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A new strong-coupling expansion for quantum field theory based on the Langevin equation
Nuclear Physics B219 (1983) 61-80 North-Holland Publishing Company
A NEW STRONG-COUPLING EXPANSION FOR QUANTUM FIELD THEORY BASED ON T H E LANGEVIN EQUATION CARL M. BENDER
Department of Physics, Washington University, St. Louis, Missouri 63130, USA FRED COOPER
Theoretical Division, Los Alamos National Laboratory, Los Alamos, New Mexico 87545, USA BARRY FREEDMAN
Department of Physics, University of Illinois, Urbana, Illinois 61801, USA Received 23 November 1982 (Final version received 24 January 1983) We review the connection between quantum field theory in d dimensions and a set of classical Langevin equations in d + 1 dimensions. We then present a new method for solving these Langevin equations in the strong-coupling domain. For the special case of supersymmetric quantum mechanics (d = 1) the Langevin equations are also in d = 1 and we use our method to calculate the strong-coupling approximation to the ground-state energy of supersymmetric quantum mechanics when the superpotential is [ ( x ) = ¼gx4. Our numerical results are accurate and they indicate that by studying the critical index of a lattice series one can determine whether the ground-state energy is zero and thereby see the signal of supersymmetry restoration or breaking in the continuum limit.
1. Introduction Recently, there has been renewed interest in the Langevin equation formulation of quantum field theory because of its connection with supersymmetry [1, 2] and because it allows a covariant quantization of gauge fields without a Gribov ambiguity [3, 4]. The Langevin equation and its relation to field theory can be traced back to early work by Ginzburg and Landau, [5] and in the context of ~b4 field theory it is known as the time-dependent Ginzburg-Landau model [6, 7]. The Langevin equation is the simplest time-dependent equation whose solution for large times relaxes to an equilibrium distribution of the Ginzburg-Landau form. The Langevin equation is a dissipative equation of motion for a classical field in the presence of external white noise. To model the effect of white noise we require that the external driving force f(~'), when time-averaged, is not self-correlated; that
is, (f(t))
= O, ( f ( s ) f ( t ) )
is proportional* t o 8(s - t), ( f ( r ) f ( s ) f ( t ) ) = O, ( f ( q ) f ( r ) f ( s ) f ( t ) }
* The proportionality constant is a multiple of Planck's constant h. This is the only place in the theory in which h appears; if it were not for this proportionality constant the theory would be completely classical. 61
62
C.M. Bender et al. / Strong-couplingexpansion
is proportional to 8 ( q - r ) 8 ( s - t ) + 8 ( q - s ) 8 ( r - t ) + 8 ( q - t ) 8 ( r - s ) , and so on. It has been proved [8] that at large times the correlation function of the product of n classical fields, obtained by computing the time average of the solution to the Langevin equation, approaches the statistical mechanics average with respect to the free energy functional of n time independent fields. Therefore, from the equivalence of statistical mechanics and euclidean quantum field theory, one can compute the n-point Green functions for a quantum field theory in d-dimensional euclidean space by (i) solving the appropriate classical Langevin equation in the presence of external sources in d + 1 dimensional space, and then (ii) averaging products of n classical fields over gaussian (white) noise. The purpose of this paper is to present a new method for solving the Langevin equation in the strong-coupling regime. This method is an outgrowth of previous ideas we have published for obtaining good approximate solutions to singular boundary layer differential equation problems [9, 11]. The method consists of introducing a lattice and then solving the discrete version of the Langevin equation for large coupling constant g as a well-defined perturbation series in inverse powers of g. We then perform the average over the white noise. Finally, we extrapolate back to the continuum*. Ordinarily, the Langevin lattice strong-coupling series will be different term by term from the usual lattice strong-coupling series obtained from the path integral [14] (the continuum limits should agree, of course). We do not yet know whether the Langevin lattice strong-coupling series is advantageous compared with the conventional lagrangian lattice strong-coupling series; such a judgment will depend upon the particular theory under examination. However, in this paper we find that the Langevin strong coupling series is extremely easy to obtain. To illustrate this procedure we have chosen a particularly simple quantum theory. By a formal argument involving functional integration, it was recently shown [ 1, 12] that the equal-time correlation functions in supersymmetric quantum mechanics (d -- 1) are identical to those in a zero-dimensional (d = 0) quantum field theory. Thus, to solve this field theory one must solve a very simple Langevin equation in d = 1 dimensions. From the lattice solution of the Langevin equation we show how to compute numerically the n-point Green functions of the theory. By obtaining the solution to this problem we can begin to answer an interesting related question. It is clear that introducing a lattice breaks the supersymmetry of a quantum theory which is supersymmetric in the continuum. Is there a way to see that the underlying continuum theory is supersymmetric? More generally, is there a signal for non-perturbative supersymmetry breaking? In this model a clear signal for supersymmetry breaking is that the ground-state energy is non-zero. Therefore, in this paper we calculate the lattice strong-coupling series for the ground-state * It is clear that this calculational method reverses the order of two procedures; the correct order is to take the continuum limit first and to perform the average over gaussian noise second. However, our numerical results in sect. 4 provide strong justificationfor the legitimacyof our method.
C.M. Bender et al. / Strong-coupling expansion
63
energy; we use the critical index for this series to show that in the continuum limit this series extrapolates to 0 and thus the supersymme:ry is restored. Specifically, we consider a supersymmetric hamiltonian of the form [13, 1] 1 2 6,3 2 H = ~ 1p 2 .+~g x t~r3gx .
(1.1)
On purely dimensional grounds we know that for this hamiltonian the ground-state energy has the form E0 = bg 1/2 ,
(1.2)
where b is a pure number. Our treatment of the Langevin equation gives a lattice strong-coupling series representation for Eo of the form
gl/2yZ/2 ~ A , y " ,
(1.3)
n~0
where
y = (a 2/g)-1/3,
(1.4)
a is the lattice spacing, and A , are numerical coefficients. The continuum limit a ~ 0 of the expression in (1.3) is obtained by finding the behavior of (1.3) as y ~ oo. For the continuum hamiltonian in (1.1) the theory is supersymmetric and Eo (and therefore b) is 0. T o find out whether the introduction of a lattice permanently breaks the supersymmetry of the continuum theory we examine the behavior of oo n ~ , =0 A , y as y --, oo. If we were to find that A,y" ~ y-1/2,
(y .-, oo) ,
(15)
n=O
then we would conclude that in the continuum limit b # 0 and supersymmetry remains broken. In fact, we find that
A , y " ~ y'~,
(1.6)
tl=O
where a < -~. Thus, in the continuum limit, supersymmetry is restored. This paper is organized as follows. In sect. 2 we review the Langevin formula. In sect. 3 we examine the connection between supersymmetric quantum mechanics and the Langevin formalism. In sect. 4 we solve the Langevin equation of quantum field theory in the strong-coupling limit and we evaluate the ground-state energy for the supersymmetric hamiltonian in (1.1). In the appendix we review the properties of the d = 0 Fokker-Planck equation, which is the master equation for the time evolution of the probability distribution determined by the Langevin equation, and show that for large times the time-dependent probability function approaches the time-independent probability relevant for quantum field theory.
64
C.M. Bender et al. / Strong-coupling expansion
2. Review of Langevin formalism
It is widely known that euclidean field theory in d dimensions is equivalent to statistical mechanics in d spatial dimensions with h -1 playing the role of ( k T ) -1. In the theory of dynamical critical phenomena time-dependent models were sought which relaxed to the equilibrium distribution. Such models were constructed by adding dissipation terms to a conservative system and coupling it to a heat bath. The probability distribution function for time-independent (equilibrium) fields is 1 P=q[4)] = ~- exp (-F0[~b ]),
(2.1)
= y ~d) exp (-Fo[~b ])
(2.2)
where Z
is a normalization constant which normalizes to 1 the functional of Peq[~b] with respect to ~b and Fo is the free energy in statistical mechanics and is the euclidean action in field theory. In euclidean field theory the objective is to compute n-point Green functions. The two-point function, for example, is given by <~b(x)~b(y)) = I ~b~b (x)~b(y)P=q[~b],
(2.3)
where x and y are points lying in d-dimensional euclidean space. The simplest equation one can write down which has the property that for large times z, Pcl[~b(x, i-)] ~ P~q[~b(x)] is the Langevin equation od)(x, r)
ar
6F o@~x, z)
= -ro,,-Z-;~-T,+f(x,
z).
(2.4)
Here, F = I da X*~E(~b),
(2.5)
where ~b is a classical field depending on the d-dimensional euclidean space variable x and an artificial time variable ~-. The function f(x, r) is a gaussian white noise source with correlations (f) = O,
(2.6)
(f(x, z)f(x', z')) = 2Fo 8(x - x') 8(T - C ) ,
and so on. We can express these correlation functions as functional integrals with respect to a gaussian probability distribution/~[f], where /~[f] = l e x p [-4-~0 1 dax d'rf2(x, "r)] ,
(2.7)
C.M. Bender et al. / Strong-coupling expansion
65
and (2.8) For example,
(f(x, r)f(x', ~')) J d~ff(x, ~)f(x', =
¢')/~[f].
(2.9)
If we take as an initial condition ~(x, 0 ) = 0 and solve (2.4) for 4~(x, r) as a functional of f, then one can show that lim (& (x, z)~b (x', ~'))f = (4~(x)4, (x')),
(2.10)
where the right side of (2.10) is the usual two-point function of the field theory described by *fie in (2.5). On a euclidean lattice where we replace x by i, the non-equilibrium probability distribution is given by
P¢,[x(i),r]=(~ 6~(i)-~b(i, r)])f
(2.11)
because now (~b(i, r)~b (/', . ) ) f = I ~f/~[[]$ (i' z)~b (], z)
= I ~ dx(k)x(i)x(/)P[x(k), ~'].
(2.12)
Thus, in the continuum limit
Pd[x, r]=(exp {8(O)I dax ln (8[X(x)-cb(x,r)])})~
(2.13)
(~(x, ~')4~(Y, r))t = ~ @gPct[X, ~']X(x)g(y) •
(2.14)
Using eqs. (2.13) and (2.14), one can show that Pc~[4~,~'] obey~ the Fokker-Planck equation
OP ~ a 8 rSP o-;=ro d
8F1
(2.15)
This equation has the form of an imaginary time functional Schr6dinger equation for P aP = aT
-liP.
(2.16)
66
C.M. Bender et al. / Strong-coupling expansion
This equation forms the basis of the proof that as r--)~, P[~b, r]~P~q[~b] (see appendix). Here we merely note that the time-independent solution to the FokkerPlanck equation satisfies 8Po ~_p 8F &b 0 ~ = constant.
(2.17)
Since P is normalizable the constant must be 0. Thus,
Po[~]:
e-F[4,]
I
~,~ e-r'[~l '
which is the equilibrium probability distribution in (2.1). Associated with ~b(x, ~-) is a path integral in d + 1 dimensions. If we want to calculate
(¢(x, ~')~b(x',~"))f= ; ~:fd:(x,~')~b(x',r')fi[~,
(2.18)
we can make a change of variablesusing (2.4)and (2.7), a~b+ 1" 8F
1
d~')/Z,
to obtain ^ 0¢
(&(x, r)~b (x', ~")), = I ~b~b (x, ~')* (x', z ) P [ ~ ' '
8r
a/
r 0 ~ - ] det I~-I"
(2.19)
In some instances, notably when d = 0 (see sect. 3), the path integral in (2.19) represents a well-defined field theory in its own right. In this case the field theory is related to a Langevin equation in the same number of dimensions.
3. Langevin equation |or supersymmetric quantum mechanics In this section we show that supersymmetric quantum mechanics is one of the systems whose associated Langevin equation involves the same number of dimensions as the original d = 1 field theory. We do this to prepare for our model calculation in sect. 4 of the ground-state energy of this system (which we know to be zero). The action for supersymmetric quantum mechanics is l . _ t -~ - ~ - , _ ~) . + ~o l _ 2 + D f , ( x ) + ½[~;" ~]/"(x)}, S = I .o& ~x. ~ + ~t~
(3.1)
where f(x) is an arbitrary function of x and is the superpotential which is invariant
C.M. Bender et al. / Strong-coupling expansion
67
under the supersymmetry transformation
iSx =e*d/e-~be,
84J =-ie*D+e*~,
8D = e ~ + ~ * e .
(3.2)
Integrating out the auxiliary field D and going to euclidean space we have
where T
S E = I o .~E dt, ,.~E = ~x 1 .2
+ ½w2(x ) + ~b*[a, - w '(x )]41,
w(x) =f'(x).
(3.4)
(3.5)
When Jr, the external source, is zero we have Z [ 0 ] = ~ e -•"r,
(3.6)
n
where E , are the energy levels of the system. In this case we can perform the path integral over the fermion degrees of freedom exactly to obtain Z = Z+ + Z _ ,
(3.7)
Z± = f ~ x exp [ - S ± ( x ) ] ,
(3.8)
T
1 , 2 +~w 1 2 (x)+~w I t S±= ~0 dt[~x (x)].
(3.9)
Z = Tr e -m" = T r e - m r + Tr e - m r ,
(3.10)
Now
where H ÷ and H _ depend only on bosonic degrees of freedom. In the energy representation we have e -Enr = ~ e -n-÷r + ~ e -n--T. rl
rl
(3.1 1)
n
The energy eigenvalues {E, } are given in terms of the combined spectrum {E, +, E,_}. When the ground-state energies of the two bosonic wells are not degenerate, with E o - < Eo+ say, one then has
Eo = Eo-. We know that a non-degenerate ground-state necessarily implies that supersymmetry is unbroken so Eo = 0. In this case, as we take T ~ oo the expectation value
C.M. Bender et al. / Strong-coupling expansion
68
of an operator Q is given by Tr (e-rrrQ)
<01OI0)= r-,~ lim Tr (e -nr)
= lim ~" ((n + [QIn +) e -~"+r + (n -[QIn-) e -E"-r r-,~ ~ , (e -E"+r + e -t~"-r)
(3.12)
=.
Thus, we obtain the correct ground-state matrix elements for the supersymmetric theory by computing with just Z_ and ignoring Z+ entirely. Suppose, w(x) = gx 3. Then H±
1 2 1 2 6.3 =~p -~g x ±~gx 2 .
(3.13)
No degeneracy is possible for this hamiltonian, so Eo = 0 and one need only consider
Integrating by parts we get
z_-- I
exp {I C-½( +
d,}
(3.15,
But, this is exactly the path integral we obtain for the Langevin system
Y¢= - w (x) + f(t) ,
(3.16)
with 1
P[~=-~exp [-½ f dtf2(t)] .
(3.17)
Note that,
(x (t)x (t'))~ = J Nfx (t)x (t')/~[fl = I ~x det l~xlX(t)x(t')P[2 +w(x)]"
(3.18)
The functional determinant is given by det I~xfl = exp f d t T r l n [ d - w ' ( x ( t ) ) ] 6 ( t - t ' ) d t The free Green function Go satisfies d
-~Go(t-t') =8(t-t') ,
'.
(3.19)
C.M. Bender et al. / Strong-coupling expansion
69
SO G o ( t - t') =
O(t - t') ,
where the constant of integration is chosen to be 0 to satisfy causality. Because of the step function 0, only one term exists in the expansion of the logarithm in (3.19) and we obtain det I~1 : e x p [-½ d t w ' ( x ) ] .
(3.20,
We conclude that Z_ corresponds to the simple one-dimensional Langevin equation in (3.16). In the appendix we show that H_ is also obtainable directly from the Fokker-Planck equation related to the Langevin equation in (3.16). We conclude this section by showing how to verify that when w ( x ) = g x 3 the ground-state energy is zero. We recall that the time-independent probability distribution function for the Langevin equation is P ( x ) = e -v(x) ,
(3.21)
where w ( x ) = ½F'(x). Thus,
e-gX4/2 (3.22)
p(x) =
f dx e -gx4/2 The ground-state energy can be calculated by the virial theorem: E o = (0[HI0) = 2g =(x 6) - 3g(x2) .
(3.23)
Using P ( x ) in (3.22) in (3.23) we have (X 6) ~. (2/g)3/21..,(7)liF(1) ,
(3.24)
(x 2) = ( 2 / g ) l / 2 F ( k ) / F ( ~ ) .
(3.25)
Thus, we verify directly that Eo = O.
4. Strong-coupling lattice expansion for the Langevin equation The method for using the Langevin equation to compute the two-point (or n-point) Green function of a quantum field theory is (i) to solve for the classical field ~b(x, ~') as a functional of the driving force f ( x , r), (ii) to use P(f) in (2.7) to evaluate (~b(x, ~-)~b(x', z))t, and (iii) to take the limit r-,oo. In weak-coupling perturbation theory this method has been shown to give the usual Feynman diagram expansion [6, 3]. Here, we present a totally new method for solving the Langevin equation based on methods we developed to solve boundary layer problems. For simplicity, consider
70
C.M. Bender et al. / Strong-coupling expansion
a (gch4)a field theory whose euclidean action is rl,0_,2+1 _4,J. S = f d a XL~ O) ~go
(4.1)
Then the Langevin equation in (2.4) with Fo = ½is &h _ ½[::]~b + g~b 3 = f .
0T
(4.2)
We are going to solve (4.2) in the large-g regime so we will treat the derivative terms 0~/0~"-~]~b as perturbations. Note that perturbing in the derivatives generally gives a singular perturbation expansion. In an earlier series of papers [9-11] we discussed a new way to solve such equations by introducing a lattice and converting the differential equation to a difference equation. On the lattice the perturbation series is no longer singular but we have had to introduce a new dimensional parameter, the lattice spacing a. We are now faced with the problem of returning to the continuum. In the continuum limit a--* 0, the lattice perturbation series diverges term by term. If we could compute all the terms in the lattice series and sum the series before taking the limit a-* 0 there would be no difficulty. However, when only a finite number of terms in the lattice series have been found Pad6-1ike extrapolation methods must be used to obtain the approximate continuum limit. In refs. [9-11] we verified these lattice extrapolation techniques by solving a number of boundary layer problems. The starting point for a strong-coupling expansion of (4.2) is to introduce a d + 1 dimensional lattice to regulate the derivative terms. We let "c = ia ,
(4.3)
x" = n ~'a,
(4.4)
where i is an integer, and
where/x : 1 ~ d. The derivatives now become differences: limb (n, i ) - ~b(n, i - 1 ) ] , O'r
(4.5)
a
d
[--12 b = ~ - 2 ~ [~b(n +l~,, i)+~b(n -~t, i)-2~.b(n, i)].
(4.6)
Next we introduce the perturbation parameters e = 1/a and e = 1/~ 2. The discrete Langevin equation now takes the form
~[4~(n, i)-~,(n, i-1)]+½8 2d~(n, i1- 2 [~(n +0. i)+4,(n - g , i1] + g~ba(n, i ) = f(n, i ) .
(4.7)
C.M. Bender et al. / Strong-coupling expansion
71
If we assume that 4, (n, i) has a formal expansion in powers of e and 6,
&(n, i)= ~ ~ cbl.,,,(n,i)et6 '' ,
(4.8)
1=0 m = 0
we find that the coefficients ~t.,,(n, i) contain increasing powers of g in the denominator, so that (4.8) is actually a strong-coupling expansion. T o lowest order (4.7) becomes
gdp3oo(n,i)=[(n, i) ,
(4.9)
so that 4,00(n, i ) = [f(n,
i)lg] 1/3 .
(4.10)
Now we must perform the average over white noise. Recall that for gaussian noise
I [ Idax d r ~, , ¢, ,2 ',x~')]
/~[f] = ~ exp -
,
(4.11)
where Z is a normalization constant which sets the functional integral o f / ~ [ f ] to 1. On the lattice (4.11) becomes exp[-a
as.f (n, i)]
.
(4.12)
n.i
In lowest order the two-point correlation function is expressible in terms of a single integral
<~oo(n, i)rkoo(m, i))f = &,,,,g-2/3 I i df ~ f 2 / 3
e-'~'~a½t2.
(4.13)
Integrals over all other lattice points give 1. Note that we obtain 0 unless n = m because fx/3 is odd and integrates to 0. Evaluating the one-dimensional integral in (4.13) gives • 2 "1/3
F(~)/~/~r.
(4.14)
This procedure is easy to carry out to any finite order. To order e t S " we find that the averaging procedure requires that we do a number of ordinary integrals all of which give gamma functions. Now let us specialize to the case d = 0 which corresponds to supersymmetric quantum mechanics. (We also replace ~b by x.) We will show how to calculate the first three terms of the series in detail and then state the results of calculating seven terms. When d = 0 (4.7) becomes e [x (i) - x (i - 1)] + gx (i)3 = [(i).
(4.15)
72
C.M. Bender et al. / Strong-coupling expansion
As in (4.8) we let l,
•
x ( i ) = ~ e xl(t),
(4.16)
/=0
and substitute (4.16) into (4.15). Collecting powers of e and solving recursively gives x (i) to any desired order:
... f xt,)=~
x/3
.
(~)+
~ rf
1/3 .
(,-)
1
3--~[ /2/36) - f
t,)j
e2rfa/3(i-1) fl/3(i-2) fl/3(i-1) f2/3(i-1)] /-4/3(i) -~f4/3(i_1) /2/3(i) f5/3(i ) j + . . . .
+~g[ If
we are interested in computing the two-point function we must square X2(i)
=
f2/3(i)+ 2e [fa/3(i-1) g2/"--"T- ~gL fv3(i)
(4.17) (4.17):
1] -
2 + oeo_~[f_2/3(i) f2/3(i-1), 2f~/3(i-2)f'/3(i) 2f-'/3(i -1)'] -p/3(i) -tp/3(i--1) -- fa/3(i) J + " ' " ~ L
(4.18) In order to calculate (x2(i))f we must average over the gaussian noise using _
/~[f]-~ e
_l/2af2(i )
a
~/~--~.
(4.19)
We interpret (_f)x/3 = _fx/3 in order to insure that (4~)~= 0 to all orders in powers of e. Keeping this in mind it is clear that all odd terms in (4.18) can be dropped [in higher order almost all terms in (4.18) are odd]. Thus, we need only consider
2/.', f2/3(i) 2 e + e 2 r l . - 2 / 3 , . \ f2/3(i--1)] x tt) .... =-~i---~g 9--g~tS tt)- p/a(i ) j + ' " .
(4.20)
Gaussian terms in (4.19) are normalized to one. All non-trivial integrals that arise when we perform the lattice gaussian average of (4.20) using/~[f] have the form
2~--~ I~ e-1/2as2$2k/3d$ =~-~ (2) k/3l"(lk + l) .
(4.21)
Thus, we find at this order (x2(n))~ is independent of n and
(x2)f = ~(2/a )l/3g-2/3F(~)_ 21/(3g) +e 2(a/2)l/3g-'*/3(F(~)/,/~" - F(~)F(-~)/~r]/9 + . " , where it is understood that the continuum limit e =
1/a --.*oo must
be taken.
(4.22)
C.M. Bender et al. / Strong-coupling expansion
73
W e then recognize that an appropriate dimensionless p a r a m e t e r is 3 2 Z ----a---~g'
(4.23)
so that lim F ( ~ ) z
z24~ " ' z" ['P(~)
F(-~)]
is valid for small z. Note that as a -* 0, z -* co so that each t e r m in the series diverges. This is how the singular nature of the continuum perturbation theory is retained in the problem. We know however [see (3.25)] that in the continuum
2 X/2 r ( ~ ) g-1/2 ( x ) f = ~ g ~ z ) = (0.4779886 . . . . ) .
(4.25)
N Thus, we must sum a series of the f o r m z ~,,,=oA,,z 2" [see (1.3)] to get a finite answer. T h e r e are several techniques for doing this. In general we have a series y ( x ) = x ~' E A , , x " ,
(4.26)
where for our case x = z 2 and a = 21-, and we have calculated a finite n u m b e r of terms K : K y r ( x ) = x ~ ~, A ~ x ' . (4.27) n=0 W e know that lim Wo(x) = V~, where Too is finite. Then we see that Xk
[ y r (x ) ]X/~ = E ~ =0 Crux m ,
(4.28)
can be m a d e equivalent to the original series for small x; that is, matching the expansion of (4.28) with (4.27) to order x r uniquely determines the coefficients Cm in terms of the An. Moreover, as x ~ o v , [y~(x)] r/~ has a finite limit. Thus, we choose W: = C r ~ / r ,
(4.29)
as the K t h approximant to 3%. The numerical value of the series for (x2)f in (4.24) is F s1 ~(1_
0.52340954z 2 + 0.24312044z 4
- 0 . 0 7 4 0 7 4 0 7 4 z 6 + 6.4196552 × 10-Sz s + 8.1172514 x 10-3z 10 - 5 . 9 8 3 8 3 0 0 x lO-3z l O + . . . ) .
(4.30)
74
C.M. Bender et al. / Strong-coupling expansion
The first six extrapolants are 1/~/g times 0.62245, 0.55236, 0.52656, 0.51281, 0.50427, 0.49847 which are converging nicely to the exact answer whose numerical value is 0.478 . . . . We are most interested in the ground-state energy E0 which, from the virial theorem is given by [see (3.23)] E0 = 292(x 6)_ 3 g ( x 2). We would like to know how well we can see that E0 = 0, that is, how obvious it is that the continuum theory is supersymmetric in our strong-coupling lattice approximation. There are two possible strategies for calculating E0. One is to separately extrapolate the series for (x 6) and (x 2) and then to examine the differences of the extrapolants and the other is to add the two lattice series first and then to carry out an extrapolation of the resulting series. The calculation of the series for (x 6) follows the same path as that for (x 2). The first few terms are 2
2
5 2/3 x 6 __ ~fO_ ~ _ ~E , .[,,~¢4/3 j o -- z~1/3.,_ ,.2/3_3j_ljo ' J - 1 J O J - r ' g~~7~k]-I 8¢1/3¢1/3 +3/'0 12_.1/3¢ ¢-2/3
3j-2 j oj -1
(4.31)
- ~fo[-~/3) + ' " .
Computing the average with respect to/~(f) using (4.21) gives
z3
r(~) 2+4r(~)z,_~z6 +
(x6), = 2-~'~ [ 1 - 3---~-z
~
"" "] .
(4.32)
In numerical form this series reads Z
3
(x6)f = 2 - ~ ( 1 - 1.0468192z 2 + 0.84913325z4- 0.55555556z 6 +0.28398382z s _ 0.10595548z lo + 0.021912058z 12 + . . . ) .
(4.33)
Recall from (3.24) that the exact answer is (x6)f= (2/g)
3/2F(7)
~=
(0.7169829...)
g-3/2.
(4.34)
The form of the series in (4.33) differs from that in (4.30) in that it starts with z 3 instead of z. The first six extrapolants a r e g-3/2 times 0.85763, 0.75975, 0.82611, 0.98872, 1.30668, 1.12249. Note that in this case we have an asymptotic sequence which at first gets close to the exact answer and then veers away. This is a typical asymptotic behavior for the extrapolants defined in (4.29) as we found in our studies of singular differential equations in refs. [9-11]. Thus, we do quite well in determining (x 2) and (x6). Our best estimates are (X2)f = 0.49847/x/g,
(4% error)
(4.35)
C.M. Bender et al. / Strong-coupling expansion (x6>~ = 0 . 7 5 9 7 5 / g 3/2,
75
(6% e r r o r ) .
(4.36)
Substituting these results into E o in (3.23) gives o u r best estimate for Eo: Eo = 0.024x/g,
(4.37)
instead of zero. Thus, o u r first strategy is reasonably successful*. T h e s e c o n d strategy outlined a b o v e consists of combining the two series in (4.24) and (4.32) to obtain a single series for Eo: 3(5/6)
Eo/4g = lim
51r3/2+3F3(~)
6,~r(~)
-----/--~-z + 2z 3
14F(~) + ~ z
5 z
7 9F3(~) x/Tr+59zr2 9 \ 108~r2 z +.").
(4.38)
T h e extrapolants for this series are - 1.3204, - 1.1434, - 1.0691, - 1.0270, - 0 . 9 9 9 6 , -0.9801 ..... a s e q u e n c e whose terms are slowly decreasing in magnitude. H o w e v e r , it is not possible to conclude with m u c h assurance that the limit of this s e q u e n c e of a p p r o x i m a n t s is 0. T h e best a r g u m e n t that the s e q u e n c e of extrapolants for (4.38) really a p p r o a c h e s 0 d e p e n d s on calculating the critical index of the series. W e assume that as z -* ~ , the right side of (4.38) goes like A z a, w h e r e A and a are constants and ot is called the critical index. If a = 0 (and A ~ 0) then Eo is a n o n - z e r o constant and s u p e r s y m m e t r y remains b r o k e n as the lattice spacing a -* 0. If a < 0 then as the lattice spacing a p p r o a c h e s 0, the right side of (4.38) a p p r o a c h e s zero as a Iaf/3 and the s u p e r s y m m e t r y is restored in the c o n t i n u u m limit. T o d e t e r m i n e a we use the identity
E~(z)
a =z ~ Eo(z ) '
(4.39)
w h e r e Eo(z) represents the r i g h t - h a n d side of (4.38). Evaluating (4.39) numerically and setting x = z 2 gives a = lim (1 - 2 . 0 9 3 6 x + 0 . 9 7 2 4 8 x 2 _ 0 . 4 3 6 9 5 x 3 + 0 . 0 6 5 3 4 5 x 4 + 0 . 1 3 5 3 2 x 5 + . . . ) . X --~oO
(4.40) T o d e t e r m i n e the value of a we find that it is best to m a p the d o m a i n 0 ~
'
(4.41)
* If we look at the sequence of differences between the extrapolants of the series in (4.30) and (4.33) we get the sequence x/g times -0.152, -0.137, +0.073, +0.439 which we see changes sign. Thus, we have crude evidence here that Eo = 0.
C.M. Bender et al. / Strong-coupling expansion
76
and to evaluate the diagonal and off-diagonal Pad6 approximants at y = 1. The series in y is ~t = l i m ( 1 - O . 5 2 3 4 y - O . 4 6 2 6 y 2 - O . 4 0 8 7 y T - O . 3 6 1 3 y 4 - O . 3 2 0 1 y 5 . . . . ).
(4.42)
y"-~ 1
The results are ao,o = 1,
a0.1 = 0.6564,
t~1,1 = - 3 . 5 0 7 , ot2.2 = - 3 . 4 9 9 7 ,
al,o = 0.4766,
al,2 = - 3 . 4 9 0 3 ,
Or2,1 =
a2,a = - 3 . 4 7 6 3 ,
-3.4903,
aa.2 = - 3 . 4 7 6 4 ,
where the subscripts are the indices of the Pad6 approximant used, The results of this Pad6 extrapolation are unambiguous; the value of a lies very close to - 3 . 5 . Hence E0 - l i m a 1.16= 0 .
(4.43)
a~0
We have thus shown that the Langevin equations provide an alternative method for performing strong-coupling expansions to the usual path integral method. Our method is simple and straightforward and avoids the necessity of diagram techniques. By applying these methods to supersymmetric quantum mechanics we were able to show that strong-coupling lattice expansions do provide a calculational tool for discussing supersymmetry breaking. By examining the index of a lattice series we can see whether in the continuum limit the series gives a zero or non-zero result for the ground-state energy. We thank the U.S. Department of Energy for partial financial support. We also thank the MIT M A T H L A B group for the use of MACSYMA. Two of us, C.M.B. and B.F., thank the Los Alamos National Laboratory for its hospitality,
Appendix In this appendix we examine the properties of the Fokker-Planck equation in d = 0 dimensions. We will show here that the time-dependent probability function which satisfies the Fokker-Planck equation approaches the time independent probability distribution of quantum field theory [8]. Consider the Langevin equation (2.4) in d = 0 dimensions [here we use t instead of ~- and write A(x) in place of F'(x)]:
Y¢(t) = -FoA (x ) + f(t).
(A. 1)
The gaussian probability distribution function is taken to be
fi[f]=exp[-I dtl'2(t)] /~, 4/"0 ] /
'
(A.2)
C.M. Bender et aL / Strong-couplingexpansion
77
Next define the function g(y, t) -~6[y -x(t)].
(A.3)
In terms of g the time-dependent probability distribution function is Pc,(Y, t) = I i6[f]g(Y' t)~f.
(A.4)
Note that from (AA) we have
I ~ f x " ( t ) P ( f ) = I dy y"Pc,(y, t). To obtain the master equation satisfied by Pcl(y, t) we write down the chain rule
at g =~x ~ = -
{g[f(t)-FoA ]},
and average this equation over/~[/] to obtain 0 (9 ~Pd-Ao~y[P~,A]=-
LOf 2 Z ~yyj ~ / g f exp [ - ~ ~ d t ] .
(A.5,
Next we use integration by parts to derive the identity
ox [q f2(t') o dt'], 8-~-lt =
8f(t)
2
= [ ~ ~- f gf exp [ - f f2(t') - - ~ °- dt']j ,
(A.6)
and note that ag 8x(t) Ox(t) 8f(t)
8g(y, t) 8f(t)
o [gsx~t)] Oy
8f(t)3 "
(A.7)
Finally, we integrate the equation
BYe(t! = $ (t - t') 8f(t') to obtain 8x (t)
= lim
0 ( t - t') =
1
(A.8)
Combining (A.6)-(A.8) with (A.5) gives the Fokker-Planck equation: 2
at
y
y
(A.9)
C.M. Benderet al. / Strong-couplingexpansion
78
From here on we set Fo = ½and recall that A = F'. Thus (A.9) reads
OPc,_lot 20yO[~-~+Pc'F'] "
(A. 10)
It is clear from (A.10) that a stationary solution (one for which OPd/Ot = 0) is Po[y ] = e -Fry] •
(A. 11)
We want to show that at large times P~[y, t]-* Po[y ]. To show this it is useful to view (A.10) as a SchrSdinger equation with imaginary time and a non-hermitian hamiltonian (non-hermitian because there is a onederivative term): ^
0
( H +~)P¢,(y, t ) = 0 ,
(A.12)
I'?I =½p2 + ipw ,
(A.13)
where w = ½A. Now if {A,} are the eigenvalues of/-t and {lug)} are the associated eigenfunctions and we assume that the set {lu,)} is sufficient to expand P¢l[y, t] we have
e¢,(x, t) = ~ C.(xIu.) e -x"' n=0
= ~ C.¢t,,(x) e -az.
(A.14)
n=0
It is clear that Po(x) is a solution to the SchrSdinger equation (A.10) with Ao = 0 so that (XlUo)=Po(x). Thus, if we are to prove that Pcl(x, t)~Po(x) as t ~ , we must show that Co = 1 and A, > 0 for n = 1, 2, 3 . . . . . To establish the positivity of the eigenvalues we relate/-t to another hamiltonian operator that is self-adjoint and has the same spectrum. Completing the square gives • 2 +~w 1 , +~w 1 2. =-½p2+ipw=½(p +tw) But,
(p+iw)tO,,(x)=pexp[-~Xw dx]tO. (x). So if we introduce a new basis
{In)}by
I n ) = e x p [ - I x w dx][u.),
(AA5)
then
luo > =
dXln),
(A.16)
C.M. Bender et al. / Strong-coupling expansion
79
and the operator on the right-hand side of (A.16), which we call H, is hermitian: H _ --e-f~axiSief~dx
1 , = -½p2 +~w
~t_lW2
(A.17)
In fact H is precisely the hamiltonian associated with supersymmetric quantum mechanics [see (3.14) for example]. The transformation in (A.17) has eliminated the imaginary magnetic field in/-I by means of a gauge transformation. The eigenvalue equation/-Ilu.) = A. ]un) becomes HIn)= X.ln),
where
In) =
e o 1/2 lu.).
(A.18)
(n'}n ) = 6,,, = (u,,IPo 1 lu.) .
(A. 19)
e/x w a X l u . ) =
Thus,
Also, the eigenfunctions {0. (x)} of the hermitian hamiltonian obey H0,, = A,,0,, (x),
(A. 20)
where 0 . = (x In I> = (x l u . > e o 1/2 .
Specializing (A.14) and (A.16) to the case n = 0 gives
0800 =
I(x lu0)12Po 1 ---eo(x,
t),
HOo = O .
Thus, Po(x) = 0*00, the square of the ground-state wave function of H corresponding to the zero eigenvalue. It is not surprising that the time-independent FokkerPlanck equation agrees with the SchrSdinger equation 0 and 0 " with E = 0. We also have 00 =exp { - I w dx) which obeys H00=0. Since H and/~r have the same spectrum it is now easy to prove that A. > 0 for n > 0. For an arbitrary state 10) we have from (A.17) 2(4, IUl,b) -- (4,1p2}4,) + (4, IwZl~>+
But since la + ib 12> 0 we know that la12+lb'2 ~>2 I m ab*.
(A.21)
80
C.M. Bender et al. / Strong-coupling expansion
Thus,
2(~blHlO) ~>0.
(A.22)
Thus all the eigenvalues are positive except for Ao, the smallest eigenvalue, which vanishes. It follows that since (y2(t)) = I Y2[P°(Y) + CI~I(y) e -xlt +" ' "] dy, where P0(y) = exp [ - F ( y ) ] we see that lina (y2(t)) = ~ y2 e-F(y) dy,
(A.23)
as stated earlier. We also notice that the rate at which the classical theory relaxes to its timeindependent limit is determined by the size of Xl, the lowest non-zero eigenvalue of H. For the speical case where A (X) = 2gx 3 the energy eigenvalues (from a pure dimensional argument)are proportional to x/g: An OCg 1/2 •
Thus, in the strong-coupling regime we expect a very fast relaxation of Ply, t] to its equilibrium value. This fast rate of convergence is also true of Monte Carlo evaluation of the functional integral for the quantum theory; it converges well at large-g but relaxes very slowly in the weak coupling domain. It is not hard to generalize the results of this appendix to the case d > 0. This generalization is sketched in ref. [3]. References [1] [2] [3] [4] [5] [6l [7] [8] [9] [10] [11] [12] [13]
F. Cooper and B. Freedman, Ann. of Phys., to be published G. Parisi and N. Sourlas, Ecole Normale Sup6rieure preprint (1982) (3. Parisi and Y.S, Wu, Scientia Sinica 24 (1981) 483 D. Zwanziger, Nucl. Phys. B192 (1981) 259 V.L. Ginzburg and LD. Landau, JETP 20 (1950) 1064 B.I. Halpern and P.C. Hohenberg, Rev. Mod. Phys. 49 (1977) 435 S.K. Ma, Modern theory of critical phenomena (Benjamin, Massachusetts, 1976) B. Muhlschlegel, in Path Integrals, ed. G. Papadopulos and I.T. de Vreese (Plenum Press, New York, 1978) pp. 39-60 C.M. Bender, F. Cooper, (3. Guralnik, E. Mjolsness, H.A. Rose and D.H. Sharp, Adv. in App. Math. 1 (1980) 22 C.M. Bender, Los Alamos Science 2 (1981) 76 C.M. Bender and D.H. Sharp, Phys. Rev. D24 (1981) 1691 C.M. Bender, F. Cooper, G.S. Guralnik, H.A. Rose and D. Sharp, J. Stat. Phys. 22 (1980) 647 E. Witten, Nucl. Phys. B15 (1981) 513
A NEW STRONG-COUPLING EXPANSION FOR QUANTUM FIELD THEORY BASED ON T H E LANGEVIN EQUATION CARL M. BENDER
Department of Physics, Washington University, St. Louis, Missouri 63130, USA FRED COOPER
Theoretical Division, Los Alamos National Laboratory, Los Alamos, New Mexico 87545, USA BARRY FREEDMAN
Department of Physics, University of Illinois, Urbana, Illinois 61801, USA Received 23 November 1982 (Final version received 24 January 1983) We review the connection between quantum field theory in d dimensions and a set of classical Langevin equations in d + 1 dimensions. We then present a new method for solving these Langevin equations in the strong-coupling domain. For the special case of supersymmetric quantum mechanics (d = 1) the Langevin equations are also in d = 1 and we use our method to calculate the strong-coupling approximation to the ground-state energy of supersymmetric quantum mechanics when the superpotential is [ ( x ) = ¼gx4. Our numerical results are accurate and they indicate that by studying the critical index of a lattice series one can determine whether the ground-state energy is zero and thereby see the signal of supersymmetry restoration or breaking in the continuum limit.
1. Introduction Recently, there has been renewed interest in the Langevin equation formulation of quantum field theory because of its connection with supersymmetry [1, 2] and because it allows a covariant quantization of gauge fields without a Gribov ambiguity [3, 4]. The Langevin equation and its relation to field theory can be traced back to early work by Ginzburg and Landau, [5] and in the context of ~b4 field theory it is known as the time-dependent Ginzburg-Landau model [6, 7]. The Langevin equation is the simplest time-dependent equation whose solution for large times relaxes to an equilibrium distribution of the Ginzburg-Landau form. The Langevin equation is a dissipative equation of motion for a classical field in the presence of external white noise. To model the effect of white noise we require that the external driving force f(~'), when time-averaged, is not self-correlated; that
is, (f(t))
= O, ( f ( s ) f ( t ) )
is proportional* t o 8(s - t), ( f ( r ) f ( s ) f ( t ) ) = O, ( f ( q ) f ( r ) f ( s ) f ( t ) }
* The proportionality constant is a multiple of Planck's constant h. This is the only place in the theory in which h appears; if it were not for this proportionality constant the theory would be completely classical. 61
62
C.M. Bender et al. / Strong-couplingexpansion
is proportional to 8 ( q - r ) 8 ( s - t ) + 8 ( q - s ) 8 ( r - t ) + 8 ( q - t ) 8 ( r - s ) , and so on. It has been proved [8] that at large times the correlation function of the product of n classical fields, obtained by computing the time average of the solution to the Langevin equation, approaches the statistical mechanics average with respect to the free energy functional of n time independent fields. Therefore, from the equivalence of statistical mechanics and euclidean quantum field theory, one can compute the n-point Green functions for a quantum field theory in d-dimensional euclidean space by (i) solving the appropriate classical Langevin equation in the presence of external sources in d + 1 dimensional space, and then (ii) averaging products of n classical fields over gaussian (white) noise. The purpose of this paper is to present a new method for solving the Langevin equation in the strong-coupling regime. This method is an outgrowth of previous ideas we have published for obtaining good approximate solutions to singular boundary layer differential equation problems [9, 11]. The method consists of introducing a lattice and then solving the discrete version of the Langevin equation for large coupling constant g as a well-defined perturbation series in inverse powers of g. We then perform the average over the white noise. Finally, we extrapolate back to the continuum*. Ordinarily, the Langevin lattice strong-coupling series will be different term by term from the usual lattice strong-coupling series obtained from the path integral [14] (the continuum limits should agree, of course). We do not yet know whether the Langevin lattice strong-coupling series is advantageous compared with the conventional lagrangian lattice strong-coupling series; such a judgment will depend upon the particular theory under examination. However, in this paper we find that the Langevin strong coupling series is extremely easy to obtain. To illustrate this procedure we have chosen a particularly simple quantum theory. By a formal argument involving functional integration, it was recently shown [ 1, 12] that the equal-time correlation functions in supersymmetric quantum mechanics (d -- 1) are identical to those in a zero-dimensional (d = 0) quantum field theory. Thus, to solve this field theory one must solve a very simple Langevin equation in d = 1 dimensions. From the lattice solution of the Langevin equation we show how to compute numerically the n-point Green functions of the theory. By obtaining the solution to this problem we can begin to answer an interesting related question. It is clear that introducing a lattice breaks the supersymmetry of a quantum theory which is supersymmetric in the continuum. Is there a way to see that the underlying continuum theory is supersymmetric? More generally, is there a signal for non-perturbative supersymmetry breaking? In this model a clear signal for supersymmetry breaking is that the ground-state energy is non-zero. Therefore, in this paper we calculate the lattice strong-coupling series for the ground-state * It is clear that this calculational method reverses the order of two procedures; the correct order is to take the continuum limit first and to perform the average over gaussian noise second. However, our numerical results in sect. 4 provide strong justificationfor the legitimacyof our method.
C.M. Bender et al. / Strong-coupling expansion
63
energy; we use the critical index for this series to show that in the continuum limit this series extrapolates to 0 and thus the supersymme:ry is restored. Specifically, we consider a supersymmetric hamiltonian of the form [13, 1] 1 2 6,3 2 H = ~ 1p 2 .+~g x t~r3gx .
(1.1)
On purely dimensional grounds we know that for this hamiltonian the ground-state energy has the form E0 = bg 1/2 ,
(1.2)
where b is a pure number. Our treatment of the Langevin equation gives a lattice strong-coupling series representation for Eo of the form
gl/2yZ/2 ~ A , y " ,
(1.3)
n~0
where
y = (a 2/g)-1/3,
(1.4)
a is the lattice spacing, and A , are numerical coefficients. The continuum limit a ~ 0 of the expression in (1.3) is obtained by finding the behavior of (1.3) as y ~ oo. For the continuum hamiltonian in (1.1) the theory is supersymmetric and Eo (and therefore b) is 0. T o find out whether the introduction of a lattice permanently breaks the supersymmetry of the continuum theory we examine the behavior of oo n ~ , =0 A , y as y --, oo. If we were to find that A,y" ~ y-1/2,
(y .-, oo) ,
(15)
n=O
then we would conclude that in the continuum limit b # 0 and supersymmetry remains broken. In fact, we find that
A , y " ~ y'~,
(1.6)
tl=O
where a < -~. Thus, in the continuum limit, supersymmetry is restored. This paper is organized as follows. In sect. 2 we review the Langevin formula. In sect. 3 we examine the connection between supersymmetric quantum mechanics and the Langevin formalism. In sect. 4 we solve the Langevin equation of quantum field theory in the strong-coupling limit and we evaluate the ground-state energy for the supersymmetric hamiltonian in (1.1). In the appendix we review the properties of the d = 0 Fokker-Planck equation, which is the master equation for the time evolution of the probability distribution determined by the Langevin equation, and show that for large times the time-dependent probability function approaches the time-independent probability relevant for quantum field theory.
64
C.M. Bender et al. / Strong-coupling expansion
2. Review of Langevin formalism
It is widely known that euclidean field theory in d dimensions is equivalent to statistical mechanics in d spatial dimensions with h -1 playing the role of ( k T ) -1. In the theory of dynamical critical phenomena time-dependent models were sought which relaxed to the equilibrium distribution. Such models were constructed by adding dissipation terms to a conservative system and coupling it to a heat bath. The probability distribution function for time-independent (equilibrium) fields is 1 P=q[4)] = ~- exp (-F0[~b ]),
(2.1)
= y ~d) exp (-Fo[~b ])
(2.2)
where Z
is a normalization constant which normalizes to 1 the functional of Peq[~b] with respect to ~b and Fo is the free energy in statistical mechanics and is the euclidean action in field theory. In euclidean field theory the objective is to compute n-point Green functions. The two-point function, for example, is given by <~b(x)~b(y)) = I ~b~b (x)~b(y)P=q[~b],
(2.3)
where x and y are points lying in d-dimensional euclidean space. The simplest equation one can write down which has the property that for large times z, Pcl[~b(x, i-)] ~ P~q[~b(x)] is the Langevin equation od)(x, r)
ar
6F o@~x, z)
= -ro,,-Z-;~-T,+f(x,
z).
(2.4)
Here, F = I da X*~E(~b),
(2.5)
where ~b is a classical field depending on the d-dimensional euclidean space variable x and an artificial time variable ~-. The function f(x, r) is a gaussian white noise source with correlations (f) = O,
(2.6)
(f(x, z)f(x', z')) = 2Fo 8(x - x') 8(T - C ) ,
and so on. We can express these correlation functions as functional integrals with respect to a gaussian probability distribution/~[f], where /~[f] = l e x p [-4-~0 1 dax d'rf2(x, "r)] ,
(2.7)
C.M. Bender et al. / Strong-coupling expansion
65
and (2.8) For example,
(f(x, r)f(x', ~')) J d~ff(x, ~)f(x', =
¢')/~[f].
(2.9)
If we take as an initial condition ~(x, 0 ) = 0 and solve (2.4) for 4~(x, r) as a functional of f, then one can show that lim (& (x, z)~b (x', ~'))f = (4~(x)4, (x')),
(2.10)
where the right side of (2.10) is the usual two-point function of the field theory described by *fie in (2.5). On a euclidean lattice where we replace x by i, the non-equilibrium probability distribution is given by
P¢,[x(i),r]=(~ 6~(i)-~b(i, r)])f
(2.11)
because now (~b(i, r)~b (/', . ) ) f = I ~f/~[[]$ (i' z)~b (], z)
= I ~ dx(k)x(i)x(/)P[x(k), ~'].
(2.12)
Thus, in the continuum limit
Pd[x, r]=(exp {8(O)I dax ln (8[X(x)-cb(x,r)])})~
(2.13)
(~(x, ~')4~(Y, r))t = ~ @gPct[X, ~']X(x)g(y) •
(2.14)
Using eqs. (2.13) and (2.14), one can show that Pc~[4~,~'] obey~ the Fokker-Planck equation
OP ~ a 8 rSP o-;=ro d
8F1
(2.15)
This equation has the form of an imaginary time functional Schr6dinger equation for P aP = aT
-liP.
(2.16)
66
C.M. Bender et al. / Strong-coupling expansion
This equation forms the basis of the proof that as r--)~, P[~b, r]~P~q[~b] (see appendix). Here we merely note that the time-independent solution to the FokkerPlanck equation satisfies 8Po ~_p 8F &b 0 ~ = constant.
(2.17)
Since P is normalizable the constant must be 0. Thus,
Po[~]:
e-F[4,]
I
~,~ e-r'[~l '
which is the equilibrium probability distribution in (2.1). Associated with ~b(x, ~-) is a path integral in d + 1 dimensions. If we want to calculate
(¢(x, ~')~b(x',~"))f= ; ~:fd:(x,~')~b(x',r')fi[~,
(2.18)
we can make a change of variablesusing (2.4)and (2.7), a~b+ 1" 8F
1
d~')/Z,
to obtain ^ 0¢
(&(x, r)~b (x', ~")), = I ~b~b (x, ~')* (x', z ) P [ ~ ' '
8r
a/
r 0 ~ - ] det I~-I"
(2.19)
In some instances, notably when d = 0 (see sect. 3), the path integral in (2.19) represents a well-defined field theory in its own right. In this case the field theory is related to a Langevin equation in the same number of dimensions.
3. Langevin equation |or supersymmetric quantum mechanics In this section we show that supersymmetric quantum mechanics is one of the systems whose associated Langevin equation involves the same number of dimensions as the original d = 1 field theory. We do this to prepare for our model calculation in sect. 4 of the ground-state energy of this system (which we know to be zero). The action for supersymmetric quantum mechanics is l . _ t -~ - ~ - , _ ~) . + ~o l _ 2 + D f , ( x ) + ½[~;" ~]/"(x)}, S = I .o& ~x. ~ + ~t~
(3.1)
where f(x) is an arbitrary function of x and is the superpotential which is invariant
C.M. Bender et al. / Strong-coupling expansion
67
under the supersymmetry transformation
iSx =e*d/e-~be,
84J =-ie*D+e*~,
8D = e ~ + ~ * e .
(3.2)
Integrating out the auxiliary field D and going to euclidean space we have
where T
S E = I o .~E dt, ,.~E = ~x 1 .2
+ ½w2(x ) + ~b*[a, - w '(x )]41,
w(x) =f'(x).
(3.4)
(3.5)
When Jr, the external source, is zero we have Z [ 0 ] = ~ e -•"r,
(3.6)
n
where E , are the energy levels of the system. In this case we can perform the path integral over the fermion degrees of freedom exactly to obtain Z = Z+ + Z _ ,
(3.7)
Z± = f ~ x exp [ - S ± ( x ) ] ,
(3.8)
T
1 , 2 +~w 1 2 (x)+~w I t S±= ~0 dt[~x (x)].
(3.9)
Z = Tr e -m" = T r e - m r + Tr e - m r ,
(3.10)
Now
where H ÷ and H _ depend only on bosonic degrees of freedom. In the energy representation we have e -Enr = ~ e -n-÷r + ~ e -n--T. rl
rl
(3.1 1)
n
The energy eigenvalues {E, } are given in terms of the combined spectrum {E, +, E,_}. When the ground-state energies of the two bosonic wells are not degenerate, with E o - < Eo+ say, one then has
Eo = Eo-. We know that a non-degenerate ground-state necessarily implies that supersymmetry is unbroken so Eo = 0. In this case, as we take T ~ oo the expectation value
C.M. Bender et al. / Strong-coupling expansion
68
of an operator Q is given by Tr (e-rrrQ)
<01OI0)= r-,~ lim Tr (e -nr)
= lim ~" ((n + [QIn +) e -~"+r + (n -[QIn-) e -E"-r r-,~ ~ , (e -E"+r + e -t~"-r)
(3.12)
=
Thus, we obtain the correct ground-state matrix elements for the supersymmetric theory by computing with just Z_ and ignoring Z+ entirely. Suppose, w(x) = gx 3. Then H±
1 2 1 2 6.3 =~p -~g x ±~gx 2 .
(3.13)
No degeneracy is possible for this hamiltonian, so Eo = 0 and one need only consider
Integrating by parts we get
z_-- I
exp {I C-½( +
d,}
(3.15,
But, this is exactly the path integral we obtain for the Langevin system
Y¢= - w (x) + f(t) ,
(3.16)
with 1
P[~=-~exp [-½ f dtf2(t)] .
(3.17)
Note that,
(x (t)x (t'))~ = J Nfx (t)x (t')/~[fl = I ~x det l~xlX(t)x(t')P[2 +w(x)]"
(3.18)
The functional determinant is given by det I~xfl = exp f d t T r l n [ d - w ' ( x ( t ) ) ] 6 ( t - t ' ) d t The free Green function Go satisfies d
-~Go(t-t') =8(t-t') ,
'.
(3.19)
C.M. Bender et al. / Strong-coupling expansion
69
SO G o ( t - t') =
O(t - t') ,
where the constant of integration is chosen to be 0 to satisfy causality. Because of the step function 0, only one term exists in the expansion of the logarithm in (3.19) and we obtain det I~1 : e x p [-½ d t w ' ( x ) ] .
(3.20,
We conclude that Z_ corresponds to the simple one-dimensional Langevin equation in (3.16). In the appendix we show that H_ is also obtainable directly from the Fokker-Planck equation related to the Langevin equation in (3.16). We conclude this section by showing how to verify that when w ( x ) = g x 3 the ground-state energy is zero. We recall that the time-independent probability distribution function for the Langevin equation is P ( x ) = e -v(x) ,
(3.21)
where w ( x ) = ½F'(x). Thus,
e-gX4/2 (3.22)
p(x) =
f dx e -gx4/2 The ground-state energy can be calculated by the virial theorem: E o = (0[HI0) = 2g =(x 6) - 3g(x2) .
(3.23)
Using P ( x ) in (3.22) in (3.23) we have (X 6) ~. (2/g)3/21..,(7)liF(1) ,
(3.24)
(x 2) = ( 2 / g ) l / 2 F ( k ) / F ( ~ ) .
(3.25)
Thus, we verify directly that Eo = O.
4. Strong-coupling lattice expansion for the Langevin equation The method for using the Langevin equation to compute the two-point (or n-point) Green function of a quantum field theory is (i) to solve for the classical field ~b(x, ~') as a functional of the driving force f ( x , r), (ii) to use P(f) in (2.7) to evaluate (~b(x, ~-)~b(x', z))t, and (iii) to take the limit r-,oo. In weak-coupling perturbation theory this method has been shown to give the usual Feynman diagram expansion [6, 3]. Here, we present a totally new method for solving the Langevin equation based on methods we developed to solve boundary layer problems. For simplicity, consider
70
C.M. Bender et al. / Strong-coupling expansion
a (gch4)a field theory whose euclidean action is rl,0_,2+1 _4,J. S = f d a XL~ O) ~go
(4.1)
Then the Langevin equation in (2.4) with Fo = ½is &h _ ½[::]~b + g~b 3 = f .
0T
(4.2)
We are going to solve (4.2) in the large-g regime so we will treat the derivative terms 0~/0~"-~]~b as perturbations. Note that perturbing in the derivatives generally gives a singular perturbation expansion. In an earlier series of papers [9-11] we discussed a new way to solve such equations by introducing a lattice and converting the differential equation to a difference equation. On the lattice the perturbation series is no longer singular but we have had to introduce a new dimensional parameter, the lattice spacing a. We are now faced with the problem of returning to the continuum. In the continuum limit a--* 0, the lattice perturbation series diverges term by term. If we could compute all the terms in the lattice series and sum the series before taking the limit a-* 0 there would be no difficulty. However, when only a finite number of terms in the lattice series have been found Pad6-1ike extrapolation methods must be used to obtain the approximate continuum limit. In refs. [9-11] we verified these lattice extrapolation techniques by solving a number of boundary layer problems. The starting point for a strong-coupling expansion of (4.2) is to introduce a d + 1 dimensional lattice to regulate the derivative terms. We let "c = ia ,
(4.3)
x" = n ~'a,
(4.4)
where i is an integer, and
where/x : 1 ~ d. The derivatives now become differences: limb (n, i ) - ~b(n, i - 1 ) ] , O'r
(4.5)
a
d
[--12 b = ~ - 2 ~ [~b(n +l~,, i)+~b(n -~t, i)-2~.b(n, i)].
(4.6)
Next we introduce the perturbation parameters e = 1/a and e = 1/~ 2. The discrete Langevin equation now takes the form
~[4~(n, i)-~,(n, i-1)]+½8 2d~(n, i1- 2 [~(n +0. i)+4,(n - g , i1] + g~ba(n, i ) = f(n, i ) .
(4.7)
C.M. Bender et al. / Strong-coupling expansion
71
If we assume that 4, (n, i) has a formal expansion in powers of e and 6,
&(n, i)= ~ ~ cbl.,,,(n,i)et6 '' ,
(4.8)
1=0 m = 0
we find that the coefficients ~t.,,(n, i) contain increasing powers of g in the denominator, so that (4.8) is actually a strong-coupling expansion. T o lowest order (4.7) becomes
gdp3oo(n,i)=[(n, i) ,
(4.9)
so that 4,00(n, i ) = [f(n,
i)lg] 1/3 .
(4.10)
Now we must perform the average over white noise. Recall that for gaussian noise
I [ Idax d r ~, , ¢, ,2 ',x~')]
/~[f] = ~ exp -
,
(4.11)
where Z is a normalization constant which sets the functional integral o f / ~ [ f ] to 1. On the lattice (4.11) becomes exp[-a
as.f (n, i)]
.
(4.12)
n.i
In lowest order the two-point correlation function is expressible in terms of a single integral
<~oo(n, i)rkoo(m, i))f = &,,,,g-2/3 I i df ~ f 2 / 3
e-'~'~a½t2.
(4.13)
Integrals over all other lattice points give 1. Note that we obtain 0 unless n = m because fx/3 is odd and integrates to 0. Evaluating the one-dimensional integral in (4.13) gives • 2 "1/3
F(~)/~/~r.
(4.14)
This procedure is easy to carry out to any finite order. To order e t S " we find that the averaging procedure requires that we do a number of ordinary integrals all of which give gamma functions. Now let us specialize to the case d = 0 which corresponds to supersymmetric quantum mechanics. (We also replace ~b by x.) We will show how to calculate the first three terms of the series in detail and then state the results of calculating seven terms. When d = 0 (4.7) becomes e [x (i) - x (i - 1)] + gx (i)3 = [(i).
(4.15)
72
C.M. Bender et al. / Strong-coupling expansion
As in (4.8) we let l,
•
x ( i ) = ~ e xl(t),
(4.16)
/=0
and substitute (4.16) into (4.15). Collecting powers of e and solving recursively gives x (i) to any desired order:
... f xt,)=~
x/3
.
(~)+
~ rf
1/3 .
(,-)
1
3--~[ /2/36) - f
t,)j
e2rfa/3(i-1) fl/3(i-2) fl/3(i-1) f2/3(i-1)] /-4/3(i) -~f4/3(i_1) /2/3(i) f5/3(i ) j + . . . .
+~g[ If
we are interested in computing the two-point function we must square X2(i)
=
f2/3(i)+ 2e [fa/3(i-1) g2/"--"T- ~gL fv3(i)
(4.17) (4.17):
1] -
2 + oeo_~[f_2/3(i) f2/3(i-1), 2f~/3(i-2)f'/3(i) 2f-'/3(i -1)'] -p/3(i) -tp/3(i--1) -- fa/3(i) J + " ' " ~ L
(4.18) In order to calculate (x2(i))f we must average over the gaussian noise using _
/~[f]-~ e
_l/2af2(i )
a
~/~--~.
(4.19)
We interpret (_f)x/3 = _fx/3 in order to insure that (4~)~= 0 to all orders in powers of e. Keeping this in mind it is clear that all odd terms in (4.18) can be dropped [in higher order almost all terms in (4.18) are odd]. Thus, we need only consider
2/.', f2/3(i) 2 e + e 2 r l . - 2 / 3 , . \ f2/3(i--1)] x tt) .... =-~i---~g 9--g~tS tt)- p/a(i ) j + ' " .
(4.20)
Gaussian terms in (4.19) are normalized to one. All non-trivial integrals that arise when we perform the lattice gaussian average of (4.20) using/~[f] have the form
2~--~ I~ e-1/2as2$2k/3d$ =~-~ (2) k/3l"(lk + l) .
(4.21)
Thus, we find at this order (x2(n))~ is independent of n and
(x2)f = ~(2/a )l/3g-2/3F(~)_ 21/(3g) +e 2(a/2)l/3g-'*/3(F(~)/,/~" - F(~)F(-~)/~r]/9 + . " , where it is understood that the continuum limit e =
1/a --.*oo must
be taken.
(4.22)
C.M. Bender et al. / Strong-coupling expansion
73
W e then recognize that an appropriate dimensionless p a r a m e t e r is 3 2 Z ----a---~g'
(4.23)
so that lim F ( ~ ) z
z24~ " ' z" ['P(~)
F(-~)]
is valid for small z. Note that as a -* 0, z -* co so that each t e r m in the series diverges. This is how the singular nature of the continuum perturbation theory is retained in the problem. We know however [see (3.25)] that in the continuum
2 X/2 r ( ~ ) g-1/2 ( x ) f = ~ g ~ z ) = (0.4779886 . . . . ) .
(4.25)
N Thus, we must sum a series of the f o r m z ~,,,=oA,,z 2" [see (1.3)] to get a finite answer. T h e r e are several techniques for doing this. In general we have a series y ( x ) = x ~' E A , , x " ,
(4.26)
where for our case x = z 2 and a = 21-, and we have calculated a finite n u m b e r of terms K : K y r ( x ) = x ~ ~, A ~ x ' . (4.27) n=0 W e know that lim Wo(x) = V~, where Too is finite. Then we see that Xk
[ y r (x ) ]X/~ = E ~ =0 Crux m ,
(4.28)
can be m a d e equivalent to the original series for small x; that is, matching the expansion of (4.28) with (4.27) to order x r uniquely determines the coefficients Cm in terms of the An. Moreover, as x ~ o v , [y~(x)] r/~ has a finite limit. Thus, we choose W: = C r ~ / r ,
(4.29)
as the K t h approximant to 3%. The numerical value of the series for (x2)f in (4.24) is F s1 ~(1_
0.52340954z 2 + 0.24312044z 4
- 0 . 0 7 4 0 7 4 0 7 4 z 6 + 6.4196552 × 10-Sz s + 8.1172514 x 10-3z 10 - 5 . 9 8 3 8 3 0 0 x lO-3z l O + . . . ) .
(4.30)
74
C.M. Bender et al. / Strong-coupling expansion
The first six extrapolants are 1/~/g times 0.62245, 0.55236, 0.52656, 0.51281, 0.50427, 0.49847 which are converging nicely to the exact answer whose numerical value is 0.478 . . . . We are most interested in the ground-state energy E0 which, from the virial theorem is given by [see (3.23)] E0 = 292(x 6)_ 3 g ( x 2). We would like to know how well we can see that E0 = 0, that is, how obvious it is that the continuum theory is supersymmetric in our strong-coupling lattice approximation. There are two possible strategies for calculating E0. One is to separately extrapolate the series for (x 6) and (x 2) and then to examine the differences of the extrapolants and the other is to add the two lattice series first and then to carry out an extrapolation of the resulting series. The calculation of the series for (x 6) follows the same path as that for (x 2). The first few terms are 2
2
5 2/3 x 6 __ ~fO_ ~ _ ~E , .[,,~¢4/3 j o -- z~1/3.,_ ,.2/3_3j_ljo ' J - 1 J O J - r ' g~~7~k]-I 8¢1/3¢1/3 +3/'0 12_.1/3¢ ¢-2/3
3j-2 j oj -1
(4.31)
- ~fo[-~/3) + ' " .
Computing the average with respect to/~(f) using (4.21) gives
z3
r(~) 2+4r(~)z,_~z6 +
(x6), = 2-~'~ [ 1 - 3---~-z
~
"" "] .
(4.32)
In numerical form this series reads Z
3
(x6)f = 2 - ~ ( 1 - 1.0468192z 2 + 0.84913325z4- 0.55555556z 6 +0.28398382z s _ 0.10595548z lo + 0.021912058z 12 + . . . ) .
(4.33)
Recall from (3.24) that the exact answer is (x6)f= (2/g)
3/2F(7)
~=
(0.7169829...)
g-3/2.
(4.34)
The form of the series in (4.33) differs from that in (4.30) in that it starts with z 3 instead of z. The first six extrapolants a r e g-3/2 times 0.85763, 0.75975, 0.82611, 0.98872, 1.30668, 1.12249. Note that in this case we have an asymptotic sequence which at first gets close to the exact answer and then veers away. This is a typical asymptotic behavior for the extrapolants defined in (4.29) as we found in our studies of singular differential equations in refs. [9-11]. Thus, we do quite well in determining (x 2) and (x6). Our best estimates are (X2)f = 0.49847/x/g,
(4% error)
(4.35)
C.M. Bender et al. / Strong-coupling expansion (x6>~ = 0 . 7 5 9 7 5 / g 3/2,
75
(6% e r r o r ) .
(4.36)
Substituting these results into E o in (3.23) gives o u r best estimate for Eo: Eo = 0.024x/g,
(4.37)
instead of zero. Thus, o u r first strategy is reasonably successful*. T h e s e c o n d strategy outlined a b o v e consists of combining the two series in (4.24) and (4.32) to obtain a single series for Eo: 3(5/6)
Eo/4g = lim
51r3/2+3F3(~)
6,~r(~)
-----/--~-z + 2z 3
14F(~) + ~ z
5 z
7 9F3(~) x/Tr+59zr2 9 \ 108~r2 z +.").
(4.38)
T h e extrapolants for this series are - 1.3204, - 1.1434, - 1.0691, - 1.0270, - 0 . 9 9 9 6 , -0.9801 ..... a s e q u e n c e whose terms are slowly decreasing in magnitude. H o w e v e r , it is not possible to conclude with m u c h assurance that the limit of this s e q u e n c e of a p p r o x i m a n t s is 0. T h e best a r g u m e n t that the s e q u e n c e of extrapolants for (4.38) really a p p r o a c h e s 0 d e p e n d s on calculating the critical index of the series. W e assume that as z -* ~ , the right side of (4.38) goes like A z a, w h e r e A and a are constants and ot is called the critical index. If a = 0 (and A ~ 0) then Eo is a n o n - z e r o constant and s u p e r s y m m e t r y remains b r o k e n as the lattice spacing a -* 0. If a < 0 then as the lattice spacing a p p r o a c h e s 0, the right side of (4.38) a p p r o a c h e s zero as a Iaf/3 and the s u p e r s y m m e t r y is restored in the c o n t i n u u m limit. T o d e t e r m i n e a we use the identity
E~(z)
a =z ~ Eo(z ) '
(4.39)
w h e r e Eo(z) represents the r i g h t - h a n d side of (4.38). Evaluating (4.39) numerically and setting x = z 2 gives a = lim (1 - 2 . 0 9 3 6 x + 0 . 9 7 2 4 8 x 2 _ 0 . 4 3 6 9 5 x 3 + 0 . 0 6 5 3 4 5 x 4 + 0 . 1 3 5 3 2 x 5 + . . . ) . X --~oO
(4.40) T o d e t e r m i n e the value of a we find that it is best to m a p the d o m a i n 0 ~
'
(4.41)
* If we look at the sequence of differences between the extrapolants of the series in (4.30) and (4.33) we get the sequence x/g times -0.152, -0.137, +0.073, +0.439 which we see changes sign. Thus, we have crude evidence here that Eo = 0.
C.M. Bender et al. / Strong-coupling expansion
76
and to evaluate the diagonal and off-diagonal Pad6 approximants at y = 1. The series in y is ~t = l i m ( 1 - O . 5 2 3 4 y - O . 4 6 2 6 y 2 - O . 4 0 8 7 y T - O . 3 6 1 3 y 4 - O . 3 2 0 1 y 5 . . . . ).
(4.42)
y"-~ 1
The results are ao,o = 1,
a0.1 = 0.6564,
t~1,1 = - 3 . 5 0 7 , ot2.2 = - 3 . 4 9 9 7 ,
al,o = 0.4766,
al,2 = - 3 . 4 9 0 3 ,
Or2,1 =
a2,a = - 3 . 4 7 6 3 ,
-3.4903,
aa.2 = - 3 . 4 7 6 4 ,
where the subscripts are the indices of the Pad6 approximant used, The results of this Pad6 extrapolation are unambiguous; the value of a lies very close to - 3 . 5 . Hence E0 - l i m a 1.16= 0 .
(4.43)
a~0
We have thus shown that the Langevin equations provide an alternative method for performing strong-coupling expansions to the usual path integral method. Our method is simple and straightforward and avoids the necessity of diagram techniques. By applying these methods to supersymmetric quantum mechanics we were able to show that strong-coupling lattice expansions do provide a calculational tool for discussing supersymmetry breaking. By examining the index of a lattice series we can see whether in the continuum limit the series gives a zero or non-zero result for the ground-state energy. We thank the U.S. Department of Energy for partial financial support. We also thank the MIT M A T H L A B group for the use of MACSYMA. Two of us, C.M.B. and B.F., thank the Los Alamos National Laboratory for its hospitality,
Appendix In this appendix we examine the properties of the Fokker-Planck equation in d = 0 dimensions. We will show here that the time-dependent probability function which satisfies the Fokker-Planck equation approaches the time independent probability distribution of quantum field theory [8]. Consider the Langevin equation (2.4) in d = 0 dimensions [here we use t instead of ~- and write A(x) in place of F'(x)]:
Y¢(t) = -FoA (x ) + f(t).
(A. 1)
The gaussian probability distribution function is taken to be
fi[f]=exp[-I dtl'2(t)] /~, 4/"0 ] /
'
(A.2)
C.M. Bender et aL / Strong-couplingexpansion
77
Next define the function g(y, t) -~6[y -x(t)].
(A.3)
In terms of g the time-dependent probability distribution function is Pc,(Y, t) = I i6[f]g(Y' t)~f.
(A.4)
Note that from (AA) we have
I ~ f x " ( t ) P ( f ) = I dy y"Pc,(y, t). To obtain the master equation satisfied by Pcl(y, t) we write down the chain rule
at g =~x ~ = -
{g[f(t)-FoA ]},
and average this equation over/~[/] to obtain 0 (9 ~Pd-Ao~y[P~,A]=-
LOf 2 Z ~yyj ~ / g f exp [ - ~ ~ d t ] .
(A.5,
Next we use integration by parts to derive the identity
ox [q f2(t') o dt'], 8-~-lt =
8f(t)
2
= [ ~ ~- f gf exp [ - f f2(t') - - ~ °- dt']j ,
(A.6)
and note that ag 8x(t) Ox(t) 8f(t)
8g(y, t) 8f(t)
o [gsx~t)] Oy
8f(t)3 "
(A.7)
Finally, we integrate the equation
BYe(t! = $ (t - t') 8f(t') to obtain 8x (t)
= lim
0 ( t - t') =
1
(A.8)
Combining (A.6)-(A.8) with (A.5) gives the Fokker-Planck equation: 2
at
y
y
(A.9)
C.M. Benderet al. / Strong-couplingexpansion
78
From here on we set Fo = ½and recall that A = F'. Thus (A.9) reads
OPc,_lot 20yO[~-~+Pc'F'] "
(A. 10)
It is clear from (A.10) that a stationary solution (one for which OPd/Ot = 0) is Po[y ] = e -Fry] •
(A. 11)
We want to show that at large times P~[y, t]-* Po[y ]. To show this it is useful to view (A.10) as a SchrSdinger equation with imaginary time and a non-hermitian hamiltonian (non-hermitian because there is a onederivative term): ^
0
( H +~)P¢,(y, t ) = 0 ,
(A.12)
I'?I =½p2 + ipw ,
(A.13)
where w = ½A. Now if {A,} are the eigenvalues of/-t and {lug)} are the associated eigenfunctions and we assume that the set {lu,)} is sufficient to expand P¢l[y, t] we have
e¢,(x, t) = ~ C.(xIu.) e -x"' n=0
= ~ C.¢t,,(x) e -az.
(A.14)
n=0
It is clear that Po(x) is a solution to the SchrSdinger equation (A.10) with Ao = 0 so that (XlUo)=Po(x). Thus, if we are to prove that Pcl(x, t)~Po(x) as t ~ , we must show that Co = 1 and A, > 0 for n = 1, 2, 3 . . . . . To establish the positivity of the eigenvalues we relate/-t to another hamiltonian operator that is self-adjoint and has the same spectrum. Completing the square gives • 2 +~w 1 , +~w 1 2. =-½p2+ipw=½(p +tw) But,
(p+iw)tO,,(x)=pexp[-~Xw dx]tO. (x). So if we introduce a new basis
{In)}by
I n ) = e x p [ - I x w dx][u.),
(AA5)
then
luo > =
dXln),
(A.16)
C.M. Bender et al. / Strong-coupling expansion
79
and the operator on the right-hand side of (A.16), which we call H, is hermitian: H _ --e-f~axiSief~dx
1 , = -½p2 +~w
~t_lW2
(A.17)
In fact H is precisely the hamiltonian associated with supersymmetric quantum mechanics [see (3.14) for example]. The transformation in (A.17) has eliminated the imaginary magnetic field in/-I by means of a gauge transformation. The eigenvalue equation/-Ilu.) = A. ]un) becomes HIn)= X.ln),
where
In) =
e o 1/2 lu.).
(A.18)
(n'}n ) = 6,,, = (u,,IPo 1 lu.) .
(A. 19)
e/x w a X l u . ) =
Thus,
Also, the eigenfunctions {0. (x)} of the hermitian hamiltonian obey H0,, = A,,0,, (x),
(A. 20)
where 0 . = (x In I> = (x l u . > e o 1/2 .
Specializing (A.14) and (A.16) to the case n = 0 gives
0800 =
I(x lu0)12Po 1 ---eo(x,
t),
HOo = O .
Thus, Po(x) = 0*00, the square of the ground-state wave function of H corresponding to the zero eigenvalue. It is not surprising that the time-independent FokkerPlanck equation agrees with the SchrSdinger equation 0 and 0 " with E = 0. We also have 00 =exp { - I w dx) which obeys H00=0. Since H and/~r have the same spectrum it is now easy to prove that A. > 0 for n > 0. For an arbitrary state 10) we have from (A.17) 2(4, IUl,b) -- (4,1p2}4,) + (4, IwZl~>+
But since la + ib 12> 0 we know that la12+lb'2 ~>2 I m ab*.
(A.21)
80
C.M. Bender et al. / Strong-coupling expansion
Thus,
2(~blHlO) ~>0.
(A.22)
Thus all the eigenvalues are positive except for Ao, the smallest eigenvalue, which vanishes. It follows that since (y2(t)) = I Y2[P°(Y) + CI~I(y) e -xlt +" ' "] dy, where P0(y) = exp [ - F ( y ) ] we see that lina (y2(t)) = ~ y2 e-F(y) dy,
(A.23)
as stated earlier. We also notice that the rate at which the classical theory relaxes to its timeindependent limit is determined by the size of Xl, the lowest non-zero eigenvalue of H. For the speical case where A (X) = 2gx 3 the energy eigenvalues (from a pure dimensional argument)are proportional to x/g: An OCg 1/2 •
Thus, in the strong-coupling regime we expect a very fast relaxation of Ply, t] to its equilibrium value. This fast rate of convergence is also true of Monte Carlo evaluation of the functional integral for the quantum theory; it converges well at large-g but relaxes very slowly in the weak coupling domain. It is not hard to generalize the results of this appendix to the case d > 0. This generalization is sketched in ref. [3]. References [1] [2] [3] [4] [5] [6l [7] [8] [9] [10] [11] [12] [13]
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