# photon propagator

I am reading Zee's book "QFT in a nutshell". I have a question on the photon propagator computation. For a massive photon, consider the Lagrangian $L = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2}m^2A_\mu A^\mu + A_\mu J^\mu$, then the path integral is $Z = \int dx ~L = \int dx ~\{ \frac{1}{2}A_\mu[(\partial^2 + m^2)g^{\mu \nu} - \partial^\mu \partial^\nu]A_\nu + A_\mu J^\mu \}$. From this we get that the photon propagator $D_{\mu \nu}$ satisfies $[(\partial^2 + m^2)g^{\mu \nu} -\partial^\mu \partial^\nu ] D_{\nu \lambda}(x) = \delta^\mu_\lambda \delta^{(4)}(x)$, and solving this, $$D_{\nu \lambda}(k) = \frac{-g_{\nu \lambda} + k_\nu k_\lambda/m^2}{k^2 - m^2}.$$

I can not see why the numerator has a term $k_\nu k_\lambda/m^2$. Any ideas?

-
Something's very wrong with your differential equation for the Green's function $D_{\mu\nu}$. Namely, the LHS is a scalar(-valued function), whereas the $\delta_\nu^\mu$ makes the RHS into a matrix(-valued function). I have taken the liberty to correct the formula; if I've screwed it up, I apologize. – Theo Johnson-Freyd Dec 1 '09 at 4:13

Just multiply it out in Fourier, where $\partial = ik$:

$$\bigl[ (-k^2 +m^2) g^{\mu\nu} + k^\mu k^\nu\bigr] \frac{ -g_{\nu\lambda} + m^{-2} k_\nu k_\lambda }{k^2 - m^2} = \frac1{k^2 - m^2} \bigl( - (-k^2 + m^2) \delta^\mu_\lambda + (-k^2 + m^2) m^{-2} k^\mu k_\lambda - k^\mu k_\lambda + k^\mu k^2 k_\lambda m^{-2} \bigr) = \delta^\mu_\lambda$$

which is a function of $k$. But converting back to position space, $1(k) = \delta(x)$.

This proves that $D$ is a solution. To be the solution, you usually have to impose boundary conditions, etc. In this case, there are no solutions to $\bigl[ (-k^2 +m^2) g^{\mu\nu} + k^\mu k^\nu\bigr] f_\nu = 0$: the corresponding equation in Fourier is $(-k^2 + m^2) g^{\mu\nu} + k^\mu k^\nu = 0$, and contracting with $g_{\mu \nu}$ gives $0 = d(-k^2 + m^2) + k^2 = dm^2 - (d-1)k^2$, where $d$ is the dimension of spacetime, so $k^2 = \frac{d}{d-1}m^2$, but $-\frac1{d-1}m^2 g^{\mu\nu} + k^\mu k^\nu$ cannot equal $0$, as $k^\mu k^\nu$ cannot be an invertible matrix. So $D$ is the only solution.

-
Unfortunately your first equation goes over the links on the right. – Paolo Ghiggini Oct 1 '12 at 14:16

Theo and David are perfectly correct. To add a bit more of a physical explanation which might help with the why part, a massive spin one particle has 3 physical degrees of freedom so there must be some condition on the four components $A_\mu$. The equation of motion for $A_\mu$ is equivalent to saying that each component of $A_\mu$ satisfies the massive Klein-Gordon equation and that in addition $\partial^\mu A_\mu=0$. This latter condition in momentum space implies that $k^\mu D_{\mu \nu}(k)=0$. So one can understand the $1/(k^2-m^2)$ from each component obeying the massive KG equation and the factor in the numerator as ensuring that $k^\mu D_{\mu \nu}(k)=0$.

-

This is just because of the algebraic matrix inversion in the momentum space. It can be seen from the identity:

$(g_{\mu\nu} - \frac{k_\mu k_\nu}{(k^2-m^2)}) (g_{\nu\lambda} - \frac{k_\nu k_\lambda}{m^2}) = g_{\mu_\lambda}$

This is a special case of a general result in linear algebra: The inverse of a matrix prportional to a weighted sum of a unit matrix and a Hermitian matrix of unit rank is also proportional to a (generally different) weighted sum of the unit matrix and the Hermitian unit rank matrix.

-