Editing Feynman diagram (section)

==== Spin 1: photons ====
The naive propagator for photons is infinite, since the Lagrangian for the A-field is:

:<math> S = \int \tfrac14 F^{\mu\nu} F_{\mu\nu} = \int -\tfrac12\left(\partial^\mu A_\nu \partial_\mu A^\nu - \partial^\mu A_\mu \partial_\nu A^\nu \right)\,.</math>

The quadratic form defining the propagator is non-invertible. The reason is the [[gauge invariance]] of the field; adding a gradient to {{mvar|A}} does not change the physics.

To fix this problem, one needs to fix a gauge. The most convenient way is to demand that the divergence of {{mvar|A}} is some function {{mvar|f}}, whose value is random from point to point. It does no harm to integrate over the values of {{mvar|f}}, since it only determines the choice of gauge. This procedure inserts the following factor into the path integral for {{mvar|A}}:

:<math> \int \delta\left(\partial_\mu A^\mu - f\right) e^{-\frac{f^2}{2} }\, Df\,. </math>

The first factor, the delta function, fixes the gauge. The second factor sums over different values of {{mvar|f}} that are inequivalent gauge fixings. This is simply

:<math> e^{- \frac{\left(\partial_\mu A_\mu\right)^2}{2}}\,.</math>

The additional contribution from gauge-fixing cancels the second half of the free Lagrangian, giving the Feynman Lagrangian:

:<math> S= \int \partial^\mu A^\nu \partial_\mu A_\nu </math>

which is just like four independent free scalar fields, one for each component of {{mvar|A}}. The Feynman propagator is:

:<math> \left\langle A_\mu(k) A_\nu(k') \right\rangle = \delta\left(k+k'\right) \frac{g_{\mu\nu}}{ k^2 }.</math>

The one difference is that the sign of one propagator is wrong in the Lorentz case: the timelike component has an opposite sign propagator. This means that these particle states have negative norm—they are not physical states. In the case of photons, it is easy to show by diagram methods that these states are not physical—their contribution cancels with longitudinal photons to only leave two physical photon polarization contributions for any value of {{mvar|k}}.

If the averaging over {{mvar|f}} is done with a coefficient different from {{sfrac|1|2}}, the two terms do not cancel completely. This gives a covariant Lagrangian with a coefficient <math>\lambda</math>, which does not affect anything:

:<math> S= \int \tfrac12\left(\partial^\mu A^\nu \partial_\mu A_\nu - \lambda \left(\partial_\mu A^\mu\right)^2\right)</math>

and the covariant propagator for QED is:

:<math>\left \langle A_\mu(k) A_\nu(k') \right\rangle =\delta\left(k+k'\right)\frac{g_{\mu\nu} - \lambda\frac{k_\mu k_\nu }{ k^2} }{ k^2}.</math>