Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Polarization vector sums in QED

  1. Nov 9, 2018 #1

    Glenn Rowe

    User Avatar
    Gold Member

    I'm working through Lahiri & Pal's book A First Book of Quantum Field Theory, Second Edition and I'm stuck on their explanation of the polarization vector in quantum electrodynamics in Chapters 8 and 9. In section 8.8, they derive a formula for the sum over the transverse polarization modes of the polarization vector ##\epsilon_{r}^{\mu}\left(k\right)##, where the subscript ##r## denotes the polarization mode (with ##r=1,2## being the transverse modes), ##\mu## is the 4-vector component and ##k## is the 4-momentum of the photon. If we expand their equation 8.87 and set ##k^2=0## for the photon, we get an equation for the sum over the transverse polarization modes:
    $$
    \begin{align}

    \sum_{r=1,2}\epsilon_{r}^{\mu}\left(k\right)\epsilon_{r}^{\nu}\left(k\right) & =-g^{\mu\nu}-\frac{k^{\mu}k^{\nu}}{\left(k\cdot n\right)^{2}}+\frac{n^{\mu}k^{\nu}+k^{\mu}n^{\nu}}{k\cdot n}

    \end{align}
    $$
    Here, ##n^\mu## is defined by Lahiri & Pal after equation 8.49 as "an arbitrary time-like vector satisfying ##n^{\mu}n_{\mu}=1## and ##n^0>0##. In practice, they usually choose ##n=\left(1,0,0,0\right)##.
    However, in equation 8.88, they state that, because the photon couples only to conserved currents, the last 2 terms in the sum don't contribute, and they then write
    $$\begin{align}
    \sum_{r=1,2}\epsilon_{r}^{\mu}\left(k\right)\epsilon_{r}^{\nu}\left(k\right)=-g^{\mu\nu}\end{align}$$which they then say is the "polarization sum of physical photons in any scattering amplitude".
    My first question is - what does this mean? How can they just throw away the last two terms?
    This is made even more mysterious in Chapter 9, where in equation 9.109 they claim that the same sum is given by
    $$\begin{align}\sum_{r}\epsilon_{ri}\left(k\right)\epsilon_{rj}^{*}\left(k\right)=\delta_{ij}-\frac{k_{i}k_{j}}{\left|\mathbf{k}\right|^{2}}\end{align}$$where the sum is over transverse polarization states only (that is, ##r=1,2##) and the indexes ##i## and ##j## refer to the spatial components 1, 2 and 3. This follows from equation (1), since the metric tensor satisfies ##g_{ij}=-\delta_{ij}## and we're taking ##n=\left(1,0,0,0\right)##. For a photon, ##k^{0}=\omega=\left|\mathbf{k}\right|## where ##\omega## is the energy. Using this formula, they derive the formula for the differential cross-section of the scattering of unpolarized photons by an electron. Their derivation makes sense up to this point, providing that we accept equation (3) as the one to use for the sum.
    Second question: Despite them saying in Chapter 8 that equation (2) above is valid for cross-section calculations, why are they now using equation (3)?
    Finally, in their Exercise 9.7 they then ask us to derive the same formula for the cross-section using equation (2) above for the sum. My last question is: how on earth can we do this, since equation (2) doesn't have any reference to the 4-momentum, which is needed to derive a cross-section formula which depends on the scattering angle of the photon?
    Thanks for any help.
     
    Last edited: Nov 9, 2018
  2. jcsd
  3. Nov 9, 2018 #2

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    That's of course imprecise, but FAPP sufficient, because due to gauge invariance you have a Ward-Takahashi identity ensuring that all on-shell amplitudes needed to evaluate the S-matrix elements by amputating the external legs and fit them with the appropriate free wave functions, do not depend on the arbitrary four-momentum longitudinal part of these wave functions, i.e., such an amplitude fulfills ##k_{\mu} A^{\mu}=0##. That's also the reason, why instead of using complicated photon propagators you can simply use the one of the Feynman gauge, ##D_{\mu \nu}(k)=-g_{\mu \nu}/k^2##.
     
  4. Nov 10, 2018 #3

    Glenn Rowe

    User Avatar
    Gold Member

    I think my main problem is that I don't know where to start to solve the problem of deriving the differential cross-section using equation (2) above rather than equation (3). In their problem statement, Lahiri & Pal say that we're not allowed to use the condition which they derive from gauge invariance, that is, we can't use the form ##\epsilon^{\mu}=\left(0,\boldsymbol{\epsilon}\right)## that is, where ##\epsilon^0=0##. In order to derive the cross-section using (2), do I need to impose some other condition on ##\epsilon^{\mu}##? If so, what condition? Do I then need to go through their whole derivation again, calculating traces and so on?
    Any hints gratefully received.
     
  5. Nov 10, 2018 #4

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Eq. (2) is obviously wrong, but it's ok when used for the sums over photon polarizations in scattering amplitudes involving photons, because the corresponding truncated N-point functions fulfill the corresponding Ward identities, i.e., you can indeed use ##-g^{\mu \nu}## as a Feynman rule for the polarization-averaged external photon leg. This is due to gauge invariance. Thus you can change the polarization matrix according to (3) which corresponds to the radiation gauge of the free photon field to
    $$\rho_{\mu \nu}'(k)=\rho_{\mu \nu} + k_{\mu} \chi_{\nu}+k_{\nu} \chi_{\mu}$$
    with an arbitrary vector field ##\chi_{\mu}##. Choosing ##\chi_{0}=-1/(2 k)##, ##\vec{\chi}=\vec{k}/(2k^2)## leads to ##\rho_{\mu \nu}'=-g_{\mu \nu}##.
     
  6. Nov 11, 2018 at 3:22 AM #5

    Glenn Rowe

    User Avatar
    Gold Member

    I haven't got far enough in QFT for this to mean much to me yet, I'm afraid. Lahiri's book does mention the Ward-Takahashi identity, but not for another 3 chapters beyond where I am now, and I haven't come across N-point functions either. I'm not sure what you mean by the 'polarization matrix'; is this just the LHS of (3)? I've come across the polarization vector which is what Lahiri calls ##\epsilon^{\mu}##, and he shows that under the transformation ##\epsilon^{\mu}\rightarrow\epsilon^{\mu}+k_{\mu}\theta## (where ##\theta## is an arbitrary constant) the Feynman amplitude is unchanged. What does ##\rho_{\mu\nu}## represent? Is it the same thing as ##\sum_{r=1,2}\epsilon_{r}^{\mu}\left(k\right)\epsilon_{r}^{\nu}\left(k\right)##? I'm guessing that by ##k## you mean ##\left|\mathbf{k}\right|##, that is, the magnitude of the 3-momentum. The symbol ##k## in Lahiri is used for the 4-momentum, so that for a photon ##k^{2}=0## which wouldn't make much sense in your definition of ##\chi##.
    Sorry for all this, but I'd really like to get to grips with this question, as I feel that I'm missing something fundamental here.
     
  7. Nov 14, 2018 at 9:38 AM #6

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    ##\rho_{\mu \nu}## is the statistical operator for polarization states in matrix representation, used to describe partially polarized (or in the extreme case unpolarized) photons.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted