By JamieBondi


2016-09-01 02:53:37 8 Comments

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 $J$ $$ S=\frac{1}{2}m\dot{x}^2-\frac{1}{2}m\omega^2x^2+Jx. $$

The transition probability would be given by $$ \int Dx\, e^{iS[x]}. $$

Define $x(t)=x_{\text{cl}}(t)+\eta(t)$, then we can Taylor expand the action about the classical path $$ S[x]=S[x_{\text{cl}}+\eta]=S[x_{\text{cl}}]+\int dt\, \eta(t)\frac{\delta S[x]}{\delta x(t)}\Big|_{x=x_{\text{cl}}}+\frac{1}{2}\int dt_1 dt_2\, \eta(t_1)\eta(t_2) \frac{\delta^2 S[x]}{\delta x(t_1)\delta x(t_2)}\Big|_{x=x_{\text{cl}}}, $$ 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:

$$ \eta\eta \delta^2 S =\eta\eta(m\delta\ddot{x}-m\omega^2\delta x), $$ where, in integrating by parts twice, we get something like

$$\frac{1}{2}\int dt_1 dt_2\, (m\dot{\eta}^2-m\omega^2\eta^2).$$

However, in the book the result is

$$\frac{1}{2}\int dt\, (m\dot{\eta}^2-m\omega^2\eta^2).$$

My confusion is how the double integration is reduced to a single.

1 comments

@OON 2016-09-01 03:06:54

You confused the action with lagrangian. $$S[x]=\int dt L\Big(x(t),\dot{x}(t)\Big)$$ As $L$ contains only simultaneous $x$ and $\dot{x}$ you'll get, $$\frac{\delta^2S}{\delta x(t_1)\delta x(t_2)}=\Big(\dots\Big)\cdot\delta(t_2-t_1)$$ $\delta$-function will eat double integration

@JamieBondi 2016-09-01 03:11:09

Right! There should be a Dirac function that eliminates one integration. The sign problem was really a stupid one, when I reviewed my post I immediately realized the sloppy thing...Thanks!!!

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