid:111) J (cid:48) ( t ) J (cid:48) ( t (cid:48) )= (cid:104) e − εm ( t − t (cid:48) ) e − im ( t − t (cid:48) ) Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) e im ( t − t (cid:48) ) Θ( t (cid:48) − t ) (cid:105) J ( t ) J ( t (cid:48) )+ (cid:104) e − εm ( t − t (cid:48) ) e im ( t − t (cid:48) ) Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) e − im ( t − t (cid:48) ) Θ( t (cid:48) − t ) (cid:105) J (cid:48) ( t ) J (cid:48) ( t (cid:48) )= e − i (1 − iε ) m | t − t (cid:48) | J ( t ) J ( t (cid:48) ) + e i (1+ iε ) m | t − t (cid:48) | J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) . (8.72)There you have the iε prescription as ordinarily understood: The frequency m gets a small,negative, imaginary part in the forward direction, and a small, positive, imaginary part inthe backward direction. Compare to Eqs. (4.25) and (4.26).86he new terms derived from Lindblad evolution are: e − εm | t − t (cid:48) | ˙ G ( t − t (cid:48) )[ − J ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J ( t (cid:48) )] + ime εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t )[ − J ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48) ( t ) J ( t (cid:48) )]+ ime εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) )[ J ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J ( t (cid:48) )] = − [ e − εm | t − t (cid:48) | ˙ G ( t − t (cid:48) ) + ime εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) − ime εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) )] J ( t ) J (cid:48) ( t (cid:48) ) − [ e − εm | t − t (cid:48) | ˙ G ( t − t (cid:48) ) − ime εm ( t − t (cid:48) ) G ( t − t (cid:48) )Θ( t (cid:48) − t ) + ime εm ( t (cid:48) − t ) G ( t (cid:48) − t )Θ( t − t (cid:48) )] J (cid:48) ( t ) J ( t (cid:48) ) = − (cid:110) e − εm ( t − t (cid:48) ) (cid:104) ˙ G ( t − t (cid:48) ) + imG ( t − t (cid:48) ) (cid:105) Θ( t − t (cid:48) ) + e − εm ( t (cid:48) − t ) (cid:104) ˙ G ( t − t (cid:48) ) + imG ( t − t (cid:48) ) (cid:105) Θ( t (cid:48) − t ) (cid:111) J ( t ) J (cid:48) ( t (cid:48) ) − (cid:110) e − εm ( t − t (cid:48) ) (cid:104) ˙ G ( t − t (cid:48) ) − imG ( t − t (cid:48) ) (cid:105) Θ( t − t (cid:48) ) + e − εm ( t (cid:48) − t ) (cid:104) ˙ G ( t − t (cid:48) ) − imG ( t − t (cid:48) ) (cid:105) Θ( t (cid:48) − t ) (cid:111) J (cid:48) ( t ) J ( t (cid:48) )= − (cid:110) e − εm ( t − t (cid:48) ) e im ( t − t (cid:48) ) Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) e im ( t − t (cid:48) ) Θ( t (cid:48) − t ) (cid:111) J ( t ) J (cid:48) ( t (cid:48) ) − (cid:110) e − εm ( t − t (cid:48) ) e − im ( t − t (cid:48) ) Θ( t − t (cid:48) ) + e εm ( t − t (cid:48) ) e − im ( t − t (cid:48) ) Θ( t (cid:48) − t ) (cid:111) J (cid:48) ( t ) J ( t (cid:48) )= − (cid:104) e i (1+ iε ) m ( t − t (cid:48) ) Θ( t − t (cid:48) ) + e i (1 − iε ) m ( t − t (cid:48) ) Θ( t (cid:48) − t ) (cid:105) J ( t ) J (cid:48) ( t (cid:48) ) − (cid:104) e − i (1 − iε ) m ( t − t (cid:48) ) Θ( t − t (cid:48) ) + e − i (1+ iε ) m ( t − t (cid:48) ) Θ( t (cid:48) − t ) (cid:105) J (cid:48) ( t ) J ( t (cid:48) )= − e i [1+sign( t − t (cid:48) ) iε ] m ( t − t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − e − i [1 − sign( t − t (cid:48) ) iε ] m ( t − t (cid:48) ) J (cid:48) ( t ) J ( t (cid:48) ) . (8.73)Compare to Eqs. (4.24) and (4.23).Comparing Eqs. (8.72) and (8.73) with the general form of the influence phase from Eq. (4.27),I infer the Feynman, Dyson, and Wightman functions produced by the Lagrangian fromEq. (8.19): G F ( t ) = i m e − i (1 − iε ) m | t | , G D ( t ) = − i m e i (1+ iε ) m | t | ,G < ( t ) = i m e i [1+sign( t ) iε ] mt , G > ( t ) = i m e − i [1 − sign( t ) iε ] mt . (8.74)That, ladies and gentlemen, is the iε prescription done right.87 Discussion
I have reviewed the Schwinger-Keldysh path integral and its generalization for OTOCs inpresumably enough detail. I will conclude this treatise, and my career, with some words.It is common to speak of the temporal “Keldysh contour,” reflecting the steps that ledto Eq. (2.37). I do not find it convenient to think that way. Instead, toward uplifting theformalism to higher-dimensional quantum field theory, I prefer to express the generatingfunction as Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) (cid:90) q (cid:48) ( t f ) = q f q (cid:48) ( t ) = q (cid:48) D q (cid:48) ( · ) e i (cid:110) S ( q,q (cid:48) )+ (cid:82) tft dt [ J ( t ) q ( t ) − J (cid:48) ( t ) q (cid:48) ( t )] (cid:111) , (9.1)with action S ( q, q (cid:48) ) = (cid:90) t f t dt (cid:2) ˙ q ( t ) − m q ( t ) − ˙ q (cid:48) ( t ) + m q (cid:48) ( t ) + ... (cid:3) , (9.2)and leave it at that. Similarly for the Larkin-Ovchinnikov path integral and whatever Idefined in Eq. (2.56). I see only one contour: A straight line from t to t f . Whether forthermal equilibrium it “could make practical calculations much easier” [51] to reformulatethe initial condition as a detour in imaginary time is beside the point. The advantage ofthe Schwinger-Keldysh formalism is to have a real-time generating function for expectationvalues in arbitrary states.The novelties are instead two flavors of field (or four, or three, etc.), one with a wrong-sign kinetic term, and a constraint between them at t f [52]. This is just the well-trodden × return to operator expressions like Eqs. (2.43),(2.67), and (2.68) instead of using them for a thing or two then throwing them away.Undoubtedly this stems from the “typical strategy in practice” of using path integrals merelyto “derive equations of motion for a given set of correlation functions” [17]—a statement thoseauthors intended as a matter of fact, but which I read as an indictment. Feynman’s paperdates to 1948, and yet fear, uncertainty, and doubt persist. Path integrals are not on “veryshaky mathematical grounds” [55]: They form the bedrock of modern physics [56, 57]. But not well-developed. I do not recall discussions of charge conjugation, parity, and time reversal, orof other staples of relativistic quantum field theory in this context; and apparently a systematic account ofrenormalization has only just begun [53, 54].
88s for why I chose to study this formalism in the first place: Black-hole evaporation.Hawking’s prediction that isolated black holes will radiate and disappear was a conceptualbreakthrough [58], but the formalism he used amounts to implicitly integrating out, insteadof tracing out, a dynamical thermal environment and inferring detailed properties of thewavefunction of the radiation [59]. That is not a sound basis for effective field theory. Butgiven the apparently unassailable assumptions behind semiclassical gravity coupled to quan-tum fields, the question is why not. If the observation scale is macroscopic compared to theanticipated regime of quantum gravity, then why exactly can the system not be decoupledfrom the bath? How exactly does effective field theory predict its own demise? That is theinformation problem. That is why it is considered a paradox, and not just one of manydifficult dynamical problems in theoretical physics.A useful comparison is to a modified history of asymptotic freedom in QCD. Supposethat you know, whether by experiment or by encyclical, that the long-distance model forthe strong force is the chiral Lagrangian, and that the short-distance model is the QCDLagrangian. But suppose you know nothing about the renormalization group—never heardof Gell-Mann and Low, let alone of Wilson. How could those two models be mutually con-sistent? They cannot coexist, they do not imply each other. One has color, the other doesnot; one has massless particles, the other does not. Both describe the real world. Not just ahard problem, but a paradox.The resolution was not to calculate the fermion condensate or nucleon masses in terms of theQCD coupling; as far as I know, that still cannot be done analytically. ’t Hooft calculatedthe meson spectrum in 1+1 dimensions [64], and Witten explained how baryons fit in [65],which demonstrated that in a toy model the picture of long-distance pions and nucleonsand of short-distance gluons and quarks was plausible. But that was not what solved theparadox. What solved the paradox was the renormalization group: The explicit demonstra-tion that the coupling of QCD shrinks as the distance scale decreases [67, 68]. That wasthe breakthrough, even though all it implies is that perturbation theory breaks down. Itdoes not prove confinement, and it does not prove that there exists an energy gap at longdistances. It simply shows that effective field theory predicts its own demise, paving the wayfor consistency between two unimpeachable models. The question of what happens when an observer falls into a black hole is a different problem. This is more or less the actual history, but with the crucial benefit of knowing that quantum field theoryis the correct mathematical framework for fundamental particles. I am not old enough to have lived thathistory—my understanding is that it cannot be overstated just how difficult it was to test not only a modelbut also the edifice on which that model was built [60, 61]. People had certainly heard of the renormalizationgroup, but they did not trust it. As an aside within an aside: I think that nowadays many people trust therenormalization group too blindly, or at least too literally. That effective field theory predicts the values ofsuperrenormalizable couplings to be of order the cutoff [62, 63] should not be interpreted as an inconsistency;instead, it should be interpreted as effective field theory saying that you have asked it a question it cannotanswer. It tells you that the most straightforward high-energy completion is incorrect, nothing more. Thehierarchy problem and the cosmological-constant problem are difficult dynamical problems but not paradoxes. See also the work on chiral-symmetry breaking by Coleman and Witten [66]. iε prescription. At minimum, I want to see ananswer to “why is the public paying us?” [71]. Whether you could be super-duper smart andposit the correct pure state for the black hole plus radiation in JT gravity makes no differ-ence, even if you enumerate all of the microstates that constitute the interior. That wouldnot solve the information problem. The SYK model is great, and I owe my career to K, butit is not gravity. It is like ’t Hooft’s toy model of QCD. It does not describe the real world. Here are some observations I have made over the last few years, which may have somerelevance given the recent work on replicas [74, 75]. First, recall that for evaporation of ablack hole formed from collapse, attempting to trace modes backward in time from futureinfinity would accelerate them to arbitrarily high energy, while tracing them forward intime from past infinity does not. Most people seem to consider that discrepancy harmless,but I had always interpreted it as evidence of a phase transition. I am pleased to see thata first-order phase transition at the Page time is now consensus [76].Second, between thinking about topological superconductors [77, 78] and gravitational shock-waves [79, 80], the first thing I thought of when seeing the commutation relation between thegravitationally backreacted ingoing and outgoing near-horizon fields is lattice dislocations. That makes sense, because that is exactly what shockwave operators are: Translational de-fect lines. When you act with two opposing shockwaves in a specified order, it sure does looklike it induces a puncture at what was formerly U = V = 0 in the Kruskal diagram [82]. Are those the Lorentzian defects that proliferate to restore the black-hole metric plus thebranching surface in the n → I thank Eva Silverstein for sharing similar thoughts during my visit to Stanford in 2017. I am well aware that near-extremal black holes have an
AdS factor [72], and that realistic black holesspin quickly [73], but I will stake my reputation on predicting that mastering JT gravity will not be goodenough to solve the real problem. If I am wrong, then hey, at least I never insisted that superpartners wouldbe discovered at the LHC. Mulling over that, I was led to the following formal observation, which I alluded to in Sec. 2. As I haveemphasized from square one, expectation values can be built from transition amplitudes, as in Eq. (2.7).Instead of pairing the density matrix with the initial field configurations and summing over a common final field configuration, I could do the reverse and thereby construct an “out-out” generating function forexpectation values like Eq. (6.28). Would that be useful? I made comments along those lines to Douglas Stanford and Nick Hunter-Jones at the Simons meetingin NYC in Dec. 2017, and to Beni Yoshida at Settlement Co. in Waterloo sometime in 2018-2019. That is an observation I shared with many people in the IQIM during 2015-2018, including Gil Refaeland probably Justin Wilson. See in particular the order-disorder commutation relations in Marino et al. [81] I have long wanted to characterize the topological nature of two opposing shockwaves by consideringsome kind of holonomy, but I have never figured out how. Alexei Kitaev gave me that idea, and I havespoken about it at length with Joe Swearngin and Alex Rasmussen. I think the pertinent connection is thetranslational gauge field of the Poincare group [83]. Remember two things: Cartan gravity is better thanEinstein gravity, and the translational gauge field is not the frame.
Acknowledgments
I thank Michael Buchhold for acquiescing to my interrogation about the Schwinger-Keldyshformalism, and I thank Beatrice Bonga for putting up with a lot. I thank the IQIM at Caltechfor letting me relive my glory days for the month of February, 2020. I thank in particular XieChen and Alexei Kitaev for always teaching me new things and for letting me vent. I thankJoe Swearngin and Justin Wilson for innumerable conversations, many of which contributedin one way or another to this project. I thank Michael Buchhold and Joe Swearngin a secondtime for feedback on a draft of this paper. I thank the CGWiki community for makingcoronavirus lockdown tolerable. Finally, I thank Leah and Leila at the Grand Surf Loungefor not ejecting me from the premises for explaining to some guy that the icosahedra he sawwhile struck by lightning were not a theory of wavefunctions. Research at Perimeter Instituteis supported in part by the Government of Canada through the Department of Innovation,Science and Economic Development Canada and by the Province of Ontario through theMinistry of Economic Development, Job Creation and Trade.91
Free-particle path integral
In this appendix I will calculate the amplitude-generating function for the free particle, Z ( q , t ; q f , t f | J ) ≡ (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) e i (cid:82) tft dt (cid:104)
12 ˙ q ( t ) + J ( t ) q ( t ) (cid:105) , (A.1)using the method of stationary phase. Then I will review how to evaluate the J = 0 versionusing time-slicing regularization. A.1 Stationary-phase method
I explained the method in Sec. 3, but here is an abbreviated retelling.Split the field into classical solutions plus fluctuations: q ( t ) ≡ ¯ q ( t ) + Q ( t ), with ¯ q ( t ) = q ,¯ q ( t f ) = q f , and Q ( t ) = Q ( t f ) = 0. The stationary-phase condition is¨¯ q ( t ) = J ( t ) . (A.2)The kinetic term produces a total derivative:˙ q = ( ˙¯ q + ˙ Q ) = ˙¯ q + 2 ˙¯ q ˙ Q + ˙ Q = ddt (¯ q ˙¯ q ) − ¯ q ¨¯ q + ddt (2 ˙¯ qQ ) − qQ + ˙ Q . (A.3)Therefore, (cid:90) t f t dt (cid:2) ˙ q ( t ) + J ( t ) q ( t ) (cid:3) = ¯ q ˙¯ q (cid:12)(cid:12) t f t = t + ˙¯ qQ | t f t = t + (cid:90) t f t dt (cid:110) ¯ q ( t ) (cid:2) − ¨¯ q ( t ) + J ( t ) (cid:3) + Q ( t ) [ − ¨¯ q ( t ) + J ( t )] + ˙ Q ( t ) (cid:111) = ¯ q ˙¯ q | t f t = t + (cid:90) t f t dt ¯ q ( t ) J ( t ) + (cid:90) t f t dt ˙ Q . (A.4)The generating function is Z ( q , t ; q f , t f | J ) = e i (cid:20) ¯ q ˙¯ q | tft = t + (cid:82) tft dt ¯ q ( t ) J ( t ) (cid:21) (cid:90) Q ( t f ) = 0 Q ( t ) = 0 D Q ( · ) e i (cid:82) tft dt
12 ˙ Q ( t ) . (A.5) A.2 Classical solution
I will offer a slightly different take on the method from Sec. 3.2.
A.2.1 Homogeneous and particular solutions
The homogeneous solution of Eq. (A.2) has the form q h ( t ) = ( t − t ) A + B , (A.6)92ith some constants A and B . The causal particular solution has the form q p ( t ) = (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) J ( t (cid:48) ) , (A.7)with conditions on the function G ( t ) to be determined by imposing ¨ q p ( t ) = J ( t ):˙ q p ( t ) = (cid:90) t f t dt (cid:48) (cid:104) δ ( t − t (cid:48) ) G ( t − t (cid:48) ) + Θ( t − t (cid:48) ) ˙ G ( t − t (cid:48) ) (cid:105) J ( t (cid:48) )= G (0) J ( t ) + (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) ˙ G ( t − t (cid:48) ) J ( t (cid:48) )= ⇒ G (0) = 0 (A.8)¨ q p ( t ) = (cid:90) t f t dt (cid:48) (cid:104) δ ( t − t (cid:48) ) ˙ G ( t − t (cid:48) ) + Θ( t − t (cid:48) ) ¨ G ( t − t (cid:48) ) (cid:105) J ( t (cid:48) )= ˙ G (0) J ( t ) + (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) ¨ G ( t − t (cid:48) ) J ( t (cid:48) )= ⇒ ˙ G (0) = 1 , ¨ G ( t ) = 0 . (A.9)That is just a way to derive the conditions on G ( t ) instead of announcing them from generalprinciples. Either way, the free-particle Green’s function is G ( t ) = t . (A.10)I will use that to express the homogeneous solution as q h ( t ) = G ( t − t ) A + ˙ G ( t − t ) B , (A.11)in line with the general expression from Eq. (3.13). Combining the homogeneous solutionwith the causal particular solution, I obtain¯ q ( t ) = G ( t − t ) A + ˙ G ( t − t ) B + (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) J ( t (cid:48) ) . (A.12) A.2.2 Boundary conditions
Since Θ( t − t (cid:48) ) = 0 within the physical interval, imposing ¯ q ( t ) = q gives B = q . (A.13)Since Θ( t f − t (cid:48) ) = 1 within that interval, imposing ¯ q ( t f ) = q f gives (let T ≡ t f − t ): q f = G ( T ) A + ˙ G ( T ) q + (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) )= ⇒ A = 1 G ( T ) (cid:20) q f − ˙ G ( T ) q − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:21) . (A.14)93 .2.3 Complete solution The classical solution is then:¯ q ( t ) = G ( t − t ) G ( T ) (cid:20) q f − ˙ G ( T ) q − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:21) + ˙ G ( t − t ) q + (cid:90) t f t dt (cid:48) Θ( t − t (cid:48) ) G ( t − t (cid:48) ) J ( t (cid:48) )= 1 G ( T ) (cid:110) G ( t − t ) q f + (cid:104) G ( T ) ˙ G ( t − t ) − G ( t − t ) ˙ G ( T ) (cid:105) q + (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( T ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J ( t (cid:48) ) (cid:27) = 1 G ( T ) (cid:26) G ( t − t ) q f + G ( t f − t ) q + (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( T ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J ( t (cid:48) ) (cid:27) . (A.15)Notice that G ( T ) ˙ G ( t − t ) − G ( t − t ) ˙ G ( T ) = T − ( t − t ) = t f − t = G ( t f − t ); trivial in thiscase, but a trigonometric identity for the oscillator, with G ( t ) = t replaced by G ( t ) = m sin( mt ). A.3 Action on classical solution
To evaluate the action on the classical solution, I will need ˙¯ q ( t ):˙¯ q ( t ) = 1 G ( T ) (cid:110) ˙ G ( t − t ) q f − ˙ G ( t f − t ) q + (cid:90) t f t dt (cid:48) (cid:104) Θ( t − t (cid:48) ) G ( T ) ˙ G ( t − t (cid:48) ) − ˙ G ( t − t ) G ( t f − t (cid:48) ) (cid:105) J ( t (cid:48) ) (cid:111) . (A.16)At t = t : ˙¯ q ( t ) = 1 G ( T ) (cid:26) q f − ˙ G ( T ) q − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:27) . (A.17)At t = t f :˙¯ q ( t f ) = 1 G ( T ) (cid:26) ˙ G ( T ) q f − q + (cid:90) t f t dt (cid:48) (cid:104) G ( T ) ˙ G ( t f − t (cid:48) ) − ˙ G ( T ) G ( t f − t (cid:48) ) (cid:105) J ( t (cid:48) ) (cid:27) . (A.18)94herefore: G ( T ) ¯ q ˙¯ q | t f t = t = q f ˙¯ q ( t f ) − q ˙¯ q ( t )= q f (cid:26) ˙ G ( T ) q f − q + (cid:90) t f t dt (cid:48) (cid:104) G ( T ) ˙ G ( t f − t (cid:48) ) − ˙ G ( T ) G ( t f − t (cid:48) ) (cid:105) J ( t (cid:48) ) (cid:27) − q (cid:26) q f − ˙ G ( T ) q − (cid:90) t f t dt (cid:48) G ( t f − t (cid:48) ) J ( t (cid:48) ) (cid:27) = ˙ G ( T )( q f + q ) − q f q + (cid:90) t f t dt (cid:48) (cid:110)(cid:104) G ( T ) ˙ G ( t f − t (cid:48) ) − ˙ G ( T ) G ( t f − t (cid:48) ) (cid:105) q f + G ( t f − t (cid:48) ) q (cid:111) J ( t (cid:48) )= ˙ G ( T )( q f + q ) − q f q + (cid:90) t f t dt (cid:48) { G ( t (cid:48) − t ) q f + G ( t f − t (cid:48) ) q } J ( t (cid:48) ) . (A.19)I will need to combine that with G ( T ) (cid:90) t f t dt ¯ q ( t ) J ( t ) = (cid:90) t f t dt [ G ( t − t ) q f + G ( t f − t ) q ] J ( t )+ (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( T ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) . (A.20)The terms linear in J add. Since J ( t ) J ( t (cid:48) ) = J ( t (cid:48) ) J ( t ), the term quadratic in J should beput into a manifestly symmetric form:Θ( t − t (cid:48) ) G ( T ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) ) = Θ( t − t (cid:48) ) [ G ( T ) G ( t − t (cid:48) ) − G ( t − t ) G ( t f − t (cid:48) )]+ Θ( t (cid:48) − t ) [ − G ( t − t ) G ( t f − t (cid:48) )]= − Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) − Θ( t (cid:48) − t ) G ( t − t ) G ( t f − t (cid:48) ) . (A.21)With that, I obtain the following form of the generating function: Z ( q , t ; q f , t f | J ) = C ( T ) e iS cl ( q ,t ; q f ,t f | J ) , (A.22)with action S cl ( q , t ; q f , t f | J ) = 12 G ( T ) (cid:26) ˙ G ( T )( q f + q ) − q f q + 2 (cid:90) t f t dt [ G ( t − t ) q f + G ( t f − t ) q ] J ( t ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) (cid:27) , (A.23)and overall factor C ( T ) ≡ (cid:90) Q ( t f ) = 0 Q ( t ) = 0 D Q ( · ) e i (cid:82) tft dt
12 ˙ Q ( t ) . (A.24)95 .4 The overall factor The overall factor C ( T ) does not depend on the external source J , so I will calculate it fromthe source-free path integral: Z ( q , t ; q f , t f |
0) = C ( T ) e iS cl ( q ,t ; q f ,t f | , (A.25)with S cl ( q , t ; q f , t f |
0) = ˙ G ( T )( q f + q ) − q f q G ( T ) . (A.26)Recall that Z ( q , t ; q f , t f |
0) = (cid:104) q f | e − i ˆ HT | q (cid:105) . That amplitude satisfies a composition law: Z ( q , t ; q , t |
0) = (cid:104) q | e − i ( t − t ) ˆ H | q (cid:105) = (cid:104) q | e − i ( t − t ) ˆ H e − i ( t − t ) ˆ H | q (cid:105) = (cid:90) ∞−∞ dq (cid:104) q | e − i ( t − t ) ˆ H | q (cid:105)(cid:104) q | e − i ( t − t ) ˆ H | q (cid:105) (cid:90) ∞−∞ dq Z ( q , t ; q , t | Z ( q , t ; q , t | . (A.27)Because that relation is nonlinear, imposing it will fix the overall normalization of the pathintegral. To calculate the integral over the intermediate field value, I will need the sum oftwo actions evaluated on classical solutions: S cl ( q , t ; q , t |
0) + S cl ( q , t ; q , t |
0) = ˙ G ( t − t )( q + q ) − q q G ( t − t ) + ˙ G ( t − t )( q + q ) − q q G ( t − t )= G ( t − t ) (cid:104) ˙ G ( t − t )( q + q ) − q q (cid:105) + G ( t − t ) (cid:104) ˙ G ( t − t )( q + q ) − q q (cid:105) G ( t − t ) G ( t − t )= G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q G ( t − t ) G ( t − t )+ (cid:104) G ( t − t ) ˙ G ( t − t ) + G ( t − t ) ˙ G ( t − t ) (cid:105) q − G ( t − t ) q + G ( t − t ) q ] q G ( t − t ) G ( t − t )= G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q G ( t − t ) G ( t − t )+ G ( t − t ) q − G ( t − t ) q + G ( t − t ) q ] q G ( t − t ) G ( t − t ) . (A.28)96o prepare for integration, complete the square in the numerator of the second term: G ( t − t ) q − G ( t − t ) q + G ( t − t ) q ] q = G ( t − t ) (cid:20) q − (cid:18) G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) q (cid:21) = G ( t − t ) (cid:34)(cid:18) q − G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) − (cid:18) G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) (cid:35) (A.29)Therefore:2 G ( t − t ) G ( t − t ) [ S cl ( q , t ; q , t |
0) + S cl ( q , t ; q , t | G ( t − t ) (cid:18) q − G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) + G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q − [ G ( t − t ) q + G ( t − t ) q ] G ( t − t )= G ( t − t ) (cid:18) q − G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) + G ( t − t ) (cid:104) G ( t − t ) ˙ G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) q (cid:105) G ( t − t ) − G ( t − t ) q + G ( t − t ) q + 2 G ( t − t ) G ( t − t ) q q G ( t − t )= G ( t − t ) (cid:18) q − G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) − G ( t − t ) G ( t − t ) q q G ( t − t )+ G ( t − t ) (cid:104) G ( t − t ) ˙ G ( t − t ) − G ( t − t ) (cid:105) q + G ( t − t ) (cid:104) G ( t − t ) ˙ G ( t − t ) − G ( t − t ) (cid:105) q G ( t − t ) . (A.30)And so, S cl ( q , t ; q , t | S cl ( q , t ; q , t |
0) = G ( t − t )2 G ( t − t ) G ( t − t ) (cid:18) q − G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) + G ( t − t ) ˙ G ( t − t ) − G ( t − t )2 G ( t − t ) G ( t − t ) q + G ( t − t ) ˙ G ( t − t ) − G ( t − t )2 G ( t − t ) G ( t − t ) q − q q G ( t − t ) . (A.31)Since G ( t − t ) ˙ G ( t − t ) − G ( t − t ) = G ( t − t ) = G ( t − t ) ˙ G ( t − t ) [I included thatfactor of 1 = ˙ G ( t − t ) for reasons that will be come clear shortly], and since G ( t − t ) ˙ G ( t − ) − G ( t − t ) = G ( t − t ) = G ( t − t ) ˙ G ( t − t ), I obtain S cl ( q , t ; q , t | S cl ( q , t ; q , t |
0) = G ( t − t )2 G ( t − t ) G ( t − t ) (cid:18) q − G ( t − t ) q + G ( t − t ) q G ( t − t ) (cid:19) + ˙ G ( t − t )( q + q ) − q q G ( t − t ) (cid:124) (cid:123)(cid:122) (cid:125) = S cl ( q , t ; q , t | . (A.32)Therefore, the integral over the intermediate field value produces: (cid:90) ∞−∞ dq Z ( q , t ; q , t | Z ( q , t ; q , t | C ( t − t ) C ( t − t ) e iS cl ( q ,t ; q ,t | (cid:90) ∞−∞ dq e iG t − t G t − t G t − t ( q − const) = C ( t − t ) C ( t − t ) e iS cl ( q ,t ; q ,t | (cid:115) πi G ( t − t ) G ( t − t ) G ( t − t ) ≡ Z ( q , t ; q , t |
0) = C ( t − t ) e iS cl ( q ,t ; q ,t | = ⇒ C ( t − t ) C ( t − t ) (cid:112) πi G ( t − t ) G ( t − t ) = C ( t − t ) (cid:112) G ( t − t ) . (A.33)The solution is C ( t ) = 1 (cid:112) πi G ( t ) . (A.34) A.5 Result
The generating function for the free particle is Z ( q , t ; q f , t f | J ) = 1 (cid:112) πi G ( T ) e iS cl ( q ,t ; q f ,t f | J ) , G ( t ) = t , T ≡ t f − t (A.35)with action S cl ( q , t ; q f , t f | J ) = 12 G ( T ) (cid:26) ˙ G ( T )( q f + q ) − q f q + 2 (cid:90) t f t dt [ G ( t − t ) q f + G ( t f − t ) q ] J ( t ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) (cid:27) . (A.36)Remark 1: For J = 0, I recover the standard result Z ( q , t ; q f , t f |
0) = 1 √ πi T e i T ( q f − q ) . (A.37)98emark 2: For J (cid:54) = 0, the coefficient of the J J term is K ( t, t (cid:48) ) ≡ − T [Θ( t − t (cid:48) )( t f − t )( t (cid:48) − t ) + Θ( t (cid:48) − t )( t − t )( t f − t (cid:48) )] . (A.38)Remark 3: The generating function and effective action for the harmonic oscillator takeexactly the same forms as in Eqs. (A.35) and (A.36), with G ( t ) = t replaced by G ( t ) = m sin( mt ). A.6 Time-slicing method
Now I will evaluate Eq. (A.35) for J = 0 (i.e., the 0-point amplitude) by time-slicing regu-larization. The defining expression is: Z ( q N , t N ) = (cid:90) q ( t N ) = q N q ( t ) = q D q ( · ) e i (cid:82) tNt dt
12 ˙ q ( t ) ≡ C N (cid:90) ∞−∞ dq N − ... (cid:90) ∞−∞ dq e iε (cid:80) N − j = 0 (cid:16) qj +1 − qjε (cid:17) . (A.39)Here I am using the notation q N ≡ q f and t N ≡ t f , and explicitly denoting the dependenceof Z only on those variables. The overall constant C N depends on N . A.6.1 Composition law
Eq. (A.39) describes the composition law for the transition amplitude: The particle propa-gates from q to q , then from q to q , and so on. Evolving forward yet another step, from( q N , t N ) to ( q N +1 , t N +1 ), would give Z ( q N +1 , t N +1 ) = C N +1 (cid:90) ∞−∞ dq N (cid:90) ∞−∞ dq N − ... (cid:90) ∞−∞ dq e iε (cid:80) Nj = 0 (cid:16) qj +1 − qjε (cid:17) = C N +1 C N (cid:90) ∞−∞ dq N e iε (cid:16) qN +1 − qNε (cid:17) Z ( q N , t N ) . (A.40)Change integration variables from q N to q ≡ q N +1 − q N : Z ( q N +1 , t N +1 ) = C N +1 C N (cid:90) ∞−∞ dq e i ε q Z ( q N +1 − q, t N ) . (A.41)The pertinent limit is always ε →
0, in which case the factor e i ε q oscillates wildly. Soonly the parts near q = 0 contribute appreciably to the integral, and Z ( q N +1 − q, t N ) can beexpanded in a series: Z ( q N +1 , t N +1 ) ≈ C N +1 C N (cid:90) ∞−∞ dq e i ε q (cid:20) Z ( q N +1 , t N ) − q ∂∂q N +1 Z ( q N +1 , t N )+ q ∂ ∂q N +1 Z ( q N +1 , t N ) (cid:21) This is the oscillatory version of the Chapman-Kolmogorov relation for Markov processes. C N +1 C N (cid:20) Z ( q N +1 , t N ) (cid:90) ∞−∞ dq e i ε q − ∂∂q N +1 Z ( q N +1 , t N ) (cid:90) ∞−∞ dq q e i ε q + ∂ ∂q N +1 Z ( q N +1 , t N ) (cid:90) ∞−∞ dq q e i ε q (cid:21) = C N +1 C N (cid:34) Z ( q N +1 , t N ) √ πiε + 0 + ∂ ∂q N +1 Z ( q N +1 , t N )( − i ) ∂∂α (cid:114) πiα (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) α = ε − (cid:35) = C N +1 C N √ πiε (cid:20) Z ( q N +1 , t N ) + iε ∂ ∂q N +1 Z ( q N +1 , t N ) (cid:21) . (A.42)Meanwhile, since t N +1 ≡ t N + ε , I can compare the above to a series expansion in time: Z ( q N +1 , t N +1 ) = Z ( q N +1 , t N + ε ) ≈ Z ( q N +1 , t N ) + ε ∂∂t N Z ( q N +1 , t N ) . (A.43)This fixes the overall factor on the right-hand side of Eq. (A.42) to be 1, in which case C N +1 C N = 1 √ πiε . (A.44)The recursive boundary condition for this is C ≡
1. (The amplitude for the particle tostart at q , t and end at q , t is by definition 1.) Therefore, C = √ πiε C = √ πiε , C = √ πiε C = √ πiε ) , and in general C N = 1( √ πiε ) N . (A.45)That fixes the overall factor. Feynman’s fundamental insight, moreover, was that matchingthe O ( ε ) parts reproduces the Schrodinger equation, i ∂∂t N Z ( q N +1 , t N ) = − ∂ ∂q N +1 Z ( q N +1 , t N ) . (A.46) A.6.2 Iterated integrals
Eq. (A.44) implies that Eq. (A.39) becomes Z ( q N , t N ) = 1( √ πiε ) N (cid:90) ∞−∞ dq N − ... (cid:90) ∞−∞ dq e iε (cid:80) N − j = 0 (cid:16) qj +1 − qjε (cid:17) . (A.47)The task is to calculate those iterated integrals. First define I ≡ (cid:90) ∞−∞ dq e i ε [ ( q − q ) +( q − q ) ] = √ πiε e i ε ( q − q ) . (A.48)100hen define I ≡ (cid:90) ∞−∞ dq e i ε ( q − q ) I = √ πiε (cid:90) ∞−∞ dq e i ε [ q − q ) +( q − q ) ]= √ πiε e i ε ( q − q ) (cid:90) ∞−∞ dq e i ε [ q − (2 q + q ) ] = (cid:114)
12 (2 πiε ) (cid:114)
23 (2 πiε ) e i ε ) ( q − q ) . (A.49)That produces a pattern even I can spot, leading to I N − = (cid:114) (2 πiε ) N − N e iN (2 ε ) ( q N − q ) . (A.50)So the free-particle path integral in Eq. (A.47) is (recall that T = N ε ) Z ( q N , t N ) = 1( √ πiε ) N I N − = 1 √ πiT e i T ( q N − q ) . (A.51) A.6.3 Free-particle factor for harmonic oscillator
Setting q N = q = 0 in Eq. (A.51) produces the missing ingredient from Eq. (3.51): Z = Z (0 ,
0) = 1 √ πiT . (A.52)With that, the overall factor for the oscillator path integral matches Eq. (3.42) for t = T : Z (0 , |
0) = Z (cid:115) TG ( T ) = 1 (cid:112) πiG ( T ) . (A.53) B Integrating out vs. tracing out
I asked two physicists the following question: “What is the difference between integratingout a field and tracing out a field?” One replied, “Why should those two have anything todo with each other?” and the other replied, “Aren’t those the same thing?” I, for one, was about halfway between the two, maximally confused. Here is what I think.Let q ( t ) and x ( t ) be independent fields, and let their combined action have the followinggeneral form: S tot ( q, x ) = S sys ( q ) + S bath ( x ) + S int ( q, x ) . (B.1)For example, S sys ( q ) = ˙ q − m q , S bath ( x ) = ˙ x − M x , and S int ( q, x ) = γqx (for somecoupling γ ); but in this section I will not work through an explicit calculation. As suggested Justin Wilson and A. Zee—I will let you guess who gave which reply. I thank both for interestingdiscussions about this. Balasubramanian et al. [87] seem to have raised the right questions but then devolved into mumblingsabout the thermal partition function instead of applying the Schwinger-Keldysh formalism. I have not triedto work through their examples. q ( t ) as the “system” field whose correlation functionsI want to calculate, and I will think of x ( t ) as the “bath” field, even though in this case it isjust an oscillator with discrete energy levels. B.1 Integrating out
The amplitude-generating function for q ( t ) is Z ( q , x ; q f , x f | J ( · )) ≡ (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) (cid:90) x ( t f ) = x f x ( t ) = x D x ( · ) e i (cid:104) S tot ( q ( · ) , x ( · ))+ (cid:82) tft dt J ( t ) q ( t ) (cid:105) . (B.2)To integrate out the bath means to calculate, possibly in some approximation, the “induced”action S ind ( x , x f | q ( · )) defined by: e iS ind ( x , x f | q ( · )) ≡ (cid:90) x ( t f ) = x f x ( t ) = x D x ( · ) e i [ S bath ( x ( · ))+ S int ( q ( · ) , x ( · ))] . (B.3)The amplitude-generating function for q ( t ) would then have the form Z ( q , x ; q f , x f | J ( · )) = (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) e i (cid:104) S eff ( q ( · ) | x ,x f )+ (cid:82) tft dt J ( t ) q ( t ) (cid:105) , (B.4)with “effective” action S eff ( q ( · ) | x , x f ) = S sys ( q ( · )) + S ind ( x , x f | q ( · )) . (B.5)Note that the effective action depends on the boundary conditions of the bath. B.2 Tracing out
Given the amplitude-generating function in Eq. (B.2), the expectation-value-generating func-tion for q ( t ) in an ensemble described by density matrix ρ is Z ( J ( · ) , J (cid:48) ( · )) ≡ (cid:90) ∞−∞ dq f dx f dq dx dq (cid:48) dx (cid:48) ρ ( q , x ; q (cid:48) , x (cid:48) ) Z ( q , x ; q f , x f | J ( · )) Z ( q (cid:48) , x (cid:48) ; q f , x f | J (cid:48) ( · )) ∗ . (B.6)To trace out the bath means to calculate, again possibly in some approximation, an “effective”density matrix ρ eff ( q , q (cid:48) ; q ( · ) , q (cid:48) ( · )) ≡ (cid:90) ∞−∞ dx f dx dx (cid:48) ρ ( q , x ; q (cid:48) , x (cid:48) ) × (cid:90) x ( t f ) = x f x ( t ) = x D x ( · ) e i [ S bath ( x ( · ))+ S int ( q ( · ) , x ( · ))] (cid:90) x (cid:48) ( t f ) = x f x (cid:48) ( t ) = x (cid:48) D x (cid:48) ( · ) e − i [ S bath ( x (cid:48) ( · ))+ S int ( q (cid:48) ( · ) , x (cid:48) ( · ))] . (B.7) In this section I will be pedantic about denoting which variables are fields and which are numbers, hencethe proliferation of parentheses. This is now reminding me of scattering from monopoles [88] and impurities [89]. q ( t ) would then have the form Z ( J ( · ) , J (cid:48) ( · )) = (cid:90) ∞−∞ dq f dq dq (cid:48) (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) (cid:90) q (cid:48) ( t f ) = q f q (cid:48) ( t ) = q (cid:48) D q (cid:48) ( · ) ρ eff ( q , q (cid:48) ; q ( · ) , q (cid:48) ( · )) × e i (cid:110) S sys ( q ( · ))+ (cid:82) tft dt J ( t ) q ( t ) − (cid:104) S sys ( q (cid:48) ( · ))+ (cid:82) tft dt J (cid:48) ( t ) q (cid:48) ( t ) (cid:105)(cid:111) . (B.8)Note that, in general, the effective density matrix depends on the entire history of the field. B.3 Remarks
Since I will not continue with an example, I should probably stop here and let the equationsspeak for themselves. But this is an appendix, after all.In a sense, both physicists were right. Integrating out means performing the path integralover the bulk fluctuations, while tracing out means integrating over the boundary conditions(against a density matrix)—two different operations that, a priori, have nothing to do witheach other. And yet, calculating the effective density matrix in Eq. (B.7) requires as inputthe induced action from Eq. (B.3).Said that way, it seems obvious: The Feynman path integral is the evolution operator, whichis used to construct the Schrodinger-picture wavefunction, and the effective Schrodinger-picture density matrix requires the effective wavefunction.
C List of results
In this appendix I will collect some key formulas and results.
C.1 Feynman path integrals
The amplitude-generating function (Feynman path integral) is Z ( q , q f | J ) ≡ (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) e i (cid:82) tft dt [ L ( q ( t ))+ J ( t ) q ( t )] . (C.1)For translationally invariant systems, Z ( q , q f | J ) will depend on t and t f only in the com-bination T ≡ t f − t . (C.2) C.1.1 Free particle
The free particle has Lagrangian L ( q ) = ˙ q , and its generating function is Z ( q , q f | J ) = 1 (cid:112) πiG ( T ) e iS cl ( q ,q f | J ) , G ( t ) = t , (C.3)103ith action S cl ( q , q f | J ) = 12 G ( T ) (cid:26) ˙ G ( T ) (cid:0) q f + q (cid:1) − q f q + 2 (cid:90) t f t dt [ G ( t − t ) q f + G ( t f − t ) q ] J ( t ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) (cid:27) . (C.4) C.1.2 Harmonic oscillator
The harmonic oscillator has Lagrangian L ( q ) = ˙ q − m q , and its generating function is Z ( q , q f | J ) = 1 (cid:112) πiG ( T ) e iS cl ( q ,q f | J ) , G ( t ) = 1 m sin( mt ) , (C.5)with action S cl ( q , q f | J ) = 12 G ( T ) (cid:26) ˙ G ( T ) (cid:0) q f + q (cid:1) − q f q + 2 (cid:90) t f t dt [ G ( t − t ) q f + G ( t f − t ) q ] J ( t ) − (cid:90) t f t dt (cid:90) t f t dt (cid:48) [Θ( t − t (cid:48) ) G ( t f − t ) G ( t (cid:48) − t ) + Θ( t (cid:48) − t ) G ( t − t ) G ( t f − t (cid:48) )] J ( t ) J ( t (cid:48) ) (cid:27) . (C.6) C.2 Schwinger-Keldysh path integrals
The expectation-value-generating function (Schwinger-Keldysh path integral) is Z ( J, J (cid:48) ) ≡ (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ . (C.7)In general it depends on the choice of density matrix ρ and on the initial time t ; for trans-lationally invariant systems, t will drop out. The results for Z ( J, J (cid:48) ) will be expressed interms of the influence phase Φ(
J, J (cid:48) ) ≡ − i ln Z ( J, J (cid:48) ) . (C.8) C.2.1 Quadratic actions
The influence phase for quadratic actions has the general formΦ(
J, J (cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) [ G F ( t, t (cid:48) ) J ( t ) J ( t (cid:48) ) − G D ( t, t (cid:48) ) J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) − G < ( t, t (cid:48) ) J ( t ) J (cid:48) ( t (cid:48) ) − G > ( t, t (cid:48) ) J (cid:48) ( t ) J ( t (cid:48) )] . (C.9)The property Φ( J, J ) = 0 implies G F ( t, t (cid:48) ) − G D ( t, t (cid:48) ) − G < ( t, t (cid:48) ) − G > ( t, t (cid:48) ) = 0 . (C.10)104he property Φ( J, J (cid:48) ) ∗ = − Φ( J (cid:48) , J ) implies G F ( t, t (cid:48) ) ∗ = G D ( t, t (cid:48) ) , G < ( t, t (cid:48) ) ∗ = − G < ( t, t (cid:48) ) . (C.11)Invariance of Φ( J, J (cid:48) ) under t ↔ t (cid:48) implies G F ( t (cid:48) , t ) = G F ( t, t (cid:48) ) , G D ( t (cid:48) , t ) = G D ( t, t (cid:48) ) , G < ( t (cid:48) , t ) = G > ( t, t (cid:48) ) . (C.12)For translationally invariant systems, G F ( t, t (cid:48) ) = G F ( t − t (cid:48) ), G D ( t, t (cid:48) ) = G D ( t − t (cid:48) ), G < ( t, t (cid:48) ) = G < ( t − t (cid:48) ), and G > ( t, t (cid:48) ) = G > ( t − t (cid:48) ). C.2.2 Harmonic oscillator in ground state
The ground-state wavefunction of the harmonic oscillator is ψ ( q ) = (cid:0) mπ (cid:1) / e − mq . Puttingthe harmonic oscillator in its ground state means setting ρ ( q, q (cid:48) ) = ψ ( q ) ψ ( q (cid:48) ) ∗ = (cid:113) mπ e − m ( q + q (cid:48) ) . (C.13)The resulting influence phase has the form in Eq. (C.9) with G < ( t ) = i m e imt , (C.14) G > ( t ) = − G < ( t ) ∗ = i m e − imt , G F ( t ) = Θ( t ) G > ( t ) + Θ( − t ) G < ( t ) = i m e − im | t | , and G D ( t ) = G F ( t ) ∗ = − i m e im | t | . C.2.3 Harmonic oscillator in thermal state
The thermal density matrix in the field basis is ρ ( q, q (cid:48) ) = (cid:113) ˙ G ( − iβ ) − iπG ( − iβ ) e i G ( − iβ ) [ ˙ G ( − iβ )( q + q (cid:48) ) − qq (cid:48) ] , (C.15)with G ( t ) = m sin( mt ). The resulting influence phase isΦ( J, J (cid:48) ) = Φ ( J, J (cid:48) ) + (cid:90) t f t dt (cid:90) t f t dt (cid:48) ∆( t − t (cid:48) ) [ J ( t ) − J (cid:48) ( t )] [ J ( t (cid:48) ) − J (cid:48) ( t (cid:48) )] , (C.16)with ∆( t ) = 1 e βm − im cos( mt ) , (C.17)where Φ ( J, J (cid:48) ) is the influence phase for the ground state. The result in Eq. (C.16) also hasthe form in Eq. (C.9), with G < ( t ) = G < ( t ) + ∆( t ) , (C.18)where G < ( t ) = i m e imt from Eq. (C.14). 105 .2.4 Quenched oscillator in ground state The quenched oscillator has Lagrangian L ( q ) = ˙ q − m ( t ) q , with m ( t > t ) = m and m ( t < t ) = m . Putting the system into its ground state amounts to using Z ( q , q f | J ) forthe oscillator with frequency m [Eqs. (C.5) and (C.6)] but calculating Z ( J, J (cid:48) ) with densitymatrix ρ ( q, q (cid:48) ) = (cid:113) m π e − m ( q + q (cid:48) ) . (C.19)The resulting influence phase has the form in Eq. (C.9) with G < ( t, t (cid:48) ) = i m ζ ( t − t ) ζ ( t (cid:48) − t ) ∗ , ζ ( t ) = ˙ G ( t ) + m G ( t ) , G ( t ) = 1 m sin( mt ) . (C.20) C.2.5 Harmonic oscillator in first excited state
The first excited state of the harmonic oscillator has wavefunction ψ ( q ) = √ m q ψ ( q ),leading to a density matrix ρ ( q, q (cid:48) ) = 2 m qq (cid:48) ψ ( q ) ψ ( q (cid:48) ) ∗ = 2 m (cid:113) mπ e − m ( q + q (cid:48) )+ln( qq (cid:48) ) . (C.21)Since ln ρ is not quadratic, the influence phase will not have the form in Eq. (C.9). Theresult isΦ( J, J (cid:48) ) = Φ ( J, J (cid:48) ) − i ln (cid:26) − m (cid:90) t f t dt (cid:90) t f t dt (cid:48) cos[ m ( t − t (cid:48) )] [ J ( t ) − J (cid:48) ( t )] [ J (cid:48) ( t ) − J (cid:48) ( t (cid:48) )] (cid:27) . (C.22) C.3 Larkin-Ovchinnikov path integrals
The OTOC-generating function (Larkin-Ovchinnikov path integral) is Z ( J, J (cid:48) , J (cid:48) , J (cid:48)(cid:48) ) ≡ (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) (cid:90) ∞−∞ dq (cid:48)(cid:48) (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:48) f ρ ( q , q (cid:48)(cid:48) ) × Z ( q , q f | J ) Z ( q (cid:48) , q f | J (cid:48) ) ∗ Z ( q (cid:48) , q (cid:48) f | J (cid:48)(cid:48) ) Z ( q (cid:48)(cid:48) , q (cid:48) f | J (cid:48)(cid:48)(cid:48) ) ∗ . (C.23)The generalized influence phaseΦ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) ≡ − i ln Z ( J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) (C.24)for quadratic actions isΦ(
J, J (cid:48) , J (cid:48)(cid:48) , J (cid:48)(cid:48)(cid:48) ) = (cid:90) t f t dt (cid:90) t f t dt (cid:48) (cid:8) G F ( t − t (cid:48) ) [ J ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )] − G D ( t − t (cid:48) ) [ J (cid:48) ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − G < ( t − t (cid:48) ) [ J ( t ) J (cid:48) ( t (cid:48) ) + J (cid:48)(cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) )] − G > ( t − t (cid:48) ) [ J (cid:48) ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )]+ G < ( t − t (cid:48) ) [ J ( t ) J (cid:48)(cid:48) ( t (cid:48) ) + J (cid:48) ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) − J ( t ) J (cid:48)(cid:48)(cid:48) ( t (cid:48) ) − J (cid:48) ( t ) J (cid:48)(cid:48) ( t (cid:48) )]+ G > ( t − t (cid:48) ) [ J (cid:48)(cid:48) ( t ) J ( t (cid:48) ) + J (cid:48)(cid:48)(cid:48) ( t ) J (cid:48) ( t (cid:48) ) − J (cid:48)(cid:48)(cid:48) ( t ) J ( t (cid:48) ) − J (cid:48)(cid:48) ( t ) J (cid:48) ( t (cid:48) )] (cid:9) . (C.25)106 .4 Lindblad-improved iε prescription The Lagrangian for a harmonic oscillator with iε prescription is L ( q, q (cid:48) ) = (1+ iε ) ˙ q − (1 − iε ) m q − (cid:2) (1 − iε ) ˙ q (cid:48) − (1+ iε ) m q (cid:48) (cid:3) − iε (cid:2) ˙ q ˙ q (cid:48) + m qq (cid:48) + im ( ˙ qq (cid:48) − ˙ q (cid:48) q ) (cid:3) . (C.26)The expectation-value-generating function is similar to Eq. (C.7), except that the integralsover forward and backward fields do not factorize: Z ( J, J (cid:48) ) = (cid:90) ∞−∞ dq f (cid:90) ∞−∞ dq (cid:90) ∞−∞ dq (cid:48) ρ ( q , q (cid:48) ) (cid:90) q ( t f ) = q f q ( t ) = q D q ( · ) (cid:90) q (cid:48) ( t f ) = q f q (cid:48) ( t ) = q (cid:48) D q (cid:48) ( · ) e i (cid:82) tft dt [ L ( q ( t ) , q (cid:48) ( t ))+ J ( t ) q ( t ) − J (cid:48) ( t ) q (cid:48) ( t )] . (C.27)The influence phase has the form in Eq. (C.9) with lesser Green’s function G < ( t ) = i m e i [1+sign( t ) iε ] mt . (C.28)The greater Green’s function is G > ( t ) = − G < ( t ) ∗ = i m e − i [1 − sign( t ) iε ] mt . The Feynmanand Dyson functions are G F ( t ) = Θ( t ) G > ( t ) + Θ( − t ) G < ( t ) = i m e − i (1 − iε ) m | t | and G D ( t ) = G F ( t ) ∗ = − i m e i (1+ iε ) m | t | . References [1] J. Schwinger. Brownian motion of a quantum oscillator.
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