The Green's function of the Klein-Gordon equation: $$\phi_s(x_\mu-y_\mu) = \int \frac{d^4k}{(2\pi)^4} \; \frac{e^{-i k^\mu (x_\mu -y_\mu)}}{-k_\mu k^\mu + m^2}$$ is the solution to the equation $$ \left(\partial_\mu \partial ^{\mu} + m^2\right) \phi_s = \delta(x_\mu-y_\mu) \, .$$ What's the corresponding equation and solution for a massless gauge field $A_\mu$? The free equation of motion reads $$\partial_\sigma ( \partial^\sigma A^\rho - \partial^\rho A^\sigma) =0 \, $$ but I'm unsure what we write here on the right-hand side to define the Green's function since there is a free index $\rho$ in the equation of motion. Moreover, although I checked several textbooks, I wasn't able to find the expression for the Green's function for photons.


1 Answer 1


In momentum space the equation of motion for photon is: $$(-p^2g_{\mu\nu}+p_\mu p_\nu)A_\mu= J_\nu$$ where $J_\nu$ is the source term. Now the term in the parentheses is not invertible since the determinant is zero (no mass term). This is also a signature of Gauge freedom of the photon fields. To remedy this, a gauge term is added to the original Lagrangian. $$\mathcal{L}= -\frac{1}{4} F_{\mu\nu}^2-\frac{1}{2\xi}(\partial_\mu A_\mu)^2 - J_\mu A_\mu.$$ Here $\xi$ is the gauge field which satisfies the equation of motions $\partial_\mu A_\mu=0$ (Lorenz gauge). With this Lagrangian, the equation of motion becomes $$\left[-p^2g_{\mu\nu}+\left(1-\frac{1}{\xi}\right)p_\mu p_\nu\right]A_\mu= J_\nu.$$ Now, one can check that the propagator is given by the following expression $$\Pi_{\mu\nu}=-\frac{g_{\mu\nu}-(1-\xi)p_\mu p_\nu/p^2}{p^2}.$$ To check that this is indeed the propagator, it should satisfy the equation $\left[-p^2g_{\mu\nu}+\left(1-\frac{1}{\xi}\right)p_\mu p_\nu\right]\Pi_{\nu\beta} = g_{\mu\beta}$.

So, the expression for propagator or Green's function is dependent on the gauge choice as it should be but all the physical observables should be independent of the gauge field $\xi$. Some most used choices of gauge fields in Quantum Electrodynamics are:

  1. Feynman gauge: $\xi=1\implies \Pi_{\mu\nu}=-\frac{g_{\mu\nu}}{p^2+i\epsilon}$.

  2. Lorenz gauge: $\xi=0 \implies \Pi_{\mu\nu}=-\frac{g_{\mu\nu}-p_\mu p_\nu/p^2}{p^2+i\epsilon}$

The $i\epsilon$ part is added to choose the causal Green's function and in the final expression of an observable $\epsilon\rightarrow 0$ limit is taken.

To read more about photon propagators you can have a look at any QED book. The discussion of the propagator should be independent of whether this is a classical or quantum field theory.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.