I've been trying to verify this result from Das' book "Field Theory: A Path Integral Approach",
Consider the action of a harmonic oscillator perturbed by a source J S=12m˙x2−12mω2x2+Jx.
The transition probability would be given by ∫DxeiS[x].
Define x(t)=xcl(t)+η(t), then we can Taylor expand the action about the classical path S[x]=S[xcl+η]=S[xcl]+∫dtη(t)δS[x]δx(t)|x=xcl+12∫dt1dt2η(t1)η(t2)δ2S[x]δx(t1)δx(t2)|x=xcl, where, by definition of the classical path, the second term vanishes. So what we need to evaluate is the third term. So far I'm basically copying from the book.
Ok, so let's now put the expression of S[x] inside to evaluate the second order variation:
ηηδ2S=ηη(mδ¨x−mω2δx), where, in integrating by parts twice, we get something like
12∫dt1dt2(m˙η2−mω2η2).
However, in the book the result is
12∫dt(m˙η2−mω2η2).
My confusion is how the double integration is reduced to a single.
Answer
You confused the action with lagrangian. S[x]=∫dtL(x(t),˙x(t)) As L contains only simultaneous x and ˙x you'll get, δ2Sδx(t1)δx(t2)=(…)⋅δ(t2−t1) δ-function will eat double integration
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