# A Path integral formula for vacuum to vacuum amplitude

1. Dec 22, 2016

### ShayanJ

I'm reading the path integral chapter of Schwartz's "Quantum Field theory and the Standard model". Something seems wrong!
He starts by putting complete sets of states(field eigenstates) in between the vacuum to vacuum amplitude:

$\displaystyle \langle 0;t_f|0;t_i \rangle=\int D\Phi_1(x)\dots D\Phi_n(x) \langle 0|e^{-i\delta t\hat H (t_n)}|\Phi_n\rangle \langle \Phi_n| \dots |\Phi_1\rangle \langle \Phi_1|e^{-i\delta t\hat H (t_0)}|0\rangle \ \ \ \ \ \ (*)$

Then he computes $\langle \Phi_{j+1}|e^{-i\delta t\hat H(t_j)}|\Phi_j\rangle$ and says that all these pieces multiply to give:

$\displaystyle \langle 0;t_f|0;t_i\rangle\propto \int D \Phi(\vec x,t) e^{iS[\Phi]}$

My problem is, not all of the factors in (*) are of the form$\langle \Phi_{j+1}|e^{-i\delta t\hat H(t_j)}|\Phi_j\rangle$. We also have $\langle 0|e^{-i\delta t\hat H (t_n)}|\Phi_n\rangle$ and $\langle \Phi_1|e^{-i\delta t\hat H (t_0)}|0\rangle$. But it seems Schwartz just ignores these two factors!
What's going on here?
Thanks

2. Dec 23, 2016

### ShayanJ

I now understand it. $|0\rangle$ is an eigenstate of both $\hat \Phi$ and $\hat{\mathcal H}$, i.e. $\hat \Phi|0\rangle=0$ and $\hat{\mathcal H}|0\rangle=E_0|0\rangle$ and this is true because the vacuum state is the same as the state with no excitation.