# Green's function for massive photon theory

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• JD_PM
In summary, the conversation discusses the 'toy' Lagrangian in Quantum Field Theory, which assumes a massive photon. The equations of motion are given by a modified version of the Klein-Gordon equation, and the goal is to find the Green's function for this equation. This is done by taking the Fourier representation and making an ansatz, resulting in the expression $\tilde{G}_{\mu \nu} = \frac{1}{p^2-m^2} (-\eta_{\mu \nu} + p_{\mu} p_{\nu}/m^2)$. This is the time-ordered propagator for vacuum QFT.
JD_PM
TL;DR Summary
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.

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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
$$\tilde{G}_{\mu \nu} = \frac{1}{p^2-m^2+\mathrm{i} 0^+} (-\eta_{\mu \nu} + p_{\mu} p_{\nu}/m^2).$$

Last edited:
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

vanhees71 said:
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.

JD_PM

## 1. What is Green's function for massive photon theory?

Green's function for massive photon theory is a mathematical tool used in quantum field theory to describe the propagation of a massive photon. It is a solution to the equations of motion for a massive photon and is used to calculate physical quantities such as scattering amplitudes and cross-sections.

## 2. How is Green's function for massive photon theory different from that of a massless photon?

The main difference between Green's function for massive and massless photons is the presence of a mass term in the equations of motion. This results in a different form of the Green's function, which takes into account the mass of the photon and its effects on its propagation.

## 3. What are the applications of Green's function for massive photon theory?

Green's function for massive photon theory has various applications in theoretical physics, particularly in quantum electrodynamics (QED). It is used to calculate scattering amplitudes, cross-sections, and other physical quantities in QED processes involving massive photons.

## 4. Can Green's function for massive photon theory be used for other particles besides photons?

Yes, Green's function can be defined for any quantum field theory, including theories that describe other particles besides photons. The specific form of the Green's function will depend on the equations of motion for the particular particle in question.

## 5. How is Green's function for massive photon theory calculated?

Green's function for massive photon theory is calculated using advanced mathematical techniques, such as perturbation theory and Feynman diagrams. These methods involve breaking down the problem into simpler components and then using mathematical tools to solve for the Green's function. The resulting expression can then be used to calculate physical quantities of interest.

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