Free Fall of a Quantum Many-Body System
Andrea Colcelli, Giuseppe Mussardo, German Sierra, Andrea Trombettoni
FFree Fall of a Quantum Many–Body System
A. Colcelli, G. Mussardo, G. Sierra, and A. Trombettoni
3, 4 SISSA and INFN, Sezione di Trieste, Via Bonomea 265, I-34136 Trieste, Italy Instituto de F´ısica Te´orica, UAM/CSIC, Universidad Aut´onoma de Madrid, Madrid, Spain CNR-IOM DEMOCRITOS Simulation Center and SISSA, Via Bonomea 265, I-34136 Trieste, Italy Department of Physics, University of Trieste, Strada Costiera 11, I-34151 Trieste, Italy
The quantum version of the free fall problem is a topic usually skipped in undergraduate QuantumMechanics courses because its discussion would require to deal with wavepackets built on the Airyfunctions – a notoriously difficult computation. Here, on the contrary, we show that the problemcan be nicely simplified both for a single particle and for general many–body systems making useof a gauge transformation of the wavefunction corresponding to a change of the reference frame,from the laboratory frame of reference to the one comoving with the falling system. Within thisapproach, the quantum mechanics problem of a particle in an external gravitational potential –counterpart of the free fall of a particle in Classical Mechanics each student is used to see fromhigh-school – reduces to a much simpler one where there is no longer gravitational potential in theSchr¨odinger equation. It is instructive to see that the same procedure applies also to many–bodysystems subjected to a two–body interparticle potential, function of the relative distances betweenthe particles. This topic provides then a handful and pedagogical example of a quantum many–bodysystem whose dynamics can be analytically described in simple terms.
I. INTRODUCTION
In Classical Mechanics one of the first paradigmatic problems that students encounter in their study isthe dynamics of a falling body, i.e. an object pulled down to the ground ( e.g. from the Pisa’s tower) by theconstant force of Earth’s gravity. However, amazingly enough, the same problem is rarely discussed in acourse of Quantum Mechanics and the reason is due to the sharp contrast of the physical simplicity of theproblem and the difficulty of its mathematical description due to how basic Quantum Mechanics coursesare structured, largely based on the solution of the time-dependent Schr¨odinger equation i (cid:126) ∂ψ∂t = Hψ for the wavefunction ψ in terms of the eigenfunctions and eigenvalues of the time-independent equation Hψ = Eψ . Indeed, in the traditional quantum mechanics approach to the problem of determining thewavefunction at time t , it is necessary to involve the Airy functions and the projection of the wavefunctionof the falling body into this set of eigenfunctions. Here we show that an alternative and easier way todeal with the quantum treatment of the problem of the falling body is both pedagogically very simple tointroduce and at the same time general enough to be applicable to the single particle case and to generalquantum many–body systems. This approach exploits the possibility to perform a gauge transformationof the wavefunction in correspondence to a change of references frames, from the inertial frame of thelaboratory to the accelerated frame of the falling body. In the new frame there is of course no longerany gravitational effect and therefore the system appears to be ”free” , i.e. non subject to the gravity.It is worth to underline that the same method can be applied to study the effect of gravitational forceon quantum many–body system where the particles interact via a generic two–body potential of theform V ( | r j − r k | ). This approach permits to easily reach some interesting results. For instance, as wediscuss in the following, the time evolution of interesting observables, such as the variance of the positionof a generic falling wavepacket, is the same as the variance of a free wavepacket: the solely effect ofgravity shows up in the behaviour of the expectation values of position (and their powers) which, onthe other hand, can be obtained instead the classical Newton’s second law of motion. This follows fromthe Ehrenfest theorem, see e.g. [1], covered in all Quantum Mechanics courses, from which we can inferthat the momentum of the wavepacket grows (for positive gravitational force, with g <
0) linearly withtime, while its position varies quadratically in time. This last fact will be valid for a generic interactingpotential in any dimensions and, in this paper, we will focus on the three– and one–dimensional casesas explanatory examples. We will also show how to get easily the expression of the energy and the totalmomentum of the falling many–body system using the basic commutation rules taught in a standardcourse of Quantum Mechanics. Finally, as a last non-trivial example, we show how to put in relation theone–body density matrix of the falling body with the corresponding one of the ”free” (although possiblyinteracting) system, and give a simple relation between the eigenvalues of the two density matrices. a r X i v : . [ phy s i c s . g e n - ph ] J u l 𝑔 𝑔 𝒙 FIG. 1: Pictorial visualization of the Einstein’s gedankenexperiment . The effect of an inertial force F = − m g inthe Einsteinium atom due to the acceleration of the rocket (right side picture) is the same as if the rocket wouldbe at rest on Earth (left side picture). II. THE QUANTUM EINSTEIN’S ROCKET
Let us begin by considering the notorious Einstein’s gedankenexperiment of a rocket of length L inan empty space ( i.e. very far from any other celestial body), and subject to an acceleration equal to g (cid:39) . ms . Suppose that inside the rocket there is a single quantum object, e.g. an Einsteinium atom,for which the relevant Schr¨odinger equation is simply i (cid:126) ∂∂t χ ( x, t ) = (cid:20) − (cid:126) m ∂ ∂x + V r ( x, t ) (cid:21) χ ( x, t ) , (1)where we’ve chosen x as the vertical direction along which the rocket is moving, while V r ( x, t ) is apotential representing the confining action of the rocket walls during the motion. If we now pass to thereference frame comoving with the rocket, by changing the spatial variable from x to (cid:101) x = x − g t i (cid:126) ∂∂t χ ( (cid:101) x, t ) = (cid:20) − (cid:126) m ∂ ∂ (cid:101) x + V r ( (cid:101) x ) + i (cid:126) g t ∂∂ (cid:101) x (cid:21) χ ( (cid:101) x, t ) , (2)where the rocket potential in the new reference is denoted by V r ( (cid:101) x ).If we now transform the wavefunction as χ ( (cid:101) x, t ) ≡ exp (cid:20) i m gt (cid:126) (cid:18)(cid:101) x + g t (cid:19)(cid:21) χ ( (cid:101) x, t ) , (3)we can get rid of the term linear in (cid:98) p , i.e. in ∂∂ (cid:101) x , in Eq. (2) and the Schr¨odinger equation becomes i (cid:126) ∂∂t χ ( (cid:101) x, t ) = (cid:20) − (cid:126) m ∂ ∂ (cid:101) x + V r ( (cid:101) x ) + m g (cid:101) x (cid:21) χ ( (cid:101) x, t ) . (4)Therefore in the comoving frame of reference, the Einsteinium atom feels the presence of a gravitationalforce F = − ∂V∂x = − m g For a discussion of the boundary conditions of the wavefunction χ see [2]. pulling it to the ground of the rocket, as sketched in Fig.1. This is essentially the Einstein’s equivalenceprinciple, which states that there is no difference between the gravitational and a fictitious force experi-enced by an observer in a non-inertial frame of reference. For a more detailed discussion on the Einstein’sequivalence principle in a Quantum Mechanics context see [3].In the following we are going to discuss how to deal with a Schr¨odinger equation of the form of (4),performing a gauge transformation on the wavefunction to wash away the gravitational potential. Noticethat the above derivation can and will be repeated as well for a many–body quantum system made ofinteracting particles. III. FREE FALL OF A QUANTUM PARTICLE
We are interested in studying the Schr¨odinger equation for a particle of mass m subject to a gravitationalpotential in one dimension (see e.g. [4]): i (cid:126) ∂∂t χ ( x, t ) = (cid:18) − (cid:126) m ∂ ∂x + m g x (cid:19) χ ( x, t ) , (5)where χ ( x, t ) is the wavefunction describing the motion of the particle under the force F = − m g , with g the gravity acceleration. This problem can be solved by Fourier transform going to momentum space,as discussed in [5], while other methods of solution, related to the Airy functions, were proposed anddiscussed in [6–9]. For the solution of the problem relative to a time dependent gravitational force seeour recent paper [2]. Gauge Transformation . Let us discuss now the method of solving Eq. (5) by means of a gauge trans-formation of the wavefunction. The starting point is indeed to perform the following gauge transformationon the wavefunction: χ ( x, t ) ≡ e iθ ( x,t ) η ( ρ ( t ) , t ) , (6)where ρ ( t ) = x − ξ ( t )and ξ ( t ) and θ ( x, t ) to be determined. Substituting (6) into (5), we see that, in order to eliminate theexternal potential term, we need to impose dξdt = (cid:126) m ∂θ∂x , − (cid:126) ∂θ∂t = (cid:126) m (cid:18) ∂θ∂x (cid:19) + m g x . (7)Assuming the validity of these equations, it is easy to see thtat η ( ρ, t ) satisfies the Schr¨odinger equationwith no external potential but with the new spatial variables, i.e. i (cid:126) ∂∂t η ( ρ, t ) = − (cid:126) m ∂ ∂ρ η ( ρ, t ) . (8)If we now make the ansatz θ ( x, t ) = m (cid:126) dξdt x + Γ( t ) , (9)and we use it in Eq. (7), we arrive to the conditions m d ξdt = − m g , (cid:126) d Γ dt = − m (cid:18) dξdt (cid:19) , (10)which determine the functions ξ ( t ) and Γ( t ) in terms of the gravitational acceleration g . Once we solvethe differential equations (10) with the trivial initial conditions ξ (0) = dξ (0) /dt = 0 and Γ(0) = 0, we getthe following expression for the gauge phase θ ( x, t ) = − m g t (cid:126) x − m g t (cid:126) , (11)while the ”translational” parameter ξ reads ξ ( t ) = − g t . (12)Eqs. (11) and (12), together with Eqs. (6) and (8), completely solves the Schr¨odinger equation (5), since η ( ρ, t ) is simply the time–dependent solution of the free Schr¨odinger equation, which is studied in anyQuantum Mechanics course, a major example being the spreading of a Gaussian wavepacket. Notice that,with our choices θ ( x,
0) = 0 and ρ (0) = x , from (6) we have χ ( x,
0) = η ( x, χ ( x, t ) = exp (cid:20) iθ ( x, t ) − i t (cid:126) (cid:98) p m (cid:21) η ( ρ,
0) = exp (cid:20) iθ ( x, t ) − i t (cid:126) (cid:98) p m − i ξ ( t ) (cid:126) (cid:98) p (cid:21) χ ( x, , (13)where we use the definition of the translation operator ψ ( x − ξ ( t ) , t ) = exp (cid:20) − i ξ ( t ) (cid:126) (cid:98) p (cid:21) ψ ( x, t ) , (14)and the free time evolution operator. In Eqs. (13) and (14) (cid:98) p refers to the momentum operator: (cid:98) p →− i (cid:126) ∂∂x . Expectation values . Using the results just discussed we can study how the expectation values of differ-ent operators, such as position and momentum, evolve during the motion for a generic initial wavepacket χ ( x, (cid:98) x are defined as (cid:10) x N (cid:11) ( t ) ≡ (cid:10) χ ( x, t ) | (cid:98) x N | χ ( x, t ) (cid:11) = ∞ (cid:90) −∞ | χ ( x, t ) | x N dx , (15)while the expectation values of powers of the momentum (cid:98) p are (cid:10) p N (cid:11) ( t ) ≡ (cid:10) χ ( x, t ) | (cid:98) p N | χ ( x, t ) (cid:11) = ( − i (cid:126) ) N ∞ (cid:90) −∞ χ ∗ ( x, t ) ∂ N ∂x N χ ( x, t ) dx (16)(where the wavefunction ξ is normalized).Assuming as initial values (cid:104) x (cid:105) (0) = x , (cid:104) p (cid:105) (0) = p , (17)we can employ the solution (13) to obtain the time evolved expectation values of these quantities. Inthe following we focus for simplicity our attention on the cases N = 1 ,
2. Since we think the calculationcan be instructive in an introductory Quantum Mechanics course, we perform them in detail so that thevarious steps can be more easily followed.
Commutation relations . Before proceeding in that direction, it is useful to study commutation rela-tions among different operators, such as the position operator with the translation and the time evolutionoperators. Let us start with (cid:104)(cid:98) x, e − i a (cid:98) p (cid:105) = ∞ (cid:88) j =0 ( − i a ) j j ! (cid:2)(cid:98) x, (cid:98) p j (cid:3) = (cid:126) a ∞ (cid:88) j =0 ( − i a ) j − ( j − (cid:98) p j − = (cid:126) a e − i a (cid:98) p , (18)where a is a generic real parameter and we used (cid:2)(cid:98) x, (cid:98) p j (cid:3) = i (cid:126) j (cid:98) p j − . (19)Then we have also (cid:104)(cid:98) x , e − i a (cid:98) p (cid:105) = ∞ (cid:88) j =0 ( − i a ) j j ! (cid:2)(cid:98) x , (cid:98) p j (cid:3) :using the commutation rules, we get (cid:2)(cid:98) x , (cid:98) p j (cid:3) = (cid:98) x (cid:2)(cid:98) x, (cid:98) p j (cid:3) + (cid:2)(cid:98) x, (cid:98) p j (cid:3) (cid:98) x = i (cid:126) j (cid:98) x (cid:98) p j − + i (cid:126) j (cid:98) p j − (cid:98) x = − (cid:126) ( j − j (cid:98) p j − + 2 i (cid:126) j (cid:98) p j − (cid:98) x , (20)where we have used (19) in the second and last equality (with j − j ). Therefore (cid:104)(cid:98) x , e − i a (cid:98) p (cid:105) = a ∞ (cid:88) j =0 ( − i a ) j − ( j − (cid:98) p j − + 2 a ∞ (cid:88) j =0 ( − i a ) j − ( j − (cid:98) p j − (cid:98) x = e − i a (cid:98) p (cid:0) a − a (cid:98) x (cid:1) . (21)Now we turn our attention to the commutation relations between the evolution operator and the positionoperator. We start again from (cid:104)(cid:98) x, e − i b (cid:98) p (cid:105) = ∞ (cid:88) j =0 ( − i b ) j j ! (cid:2)(cid:98) x, (cid:98) p j (cid:3) , where b is some real parameter that will be fixed in the calculations. Using (19) with 2 j instead of j , wefind (cid:104)(cid:98) x, e − i b (cid:98) p (cid:105) = 2 (cid:126) b ∞ (cid:88) j =0 ( − i b ) j − ( j − (cid:98) p j − = 2 (cid:126) b e − i b (cid:98) p (cid:98) p . (22)We then evaluate (cid:104)(cid:98) x , e − i b (cid:98) p (cid:105) = (cid:98) x (cid:104)(cid:98) x, e − i b (cid:98) p (cid:105) + (cid:104)(cid:98) x, e − i b (cid:98) p (cid:105) (cid:98) x = 2 (cid:126) b (cid:98) x e − i b (cid:98) p (cid:98) p + 2 (cid:126) b e − i b (cid:98) p (cid:98) x (cid:98) p == 2 (cid:126) b e − i b (cid:98) p (cid:98) p (cid:98) x + (2 (cid:126) b ) e − i b (cid:98) p (cid:98) p + 2 (cid:126) b e − i b (cid:98) p (cid:98) x (cid:98) p == 4 (cid:126) b e − i b (cid:98) p (cid:98) p (cid:98) x + (2 (cid:126) b ) e − i b (cid:98) p (cid:98) p + 2 i (cid:126) b e − i b (cid:98) p , (23)where we used (22) in the second and third equality, and we used the canonical commutation relation[ (cid:98) x, (cid:98) p ] = i (cid:126) in the last equality.Let us now consider the momentum operator. It obviously commutes with the translation and timeevolution operators, but it will be useful to evaluate its commutation relation with the gauge phase term: (cid:2)(cid:98) p, e i θ ( (cid:98) x,t ) (cid:3) . Writing θ ( (cid:98) x, t ) ≡ (cid:98) xA + B , where in our case A = − m g t (cid:126) , , B = − m g t (cid:126) , we have (cid:104)(cid:98) p, e i θ ( (cid:98) x,t ) (cid:105) = e i B ∞ (cid:88) j =0 ( i A ) j j ! (cid:2)(cid:98) p, (cid:98) x j (cid:3) = (cid:126) A e i B ∞ (cid:88) j =0 ( i A ) j − ( j − (cid:98) x j − = (cid:126) A e i ( (cid:98) xA + B ) ≡ (cid:126) A e i θ ( (cid:98) x,t ) , (24)where we used (cid:2)(cid:98) p, (cid:98) x j (cid:3) = − i (cid:126) j (cid:98) x j − in the second equality. Finally we need to calculate (cid:104)(cid:98) p , e i θ ( (cid:98) x,t ) (cid:105) = (cid:98) p (cid:104)(cid:98) p, e i θ ( (cid:98) x,t ) (cid:105) + (cid:104)(cid:98) p, e i θ ( (cid:98) x,t ) (cid:105) (cid:98) p = (cid:126) A (cid:98) p e i θ ( (cid:98) x,t ) + (cid:126) A e i θ ( (cid:98) x,t ) (cid:98) p == ( (cid:126) A ) e iθ ( (cid:98) x,t ) + 2 (cid:126) A e iθ ( (cid:98) x,t ) (cid:98) p , (25)where we used the usual commutation relation in the first equality, and the result (24) in the second andlast equality. Time evolution of (cid:98) x’s operators . Now we have all quantities we need to evaluate expectation valuesof the state χ ( x, t ) in (13). Let us start with (cid:104) χ ( x, t ) | (cid:98) x | χ ( x, t ) (cid:105) = (cid:28) η ( ρ, (cid:12)(cid:12)(cid:12) exp (cid:20) i t m (cid:126) (cid:98) p (cid:21) (cid:98) x exp (cid:20) − i t m (cid:126) (cid:98) p (cid:21) (cid:12)(cid:12)(cid:12) η ( ρ, (cid:29) , (26)where we used the fact that (cid:98) x commutes with e i θ ( (cid:98) x,t ) . Using (22) with b = t m (cid:126) , we may rewrite theexpectation value as (cid:104) χ ( x, t ) | (cid:98) x | χ ( x, t ) (cid:105) = tm (cid:104) η ( ρ, | (cid:98) p | η ( ρ, (cid:105) + (cid:104) η ( ρ, | (cid:98) x | η ( ρ, (cid:105) == tm (cid:104) η ( x, | (cid:98) p | η ( x, (cid:105) + (cid:28) η ( x, (cid:12)(cid:12)(cid:12) exp (cid:20) i ξ ( t ) (cid:126) (cid:98) p (cid:21) (cid:98) x exp (cid:20) − i ξ ( t ) (cid:126) (cid:98) p (cid:21) (cid:12)(cid:12)(cid:12) η ( x, (cid:29) , (27)where we used the definition of the translation operator (14) in the last equality, since ρ ( t ) = x − ξ ( t ).Next using the commutation relation (18) with a = ξ ( t ) (cid:126) , we get (cid:104) χ ( x, t ) | (cid:98) x | χ ( x, t ) (cid:105) = tm p + ξ ( t ) (cid:104) η ( x, | η ( x, (cid:105) + (cid:104) η ( x, | (cid:98) x | η ( x, (cid:105) = tm p + ξ ( t ) + x , (28)where we employed normalization condition (cid:104) η ( x, | η ( x, (cid:105) = (cid:104) χ ( x, | χ ( x, (cid:105) = 1 , (29)and the definitions for the initial expectation values of momentum and position in Eq. (17). We canalso evaluate the expectation value of (cid:98) x using the same procedure but employing Eqs.(21) and (23), thistime with a = ξ ( t ) / (cid:126) and b = t/ (2 m (cid:126) ) respectively, and we find (cid:10) χ ( x, t ) (cid:12)(cid:12) (cid:98) x (cid:12)(cid:12) χ ( x, t ) (cid:11) = ξ ( t ) − ξ ( t ) tm p − ξ ( t ) x + (cid:10) x (cid:11) free ( t ) , (30)where we have defined (cid:10) x (cid:11) free ( t ) ≡ (cid:10) η ( x, t ) (cid:12)(cid:12) (cid:98) x (cid:12)(cid:12) η ( x, t ) (cid:11) , (31)which is the expectation value of x evaluated on the free Schr¨odinger equation solution η ( x, t ), preparedin the initial state η ( x,
0) = χ ( x, x ∆ x ( t ) = (cid:113) (cid:104) x (cid:105) ( t ) − (cid:104) x (cid:105) ( t ) , (32)using the results (28) and (30), from which we get∆ x ( t ) = (cid:115) (cid:104) x (cid:105) free ( t ) − (cid:18) tm p + x (cid:19) = (cid:113) (cid:104) x (cid:105) free ( t ) − (cid:104) x (cid:105) free ( t ) ≡ ∆ x free ( t ) , (33)where we have rewritten (cid:104) x (cid:105) free ( t ) ≡ tm p + x , which is indeed (cid:10) η ( x, t ) (cid:12)(cid:12) (cid:98) x (cid:12)(cid:12) η ( x, t ) (cid:11) with η ( x, t ) thesolution of the free Schr¨odinger equation. Time evolution of (cid:98) p’s operators . The same computations can be done for the expectation valuesinvolving the momentum. Let’s start with (cid:104) χ ( x, t ) | (cid:98) p | χ ( x, t ) (cid:105) = (cid:68) η ( ρ, t ) | e − i θ ( (cid:98) x,t ) (cid:98) p e i θ ( (cid:98) x,t ) | η ( ρ, t ) (cid:69) = − m g t (cid:104) η ( ρ, t ) | η ( ρ, t ) (cid:105) + (cid:104) η ( ρ, t ) | (cid:98) p | η ( ρ, t ) (cid:105) , where we used Eq. (24), with A = − ( m g t ) / (cid:126) , in the second equality. Now the computation is very simple,since the momentum operator commutes with the translation and time evolution operators: therefore,using the normalization condition (29) and the initial conditions (17), we get (cid:104) χ ( x, t ) | (cid:98) p | χ ( x, t ) (cid:105) = − m g t + p = − m g t + (cid:104) p (cid:105) free (0) , (34)where obviously (cid:104) p (cid:105) free ( t ) ≡ (cid:104) η ( x, t ) | (cid:98) p | η ( x, t ) (cid:105) = (cid:104) η ( x, | (cid:98) p | η ( x, (cid:105) = p ≡ (cid:104) p (cid:105) free (0). For the timeevolution of the expectation value of (cid:98) p we have (cid:10) χ ( x, t ) | (cid:98) p | χ ( x, t ) (cid:11) = m g t − m g t p + (cid:10) p (cid:11) free (0) , (35)where we use the commutation relation (25) with A = − ( m g t ) / (cid:126) , and we defined (cid:10) p (cid:11) free ( t ) ≡ (cid:10) η ( x, t ) | (cid:98) p | η ( x, t ) (cid:11) = (cid:10) η ( x, | (cid:98) p | η ( x, (cid:11) ≡ (cid:10) p (cid:11) free (0) . Finally, let’s compute the variance of (cid:98) p using Eqs. (34) and (35), so that∆ p ( t ) = (cid:113) (cid:104) p (cid:105) ( t ) − (cid:104) p (cid:105) ( t ) = (cid:112) (cid:104) p (cid:105) (0) − ( p ) ≡ ∆ p (0) (36)= (cid:113) (cid:104) p (cid:105) free ( t ) − ( (cid:104) p (cid:105) free ) ( t ) ≡ ∆ p free ( t ) , (37)According to this result, we see that the variance of the momentum remains equal to its initial value at t = 0 while Eq. (37) shows that the evolution of the variance of (cid:98) p for a falling wavepacket is exactlythe same as the one of a free expanding wavepacket. As evident from Eq. (33), also the variance of theposition of the falling wavepacket behaves as in the free expanding ( i.e. no gravity) case. A simple check . We can check the results obtained above in the case of the time evolution of an initialGaussian wavepacket subject to a gravitational force. Since the spreading of the wavepacket in absenceof external potential is done practically in any introductory course of Quantum Mechanics, we think isinstructive to explicitly see the same problem in presence of the gravity, i.e. of a linear potential.We prepare a Gaussian wavepacket centered in x with variance σ , and with initial momentum k asinitial state: χ ( x,
0) = 1 √ πσ exp (cid:20) i k x − ( x − x ) σ (cid:21) . (38)In order to find the evolved state χ ( x, t ) we could expand the initial wavepacket with respect to the basisof eigenfunctions of the Schr¨odinger equation − (cid:126) m ∂ χ∂x + m g x χ ( x, t ) = E χ ( x, t ) . (39)From (13) it is clear that the right basis to use is χ basis ( x, t ) = 1 √ π exp (cid:20) i θ ( x, t ) − i t (cid:126) k m − i ξ ( t ) k + i k x (cid:21) , with k = (cid:114) m E (cid:126) , while θ ( x, t ) and ξ ( t ) are given by Eqs. (11) and (12) respectively. Making thisexpansion according to the standard methods of Quantum Mechanics textbooks (see for instance [1]), wefind χ ( x, t ) = 1 √ π σ e i θ ( x,t ) (cid:114) i (cid:126) t m σ exp − [ x − x − ξ ( t )] + 4 i k σ [ x − x − ξ ( t )] + 2 i (cid:126) t ( k σ ) /m (cid:18) σ + i (cid:126) t m (cid:19) , (40)which has to be compared with the one obtained from the free evolution expansion of the initial wavepacket(38), under the same Schr¨odinger equation but with g = 0, i.e. with χ ( x, t ) free = 1 √ π σ (cid:114) i (cid:126) t m σ exp − ( x − x ) + 4 i k σ ( x − x ) + 2 i (cid:126) t ( k σ ) /m (cid:18) σ + i (cid:126) t m (cid:19) . (41)It is evident that the spreading of these wavefunctions as a function of time is the same and coincideswith the expecting value coming from Eq. (33)∆ x ( t ) = (cid:114) σ + (cid:126) t m σ , (42)while the motion of their centers of mass coincides with the expected value given in Eq. (28). Noticethat the motion of the center of the wavepacket in (40) is the same as the motion of a one–dimensionalclassical particle subjected to the gravitational force, i.e. it is an accelerated motion, and this followsfrom the Ehrenfest theorem. The analysis can be performed also for the momentum variables and oneeasily find the results reported in equations (34), (36) and (37). Three–dimensional case . It is simple to extend the analysis which we presented above to the case of asingle particle falling along the x–direction (once the axis are opportunely chosen) in a three–dimensionalspace. In this case the Schr¨odinger equation reads i (cid:126) ∂∂t χ ( r , t ) = i (cid:126) ∂∂t χ ( x, y, z, t ) = (cid:18) − (cid:126) m ∇ x + m g x (cid:19) χ ( r , t ) , (43)where the vector position r is expressed in Cartesian coordinates in the second equality, and we havedenoted with ∇ x ≡ ∂ ∂x + ∂ ∂y + ∂ ∂z the Laplacian. Proceeding in the same way as for the 1 D case, we perform a gauge transformation onthe wavefunction χ ( r , t ) = e i θ ( x,t ) η ( ρ ( t ) , y, z, t ) , (44)where ρ ( t ) = x − ξ ( t ), and the gauge phase θ ( x, t ) and the translational parameter ξ ( t ) satisfy Eqs. (7).Within these conditions, the Schr¨odinger equation (43) is reduced to the free Schr¨odinger equation for η ( ρ, y, z, t ): i (cid:126) ∂∂t η ( ρ, y, z, t ) = − (cid:126) m ∇ ρ η ( ρ, y, z, t ) , (45)where we define ∇ ρ ≡ ∂ ∂ρ + ∂ ∂y + ∂ ∂z . Analogously to (13), choosing θ ( x, t ) to be (11) and ξ ( t ) given by Eq. (12), we can rewrite (44) withrespect to the evolution and translation operators as χ ( r , t ) = exp (cid:26) iθ ( x, t ) − i t (cid:126) (cid:98) p m − i ξ ( t ) (cid:126) (cid:98) p x (cid:27) χ ( r , , (46)where we have defined (cid:98) p = (cid:98) p x + (cid:98) p y + (cid:98) p z , with the momentum operators acting on different Cartesian coordinates as (cid:98) p α → − i (cid:126) ∂∂α . We are nowable to study how expectation values of different physical quantities evolve. First we redefine expectationvalues of position operator and its powers as (cid:10) α N (cid:11) ( t ) ≡ (cid:10) χ ( r , t ) | (cid:98) α N | χ ( r , t ) (cid:11) = ∞ (cid:90) −∞ dx ∞ (cid:90) −∞ dy ∞ (cid:90) −∞ dz | χ ( r , t ) | α N , (47)where α can be the x , y or z coordinate, while for the powers of the momentum (cid:10) p N α (cid:11) ( t ) ≡ (cid:10) χ ( r , t ) | (cid:98) p N α | χ ( r , t ) (cid:11) = ( − i (cid:126) ) N ∞ (cid:90) −∞ dx ∞ (cid:90) −∞ dy ∞ (cid:90) −∞ dz χ ∗ ( r , t ) ∂ N ∂α N χ ( r , t ) , (48)where α labels the x , y or z component. From Eq. (46) it is straightforward to perform the calculationof expectation values of different coordinates, and since operators acting on different spaces commutes(like (cid:98) x and (cid:98) p y or (cid:98) p y and (cid:98) p z and so on) then the motion on y and z directions is trivially evaluated tobe the free one ( g = 0), while for the x component one relies on the results presented previously for theone–dimensional case.Notice that by writing the vector position in three dimensions as r = x · e x + y · e y + z · e z , (49)where e x , e y and e z are the usual unit vectors in Cartesian coordinate system, i.e. e x = , e y = , e z = , (50)then the expectation value of the vector position r can be written in terms of its single componentsexpectation values (cid:104) r (cid:105) ( t ) = e x · (cid:104) x (cid:105) ( t ) + e y · (cid:104) y (cid:105) ( t ) + e z · (cid:104) z (cid:105) ( t ) , (51)and similarly can be done for the squared vector position r = r · r , which will read (cid:10) r (cid:11) ( t ) = (cid:10) x (cid:11) ( t ) + (cid:10) y (cid:11) ( t ) + (cid:10) z (cid:11) ( t ) . (52)The same decomposition in terms of the single components expectation values obviously holds also forthe momentum operators.So, in summary, the motion of a wavepacket in three–dimensions under the action of gravity is describedby a spreading which is completely analogous to the spreading of a free (no gravity) expansion in alldirections, while its center of mass moves along the direction of the gravitational force as a classicalparticle would do. IV. FREE FALL OF TWO INTERACTING PARTICLES
We now study a three–dimensional system made of 2 interacting particles subject to gravity. TheSchr¨odinger equation reads i (cid:126) ∂∂t χ ( r , r , t ) = (cid:20) − (cid:126) m (cid:0) ∇ r + ∇ r (cid:1) + V ( | r − r | ) + g ( x + x ) (cid:21) χ ( r , r , t ) , (53)where ∇ r j ≡ ∂ ∂x j + ∂ ∂y j + ∂ ∂z j , (54)for j = 1 ,
2, and V ( | r − r | ) describes the interaction among particles and depends only on the distancebetween them | r − r | = (cid:113) ( x − x ) + ( y − y ) + ( z − z ) . (55)In order to solve the Schr¨odinger equation, we employ the same method outlined in the previous Section:We perform a gauge transformation on the wavefunction χ ( r , r , t ) = e i [ θ ( x ,t )+ θ ( x ,t )] η ( (cid:37) ( t ) , (cid:37) ( t ) , t ) , (56)where (cid:37) j ( t ) = ( ρ j , y j , z j ), with ρ j ( t ) = x j − ξ ( t ), while θ ( x j , t ) and ξ ( t ) obey Eqs. (7) for x = x j andwith j = 1 ,
2. Notice that because the interacting potential depends on distance among the particles, itwill remain of the same form after the definition of the new spatial variables ρ j ( t ) and (cid:37) j ( t ). Using theansatz (9) and by choosing ξ (0) = dξ (0) /dt = 0 and Γ(0) = 0 we have that θ ( x j , t ) is given by (11), while ξ ( t ) is given by (12). Under these conditions, η ( (cid:37) , (cid:37) , t ) will satisfy the free Schr¨odinger equation fortwo interacting particles i (cid:126) ∂∂t η ( (cid:37) , (cid:37) , t ) = (cid:20) − (cid:126) m (cid:0) ∇ (cid:37) + ∇ (cid:37) (cid:1) + V ( | (cid:37) − (cid:37) | ) (cid:21) η ( (cid:37) , (cid:37) , t ) , (57)0with ∇ (cid:37) j ≡ ∂ ∂ρ j + ∂ ∂y j + ∂ ∂z j , (58)for j = 1 ,
2. Therefore if one knows how to solve Eq. (57), then the complete solution of (53) reads χ ( r , r , t ) = exp (cid:20) − i m g t (cid:126) (cid:18) g t x + x (cid:19)(cid:21) η (cid:18) x + g t , y , z ; x + g t , y , z ; t (cid:19) , (59)where we have used the solution of the original Schr¨odinger equation with a gravitational force term,with respect to the free ( g = 0) solution of (57).We can now ask the same questions as before: If we start from a generic wavepacket χ ( r , r ,
0) and welet it evolve under the action of gravity, how do its variances and expectation values of powers of positionbehave? Let’s define as usual (cid:10) α N j (cid:11) ( t ) ≡ (cid:10) χ ( r , r , t ) | (cid:98) α N j | χ ( r , r , t ) (cid:11) = (cid:90) dr (cid:90) dr | χ ( r , r , t ) | α N j (60)with α which can be either x , y or z , while j = 1 , (cid:98) p α j for j = 1 , (cid:68) p N α j (cid:69) ( t ) ≡ (cid:68) χ ( r , r , t ) | (cid:98) p N α j | χ ( r , r , t ) (cid:69) = ( − i (cid:126) ) N (cid:90) dr (cid:90) dr χ ∗ ( r , r , t ) ∂ N ∂α N j χ ( r , r , t ) (61)with (cid:82) dr j = (cid:82) ∞−∞ dx j (cid:82) ∞−∞ dy j (cid:82) ∞−∞ dz j . For the initial conditions we take ( j = 1 , (cid:104) α j (cid:105) (0) = α ( j )0 , (cid:10) p α j (cid:11) (0) = p ( j ) α , (62)It is actually very simple to prove that the same results for the position variables of the one particlecase will hold, that is to say the variances of positions of the particles will behave as the free expandingcase, while the expectation values of powers of the x component for positions have the same expressionsof the one body case, see Eqs. (28) and (30), with an additional index j = 1 , y and z components instead one has as result the formulas referring to g = 0, since the gravitationalpotential acts only along the x direction. The simplicity of this result comes from the fact that thecommutators among operators acting on different particles vanish, therefore (cid:104)(cid:98) x j , e − i a (cid:98) p xk (cid:105) = (cid:126) a e − i a (cid:98) p xk δ j,k , (cid:104)(cid:98) x j , e − i a (cid:98) p xk (cid:105) = e − i a (cid:98) p xk (cid:0) a − a (cid:98) x j (cid:1) δ j,k , (cid:104)(cid:98) x j , e − i b (cid:98) p xk (cid:105) = 2 (cid:126) b e − i b (cid:98) p xk (cid:98) p x k δ j,k , (cid:104)(cid:98) x j , e − i b (cid:98) p xk (cid:105) = (cid:104) (cid:126) b e − i b (cid:98) p xk (cid:98) p x k (cid:98) x j + (2 (cid:126) b ) e − i b (cid:98) p xk (cid:98) p x k + 2 i (cid:126) b e − i b (cid:98) p xk (cid:105) δ j,k , where δ j,k is the Kronecker delta and we have rewritten θ ( (cid:98) x j , t ) = (cid:98) x j A + B . We can rewrite (56) as χ ( r , r , t ) = exp (cid:26) i [ θ ( (cid:98) x , t ) + θ ( (cid:98) x , t )] − i t (cid:126) (cid:20) (cid:98) p + (cid:98) p m + V ( | (cid:98) (cid:37) − (cid:98) (cid:37) | ) (cid:21)(cid:27) η ( (cid:37) , (cid:37) ,
0) = (63)= exp (cid:26) i [ θ ( (cid:98) x , t ) + θ ( (cid:98) x , t )] − i t (cid:126) (cid:98) H − i ξ ( t ) (cid:126) ( (cid:98) p x + (cid:98) p x ) (cid:27) χ ( r , r , , where we have defined (cid:98) H ≡ (cid:98) p + (cid:98) p m + V ( | (cid:98) (cid:37) − (cid:98) (cid:37) | ) . We report for convenience only the commutators on the x components, but the same commutator rules will be valid alsofor y and z . (cid:104) x j (cid:105) ( t ) = tm p ( j ) x + ξ ( t ) + x ( j )0 , (64) (cid:10) x j (cid:11) ( t ) = ξ ( t ) − ξ ( t ) tm p ( j ) x − ξ ( t ) x ( j )0 + (cid:10) x j (cid:11) free ( t ) , (65)∆ x j ( t ) = (cid:113)(cid:10) x j (cid:11) ( t ) − (cid:104) x j (cid:105) ( t ) = (∆ x j ) free ( t ) , (66)where as usual we label with the subscript ”free” the expectation values evaluated on the wavefunction η ( r , r , t ) of the free expanding problem . The same expressions, but with g = 0, are valid for theexpectation values on the y and z components.We therefore conclude that our gauge transformation can be also used to reduce the initial Schr¨odingerequation describing the dynamic of two falling interacting particles to the simpler Schr¨odinger equationwhere no gravitational potential is present, and with the same interacting potential among the particles.The fundamental requirement is that the two–body potential depends only on the relative distancebetween the particles, after all a typical feature for a many–body system. When this condition holds, itis straightforward to generalize the results presented so far for a quantum many–body system subject toa gravitational force. V. FREE FALL OF A QUANTUM MANY–BODY SYSTEM
In the literature the problem of describing the motion of a ”structured”, many–body quantum systemunder the action of a gravitational potential has been addressed for various situations ranging from Bose–Einstein condensates [10–12] to one–dimensional integrable systems [13, 14]. Nevertheless the case of ageneral three–dimensional many–body system subject to gravity can be explicitly addressed using themethod described in the previous Sections, even though it should be now clear how to approach it.Let’s then focus our attention on the Schr¨odinger equation of N (for simplicity spinless) interactingparticles subject to gravity along the x direction i (cid:126) ∂∂t χ ( r , . . . , r N , t ) = − (cid:126) m N (cid:88) j =1 ∇ r j + (cid:88) j
0, as one would expect.Thanks to the simple rewriting of the many–body wavefunction in Eq. (68), we are also able to writethe one–body density matrix of the falling system in terms of the free, free non-falling system. Theone–body density matrix is defined as [15]: ρ ( r , r (cid:48) , t ) = N (cid:90) d r . . . d r N χ ∗ ( r , r , . . . , r N , t ) χ ( r (cid:48) , r , . . . , r N , t ) . (74)Therefore using Eq. (68) we can rewrite the density matrix as: ρ ( r , r (cid:48) , t ) = N e i [ θ ( x (cid:48) ,t ) − θ ( x,t ) ] (cid:90) d (cid:37) . . . d (cid:37) N η ∗ ( (cid:37) , (cid:37) , . . . , (cid:37) N , t ) η ( (cid:37) (cid:48) , (cid:37) , . . . , (cid:37) N , t ) , (75)since d r j = d (cid:37) j for every j , while (cid:37) ( t ) = r − ξ ( t ), (cid:37) (cid:48) ( t ) = r (cid:48) − ξ ( t ), and with x and x (cid:48) being the x -components of r and r (cid:48) respectively. So finally: ρ ( r , r (cid:48) , t ) = e i [ θ ( x (cid:48) ,t ) − θ ( x,t ) ] ρ free ( (cid:37) , (cid:37) (cid:48) , t ) , (76)where ρ free ( (cid:37) , (cid:37) (cid:48) , t ) is defined in terms of the wavefunction η solution of the Schr¨odinger equation withoutgravitational field.For a translational invariant system, the above equation may be further simplified by writing everythingin terms of the relative coordinate R ≡ r − r (cid:48) . In this case, since it is also true that R = (cid:37) − (cid:37) (cid:48) , thenEq. (76) may be rewritten as: ρ ( R , t ) = e i m g t X/ (cid:126) ρ free ( R , t ) , (77)where Eq. (11) has been used and X is the x -component of the R vector position.We may further analyse the eigenvalues of the one–body density matrix for a translational invariantsystem. In the static case, the one–body density matrix satisfies the eigenvalue equation [15] (cid:90) ρ ( r , r (cid:48) ) φ i ( r (cid:48) ) d r (cid:48) = λ i φ i ( r ) , (78)3where λ i is the occupation number of the i –th natural orbital eigenvector φ i ( r ). The λ i are such that (cid:80) i λ i = N . When the Galilean invariance is not broken, the quantum number labeling the occupationof the natural orbitals is the wavevector k , and for an homogeneous system the effective single particlestates are simply plane waves, i.e. ϕ k ( r ) = 1 √ L e i k · x − i t (cid:126) k m , with L denoting the size of the system andwe have considered the free time evolution of the state. Therefore we may write Eq. (78) for a fallingtranslational invariant many–body system as: λ k ( t ) = (cid:90) ρ ( R , t ) e i k · R d R . (79)Now, thanks to Eq. (77), we can write the following relation between the natural orbitals occupationnumbers of the falling system with those of the free non-falling one: λ k ( t ) = λ free (cid:101) k ( t ) , (80)where (cid:101) k = ( k x + m g t/ (cid:126) ) · e x + k y · e y + k z · e z , and we have defined the occupation numbers of the systemwithout gravity ( g = 0) as: λ free k ( t ) = (cid:90) ρ free ( R , t ) e i k · R d R . (81)From the above relations, one may observe that there is only a time–dependent translation over the x -component of the momentum wavevector which identifies the occupation numbers of the falling systemwith respect to the ”free” case with no gravitational potential. VI. CONCLUSIONS
In this paper we have shown that the quantum description of the free fall motion in a gravitational fieldcan be nicely simplified making use of a gauge transformation of the wavefunction, which correspondsto a change of the reference frame for the system, from the laboratory reference frame to the one thatmoves within the falling body. We have also discussed the time evolution relative to a generic three–dimensional quantum many–body system subject to a gravitational potential and we have shown that itcan be described in terms of the free time evolution.The method of gauge transformation appears to be highly versatile and easily applicable, since theexpectation values of relevant physical quantities can be computed in terms of the free expanding results.In particular we have shown that the variances of the initial wavepacket are exactly the same as if thesystem doesn’t feel any gravitational force at all. Regarding the application of the presented method toother systems we mention that it could be pedagogically interesting to apply it to the Dirac equation ina linear potential.A comment on an aspect of the presentation can be useful: we referred to the case in which the quantumsystem evolves without gravity ( g = 0, i.e. absence of the linear potential) as the free case, with ”free”referring to the non falling case. Since the topic of the motion of an object in a linear potential such thegravity ( g (cid:54) = 0) is traditionally called the ”free fall”, one may ask if the choice of referring to the nonfalling case as ”free” is convenient. However, the derivations explicitly presented clearly show that, ingeneral, really the ”free fall” is ”free”, since all the physical observables and quantities (such the one–body density matrix) during the falling dynamics are related to the corresponding ones of the non fallingsystems, and actually are the same if measured in the comoving system. Finally, it is worth stressingthat all the calculations we presented requires only basic knowledge of Quantum Mechanics, available tostudents who attended the first undergraduate courses on this topic. Acknowledgements:
A.T. acknowledges discussions with A. P. Polychronakos during the conference”Mathematical physics of anyons and topological states of matter” in Nordita, Stockholm (March 2019).Both numerical and analytical exercises on the topics presented here were done and discussed duringcourses the authors taught during the years, and we acknowledge feedback and suggestions from thestudents of these courses. [1] D.J. Griffiths,