Deriving the Propagator: Understanding Zee's Book and the Fourier Transform

[BWV]

This is more of a math question than a physics one, but following the discussion of the propagator in Zee's book:

-(∂²+m²)D(x-y)=δ(x-y)

he then gets, by taking the Fourier transform of the Dirac delta and dividing through:

D(x-y) = $\int\frac{d^4k}{2π^4} \frac{e^{ik(x-y)}}{k^2-m^2+iε}$

I get the FT and adding iε to avoid a pole, but not how you take

D(x-y)= -(∂²+m²)^-1$\int\frac{d^4k}{2π^4} e^{ik(x-y)}$

and change the differential operator outside the integral to $1/ (k^2-m^2)$ inside it

 

[Chopin]

The differential is, properly speaking, with respect to x. Because of that, you are allowed to move it inside the integral. Then, just evaluate the derivative against the exponential, and you should get $-k^2$, so you can replace it with that. You can go through all of the calculus mechanics of evaluating a derivative in the denominator, or you can just imagine expanding the fraction in a Taylor expansion, performing all of the derivatives that way, and then recognizing that they slurp back together to form $1/(k^2-m^2)$. Note that this only works because you're hitting the derivative against an exponential, which is special because you can differentiate it as many times as you want, and it keeps giving you back itself times a factor.

Doing crazy things like this with operators inside of equations is something you'll do a lot in QFT equations, so it's probably a good idea to get used to it now.

 

[The_Duck]

I get the FT and adding iε to avoid a pole, but not how you take

D(x-y)= -(∂²+m²)^-1$\int\frac{d^4k}{2π^4} e^{ik(x-y)}$

and change the differential operator outside the integral to $1/ (k^2-m^2)$ inside it

I think you are doing things out of order. Instead of "dividing through" by the differential operator, which is a slippery formal manipulation, just write D in the original equation in terms of its Fourier transform, and also write the delta function in terms of its Fourier transform:

$$-(\partial^2 + m^2) \int\frac{d^4k}{2π^4}\tilde{D}(k) e^{ik(x-y)} = \int\frac{d^4k}{2π^4} e^{ik(x-y)}$$

Now apply the differential operator on the left. Then, from the fact that the two sides are equal, deduce what form D-twiddle must have.

 

[vanhees71]

The correct use of the $\mathrm{i} \epsilon$ description is of utmost importance here. Do not omit this and make youself clear, when which description and thus which propagator is applied in physical problems.

E.g. in vacuum perturbation theory you work with the time-ordered propagator. For a free hermitean scalar field it is defined by
$$\mathrm{i} D(x-y)=\langle \Omega|T_c \phi(x) \phi(y)| \Omega \rangle.$$
Here, $|\Omega \rangle$ is the vacuum state and $T_c$ the time-ordering operator, which orders the field operators from right to left according to increasing time arguments. We have used the temporal and spatial translation invariance of the vacuum to the effect that the propagator can only depend on the difference of the space-time arguments.

In energy-momentum representation the time-ordering operator implies uniquely the correct $\mathrm{i} \epsilon$ "prescription", which thus is not a "prescription" but derived from the principles of quantum field theory. The correct way for the time-ordered propagator turns out to be
$$\tilde{\Delta}(p)=\lim_{\epsilon \rightarrow 0^+} \frac{1}{p^2-m^2 + \mathrm{i} \epsilon}.$$
The limit is to be read as a weak limit, i.e., the limit has to be taken at the very end of the calculation, i.e., after performing all integrals in perturbation theory.