- #1

JD_PM

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- 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.

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

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.