photon propagator - MathOverflow

photon propagator

2009-12-01T03:30:18Z

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?

Answer by David Bar Moshe for photon propagator

2009-12-01T04:28:08Z

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

(g_mu_nu - k_mu * k_nu/(k^2-m^2)) *(g_nu_lambda - 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.

Answer by Theo Johnson-Freyd for photon propagator

2009-12-01T04:43:16Z

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.

Answer by Jimiras for photon propagator

2012-09-30T15:12:27Z

Sir Theo Johnson-Freyd had not answer "how" the kνkλ/m2 term came up in numerator. If plugging directly the answer which A.Zee gave into the differential equation, it just confirm the massive photon propagator has that form.

Answer by Jeff Harvey for photon propagator

2012-10-01T12:16:10Z

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 $k^\mu D_{\mu \nu}(k)=0$.