Why Is the Path Integral W[J] a Vacuum-to-Vacuum Amplitude?

[PGaccount]

In the functional integral approach to quantum field theory, we have

W[J] = ⟨0₊|0_-⟩_J = ∫ D[Φ] e^{iS + ∫ JΦ}

Can someone give me some insight into why this path integral is a vacuum-to-vacuum amplitude. How and also why does this path integral W[J] become a vev?

 

[MathematicalPhysicist]

Isn't it discussed in Sredinicki's?

 

[vanhees71]

It is important to carefully take into account the "$\mathrm{i} \epsilon$ prescription to understand, why this is the "vacuum-to-vacuum transition amplitude under influence of the external source $J$". See my QFT manuscript,

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf
In Sect. 1.10 it's explained for ordinary non-relativistic QT of a single particle. It translates easily to the QFT case, which is treated in Sect. 3.5 ff.

 

[PGaccount]

I guess, because this is the partition function, $\langle \mathrm{O} \rangle = \int e^{iS} \mathrm{O}$ similar to $\langle \mathrm{O} \rangle = \mathrm{tr \ O \rho}$, we get

$$\langle 0 | 0 \rangle = \int e^{iS}$$

@vanhees71 how does this fit in with what you said about iε?

 

[PGaccount]

Also, is there a way to interpret the infinite gaussian integral

$$\int dx \ e^{-x^T A x}$$

as a trace? I mean, is it possible to think of $x^T A x$ as a matrix element of A, i.e. $A_{xx} = \langle x | A | x \rangle$? Then the integral can be written

$$\int dx \ e^{- A_{xx}} = tr \ e^{-A} = \left( \frac{1}{\det \mathrm{A}} \right)^{\frac{1}{2}} $$

Then for free field theories, $\int e^{iS}$ can be written as a trace $tr \ e^{-A}$ where A is the inverse of the propagator. This would make the analogy with $\langle \ \rangle = tr \rho$ and $\langle \mathrm{O} \rangle = \mathrm{tr O\rho}$ exact