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Consider some quantity from QFT $\langle 0| \hat{F}|0\rangle$. This can be written in terms of path integrals as:

$$\frac{\int F[\phi] e^{iS[\phi]} D\phi}{\int e^{iS[\phi]} D\phi}$$

if the limits are taken carefully.

Now consider a theory (such as string theory) with multiple vacuums $|0_1\rangle$, $|0_2\rangle$ etc. Now I would assume that the depending on which vacuum the Universe has settled in we would get different values for $\langle 0_i| \hat{F}|0_i\rangle$. And yet it seems to be written identically in terms of path integrals. Does this mean path integrals are many-valued. (In the same way as a many-valued function such as $\pm\sqrt{x}$ ?)

Does this mean either:

(1) My reasoning is wrong

(2) Path integral expressions can be multi-valued (depending on how the limits are taken?)

(3) The value of $\langle F\rangle$ gives the same value for any vacuum?

(4) Actions don't really have multiple vacuums.

(5) The operator $\hat{F}$ is different depending on the vacuum.

(6) Something else.

Edit

From a previous question I think the above path integral expression should be equal to a sum of multiple degenerate vaccua:

$$\sum_n \alpha_n\langle 0_n |\hat{F} |0_n\rangle$$

which makes me wonder about a couple of things: Is there a path integral expression which singles out the value for a particular vacuum? Are all the coefficients $\alpha_n$ the same? What meaning does this expression have? It seems to sum over all values of $F$ for all different vaccua which doesn't seem useful!

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The path integral as you've written it is defined with the $i\epsilon$ prescription. The role of this is to pick out the lowest energy state in the theory as the state which appears in the expectation value on the left hand side.

Perturbatively, you can expand around a metastable state that is not truly the ground state of the system. If you treat the path integral perturbatively by expanding around a given vacuum, you can

If you want to describe the expectation value in a specific state, you should include the wavefunction for that state in the path integral. (The $i\epsilon$ prescription is a trick that has the same effect as including the ground state wavefunction). This is discussed in Weinberg's book. Or also on stack exchange: https://physics.stackexchange.com/a/589493/27732

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  • $\begingroup$ Am i correct in saying there is only one "ground state wavefunction" but there may be many metastable vaccua? $\endgroup$
    – user84158
    Commented Jun 29, 2022 at 4:45
  • $\begingroup$ For another state $\Psi$ (not the ground state) I'd assume you'd just write $\int \Psi^*(\phi)F[\phi]\Psi(\phi) D\phi$ $\endgroup$
    – user84158
    Commented Jun 29, 2022 at 4:47
  • $\begingroup$ @zooby There's not necessarily only one ground state wavefunction, since can have multiple degenerate vacuum states (eg the Higgs potential), in which case I am not actually completely sure what the normal $i\epsilon$ prescription will do non-perturbatively; some care is needed in that case. But, metastable vacuaa are certainly not the lowest energy states. For another state $\Psi$, your path integral is right, as described for example in Weinberg's book or the stack exchange answer I linked above. $\endgroup$
    – Andrew
    Commented Jun 29, 2022 at 12:49

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