Green's function for massive photon theory

[JD_PM]

TL;DR

I want to understand how to derive the Green's function for the massive photon 'toy' theory, which I studied from section 1.5, Quantum Field Theory In a Nutshell by A. Zee

I am studying the 'toy' Lagrangian (Quantum Field Theory In a Nutshell by A.Zee).

$$\mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} + \frac{m^2}{2}A_{\mu}A^{\mu}$$

Which assumes a massive photon (which is of course not what it is experimentally observed; photons are massless).

The equations of motion (E.O.M.) of the theory are given by

$$\Box A^{\mu} - \partial^{\mu}(\partial_{\nu}A^{\nu}) + m^2A^{\mu} = 0 \tag{1}$$

We can rewrite $(1)$ as follows

$$\left[ (\Box + m^2)\eta^{\mu \nu} - \partial^{\mu}\partial^{\nu} \right]A_{\nu}= 0 \tag{2}$$

Now to find the Green's function $D_{\nu \rho}(x)$ we 'have to find the inverse of the differential operator in the square bracket' i.e. we have to find $D_{\nu \rho}(x)$ so that the following equation is satisfied

$$\left[ (\Box + m^2)\eta^{\mu \nu} - \partial^{\mu}\partial^{\nu} \right]D_{\nu \rho}(x)= \delta_{\rho}^{\mu} \delta^{(4)}(x) \tag{3}$$

I see that taking $(3)$ to momentum space (where I used $\partial_{\mu} \to ik_{\mu}$) yields

$$\left[ -(k^2 - m^2)\eta^{\mu \nu} + k^{\mu}k^{\nu} \right]D_{\nu \rho}(k)= \delta_{\rho}^{\mu} \tag{4}$$

But then Zee simply states the result

$$D_{\nu \rho}(k) = \frac{-\eta_{\nu \rho} + k_{\nu}k_{\rho}/m^2}{k^2 - m^2} \tag{5}$$

Might you please shed light on how to get $(5)$?

Thanks!

PS: I attached A. Zee's relevant pages.

 

[vanhees71]

You get it as any free propgator from solving for the Green's function. The Green's function in this case is a tensor $G_{\mu \nu}(x-y)$ and it obeys by definition the equation of motion
$$(-\Box G_{\mu \nu} + \partial^{\mu} \partial^{\rho} G_{\rho \nu}-m^2 G^{\mu \nu}=-\eta_{\mu \nu} \delta^{(4)}(x-y).$$
All the derivatives apply to to the four-vector $x$.

Now consider the Fourier representation
$$G_{\mu \nu}(x-y)=\int_{\mathbb{R}^4} \mathrm{d}^4 p \frac{1}{(2 \pi)^4} \tilde{G}_{\mu \nu}(p) \exp[-\mathrm{i} p \cdot (x-y)].$$
Plugging this into the equation of motion yields
$$(p^2-m^2) \tilde{G}_{\mu \nu} -p_{\mu} p^{\rho} \tilde{G}_{\rho \nu} = -\eta_{\mu \nu}. \qquad (*)$$
Of course $\tilde{G}_{\mu \nu}$ must be built with $p_{\mu} p_{\nu}$ and $\eta_{\mu \nu}$. To solve the equation we thus make the ansatz
$$\tilde{G}_{\mu \nu}=\tilde{G}_{1}(p^2) \eta_{\mu \nu}+ \tilde{G}_{2}(p^2) p_{\mu} p_{\nu}/p^2.$$
Plugging this into (*) you get [EDIT: Corrected in view of #2]
$$\eta_{\mu \nu} (p^2-m^2)\tilde{G}_1 - p_{\mu} p_{\nu} (\tilde{G}_1+m^2/p^2 \tilde{G}_2)=-\eta_{\mu \nu}.$$
From this we get
$$\tilde{G}_1=-\frac{1}{p^2-m^2}, \quad \tilde{G}_2=-p^2/m^2 \tilde{G}_1,$$
i.e., finally
$$\tilde{G}_{\mu \nu} = \frac{1}{p^2-m^2} (-\eta_{\mu \nu} + p_{\mu} p_{\nu}/m^2).$$
Of course, here I was as sloppy as Zee concerning the denominator (which didactically is a no-go to be honest!). You have to also know which propgator you want. In perturbation theory for vacuum QFT you need the time-ordered propgator. So the correct location of the poles is given by writing
$$\tilde{G}_{\mu \nu} = \frac{1}{p^2-m^2+\mathrm{i} 0^+} (-\eta_{\mu \nu} + p_{\mu} p_{\nu}/m^2).$$

 

[JD_PM]

Thank you for your reply @vanhees71. I understand how to derive Green's function now.

Just a small comment that does not modify the final answer

Plugging this into (*) you get
$$\eta_{\mu \nu} (p^2-m^2)\tilde{G}_1 + p_{\mu} p_{\nu} (\tilde{G}_1+m^2/p^2 \tilde{G}_2)=-\eta_{\mu \nu}.$$

I get

$$\eta_{\mu \nu} (p^2-m^2)\tilde{G}_1 - p_{\mu} p_{\nu} (\tilde{G}_1+m^2/p^2 \tilde{G}_2)=-\eta_{\mu \nu}.$$

 

[vanhees71]

Sure, that's a typo. I've corrected it in the original posting.