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I am reading from Schwarz book on QFT the Path Integral chapter and I am confused about something. I attached a SS of that part. So we have $$<\Phi_{j+1}|e^{-i\delta H(t_j)}|\Phi_{j}>=N \exp(i\delta t \int d^3x L[\Phi_j,\partial_t \Phi_j]).$$ What happens when we have the left and right most terms i.e. $$<\Phi_{1}|e^{-i\delta H(t_0)}|0>$$ and $$<0|e^{-i\delta H(t_n)}|\Phi_{n}>~?$$ Another thing that I am confused about, where are we using the fact that the state in the beginning and the end is $|0>$? I see we are using the boundaries on time, but I can't see where we use explicitly the fact that we start and end with vacuum i.e. if we had any $<f|i>$ where would we have a different term? Also in this case, if we are in the free theory (which I assume is the case as the vacuum state is $|0>$, if we start in the vacuum state won't we always end up in the vacuum state? What can happen in a non-interaction theory that can disturb the system from the vacuum?

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It is advantageous to take the vacuum $|0\rangle$ since its time-evolution is trivial:

$ e^{ -i \delta t H} |0\rangle = |0\rangle . $

Hence one only needs to evaluate the overlap

$ \langle \Phi | 0 \rangle \ . $

This is apparently ignored in the book you are reading, but in Weinbergs book (The Quantum Theory of Fields, Vol. 1, Chapter 9.2) you may find a calculation of the "wave function of the vacuum" for the case of asymptotically decaying interactions.

Some comments on why one may call it a wave function: in deriving the path integral formula one inserts the identities

$1 = \int \mathcal{D}\Phi |\Phi\rangle \langle \Phi|$

This looks like in quantum mechanics, were one often inserts identities of the form

$1 = \int d x | x \rangle \langle x | $

And then has to evaluate overlaps like $\langle x | \psi \rangle$, which is then called the wave-function of $|\psi\rangle$.

Note that taking general matrix elements would involve calculating overlaps with your basis vectors with general vectors, which may or may not be very hard. There are some remarks on that in the above quoted chapter of Weinberg.

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