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Photon Polarization Sum

  1. Mar 15, 2010 #1

    Hepth

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    Gold Member

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

    and you can replace the sum over polarizations with a [tex]-g_{\mu \nu}[/tex]

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

    [tex]M=\epsilon_{\mu \alpha \beta \sigma} \epsilon^{* \alpha} q^\beta p^\sigma + A\epsilon^{*}_\mu[/tex]

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

    [tex]\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)[/tex]


    [tex]\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)[/tex]

    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?
     
  2. jcsd
  3. Mar 16, 2010 #2
    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 [itex]\sum_\epsilon \epsilon^*_\mu\epsilon_\nu = -g_{\mu\nu}[/itex].

    For the particular case you're asking about, you're overcomplicating things. Remember that any object with a single Lorentz index obeys [itex]a_\mu = g_{\mu\nu}a^\nu[/itex]. So, in your case,
    [tex]\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}[/tex].

    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.
     
    Last edited: Mar 16, 2010
  4. May 5, 2010 #3
    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?
    Thanks in advance
     
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