# Photon Polarization Sum - Simplifying Amplitudes with Photon Polarizations

• Hepth
In summary, for a given amplitude, if it can be written as:M=M^{\mu} \epsilon^{*}_{\mu}}then\sum_\epsilon |\epsilon^{*}_\mu M^\mu |^2 = \sum_\epsilon \epsilon^{*}_\mu epsilon\nu M^\mu M^{* \nu}andyou can replace the sum over polarizations with a -g_{\mu \nu}But what if you cannot separate it out? Say your M is of the form:M=

#### Hepth

Gold Member
When summing over photon polarizations for a given amplitude if it can be written as:
$$M = M^{\mu} \epsilon^{*}_{\mu}}$$
then
$$\sum_\epsilon |\epsilon^{*}_\mu M^\mu |^2 = \sum_\epsilon \epsilon^{*}_\mu epsilon\nu M^\mu M^{* \nu}$$

and you can replace the sum over polarizations with a $$-g_{\mu \nu}$$

But what if you cannot separate it out? Say your M is of the form:

$$M=\epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma + A\epsilon^{*}_\mu$$

Do you square it out, but then the first term will be a $$-g^{\alpha \alpha'}$$ so each term gets summed over different indices?

$$\sum_\epsilon |M|^2 =\left( \epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma + A\epsilon^{*}_\mu\right)\left(\epsilon_{\nu \alpha' \beta' \sigma'} \epsilon^{\alpha'} q^{\beta'} p^{\sigma'} + A\epsilon_\nu \right)$$

$$\sum_\epsilon |M|^2 =\left( \epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma \epsilon_{\nu \alpha' \beta' \sigma'} \epsilon^{\alpha'} q^{\beta'} p^{\sigma'} + A\epsilon^{*}_\mu\epsilon_{\nu \alpha' \beta' \sigma'} \epsilon^{\alpha'} q^{\beta'} p^{\sigma'} + \epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma A\epsilon_\nu+ A\epsilon^{*}_\mu A\epsilon_\nu\right)$$

So my question is, for every pair of polarization vectors do I make the replacement to the metric tensor? Or do I multiply the entire thing by g mu nu?

Generically, the idea is that you can write the amplitude as the inner product of a polarization-independent 4-vector and a polarization 4-vector. Then, you can pull the polarization-independent part out of the polarization sum and use the fact that $\sum_\epsilon \epsilon^*_\mu\epsilon_\nu = -g_{\mu\nu}$.

For the particular case you're asking about, you're overcomplicating things. Remember that any object with a single Lorentz index obeys $a_\mu = g_{\mu\nu}a^\nu$. So, in your case,
$$\epsilon_{\mu\alpha\beta\sigma}\epsilon^{*\alpha}q^\beta p^\sigma+A\epsilon^*_\mu = \left(\epsilon_{\mu\alpha\beta\sigma}q^\beta p^\sigma+Ag_{\mu\alpha}\right)\epsilon^{*\alpha}$$.

That said, the object you're asking about still has a free Lorentz index; so, it can't, by itself, be an amplitude, since it isn't a Lorentz scalar.

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there are a simmilar formula for massive bosons.
(k^{\mu} k^{\nu})/k^2-g^{(\mu \nu)}.

How can i get only the transverse polarization sum?