Functional integrals 467to imaginary time usings=−iτ, then the equation of motion becomesmd^2 x/ds^2 −→
md^2 x/d(−iτ)^2 =−md^2 x/dτ^2 =−∂U/∂x, which is just eqn. (12.4.14). Thus, domi-
nant paths are solutions to eqn. (12.4.14) subject to the endpointconditionsx(0) =x
andx(β ̄h) =x′.
We can now use eqn. (12.4.12) to construct a functional integral expression for the
partition functionQ(β). Since
Q(β) =∫
dx ρ(x,x;β), (12.4.15)we begin by taking the diagonal element of ˆρin eqn. (12.4.12):
ρ(x,x;β) =∫x(β ̄h)=xx(0)=xDx(τ) exp{
−
1
̄h∫β ̄h0dτ[
1
2
mx ̇^2 (τ) +U(x(τ))]}
=
∫x(β ̄h)=xx(0)=xDx(τ) exp{
−
1
̄h∫β ̄h0dτΛ(x(τ),x ̇(τ))}
=
∫xxDxe−S[x]/ ̄h. (12.4.16)Note that although the upper and lower limits of integration on the functional measure
Dxare the same, the integral does not vanish as would be the case foran ordinary
integral. Rather, eqn. (12.4.16) denotes an integral over all paths that begin and end
at the same pointx(x(0) =x(β ̄h) =x). In order to construct the partition function,
we must integrate over allx, which gives
Q(β) =∫
dx∫x(β ̄h)=xx(0)=xDx(τ) exp{
−
1
̄h∫β ̄h0dτ[
1
2
mx ̇^2 (τ) +U(x(τ))]}
=
∫
dx∫x(β ̄h)=xx(0)=xDx(τ) exp{
−
1
̄h∫β ̄h0dτΛ(x(τ),x ̇(τ))}
=
∫
dx∫xxDxe−S[x]/ ̄h≡
∮
Dxe−S[x]/ ̄h. (12.4.17)The
∮
symbol in last line of eqn. (12.4.17) indicates that the functional integral is
to be taken over all paths that satisfy the conditionx(0) =x(β ̄h). These paths are
periodic in imaginary time with periodβ ̄h. Consequently, the dominant contribution
to the path integral for the partition function are paths near thesolutions to eqn.
(12.4.14) that satisfyx(0) =x(β ̄h).
12.4.1 Example: The harmonic oscillator
In order to illustrate the use of the functional integral formalism,consider a simple
harmonic oscillator of massmand frequencyωdescribed by the Hamiltonian of eqn.
(9.3.18). The functional integral for the full density matrix is