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I have two questions about Vacuum to vacuum transition amplitude.

  1. Can any particle stay in $|0\rangle$?
  2. I was studying this topic from Srednicki's QFT book. He writes in eq.$(6.22)$ $$\langle0|0 \rangle_{f,h} = \int \mathcal{D}p \mathcal{D}q \, \exp \Big[i\int_{-\infty}^\infty dt (p \dot{q} - H_0(p,q) - H_1(p,q) + fq + hp \Big]$$ $$ =\exp \Big[-i \int_{-\infty}^\infty dt H_1 \Big(\frac{1}{i} \frac{\delta}{\delta h(t)}, \frac{1}{i} \frac{\delta}{\delta f(t)} \Big) \Big] \times \int \mathcal{D}p \mathcal{D}q \,\exp \Big[i\int_{-\infty}^\infty dt (p \dot{q} - H_0(p,q) + fq + hp \Big] $$

Here in the second line the arguments of $H_1$ suddenly changes from $p$,$q$ to $\frac{1}{i} \frac{\delta}{\delta h(t)}, \frac{1}{i} \frac{\delta}{\delta f(t)}$. How can I derive this line from the first line?

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Srednicki explains the second part in the two sentences following eq (6.22). The functional derivatives act on the $fq+hp$ parts and pull out the required $p$ and $q$ factors, leaving you effectively with $H_1(p,q)$ instead of $H_1(\frac{1}{i}\frac{\delta}{\delta h(t)}, \frac{1}{i}\frac{\delta}{\delta f(t)})$ –  twistor59 Apr 2 '13 at 15:58
@twistor59:Here I don't know the explicit form of $H_1(\frac{1}{i} \frac{\delta}{\delta h(t)}, \frac{1}{i} \frac{\delta}{\delta f(t)})$. How can I operate this on the term next to it? –  omehoque Apr 2 '13 at 16:35
The idea is that no matter what the functional form of $H_1$ is, you can imagine inserting the functional derivatives as its arguments. If you do that, then, whilst operating on the integrand with the resulting object, the functional derivatives will ignore everything except the $h$ and $f$, and when they operate on $h$ and $f$ they'll pull out the respective coefficients. So what you end up with effectively is their coefficients (namely $p$ and $q$) inserted into the arguments of $H_1$. –  twistor59 Apr 2 '13 at 16:48
To add to what @twistor59 says, imagine for example that $H_1$ is a polynomial in $p$ and $q$ and interpret a functional derivative raised to a power $n$ as the $n$-fold functional derivative. –  joshphysics Apr 2 '13 at 18:37
@joshphysics: Can you please write your comment in equation ? :P –  omehoque Apr 2 '13 at 21:04

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