CComplex Langevin: Boundary Terms at Poles
Erhard Seiler a ∗ a Max-Planck-Institut f¨ur Physik (Werner-Heisenberg-Institut)M¨unchen, Germany
July 21, 2020
Abstract
We discuss the problem of possible boundary terms at poles of thedrift in the complex Langevin method, which spoil correctness of themethod. For the simplest, however paradigmatic cases we can findcomplete answers. Lessons for more generic cases as well as openmathematical problems are discussed. ∗ email: [email protected] a r X i v : . [ h e p - l a t ] J u l ontents β = 0 n p = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 74.1.1 The L c evolution . . . . . . . . . . . . . . . . . 74.1.2 Comparison with the FPE evolution . . . . . . 84.1.3 Interpolating function and boundary term . . . 94.2 Remarks on general n p > β > and z p = 0 n p = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 125.2 General n p > n p > , β > , z p (cid:54) = 0 τ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Introduction
The complex Langevin (CL) method has been studied for almost fortyyears [1, 2], but it still has unsolved mathematical aspects. We havegiven a formal justification of the method in [3, 4], pointing out al-ready the possible failure of the justification due to unwanted bound-ary terms. In [5] and [6] we studied these boundary terms in greatdetail for some models, leading to verifiable criteria for correctnessand even to the computation of corrections in cases of failure. Allthese were boundary terms at infinity, resulting from slow decay ofthe probability at infinity, since we were dealing with holomorphicdrifts.A particularly thorny issue is the problem of meromorphic drift arisingfrom zeroes of the complex density defining the models. We havepresented a detailed study of this issue in [7], with the emphasis onnumerical analysis of models, from the simplest one-dimensional caseto QCD.In this note I am beginning a mathematical analysis of the boundaryterms at poles of the drift, focusing on the simplest model alreadystudied in [7], the so-called one-pole model. In fact, I start with afurther simplification of that model in which only the pole term of thedrift is kept; this allows to clarify the issue by carrying out explicitcalculations. A heuristic justification for this simplification is the real-ization that near the pole, this term dominates the drift, so neglectingthe rest of the drift should be a good approximation of the CL pro-cess while it is spending time in the neighborhood of the poles. In thissimple approximation the linkage of the failure of the CL method withthe appearance of boundary terms at the poles becomes manifest.In [7] it was incorrectly claimed that those boundary terms had beenfound in the long time equilibrium limit; this error was pointed outby L. L. Salcedo; see the erratum to [7]. So here we move away fromequilibrium and consider the short time evolutions, where we do findindeed the sought for boundary terms.For the benefit of the reader we briefly recapitulate the general idea ofthe formal justification of the CL method and show where boundaryterms may arise at poles of the drift, invalidating the formal justifica-tion.We we consider a complex density on R ρ ( x ) = exp( − S ( x )) , (cid:90) dxρ ( x ) = 1 , (1) here ρ extends to an entire analytic function, but with possible zeroes(in which case S is of course multivalued).The complex Langevin equation (CLE) in the form used here is dx = K x dt + dw,dy = K y dt , (2)where dw is the Wiener process normalized as (cid:104) dw (cid:105) = 2 dt (3)and the drift is given by K = − S (cid:48) = ρ (cid:48) ρ ; K x = Re K K y = Im K . (4)The drift thus is univalued but has simple poles at the zeroes of ρ .The average of a generic holomorphic observable O is denoted by (cid:104)O(cid:105) t,z ≡ (cid:90) dxdyP z ( x, y ; t ) O ( x + iy ) ; (5)where P z is the probability density on C produced by the CL processstarting at z = x + iy and running for time t ; the time evolutionof P is given by the Fokker-Planck equation (FPE). We say that theCL process yields correct results if (cid:104)O(cid:105) ∞ ,z agrees with the ‘correct’expectation value of the same observable defined as (cid:104)O(cid:105) c = (cid:90) dx O ( x ) ρ ( x ) , (6)i. e. if (cid:104)O(cid:105) ∞ ,z = (cid:104)O(cid:105) c . (7)In [3, 4] it was shown that correctness is assured if the so-called inter-polating function F O ( t, τ ) = (cid:90) ∞ dxP z ( x ; t − τ ) O ( x ; τ ) (8)is independent of the parameter τ ∈ [0 , t ]. Here O ( z ; τ ) is the solutionof the initial value problem ∂∂τ O ( z ; τ ) = L c O ( z ; τ ) , O ( z ; 0) = O ( z ) ; L c = ( ∂ z + K ( z )) ∂ z , (9) he interpolation property follows from F O ( t,
0) = (cid:104)O ( x + iy ) (cid:105) t,x ; F O ( t, t ) = O ( x + iy ; t ) , (10)so ∂ τ F O ( t, τ ) = 0 = ⇒ (cid:104)O(cid:105) t,z = (cid:104)O ( z ; t ) (cid:105) ∀ t > , (11)from which correctness (7) can be deduced. The left hand side of(11), via integration by parts, is equal to a boundary term, arisingfrom possible slow decay at infinity as well as from poles of the drift.Details are found for instance in [3, 7]. In [5, 6] we found boundary terms at infinity by considering the equi-librium distributions. But we could not find boundary terms nearpoles that way, because the equilibrium distribution P ( x, y ; t = ∞ )of the probability density was always found to vanish at least linearlyat the poles of the drift, so holomorphic observables could not lead toboundary terms at the pole, as we will see.The argument goes as follows: for simplicity let’s assume that there isa pole at the origin; for the boundary term arising in equilibrium andfor τ = 0 (see [7]) consider (cid:90) x + y ≤ δ dx dyP ( x, y ; t = ∞ ) L c O ( x + iy ) . (12)Using integration by parts and the Cauchy-Riemann equations (12) is (cid:90) x + y ≤ δ dx dy O ( x + iy )( L T P )( x, y ; t = ∞ ) + B δ = B δ (13)(where L T is the Fokker-Planck operator, see [3]), since the first termof the left hand side vanishes in equilibrium. B δ is a boundary term.Now, since O is holomorphic, L c O has at most a simple pole at theorigin, stemming from the pole in the drift. Since P vanishes linearlyat the origin, the integrand of (12) is bounded in the region of inte-gration, hence the boundary term vanishes for δ →
0. If we considerthe time evolution for finite time t , the boundary term now is givenby B δ = (cid:90) x + y ≤ δ dx dy (cid:8) O ( x + iy ) L T P z ( x, y ; t ) − P z ( x, y ; t ) L c O ( x + iy ) (cid:9) (14) nd the first term of this expression is no longer zero. In fact, belowwe will give an example where the second term of (14) vanishes, butthere is a boundary term arising solely from the first term of (14).The main difficulty is now to understand the L c evolution (9) of ob-servables in the presence of poles. We focus on the simplest model,dubbed one-pole model in [7]. Since the equilibrium distribution doesnot lead to a boundary term, in this note we focus on the short timeevolution, and we do indeed find boundary terms there. The action for the one-pole model can be written (after shifting thecontour of integration) as S = − ln ρ ( x ) = − n p ln x + β ( x + z p ) , (15) ρ ( x ) = x n p exp( − β ( x + z p ) ) (16)with n p a positive integer. The drift of the CL process is then givenby the real and imaginary parts of K ( z ) = ρ (cid:48) ρ = n p z − β ( z + z p ) (17)and the complex Langevin operator determining the evolution of holo-morphic observables is L c = ( D z + K ) D z = ( D z + n p z − β ( z + z p )) D z . (18)where we wrote D z for ddz (19)and we later use the same symbol for the partial derivative. β = 0 Since we are not analyzing the equilibrium distribution we have thefreedom to study systems that do not possess one; this leads to theconsideration of the one-pole model for β = 0. This is the absolutelysimplest model having a pole in the drift. Since z p plays no role, wealso set if equal to zero. e compare the finite time evolution of the probability density underthe CL process with the evolution of the observables under (9) orequivalently the semigroup exp( tL c ). The two evolutions should beconsistent if there are no boundary terms (see [3, 4, 7]).For β = 0 the system could be treated by a real Langevin process. Wewill nevertheless study the system in the complex domain by choosinga complex starting point for the Langevin process. n p = 2 L c evolution The Langevin operator for this special case is L c = D z + 2 z D z . (20)For n p = 2 there is a simplification, pointed out already in [7]: L c is related to the Langevin operator with zero drift by a similaritytransformation: L c = 1 z D z z (21)and hence exp( tL c ) = 1 z exp( tD z ) z , (22)with the integral kernelexp( tL )( z, x (cid:48) ) = x (cid:48) z √ πt exp (cid:18) − ( z − x (cid:48) ) t (cid:19) . (23)Here x (cid:48) is to be understood as an integration variable along the realaxis. The evolution of a holomorphic observable O ( z ) is thus given by O ( z ; t ) = 1 √ πtz (cid:90) ∞−∞ dx (cid:48) x (cid:48) exp (cid:18) − ( z − x (cid:48) ) t (cid:19) O ( x (cid:48) ) . (24)For the observables O k ( z ) ≡ z k , k = 1 , . . . k = − O ( z ; t ) = z + 2 tz , (25) O ( z ; t ) = z + 6 t , (26) O ( z ; t ) = z + 12 tz + 12 t z , (27) O ( z ; t ) = z + 20 tz + 60 t (28) nd O − ( z ; t ) = 1 z . (29)More generally, it is easy to prove inductively that for any k ≥ − O k ( z ; t ) is a polynomial in t ; for even k it is holomorphic, while forodd k it is meromorphic with a simple pole at z = 0. So exp( tL c )applied to polynomial observables is indeed given by the exponentialseries, which actually terminates after finitely many terms. To study the FPE evolution, we have to resort to numerical sim-ulation. We proceed by running 10000 trajectories of the CL pro-cess all with a fixed starting point z , stopping after Langevin times t = 0 . , . , . , . L c evolution, we findthere is good agreement for the even powers but drastic disagreementfor the odd ones, except for very small times.In Fig.1 we show two plots comparing the evolution of the even pow-ers O ( z ; t ) and O ( z ; t ) with the corresponding results (cid:104)O (cid:105) t,z and (cid:104)O (cid:105) t,z based on the FPE, for starting points z = 0 . i , z = 0 . . i and z = 1 . . i . Figure 1: Comparison of O k ( z ; t ) with (cid:104)O k (cid:105) t,z for k = 2 (left) and k = 4(right) for β = 0. The agreement between the FPE and L c evolutions for even powers orresponds to the fact that in this case there are no 1 /z terms ap-pearing, hence no boundary terms at the origin.The opposite is true for the odd powers, there is strong disagreement,indicating the presence of a boundary term. As an example we show inFig.2 the comparison of the FPE and L c evolutions for the observable O − and the same three starting points z = 0 . i , z = 0 . . i and z = 1 . . i ; some data of the comparison are compiled in Table 1. Figure 2: Comparison of O k ( z ; t ) with (cid:104)O k (cid:105) t,z for k = − β = 0. Solidlines: correct results, dashed lines: CL results, connected to guide the eye. From the failure of agreement for the odd powers it is easy to see thatthe interpolating function F O k ( t, τ ) is not independent of τ when k isodd. We choose as the simplest case O − = 1 /z ; according to (29) O − ( z ; τ ) = 1 /z , independent of τ , so according to (8) F O − ( t, τ ) = (cid:90) P z ( x, y ; t − τ ) 1 x + iy dxdy . (30) (cid:104)O (cid:105) t,z O ( z ; t ) 0.46I 0.1i -1.5i -7.5i (cid:104)O (cid:105) t,z -0.18906(3) 0.3435(35) 2.727(21) 12.0794(10) O ( z ; t ) -0.19 0.35 2.75 11.75 (cid:104)O (cid:105) t,z -0.06623(43)i 0.2356(34)i 0.650(11)i 1.284(27)i O ( z ; t ) -0.0674i 0.235i -3.125i -84.125i (cid:104)O (cid:105) t,z O ( z ; t ) 0.0185 0.1625 12.5625 230.063 (cid:104)O − (cid:105) t,z -1.9957(21)i -0.6297(86)i -0.03144(58)i -0.00356(8) i O − ( z ; t ) -2i -2i -2i -2iTable 1: Comparison of CL results with O k ( z ; t ) for z = 0 . i , n p = 2. From Fig.2 and Table 1 it is clear that this is not independent of τ ,and ddτ F O − ( t, τ ) = (cid:90) ( L T P z )( x, y ; t − τ ) 1 x + iy dxdy (cid:54) = 0 . (31)As discussed in Section 2, this is a boundary term, and it can onlybe due to the pole at z = 0, because for finite time P z shows strong(Gaussian) decay. It could be evaluated also directly as a boundaryterm, but this is not necessary.We can also see that there is no boundary term for even observables,such as O ( z ) = z . While it is difficult to evaluate the τ derivativedirectly because it involves L T P ( x, y : t − τ ), we can compute theinterpolating function for different values of τ to see that it is constant:Let’s take for instance t = 2 . and t − τ = 0 . , . , . , .
0. We thenfind, using (26,28) F O (2 , τ ) = 6 τ + (cid:104)O (cid:105) t − τ,z F O (2 , τ ) = 60 τ + 20 τ (cid:104)O (cid:105) t − τ,z + (cid:104)O (cid:105) t − τ,z , (32)In Table 2 we present the values of these quantities, showing indepen-dence of τ within the errors. = 0 . i : τ F O (2 , τ ) 11.75(0) 11 . . . . F O (2 , τ ) 230.0625(0) 230 . . . . . z = 0 . . i : τ F O (2 , τ ) 12.0(0) 12 . . . . F O (2 , τ ) 0.5(0) 0 . . . . F O (2 , τ ) 239.75(0) 239 . . . . . F O (2 , τ ) 20.0(0) 20 . . . . τ and startingpoints z = 0 . i and z = 0 . . i . n p > We have L c = D z + n p z D z . (33)This operator still leaves the linear space spanned by the even non-negative powers z (cid:96) , (cid:96) ≥ n p >
0. The oddpowers z (cid:96) − for (cid:96) ≥ n p / n p odd, iterating the application of L c to O k will not termi-nate. For n p even, however, it does terminate at the power z − n p ; theobservable O − n p ( z ) ≡ z − n p (34)is an eigenvector with eigenvalue 0. As for n p = 2, for k even (cid:104)O k (cid:105) t,z and O k ( z ; t ) agree , whereas for the odd powers they disagree. Like-wise we find that exp( tL c ) O k is a polynomial in t ; for k = 2 (cid:96) it will bea polynomial in z , whereas for k = 2 (cid:96) − z with the lowest negative power being z − n p .As an example, we consider three observables for the case n p = 4: wefind by a simple calculation O ( z ; t ) = z + 2(1 + n p ) t , (35) O ( z ; t ) = z + 4(3 + n p ) tz + 4(1 + n p )(3 + n p ) t , (36) O − n p ( z ; t ) = z − n p . (37) ome data comparing O k ( z ; t ) with (cid:104)O k (cid:105) t,z for the case n p = 4 arecompiled in Table 3. t (cid:104)O (cid:105) t,z -0.14981(30) 0.7528(50) 4.731(30) 9.776(61) 19.56(12) O ( z ; t ) -0.15 0.75 4.75 9.75 19.56 (cid:104)O (cid:105) t,z O ( z ; t ) -0.065 0.7625 31.5625 133.063 546.063 (cid:104)O (cid:105) ,z t O − ( z ; t ) 8i 8i 8i 8i 8iTable 3: Comparison of CL results with O k ( z ; t ) for z = 0 . i , n p = 4. β > and z p = 0 n p = 2 The Langevin operator is now L c = D z + (cid:18) z − βz (cid:19) D z (38)with D z = ddz . (39)As pointed out in [7], the similarity transformation,exp( − S/ L c exp( S/ ≡ − H F P = ( D z + 12 K )( D z − K ) (40)after restricting to the real axis yields in this case essentially theHamiltonian of a harmonic oscillator H F P = − D x + β x − β . (41)(40) implies the relation for the semigroupsexp( − S/
2) exp( tL c ) exp( S/
2) = exp( − tH F P ) . (42)Note that H F P is not positive on L ( R ); it has exactly one negativeeigenmode: ψ ( x ) ∝ exp (cid:18) − β x (cid:19) (43) n [7] it is explained that this problem disappears if one considers H F P on L ( R + ) with Dirichlet boundary conditions at 0; then only the oddeigenvectors contribute. Because ρ ( x ) vanishes at x = 0 the (real)Langevin process avoids the origin.In any case, because of (42) we can use Mehler’s formula [8]exp( − tH F P )( x, y ) = − (cid:115) βπ (1 − e − βt ) × exp (cid:20) − β ( x + y )2 tanh( βt ) (cid:21) exp (cid:18) βxy sinh( βt ) (cid:19) exp(2 βt ) . (44)to obtain the kernel for exp( tL c ):exp( tL c )( x, y ) =2 yx exp (cid:18) β x − y ) (cid:19) exp(2 βt ) (cid:115) βπ (1 − e − βt ) × exp (cid:20) − β ( x + y )2 tanh(2 βt ) (cid:21) exp (cid:18) βxy exp(2 βt ) (cid:19) . (45)We define b ≡ β sinh(2 βt ) ; σ ≡ β (coth(2 βt ) + 1) = 1 − exp( − βt )2 β , (46)so that bσ = exp( − βt ) . (47)Thus (45) becomesexp( tL c )( x, y ) = A ( x ; t ) y exp (cid:18) − y σ (cid:19) exp( bxy ) (48)with A ( x ; t ) = 2 x (cid:115) βπ (1 − e − βt ) exp(2 βt ) exp (cid:18) − x b σ (cid:19) . (49)We now replace x by z , considering it as a complex variable by analyticcontinuation. We consider again the observables O k ( z ) ≡ z k , k =1 , . . . k = −
1; the integrals can be carried out analytically andyield O ( z ; t ) = 1 bz + bσz , (50) O ( z ; t ) = 3 σ + b σ z , (51) ( z ; t ) = 3 σbz + 6 bσ z + b σ z , (52) O ( z ; t ) = 15 σ + 10 b σ z + b σ z (53)and O − ( z ; t ) = (cid:104) /z (cid:105) ρ ( t ) = 1 bσz . (54)As for β = 0, the expressions for the odd powers are now meromorphicfunctions of z , with simple poles at z = 0. For the even powers wehave polynomials in z .Consistency of these results with the earlier ones for β → β → b = 12 t , lim β → σ = 2 t . (55)We can now also consider the limit t → ∞ , usinglim t →∞ b = 0 , lim t →∞ σ = 12 β . (56)For k odd O k ( z ; t ) grows exponentially in t , whereas for k evenlim t →∞ O ( z ; t ) = 3 σ = 32 β (57)and lim t →∞ O ( z ; t ) = 15 σ = 154 β , (58)which are the correct expectation values.We can again compare O k ( z ; t ) and (cid:104)O k (cid:105) t,z for finite times; qualita-tively the situation is not different from the case β = 0, see Fig. 3,except that now the limit t → ∞ can be considered. We find againagreement for k even and disagreement for k odd; for t = 2 the expec-tation values of the even powers have almost reached the infinite timelimit.As before, for k odd there is strong disagreement between O k ( z ; t )(which has a finite limit for t → ∞ ) and (cid:104)O k (cid:105) t,z (which grows expo-nentially); this implies that the interpolating function has a nonzeroslope, signaling a boundary term. n p > For this case we can still give a rather complete analysis, even thoughwe do not have the benefit of the Mehler fomula. We have L c = ( D z + K ) D z = ( D z + n p z − βz ) D z ; (59) O k ( z ; t ) with (cid:104) k (cid:105) t,z for k = 2 , β = 1. so the even and odd subspaces are still invariant under L c . For aholomorphic observable O ( z ), given by a convergent power series O ( z ) = ∞ (cid:88) n =0 a n z n (60)we find ( L c O )( z ) = ∞ (cid:88) n =0 a n ( L c z n ) ≡ ∞ (cid:88) n = − ( L Tc a ) n z n , . (61)The dual action on the coefficients is thus( L Tc ) n = ( n + 2)( n + 1 + n p ) a n +2 − βna n . (62)Looking for eigenvalues of L Tc we find( n + 2)( n + 1 + n p ) a n +2 − βna n = λa n ; (63)writing λ = 2 βk, k ∈ Z , (64)this can be rewritten as an upward recursion a n +2 = 2 β ( n + k )( n + 2)( n + 1 + n p ) a n . (65)There are two choices to start the recursion:(a) at n = 0 b) at n = 1 − n p and the recursion will stop at n = − k . So for fixed k only finitelymany a n will be different from 0 in both cases.The semigroup exp( tL c ) applied to z k , k = 1 − n p , . . . , , , . . . will thusbe given by a polynomial in z and 1 /z . We give a simple example: O ( z ; t ) = 1 + n p β (cid:16) − e − βt (cid:17) + z e − βt , (66)which generalizes (51) to general n p . We now distinguish two cases: • n p even:Let 0 < n p = 2 (cid:96) , (cid:96) integer. The eigenfunctions of L c are in case(a) even polynomials in z and in case (b) odd polynomials in z and 1 /z , i.e. rational functions.For the choice (a) the eigenvalues are nonpositive, correspond-ing to k = 0 , − , − , . . . and the semigroup exp( tL c ) applied to z (cid:96) , (cid:96) = 0 , , , . . . will be given by even polynomials in z .For choice (b) there are positive and negative eigenvalues, cor-responding to k = . . . , − , − , , . . . n p −
1. The semigroupexp( tL c ) applied to z (cid:96) +1 , (cid:96) = − n p . . . , , , , . . . will be givenby odd polynomials in z and z − , with largest negative power z − n p . In particular z − n p is an eigenfunction with the positiveeigenvalue 2 β ( n p − • n p odd:Let 0 < n p = 2 (cid:96) + 1, (cid:96) integer.Again for choice (a) the eigenfunctions of L c are even polynomialsin z . For (b) we obtain even polynomials in z and 1 /z . But thelinear space of the latter contains the polynomials in z arisingfrom case (a). There is only a finite ( (cid:96) − )dimensional space ofpolynomials in 1 /z with the highest negative power being z − (cid:96) ,containing all the eigenvectors with positive eigenvalues. Thereare no odd eigenfunctions. n p > , β > , z p (cid:54) = 0 This case can no longer be solved analytically, whether z p is real or not.It was studied numerically in [7]. Here we discuss some mathematical ubtleties arising in this case.First we want to formulate a mathematical conjecture that might seemplausible, but is in general not correct: Conjecture 1:
Let K ( z ) be meromorphic in C , holomorphic in a do-main G ⊂ C and O ( z ) also holomorphic in G . Then there is a solutionto the initial value problem (9), which is jointly holomorphic in ( t, z ) for z in any simply connected subset of G and t in a neighborhood of R + = { t | t > } . But Conjecture 1 is wrong even for the case of K being an entirefunction, in fact even for K = 0. The following counterexample is dueto Sofya Kovalevskaya (1875): it was found in Wikipedia: Counterexample:
Let K = 0, i. e. L c = D z and O ( z ) = 1 / (1 + z ).Then O ( z ; t ) is not analytic in ( t, z ) at (0 , Proof: O ( z ; t ) is given, using the heat kernel, by O ( z ; t ) = 1 √ πt (cid:90) ∞−∞ dy exp (cid:18) − ( z − y ) t (cid:19) ( y + 1) − . (67)A closed analytic form of this could be given in terms of error func-tions, but it is not needed. The non-analyticity can be seen eitherby looking at a power series ansatz in t and z for the solution, whichdiverges for any t (cid:54) = 0, or by looking at the result for z = 0, which is (cid:114) π t exp (cid:18) t (cid:19) Erfc(1 / √ t ) . (68)This is clearly not analytic in t at t = 0. (cid:3) On the other hand we can express O ( z ; t ), using Fourier transforma-tion, as O ( z ; t ) = 12 π (cid:90) ∞−∞ dk exp( − tk − | k | ) exp( ikz ) (69)showing that for any t ∈ C with Re t > O ( z ; t ) is an entire functionof z .Returning to our one-pole model, − L c is still conjugate to the Hamil-tonian H F P : H F P = − ( D x + 12 K )( D x − K ) = − D x − S (cid:48)(cid:48) + 14 ( S (cid:48) ) , (70)via the similarity transformation (40). Inserting S (cid:48) = − n p x + 2 β ( x + z p ) ; S (cid:48)(cid:48) = n p x + 2 β (71) he Hamiltonian becomes H F P = − D x + n p x ( n p −
2) + β ( x + z p ) − βn p z p x − β ( n p + 1)= H even + H odd . (72)with H even = − D x + n p ( n p − x + β x + β z p − β ( n p + 1) (73)and H odd = − βn p z p x + 2 β xz p (74)For nonreal z p this is a non-hermitian operator and we do not knowmuch about its spectrum; in fact to give the term a precise meaningwe would first have specify the space in which H F P operates. A simplechoice is L ( | ρ ( x ) | dx ).But let us try to find the action of the semigroup exp( tL c ) on powersof z . We have L c z n = n ( n + n p − z n − − βnz n − βnz p z n − (75)and hence the dual action on the Taylor coefficients a n of O is( L Tc a ) n = ( n + 2)( n + 1 + n p ) a n +2 − βna n − β ( n + 1) a n +1 z p . (76)The z p term mixes the even and odd subspaces.(76) has the structure of a downward recursion; iterating this recursionwill produce nonvanishing coefficients a n with n arbitrarily large nega-tive. Unfortunately this will lead to coefficients a n growing factoriallyfor n → ∞ .Formally, (exp( tL c ) O )( z ) will be given by a Laurent series(exp( tL c ) O k ) ( z ) = k (cid:88) n = −∞ a n ( t ) z n , (77)with the coefficients produced by exponentiating the recursion (76).We show in Fig. 4 the expressionlog | a − n ( t ) | log( n ) (78)for the observable O ( z ) = z and the parameters given in the caption.It is seen clearly that there is a linear increase, indicating that | a − n ( t ) | grows roughly like n n/ , so the Laurent expansion diverges for any z . | a − n | ) /n for n = 1 , . . . ,
100 with parameters β = 1 , z p = i, n p =2 , t = 0 . We interprete this situation as follows: it is analogous to the one in thecounterexample above: the semigroup exp( tL c ) applied to the powers z k is not analytic in t at t = 0, so it cannot be constructed via theexponential series. In the counterexample there was a simple way outof this dilemma: just avoid observables with poles. Here there is alsoa subspace of observables that produces a convergent Laurent series(see below). But for pure powers we cannot construct the semigroupby means of the exponential series. We expect that the solutions tothe initial value problem (9) exist and are meromorphic in z , but ingeneral nonanalytic in t at t = 0, just as in the counterexample above.So the analyticity properties of the solution to the initial value problemfor a general holomorphic observable, where it exists, are not as simpleas Conjecture 1 would suggest, undermining the formal justificationof the CL method.But as stated above, it is possible to find a linear subspace of observ-ables that do not suffer from this disease; this subspace is obtained asa deformation of the even subspace for z p = 0. It can be obtained asthe linear span of the eigenvectors of L c to nonpositive eigenvalues,which in turn are deformations of the eigenvectors obtained for z p =0.The dual action L Tc (76) on the coefficients leads to the eigenvalueequation( n + 2)( n + 1 + n p ) a n +2 − βna n − β ( n + 1) a n +1 z p = λa n . (79) his can again be rewritten as an upward recursion a n +2 = 2 β ( n + 2)( n + 1 + n p ) [( n + k ) a n + ( n + 1) z p a n +1 ] , (80)with λ = 2 βk , k = 0 , − , − , . . . . (81)The recursion has to start at n = 0; the lowest coefficients are a = 0 , a = βk n p a . (82)For z p (cid:54) = 0 the recursion no longer terminates, but it produces asequence decaying roughly like 1 / Γ( n/ z : it is not hard to prove by induction a bound of the form | a n | < n/ | a | (2 | β | (1 + | z p | )) n/ (83)For observables from this space we do not expect any boundary termsand the CL method should work.We checked the correctness of the CL method numerically for theexample O , the eigenvector of L c obtained for k = − n = 50, whenthe coefficients a n are below 10 − . In Table 4 we compare the CLresults with those of the L c evolutions. The parameters are given inthe caption. t (cid:104)O (cid:105) t,z exact0 .
00 1.14289(0) 1.142890 .
01 1.09798(12) 1.098070 .
10 0.7664(18) 0.766100 .
50 0.1596(79) 0.154671 .
00 -0.015(12) 0.020932 .
00 -0.036(12) 0.00038Table 4: CL results for the time evolution of the observable O , which is aneigenvector with eigenvalue − β for n p = 2, β = 1, z p = i ; the exact resultsare obtained as exp( − βt ) O ( z ) with z = 0 . i . The table shows that the CL results are correct, possibly with a smalltruncation error for t = 2.It should be noted that the recursion (80) works for any k ∈ R , leadingto an entire function of z . Of course k > t does not lead to convergence fore t → ∞ . But does this mean thatthe whole positive real axis belongs to the spectrum? This questionis not well posed without specifying the space (Hilbert space, Banachspace or a more general topological vector space) in which the problemis posed.In a slightly different way, a subspace of entire functions invariantunder L c is obtained by forming linear combinations of observables forvarious values of k . This means that the second condition of (82) is nolonger enforced, but the condition a = 0 still holds. The invarianceunder L c requires that a is given as a = 2 βz p n p + 2) a ; (84)to preserve this relation under L c enforces a similar linear relationbetween a and a . Continuing this kind of reasoning, we learn thatfor any (cid:96) > a (cid:96) +2 is a fixed multiple of a (cid:96) with a factor that goesto zero at least linearly with z p . The coefficients still obey the bound(83), so this defines the subspace V + of entire functions invariant under L c ; the elements of this subspace will not give rise to boundary termsand the CL process produces correct results for them.It would be nice to find a similar invariant subspace, consisting offunctions holomorphic in C \ { } , and which reduces to the odd sub-space for z p = 0. We could not find such a space because of the theconvergence problems of the Laurent expansion discussed above. Butit is clear that for observables O / ∈ V + the formal justification of theCL method fails and boundary ptoblems are to be expected.. τ = 0 Even though we cannot solve the evolution of holomorphic observablesin the general case, we can still study directly the boundary term B δ (see (13)) at the pole numerically. Applying integration by parts inthe form of Gauss’s and Green’s theorems twice to (14) one finds B δ = − (cid:73) ∂G δ (cid:126)K · (cid:126)n P z ( x, y ; t ) O ( x + iy ) ds + o ( δ ) , (85)with (cid:126)n the outer normal to ∂G δ and (cid:126)K = ( K x , K y ); see [7] for details.The terms lumped together in o ( δ ) can be seen to go to zero for δ → B δ →
0. Omitting furthercontributions which are o ( δ ) and omitting multiplicative constants, we an replace the observable O by its value at the pole, replace this by1 and replace K by 1 / ( z − z p )So we are reduced to considering finally B δ ≡ (cid:73) ∂G δ ( x − x p ) − ( y − y p ) r P z ( x, y ; t ) ds . (86)with r = (cid:112) ( x − x p ) + ( y − y p ) . (In our reasoning we assumed O ( z p ) (cid:54) = 0. If O ( z p ) = 0), one has to consider higher derivativesof O , which cannot all vanish, but this would lead too far afield).To estimate (86) numerically, we consider a sequence of rings aroundthe pole, given by δ (1 − η ) < | z − z p | < δ (1 + η ) , δ = 0 . , . , . , .
01 ; (87)with η = 0 . η = 0 . z p = − . i and the startingpoint z = 0 . i . The first thing to note is that the CL process needsa certain amount of time t min before it reaches the line y = y p . Sincethe motion in the y direction is deterministic, we can estimate thisminimal time by moving along the y axis and find t min ≈ . . (88)Secondly, it is difficult to achieve sufficient statistics near the pole,because the probability density P vanishes linearly at the pole, where | (cid:126)K · (cid:126)n | is maximal, creating an ‘overlap problem’. We ran 10 in-dependent trajectories, but still obtained only about 50 hits for δ =0 . , η = 0 . t = 0 .
24 and 0 .
25, and even fewer for other valuesof t .Therefore our results are not precise enough for a reliable extrapolationto δ = 0. But we think it is fair to estimate B ≈ .
15 for t = 0 . t = 0 . B δ for t ≤ .
22 explained above is confirmed by thesimulation. In Fig. 5 we present the results for the Langevin times t = 0 . , . , , , . , . t = 1250).So the numerics indicate that beginning at ≈ .
23 indeedlim δ → B δ (cid:54) = 0 , (90)decreasing again with increasing t and vanishing in the long time (equi-librium) limit. B δ for η = 0 . The first conclusion is that to find boundary terms at poles, it is notsufficient to look at the equilibrium distribution; it is necessary tostudy the short time evolution.The second point is that quite likely the Conjecture 1 (analyticityin t at t = 0) is in general not correct for the simple observableslike powers of z . It does seem to hold, however, for a subspace ofholomorphic observables; for the one-pole model such a subspace hasbeen constructed in Section 6; unfortunately for lattice models it seemsvery hard imitate this construction.Finally, for the pure pole model we established explicitly the existenceof boundary terms at the pole by analyzing the short time evolution.Since in the vicinity of the pole the pole term alway dominates, thepure pole model should give a good approximation of the CL pro-cess for more general models; thus we expect such boundary terms enerally, provided the CL process comes arbitrarily close to the pole.An open question concerns the analyticity properties that can be ex-pected for the solutions of the general initial value problem (9), giventhe analyticity properties of the drift and the initial value O ( z ; 0).Since Conjecture 1 failed, inspired by Kovalevskaya’s counterexample,we formulate something weaker: Conjecture 2:
Let K ( z ) be meromorphic in C , holomorphic in adomain G ⊂ C and O ( z ) also holomorphic in G . Then there is asolution O ( z ; t ) to (9) which for t > is holomorphic in z for z inany simply connected subset of G .Remark: Conjcture 2 of course implies that O ( z ; t ) can only have iso-lated singularities at the poles of K ( z ); these may be poles or essentialsingularities, but could also be branch points.Conjecture 2 has implicitly been assumed to be true for instance in[7]. It certainly would be worth knowing if it can be converted into atheorem.Our experience in the previous section unfortunately suggests the fol-lowing: Conjecture 3:
Let K ( z ) be meromorphic in C , holomorphic in adomain G ⊂ C and O ( z ) also holomorphic in G . Then ‘generically’there is no solution O ( z ; t ) to (9) that is holomorphic jointly in ( t, z ) for z in simply connected subsets of G and t in a neighborhood of R + . The trouble is the lack of analyticity in t at t = 0, just as found in theexample above. Of course the term ‘generically’ is a bit vague; in ourmodel we found that for Conjecture 3 holds for z p (cid:54) = 0 and O / ∈ V + ,but not for z p = 0 and not for z p (cid:54) = 0, O ∈ V . Acknowledgment:
I would like tho thank Gert Aarts, Manuel Scherzer,Denes Sexty and Nucu Stamatescu for the long and fruitful collabo-ration on the CL method. Clearly this note is an outgrowth of thatcollaboration. I am also grateful to Denes Sexty for useful commentson this manuscript.
A The importance of using the cor-rect function space
We want to highlight a subtlety of defining the the semigroup exp( tL c )or equivalently the initial value problem (9) by looking at a simple xample. What we said about the spectrum is equally valid for thesemigroup: without fixing the space in which we search for solutions,the initial value problem is not well posed.As an example consider the observable O − ( z ; 0) = 1 /z for the purepole model with β = 0 , z p = 0 , n p = 2.In (29) we gave the solution as O − ( z ; t ) = 1 /z . (91)But there is a second solution: O − ( z ; t ) = 1 z Erf (cid:18) z √ t (cid:19) . (92)The difference lies in the analyticity properties in t : while for any fixed t ∈ R + (92) is holo morphic in z in the whole complex plane C , it hasan essential singularity in t at t = 0 for any fixed z . Furthermore, for | Re z | < | Im z | , the limit t → | Re z | < | Im z | , for the solu-tion (92) the argument for correctness fails. On the other hand, the real Langevin process, occurring for real starting points, never movesinto the dangerous region and reproduces the second solution for ρ ( x )restricted to the half line x ≥ mero morphicin z for fixed t , but holomorphic in t for fixed z , and solves the initialvalue problem correctly; here it is the pole at the origin that invalidatesthe formal argument by giving rise to a boundary term.So it is essential to specify at least the analyticity domains of thesolutions when attempting to construct the solution of the initial valueproblem (9). The real and complex cases demand different spaces. References [1] G. Parisi,
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