A Time-ordered products derivation in "QFT and the SM" by Schwartz

[Hill]

TL;DR

Replacing a field eigenstate by the field operator

This question is not crucial, but I'd like to understand better the equation (14.35) in this derivation:

Here $\Phi$ is an eigenvalue of $\hat \phi$, i.e., $\hat \phi (\vec x ) |\Phi \rangle = \Phi (\vec x) |\Phi \rangle$.

First, I think that there is a typo in (14.35): the Hamiltonian should be evaluated at time $t_{j+1}$ rather than $t_n$. Is it right?

But the question is, why the exponential is included in (14.35)? Wouldn't it be correct just to write, $$\int \mathcal D \Phi_j(\vec x) \, |\Phi_j \rangle \Phi_j (\vec x_j) \langle \Phi_j| = \hat \phi (x_j) \int \mathcal D \Phi_j(\vec x) \, |\Phi_j \rangle \langle \Phi_j|$$?

 

[Demystifier]

Your formula without the exponential is correct as well. A formula with an exponential is studied because that's what one needs in (14.34).

 

[Hill]

Your formula without the exponential is correct as well. A formula with an exponential is studied because that's what one needs in (14.34).

Thank you for the clarification.

 

[fcoemmanoel]

The scalar fields depends of \phi_{j}=\phi(x,t), so its reasonable to separate the variables as \phi(x,t)=\phi(x)\exp(-iH\delta{t}), since j its fixed u can take a element of time \delta{t} in neiborhood

 

[WWGD]

The scalar fields depends of \phi_{j}=\phi(x,t), so its reasonable to separate the variables as \phi(x,t)=\phi(x)\exp(-iH\delta{t}), since j its fixed u can take a element of time \delta{t} in neiborhood

Could you please wrap it in ##'s to render the Latex, make it easier to read?

 

[fcoemmanoel]

Could you please wrap it in ##'s to render the Latex, make it easier to read?

I see again with more carefully, in truth he takes one of the pieces of (14.34) as in (14.27) because similarly with quantum mechanics:
\begin{equation}
A<B|C>=A<B|A><A|C>=<B|Â|A>\delta_{AC}=<B|Â|C>
\end{equation}
its intuitive.