Computational Physics

12.4 Path-integral Monte Carlo 405

Note that the numerator differs from the path integral (which occurs in the denom-
inator) in the absence of the integration overR 0. Removing all the unit operators
we obtain

P(R 0 )=

∫ 〈R^0 |e−τH|R^0 〉

dR 0 〈R 0 |e−τH|R 0 〉

. (12.77)

Largeτis equivalent to low temperature. But ifτis large indeed, then the operator
exp(−τH)projects out the ground stateφG:

e−τH≈|φG〉e−τEG〈φG|, largeτ. (12.78)

Therefore we have

P(R 0 )=

1

Z

e−τEG|〈φG|R 0 〉|^2 , largeτ. (12.79)

Because of the periodic boundary conditions in theτdirection we obtain the same
result for each time slicem. To reduce statistical errors, the ground state can be
therefore obtained from theaveragedistribution over the time slices via a histogram
method.
The expectation value of a physical quantityAfor a quantum system at a finite
temperature is found as

〈A〉β=

Tr(Ae−βH)

Tr e−βH

. (12.80)

The denominator is the partition functionZ. We can use this function to determine
the expectation value of the energy

〈E〉β=

Tr(He−βH)

Z

=−

∂

∂β

lnZ(β). (12.81)

If we apply this to the path-integral form ofZ, we obtain for the energy per particle
(in one dimension):

〈

E

N

〉β=

M

2 β

−

1

N

(〈K〉−〈V〉). (12.82)

with

K=

M∑− 1

m= 0

(Rm−Rm+ 1 )^2

2 β^2

(12.83)

andVis the potential energy (see also Problem 12.1). The first term in(12.82)
derives from the prefactor 1/

√

2 π
βof the kinetic Green’s function. The angular
brackets in the second and third term denote expectation values evaluated in the
classical statistical many-particle system. It turns out that this expression for the
energy is subject to large statistical errors in a Monte Carlo simulation. The reason