Outline - heuristic derivation of the Ward identity from the requirement of the Lorentz invariance

Suppose we have the free quantized gauge theory (with quanta called photons) with the Hilbert space restricted to transversal polarizations only. Next, suppose interactions with the matter, and assume the matrix element $M$ with (at least) one external photon line having the polarization vector $\epsilon^{\mu}_{\perp}(p)$. It has the form $$ \tag 1 M = \epsilon^{\mu}_{\perp}(p)M_{\mu} $$ Assume $M_{\mu} \sim p_{\mu}$. Next, perform the massless Lorentz orbit little group transformation $\Lambda$ defined by $$ \tag 2 (\Lambda p) = p, \quad (\Lambda \epsilon_{\perp_{1}}) = \epsilon_{\perp_{1}} +cp, \quad (\Lambda \epsilon_{\perp_{2}}) = \epsilon_{\perp_{2}} $$ and apply it to transformed squared averaged matrix element $$ |\bar{M}|^{2} \equiv \sum_{\perp_{1,2}}|M|^{2} $$ Since it must be invariant in the Lorentz-invariant gauge theory, we obtain $$ |\bar{M}|^{2} = |\bar{M}^{2}| + c(p^{\mu}\epsilon_{\perp_{1}}^{*\nu} + p^{\nu}\epsilon_{\perp_{1}}^{\mu})M_{\mu}M^{*}_{\nu}, $$ from which follows the requirement $$ \tag 3 p^{\mu}M_{\mu} = p^{\nu}M_{\nu}^{*} = 0, $$ i.e., the Ward identity.

My question

1) What is the physical sense of the transformation $(2)$? Precisely, may I state that it generates the unphysical "longitudinal" state from the physical transverse one?

2) Next, does from the derivation above follow that the unitarity (a-la optical theorem) is broken as long as $(3)$ isn't valid? Precisely, being inserted as the sub-diagram the amplitude $M_{\mu}$ will distinguish physical and unphysical polarizations, and that's enough for proving the non-unitarity (see this related question)? For example, one may calculate the optical theorem (with $A$ denoting the state containing the photon) $$ 2\text{Im}M_{A \to A} = (2\pi)^{4}\sum_{n}|M_{A \to n}|^{2}, $$ where for the given order of perturbation theory $M_{A \to A}$ doesn't respect the Ward identity while $M_{A \to n}$ respects it. Then one may perform the transformation $(2)$ and deduce that there is indeed the violation of the unitarity.

3) Finally, does this relate directly the Lorentz invariance to the unitarity of the gauge theory?

  • $\begingroup$ I have a question about your question: why do you need to square the amplitude and sum over the polarizations to get the Ward identity? Why isn't the fact that eq.2 maps the state into an equivalent one but shifting of $\epsilon$, is not enough to implies eq.3, without going through the squaring and averaging? $\endgroup$
    – TwoBs
    Commented Jan 9, 2017 at 21:42
  • $\begingroup$ @TwoBs : the reason is formal: the amplitude $M$ is not in general required to be the Lorentz-invariant quantity, and in order to avoid any problems I decide to impose the requirement of the Lorentz invariance on the quantity $|\bar{M}|^{2}$, which has to be Lorentz-invariant by the definition (at least for correctly constructed representations of the Poincare group). $\endgroup$
    – Name YYY
    Commented Jan 9, 2017 at 22:06
  • $\begingroup$ yes, the amplitude is generically transforming with the Wigner rotation associated to the little group. But then you should not summing only over the $\epsilon$ but over all the other states pol. in $M_\mu$, changing your conclusion. I was instead wondering if for your choice of $\Lambda$, itself an element of the little group, the Wigner rotation was actually trivial and there is perhaps non need to sum the polarizs. It seems it is in fact of a special type, $L(p)^{-1}\Lambda L(p)$ where $L(p)$ sends the reference victor to $p$. If this lives in the translations of iso(2)... $\endgroup$
    – TwoBs
    Commented Jan 10, 2017 at 6:18
  • $\begingroup$ ...the states would be invariant and the Ward identity would follow. $\endgroup$
    – TwoBs
    Commented Jan 10, 2017 at 6:30
  • $\begingroup$ Actually, I take back what I have said in my comment above about the need of summing over all other particles' polarizations. I see the point more clearly now. Nice trick this way of proving the Ward identity. $\endgroup$
    – TwoBs
    Commented Jan 10, 2017 at 8:39

1 Answer 1


1) see Weinberg's book.

2) You (and some introductory texts) have the logic inverted. Your "derivation" of $(3)$ is not a derivation of the Ward identity; it only serves to explain why the Ward identity is important. A quick proof of the Ward identity can be found in Srednicki's book.

3) If you use the Coulomb gauge then the Ward identity guarantees that the theory is covariant. If it is not satisfied, then covariance is broken but the theory is still unitary: there are no negative norm states.

If you use the $R_\xi$ gauge then the Ward identity guarantees that the theory is unitary. If it is not satisfied, then unitarity is broken but the theory is still covariant: the polarisation vectors are all true vectors, but the longitudinal ones are not decoupled.

This means that the Ward identity is related to both covariance and unitarity, but not at the same time. If the Ward identity becomes anomalous you either lose covariance or unitarity, but not both. And you must decide which one you want to give up (the choice is irrelevant anyway because if one of these properties isn't satisfied, then the whole theory becomes meaningless).

  • $\begingroup$ Thank You for the response! But let me write few comments. For Your second paragraph: I wrote "from which follows the requirement", which should be interpreted as "the Lorentz invariance requires...". For Your third paragraph: maybe I don't agree with You. It seems that the propagator numerator in the Coulomb gauge doesn't coincide with the transverse polarizations sum rule. They are different up to terms proportional to $k^{\mu}$. Therefore as long as the Ward identities doesn't hold, the optical theorem (and hence the perturbative unitarity) is broken. $\endgroup$
    – Name YYY
    Commented Jan 8, 2017 at 20:58
  • $\begingroup$ @NameYYY the numerator of the propagator in the Coulomb gauge does coincide with the polarisation sum. In fact, it is possible to define the propagator through polarisation sums. The problem is that in the Coulomb gauge there are two kind of vertices, the standard $-ie\boldsymbol \gamma$ and a non-local one. We can formally regard the latter as a modification of the propagator, but this doesn't mean that the true propagator is different than the polarisation sum. It is an "effective" propagator so to speak. (1/2) $\endgroup$ Commented Jan 8, 2017 at 21:30
  • $\begingroup$ (2/2) But if you check the unitarity (through the optical theorem) with the true propagator and the two kinds of interactions, everything works out just fine. In the end, there are no negative norm states and therefore there is no reason to expect any break down of the unitarity (unless there is some other anomaly). $\endgroup$ Commented Jan 8, 2017 at 21:30
  • $\begingroup$ Thank You again. Also I have the question about Your last paragraph. What's about the gauge $A_{0} = 0$, when it seems that both unitarity and Lorentz covariance suffer (the theory isn't Lorentz covariant and the gauge field propagator seems to contain the longitudinal polarization)? $\endgroup$
    – Name YYY
    Commented Jan 9, 2017 at 10:46
  • $\begingroup$ @NameYYY oh, sorry I forgot to answer. The Coulomb gauge is "natural" because it follows from the properties of the representations of the Poincaré Group. $R_\xi$ gauges are "natural" because they represent the massless limit of a Stückelberg field. But axial gauges $n\cdot A=0$ are pretty "unnatural": there is no physical mechanism that could lead to them. Only if the theory is gauge invariant the choice of gauge is irrelevant. But in the absence of the Ward identity axial gauges make no sense. (1/2) $\endgroup$ Commented Jan 9, 2017 at 21:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.