460 The Feynman path integral
Suppose, next, thatAˆis a function of just the momentum operator:Aˆ=Aˆ(ˆp). In
this case, it is no longer possible to expressAˆin terms of the diagonal elements of the
density matrix. Hence, starting with eqn. (12.3.2),|x〉is no longer an eigenvector of
Aˆ(ˆp) and cannot be brought outside the matrix element〈x|Aˆ(ˆp) exp(−βHˆ)|x〉. How-
ever, if we insert an identity operator in the form of eqn. (12.2.10) betweenAˆand
exp(−βHˆ), then we have a product of two matrix elements:
〈Aˆ〉=
1
Q(L,T)
∫
dxdx′〈x|Aˆ|x′〉〈x′|e−β
Hˆ
|x〉. (12.3.14)Eqn. (12.3.14) requires diagonal and off-diagonal elements of the density matrix. Sub-
stituting eqn. (12.2.22) into eqn. (12.3.14) gives a path integral expression for〈Aˆ〉:
〈Aˆ〉=
1
Q(L,T)
lim
P→∞(
mP
2 πβ ̄h^2)P/ 2 ∫
dx 1 ···dxP+1〈x 1 |Aˆ|xP+1〉×exp{
−
1
̄h∑P
k=1[
mP
2 β ̄h(xk+1−xk)^2 +β ̄h
2 P(U(xk+1) +U(xk))]}
. (12.3.15)
Note that the paths in eqn. (12.3.15) are no longer cyclic, andx 16 =xP+1. In gen-
eral, a sum over open paths is more difficult to evaluate than a sum over closed,
cyclic paths because of the large fluctuations in the endpoints and quantities such as
〈x 1 |Aˆ(ˆp)|xP+1〉that depend on them. According to eqn. (4.5.33), the distribution of
the end-to-end distance for a free particle is a Gaussian whose width grows asT→0.
An interesting example of a quantity that requires such off-diagonal elements is the
momentum distributionn(p), which is obtained by takingAˆ(ˆp) =δ(ˆp−p′Iˆ), wherep′
is a pure number, so that
n(p′) =〈δ(ˆp−p′Iˆ)〉=1
2 π ̄h∫
dxdx′eip′(x−x′)
〈x′|e−β
Hˆ
|x〉. (12.3.16)This distribution can be measured in neutron Compton scattering experiments and
can be computed using an algorithm introduced by Morroneet al.(2007).
Expectation values of operator functions that depend on both position and mo-
mentum can be equally difficult to evaluate depending on how the operators ˆxand ˆp
appear inAˆ(ˆx,pˆ) (see eqn. (9.2.51)). The thermodynamic functions in the canonical
ensemble are exceptional in that they can be evaluated using cyclic path integrals, as
we will now demonstrate.
Consider first the evaluation of the average energy
E=〈Hˆ〉=
〈
ˆp^2
2 m+U(ˆx)〉
. (12.3.17)
Although the Hamiltonian is a function of both position and momentum,and it would,
therefore, seem that both closed and open paths are needed to evaluate〈Hˆ〉, we can
evaluateEstraightforwardly via the thermodynamic relation