# Transition amplitude computed by the path integral

I would like to understand the transition amplitude given by the path integral as it is presented in Srednicki's book in formula (8.3):

$<0|0>_J = \int D\phi \exp[i \int d^4x [L_0 + J\phi]$. (8.3)

Apparently it is not equal to 1 as it is shown only a couple of lines later. However, only 19 pages before the normalization of ground state stated to be:

$<0|0>=1.$ (5.4).

Unfortunately Srednicki does not clearly say if $|0>$ represents a ground state of an interacting QFT, however, as the normalization of $|0>$ is stated at the beginning of the explanation of LSZ, I strongly assume that $|0>$ represents the ground state of an interacting QFT. The consultation of Peskin &Schroeder (P&S) also supports this. So if this is the case, I consider $L=L_0 + J\phi$ as the Lagrangian of an interacting QFT, so I don't see a reason why $<0|0>_J$ should not be equal 1. In particular if the considered QFT is QED, the corresponding current $J^\mu$ has clear physical meaning as Dirac current, for this well-known interacting QFT $<0|0>_J$ should be equal $=<0|0>=1$ according to (5.4).

Unfortunately, it comes even worse. According to the middle of p.55 of Srednicki $<0|0>_{J=0}=1$. So for a free QFT the ground state seems to be indeed normalized. However, consulting P&S on the subject, I found the following:

the first = sign according to Srednicki, the rest according to P&S:

$<0|0>_{J=0}=\int D\phi\, e^{S_0} =\cdots = \Pi_{all\, k_n} \sqrt{\frac{-i\pi V}{m^2-k_n^2}}$ (9.23) with $\,\,S_0= \int d^4x [\frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2\,\,$] P&S(9.19)

(P&S gets this result by a discretization of the path integral.)

which is a obviously a free QFT, but the transition amplitude is not 1. Definition of $k_n$: $k_n=\frac{2\pi n^\mu}{L}$ with $n^\mu$ an integer $|k^\mu|<\pi/\epsilon$ and $V=L^4$.

May be Srednicki wants to say $<0( T=\infty)| 0 (T=-\infty)>=\int D\phi e^{S_0}$ when he writes $<0|0>_{J=0}=\int D\phi\, e^{S_0}$ ?

However, is the ground state supposed to change in time ? My intuition tells me no. So for me the whole concept is rather shaky and confusing. I really would appreciate if somebody shows me the real meaning of $<0|0>_{J}$ and its difference to $<0|0>=1$ and even show why $<0|0>_{J=0} \neq 1$ apparently sometimes.

## 1 Answer

Suppose you have a vacuum state $|0>$ at some time $t$, i.e. $|0>=|0,t>$. Then, clearly you have the usual normalization condition

$<0,t|0,t>=1$.

In the case $t' \neq t$ however, it holds

$A=<0,t'|0,t> = <0,t'|e^{iH(t'-t)}|0,t'> \neq 1$.

The reason is that the Quantum vacuum is not stable. The vacuum persistence amplitude I have denoted by $A$. It differs from 1 in the case if particle-antiparticle pairs are created. The reason is that the particle Hamilton Operator, which is proportional to $\psi^\dagger \psi$ (this Looks normal-ordered) can have non-normal-ordered Terms if you write out the particle field Operators as follows:

$\psi = \int d^3k \frac{1}{\sqrt{2 \omega_k}}(a_k e^{i (\omega_kt - kx)}+b_k^\dagger e^{-i (\omega_kt - kx))}$.

Here, $a_k$ is the particle Annihilation Operator and $b_k^\dagger$ the antiparticle creation Operator. Therefore, the Hamiltonian operator will create and annihilate pairs of particles and their antiparticles, such that time Evolution of the vacuum will lead to other states than the vacuum state.

Srednicki probably means the above mentioned vacuum-to-vacuum Amplitude (or vacuum persistence amplitude), when the inner product of vacuum states is

$\int D \phi e^{S_0}$.

Only for equal times the inner product of the vacuum is 1.