Computational Physics

12.2 The variational Monte Carlo method 385

We again separate the exponent into two terms, one containingˆxand the otherpˆ,at
the expense of anO(
t^2 )error. Calculating Gaussian Fourier transforms as before,
we obtain the result:

G(x,y; t)=

1

√

2 π

t

e−[y−x−F(x)

t/^2 ]

(^2) /( 2
t)

. (12.45)

Note that this expression is a first order approximation in tof the exact Green’s
function. This is normalised, and we can therefore use it again for constructing a
Markov chain. This is done by moving the random walker first from its old position
xto the positionx+F(x)
t/2 and then adding a random displacementη

√

t,
whereηis drawn from a Gaussian distribution with a variance 1 (see Eq. (12.29)).
In formula, the method reads

x(t+ t)=x(t)+ tF[x(t)]/ 2 +η

√

t, (12.46)

so it is a discrete Langevin equation with ‘force’F.
We end this section with a few remarks. First, all results can be extended straight-
forwardly to higher dimensions. Using a 3N-dimensional variableRinstead of
the one-dimensional variablex(Rdenotes the positions of a set of particles in
three dimensions as usual), the Green’s function of the simple diffusion equation
Eq. (12.23)withγ= 1 /2is

G 3 N(R,R′;t)=

1

( 2 πt)^3 N/^2

e−(R

′−R) (^2) /( 2 t)

. (12.47)

The Green’s function of the Fokker–Planck equation(12.42)becomes

G 3 N(R,R′; t)=

1

( 2 π

t)^3 N/^2

e−[R

′−R− tF(R)/ 2 ] (^2) /( 2
t)
, (12.48)
whereF(R)is a three-dimensional vector, given by
F(R)=∇Rρ(R)/ρ(R). (12.49)
You might have been surprised by the way in which the exponential containing
noncommuting operators was split inEq. (12.35). After all, the following splitting
e−
τ (V+K)=e−
τV/^2 e−
τKe−
τV/^2 +O(
τ^3 ) (12.50)
is more accurate: you can check that the first order CBH commutator vanishes,
hence theO(
τ^3 )error. The reason we use the simpler splitting(12.35)is that
diffusion steps are carried outsuccessively, hence the rightmost term in the right
hand side of(12.50)at one step combines with the leftmost term at the next step,
so that the total effect of the more accurate splitting is reduced to a different first
and final step. This difference is, however, of the same order of magnitude as the
accumulated error of the sequence of steps, and therefore it does not pay to use
(12.50).