I'm working through Lahiri & Pal's book(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# A Polarization vector sums in QED

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