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

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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:

1710605797266.png

1710605837206.png

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, $$\int \mathcal D \Phi_j(\vec x) \, |\Phi_j \rangle \Phi_j (\vec x_j) \langle \Phi_j| = \hat \phi (x_j) \int \mathcal D \Phi_j(\vec x) \, |\Phi_j \rangle \langle \Phi_j|$$?
 
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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).
 
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Demystifier said:
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.
 
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
 
fcoemmanoel said:
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?
 
WWGD said:
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.
 
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