468 The Feynman path integral
ρ(x,x′;β) =∫x(β ̄h)=x′x(0)=xDx(τ) exp[
−
1
̄h∫β ̄h0dτ(
1
2
mx ̇^2 +1
2
mω^2 x^2)]
. (12.4.18)
As we have already seen, paths in the vicinity of the classical path onthe inverted
potential dominate the functional integral. Thus, in order to perform the functional
integral, we utilize a technique known asexpansion about the classical path. Suppose
we are able to solve eqn. (12.4.14) for a classical pathxcl(τ) satisfyingxcl(0) =xand
xcl(β ̄h) =x′. Given this path, we perform a “change of variables” in the functional
integral; that is, we change the function of integrationx(τ) to a new functiony(τ) via
the transformation
x(τ) =xcl(τ) +y(τ). (12.4.19)Eqn. (12.4.19) is similar to a change of variables of the formx=a+yin an ordinary
integral
∫
f(x)dx, whereais a constant, so that dx= dy. Here, sincexcl(τ) is a single
function, it is analogous to the constanta, andDx(τ) =Dy(τ). For the harmonic
oscillator,xcl(τ) satisfies the classical equation of motion on the inverted potential
surface−U(x) =−mω^2 x^2 / 2
mx ̈cl=mω^2 xcl, (12.4.20)withxcl(0) =xandxcl(β ̄h) =x′. Consequently,y(0) =y(β ̄h) = 0.
Substitution of this change of variables into the action integral yields
S=
∫β ̄h0dτ[
1
2
mx ̇^2 +1
2
mω^2 x^2]
=
∫β ̄h0dτ[
1
2
m( ̇xcl+ ̇y)^2 +1
2
mω^2 (xcl+y)^2]
=
∫β ̄h0dτ[
1
2
mx ̇^2 cl+1
2
mω^2 x^2 cl]
+
∫β ̄h0dτ[
1
2
my ̇^2 +1
2
mω^2 y^2]
+
∫β ̄h0dτ[
mx ̇cly ̇+mω^2 xcly]
. (12.4.21)
The last line of eqn. (12.4.21) contains cross terms betweenxcl(τ) andy(τ), but these
terms can be shown to vanish using an integration by parts:
∫β ̄h0dτ[
mx ̇cly ̇+mω^2 xcly]
=mx ̇cly∣
∣
∣
∣
β ̄h0+
∫β ̄h0dτ[
−mx ̈cl+mω^2 xcl]
y= 0. (12.4.22)
The boundary term vanishes becausey(0) =y(β ̄h) = 0, and the second term vanishes
becausexcl(τ) satisfies eqn. (12.4.20).