Peskin and Schroeder say that the Ward Identity of QED proves that non-transverse photon polarizations can be consistently ignored, but I'm confused about the details.


One starts by considering some process with an external photon whose momentum is chosen to be $k^\mu=(k,0,0,k)$ and let the two transverse polarization vectors be $\epsilon_1^\mu=(0,1,0,0)$ and $\epsilon_2^\mu=(0,0,1,0)$. The Ward identity tell us that if the amplitude for the process is $\mathcal{M}=\mathcal{M}^\mu\epsilon_\mu(k)$, where we've factored out the polarization vector for the external photon in consideration, then the amplitude obeys $\mathcal{M}^\mu k_\mu=0$, on shell. With our setup, this simply tells us $\mathcal{M}^0=\mathcal{M}^3$. If we then calculate the square of the amplitude and sum over external polarizations we'd find $|\mathcal{M}|^2=\sum_{i\in\{1,2\}}\epsilon_{i\mu}\epsilon_{i\nu}^*\mathcal{M}^\mu\mathcal{M}^{*\nu}=|\mathcal{M}^1|^2+|\mathcal{M}^2|^2$. Due to the Ward identity, this is equal to $-\eta_{\mu\nu}\mathcal{M}^\mu\mathcal{M}^{*\nu}$ and so we can make the replacement $\sum_{i\in\{1,2\}}\epsilon_{i\mu}\epsilon_{i\nu}\to -\eta_{\mu\nu}$. Peskin and Schroeder claim (pg 160-161) that this is proof that non-transverse photons can be consistently ignored.

Multiple Questions

1) P&S appear to be claiming that had we also summed over non-transverse polarizations we would find that the polarization sum just turns into $-\eta_{\mu\nu}$. However, if I use the two vectors $\alpha_1^\mu=(1,0,0,0)$ and $\alpha_2^\mu=(0,0,0,1)$ as the basis for the non-transverse polarizations then it would appear that the polarization sum would turn into $\delta_{\mu\nu}$ rather than $-\eta_{\mu\nu}$. How does one know that there should be an extra minus sign so that $|\mathcal{M}^0|^2$ comes in with a minus sign relative to all the other squared amplitudes? That is, I'd somehow know that the proper calculation is given by $\sum_{\rm all\ polarizations}|\mathcal M|^2=\mathcal{M}^\mu\mathcal{M}^{*\nu}(\epsilon_{1\mu}\epsilon^*_{1\nu}+\epsilon_{2\mu}\epsilon_{2\nu}^*-\alpha_{1\mu}\alpha_{1\nu}^*+\alpha_{2\mu}\alpha_{2\nu}^*)$, but I don't see where that minus sign would arise from. One could guess that you'd have to end up with something proportional to $\eta_{\mu\nu}$ due to Lorentz-Invariance, but that's not very satisfactory. Finally, the polarization sum is naively a sum of manifestly positive numbers, yet the P&S argument depends on some kind of cancellation between these number, so how is this possible?

2) What about scenarios where one doesn't perform the polarization sum? I thought that the polarization sum was only performed when the detector is insensitive to polarization, which is not always the case at hand. I would have thought that one could show that non-transverse polarization are unphysical without having to do a polarization sum. For example, given $\mathcal{M}^\mu$ I would have thought that we could have contracted this with one of the non-traverse polarization vectors, say $\alpha_2^\mu$, and we should find that the amplitude for this process vanishes by itself.

3) Shouldn't we be showing that we can also ignore non-transverse states in all parts of the diagram, not just on external legs? If non-transverse states can run in loops then they need to be included in physical initial and final states, due to the optical theorem, which needs to be avoided. Or are P&S claiming that they've proved that they've shown that non-transverse polarizations can be ignored in the initial and final states, so by the optical theorem they can be ignored in loops, too?

  • $\begingroup$ About 2) The terms with the unphysical $\alpha_1, \alpha_2$ sandwiched with the $M_\mu$ give zero (in fact, in the axial gauge, for instance, you could see that the terms $\alpha_1, \alpha_2$ are proportionnal to $k_\mu$ or to $k_\nu$ or to the both, so they vanished when sandwiched with the $M_\mu$). So, in fact, you may keep one term ($\epsilon_1$) or the other term ($\epsilon_2$), or both terms $\epsilon_1, \epsilon_2$, depending on which experiment you are doing (polarized or unpolarized) $\endgroup$
    – Trimok
    Commented Aug 12, 2013 at 10:59
  • $\begingroup$ @Trimok: My understanding is that the unphysical polarization vectors are gauge artifacts. You only introduce both $\alpha_1$ and $\alpha_2$ in a covariant gauge, there they don't separately cancel but only in the covariant combination $k_\mu$. In axial gauge you simply don't introduce the polarization vector for the direction you are setting to zero, and so in that case there is one unphysical polarization vector $\alpha$ which cancels on its own. Do you agree, or are you saying that $\alpha_1$ on its own in a covariant gauge gets killed by $M_\mu$? (not a challenge, I'm just trying to learn) $\endgroup$
    – Andrew
    Commented Aug 12, 2013 at 15:27
  • $\begingroup$ @Andrew : Yes, the unphysical polarizations $\alpha_1, \alpha_2$ are a whole, and are not separable (so I was not enough precise, my fault..). This is true for any gauge. Of corse we may think with something specific with the Coulomb gauge or the axial gauge, but there are unphysical degrees of freedom, so maybe it is better to consider it as a whole. The idea is that the term $\eta_{ij}\epsilon^{(i)}_\mu \epsilon^{(j)}_\nu$ , for all polarizations, is always equals, by definition, to $\eta_{\mu\nu}$, $\endgroup$
    – Trimok
    Commented Aug 12, 2013 at 18:19
  • $\begingroup$ @Andrew :... while the same term for physical polarizations are related to the expression of the propagator in some gauge $\frac{1}{k^2}(\eta_{\mu\nu} + X_{\mu\nu})$. So the unphysical polarizations correspond to $-X_{\mu\nu}$. Now, the interest of the axial gauge, is that $-X_{\mu\nu}$ is proportionnal to $k_\mu$ or $k_\nu$ or both. So, sandwiched with the $M_\mu$, this gives zero. $\endgroup$
    – Trimok
    Commented Aug 12, 2013 at 18:20

1 Answer 1


(1) The completeness relationship for a basis of vectors orthonormal with respect to $\eta_{\mu\nu}$ is \begin{equation} \eta_{ij}\epsilon^{(i)}_\mu \epsilon^{(j)}_\nu = \eta_{\mu\nu} \end{equation} This normalization convention is picked for Lorentz invariance... I know you said you didn't want that answer but the point is that the normalization of these vectors is a matter of convention and it's best to pick a Lorentz invariant one. One advantage of choosing a L.I. normalization is that we don't need to specify the argument: the $\epsilon$ depend on the momentum, but these normalization conditions do not. The $\eta_{ij}$ provides the minus sign you are missing. Also here you see the basic problem that the gauge symmetry fixes: one of the polarization vectors necessarily has a negative norm.

(2) Having said that, $\epsilon_\mu^{0}$ and $\epsilon_\mu^{3}$ are not valid on shell quantities. They are a convenient mathematical fiction, needed to make an orthonormal basis, which allows things to be written in a nice, Lorentz invariant way. But the external legs of Feynman diagrams must be on shell, and as a result you can only put real honest on shell polarization vectors there, and so you aren't allowed to put $\epsilon^{(0,3)}$ there at all. Put another way, you can't satisfy the equations of motion for the photon with the longitudinal and time like modes, but the LSZ formula picks out the external wave functions that satisfy the classical equations of motion. However, since $k_\mu \mathcal{M}^\mu=0$, you could add $0$ in the funny combination $\left(\epsilon^{(0)}_\mu-\epsilon^{(3)}_\mu\right)\mathcal{M}^\mu$, which you can then add to your other basis vectors to form $\eta_{\mu\nu}\mathcal{M}^{*\mu}\mathcal{M}^\nu$ when you square to form the probablity. If the hypocricy of this angers you, that is a natural reaction, you'll eventually just accept it. (Welcome to gauge theory).

(3) EXCELLENT question. You need the off shell formulation of the Ward identity to give a real answer to this, that's in chapter 7 of P&S. Basically there's more to it than just "replace the external polarization vector by $k_\mu$", you can really show that the parts of the propagator proportional to $k_\mu k_\nu$ never matter even in loops. However, in Yang Mills theories the corresponding statement is not true! So your question is exactly on the money for Yang Mills theories, you get contributions in loops from the longitudinal and timelike modes, and by the optical theorem this taken at face value would lead to the production of unphysical particles. The fix is to add yet more unphysical particles to the theory to cancel out these parts of the loop diagrams, they are called Fadeev Popov ghosts.

After flipping through Peskin and Schroder to answer this question, I have to say that they are proving things in a very roundabout way. It's good that it teaches how to think about Feynman diagrams in a very detailed way... But there are other, less painful ways to prove and think about the Ward Identity (such as using the path integral).

  • 1
    $\begingroup$ Excellent answer. Historical remark: (I haven't read the original work, but) I believe that the roundabout argument in (3) is what originally led Feynman to introduce ghosts, and it was only later that it was understood directly through the Fadeev-Popov determinant. So the OP is retracing Feynman's steps as it were. $\endgroup$
    – Michael
    Commented Aug 12, 2013 at 10:49
  • $\begingroup$ You are right about this. Although the path integral gets you that answer much faster! $\endgroup$
    – Andrew
    Commented Aug 12, 2013 at 12:48
  • $\begingroup$ Thanks for the response! You seem to often end up answer my questions or at least commenting on them and I appreciate it. A few questions of mine in response: 1) I get that the polarization vectors obey that completeness relation, but I don't see exactly why that completeness relation enters into the calculation. I'd think you'd still in the end adding up positive definite quantities like $|\epsilon_i\cdot \mathcal{M}_i|^2$. 2) I thought the point was that even if you were silly and tried to contract $\mathcal{M}^\mu$ with unphysical polarization vectors, you'd find that these (continued) $\endgroup$
    – user26866
    Commented Aug 12, 2013 at 21:29
  • $\begingroup$ (continued) contributions cancel. What if I tried to add $\left(\epsilon_\mu^{(0)}+\epsilon_\mu^{(3)}\right)\mathcal M^\mu$? $\endgroup$
    – user26866
    Commented Aug 12, 2013 at 21:38

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