A Path Integral Approach To Derive The KG Propagator

[Woolyabyss]

I'm having trouble understanding a specific line in my lecturers notes about the path integral approach to deriving the Klein Gordon propagator. I've attached the notes as an image to this post. In particular my main issue comes with (6.9). I can see that at some point he integrates over x to get a Dirac delta, which is how he gets the minus p's. I'm just not certain how he gets the desired dirac dellta function to make this expression? When I do the calculation I end up with 𝛿 (2p). Any help would be appreciated.

 

[DarMM]

Try to replace the current $J$ and the field $\phi$ by the Fourier transforms in 6.8 (use separate momentum space variables). You should get the delta you need.

Same for the kinetic term.

Then for the source term, remember the fields are real valued.

 

[Avodyne]

When you plug in the right-hand side of the 2nd equation in (6.8) for $\phi(x)$, you need to use two different dummy integration 4-momenta for each of the two factors of $\phi(x)$ in each term; call these $p$ and $p'$. Then when doing the $d^4x$ integral, you will get $(2\pi)^4\delta^4(p+p')$.

Also, note that there should be a factor of $i$ before the $\int J\phi$ term on the left-hand side of (6.9).

 

[Woolyabyss]

Thanks for the replies, I've managed to get out (p^2 -m ) term of (6.9) but am still unsure for the second term. It appears as though its being split in two and they are flipping the momentum variables, But why?

 

[vanhees71]

One should note that (6.10) misses completely the most important point about this entire calculation. It's a nonsensensical expression. You have to specify the propagator you want to calculate, and both operator perturbation theory as well as the path-integral approach show that you need the Feynman propagator here, i.e.,
$$\Delta(p)=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.$$
The crucial point is the $\mathrm{i} 0^+$ term here.

For its derivation in non-relativistic first-quantization path integrals see Sect. 1.10 in

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
The derivation for the QFT case is completely analogous to this.