Photon propagator in path integral vs. operator formalism

Question

I am self-studying the book "Quantum field theory and the standard model" by Schwartz, and I am really confused about the derivation of the Photon propagator on page 128-129.

He starts saying: "we are going to need to know its propagator $\Pi^{\mu\nu}$, defined by:"

$$ \langle0|T\{A^\mu(x)A^\nu(y)\}| 0\rangle = i \int {{d^4p}\over{(2\pi)^4}}e^{ip(x-y)}\Pi^{\mu\nu}(p).\tag{8.92} $$

It's not really clear this definition, which is the Photon propagator under definition? Is it $\Pi^{\mu\nu}(p)$ or $\langle0|T\{A^\mu(x)A^\nu(y)\}| 0\rangle$? I guess this definition actually assumes the Photon propagator can be expressed as the Fourier transform of $\Pi^{\mu\nu}(p)$, so they can be "identified".

Anyway, let's move on. Initially he gets the classical Green's function (aka propagator in this book) by using the equation of motion from the Lagrangian, the results is:

$$\Pi_{\mu\nu} = - {{g_{\mu\nu}-(1-\xi){{p_\mu p_\nu}\over{p^2}}}\over{p^2}}.\tag{8.100}$$

Then he merely states that "the time-ordered Feynman propagator for a photon can be derived just as for scalar field (problem 8.4) with the result"

$$i\Pi^{\mu\nu}(p) = - {{i}\over{p^2+i\epsilon}}\left[g^{\mu\nu}-(1-\xi){{p^\mu p^\nu}\over{p^2}}\right].\tag{8.102}$$

When he did the calculation for scalar field (page 75-77) he starts from $\langle0|T\{\phi_o(x_1)\phi_0(x_2)\}| 0\rangle$ then he uses the standard definition of $\phi_o(x)$ as:

$$ \phi_o(x) = \int {{d^3p}\over{(2\pi)^3}} {{1}\over{\sqrt{2\omega _p}}}\left(a_pe^{-ipx} + a_p^{\dagger} e^{ipx} \right)\tag{6.23} $$

substitute in $<0|T\{\phi_o(x_1)\phi_0(x_2)\}| 0>$ and after some simple calculation and easy complex analysis results he gets the final expression.

For the Photon propagator it is not clear what I am supposed to do. Should I use the expression for the $A_\mu (x)$ field on page 126, substitute in $\langle0|T\{A^\mu(x)A^\nu(y)\}| 0\rangle$ and do all the calculation?

If that is the case I don't get why he calculated initially the the classical Green's function.

Moreover in this link for the book:

https://schwartzqft.fas.harvard.edu/fourth-edition-corrections

you can read:

"Problem 8.4: This problem is extremely difficult. See Greiner and Reinhardt “Field Quantization” chapter 7 for a solution."

In Greiner and Reinhardt book that calculation takes about 10 pages.

So I am really confused about how I can get the Photon propagator and why he calculated initially the the classical Green's function.

My guess is that the "rigorous" calculation is the one requested in problem 8.4 and performed through 10 pages in Greiner and Reinhardt book, but Schwartz heuristically get it by an analogy with the scalar case.

The analogy goes as below:

In scalar field the Feynman propagator is just like the Green's function for the classical Klein-Gordon equation, we merely have to add the $i\epsilon$ in the denominator, which is introduced by the time ordering of creation operators. By analogy, for photon we get the Green's function for the equation of motion of classical electromagnetic field (in presence of current), then to get the Photon propagator we merely add the $i\epsilon$ in the denominator.

Answer 1

Right off the bat, to answer your first question $\Pi^{\mu\nu}(p)$ is the Fourier transform of $\langle 0| T\{A_{\mu}(x) A_{\nu}(x') \} |0 \rangle$. Usually $\Pi^{\mu\nu}(p)$ would be referred to as the momentum space propagator, while $\langle 0| T\{A_{\mu}(x) A_{\nu}(x') \} |0 \rangle$ would be referred to as the position space propagator.

Now I will address the deeper question. Technically what is usually referred to as the "propagator" is the thing that appears on internal lines of Feynman diagrams. For scalar fields, the propagator is the same as the time-ordered correlation function. This is not always the case for more complicated theories. For example for a massive vector field we have (I use the massive vector propagator because it is a little more straightforward than the massless case), $$\langle 0| T\{A_{\mu}(x) A_{\nu}(x') \} |0 \rangle = -i\int \frac{d^4 p}{(2\pi)^4} e^{ip(x-x')} \frac{g_{\mu\nu} - \frac{p_\mu p_\nu}{m^2}}{p^2-m^2 + i \epsilon} - i \frac{\delta^0_\mu\delta^0_\nu}{m^2} \delta^4(x-x') $$ We see the presence of noncovariant terms showing up in the creation function. However, if we remember the Feynman rules in the canonical formalism come from the Dyson series which is defined in terms of the interaction Hamiltonian not the interaction Lagrangian. So, we must consider what happens when we take the Legendre transform. In fact, in doing so you will find constraints which make $A^0$ no longer a dynamical field which leads to a new interaction term in the interaction Hamiltonian. After a long calculation we will find, $$ H_{\text{int}} = - L_{\text{int}} - \frac{J_0^2}{2m^2} $$ This extra term exactly cancels with the extra terms introduced in the correlation function. Leaving us with the propagator, $$-i\int \frac{d^4 p}{(2\pi)^4} e^{ip(x-x')} \frac{g_{\mu\nu} - \frac{p_\mu p_\nu}{m^2}}{p^2-m^2 + i \epsilon}$$ Which is just the classical propagator (with the $+i\epsilon$ prescription) for a massive vector field.

The reason no one talks about this is because if you work in the Feynman path integral formalism you work directly with the lagrangian. None of these extra terms show up in the derivation of the Feynman rules, and you find that the propagators are just the classical propagators.

So to derive the propagators in the canonical formalism the "correct" derivation is to follow the long calculation in Greiner and Reinhardt. However, the propagator must always agree with the one calculated in the path integral formalism, which will be the classical green's function.

A good (but not easy) resource that explains this is Weinberg's book Quantum theory of Fields Vol I.