I have read that the photon is protected from acquiring a mass due to gauge invariance. However, I was under the impression that gauge invariance is not a true physical symmetry, but rather a redundancy in description of a given physical theory?

Is it simply that gauge invariance is required such that one can remove the two unphysical degrees of freedom in the photon field. If the theory weren't gauge invariant then one would not be able to do this. Accordingly, requiring gauge invariance prevents one from writing down a mass term for the photon field sinc such a term is not gauge invariant?!

If this is true, however, I have read that without the Higgs mechanism all standard model particles are massless and that this is due to the gauge symmetry, so what is the general reason for why this is the case if gauge symmetry is simply a redundancy in our description?

  • $\begingroup$ I'd drop "custodial" from your question. It is not a gauge symmetry--it is just the SO(4) global symmetry of the Higgs potential. Fermions are not protected against getting a mass by a gauge symmetry, but by their EW Left handed SU(2). They do get a mass without the Higgs if they interact strongly, since QCD breaks chiral symmetry dynamically, and it is a gauge theory, to boot. You have conflated 3 points so impossibly that it is all-but impossible to parse them out, when their logical connections are so exiguous. $\endgroup$ Jan 12, 2017 at 1:27
  • $\begingroup$ @CosmasZachos Why is it so often stated in introductory texts that photons (and other SM particles without the Higgs mechanism) are protected from acquiring a mass by gauge symmetry? Is this just an abuse of terminology from the fact that they are in gauge groups (unbroken before spontaneous symmetry breaking) that have SU(2) and U(1) symmetries? $\endgroup$
    – user35305
    Jan 12, 2017 at 8:21
  • $\begingroup$ No abuse of terminology: Gauge invariance prevents gauge boson masses, and ensures that quantum corrections preserve it, G.I., so mass effects cannot appear in perturbation theory. That is all what is meant by "protection". Mentally insert a "quantum" before it, and let it go. $\endgroup$ Jan 12, 2017 at 13:48
  • $\begingroup$ @Cosmos Zachos What I find confusing though is that gauge invariance isn't a physical symmetry, but a redundancy in our description, so how can it protect against quantum corrections?! Is it simply that we demand physical theories to gauge invariant and by requiring this we cannot have mass terms appearing in the Lagrangian describing the theory?! $\endgroup$
    – user35305
    Jan 12, 2017 at 16:28
  • $\begingroup$ Yes, just that. I think you misunderstand "not physical". It is physical in the sense it is impossible without an underlying global/rigid symmetry enabling it. The only "unphysical" part is failure to rotate components of a nonexistent full multiplet for gauge bosons. $\endgroup$ Jan 12, 2017 at 16:36

1 Answer 1


However, I was under the impression that gauge invariance is not a true physical symmetry, but rather a redundancy in description of a given physical theory?!

This is true. Instead of the symmetry transformation $$ U|\Psi\rangle \to |\Psi'\rangle $$ the unitary gauge transformation operator $U$ with the generator $G$ acts on physical state $|\Psi\rangle$ as $$ \tag 1 U(G)|\Psi\rangle = |\Psi\rangle \leftrightarrow G(\mathbf x)|\Psi\rangle = 0 $$ Analogically, in theory with interactions there is another "do-nothing transformation", called the Ward identity. It states that any amplitude $M$ with at least one external photon line, i.e., $M = M_{\mu}\epsilon^{\mu}(p)$, is transversal: $$ \tag 2 p^{\mu}M_{\mu} = 0 $$
It is the analog of gauge current conservation in the classical theory.

Together, $(1)$ and $(2)$ can be called as "the gauge invariance of the quantum field theory", but in fact they are just the description of the gauge redundancy.

It turns out that typically $(2)$ forbids the mass of the photon. Precisely, it can be shown that in most cases it prevents the position of the photon propagator pole (which defines its mass) to be shifted from zero.

However, in some particular cases we can write down the gauge field mass without violating the gauge invariance (this statement in fact is not exact, see below, however typically people say it as "mantra").

Really, classically one can write down the gauge-invariant field $$ \tag 3 V_{\mu} = A_{\mu} - \partial_{\mu}\frac{1}{\square}\partial_{\nu}A^{\nu} $$ Corresponding mass term $m^{2}V_{\mu}V^{\mu}$ doesn't violate the gauge invariance, although it looks cumbersome. The second summand in $(3)$ looks bad, but the situation becomes good when there is some scalar degree of freedom $\varphi$ which parametrizes it: $$ \frac{1}{\square}\partial_{\nu}A^{\nu} \to \varphi $$ In the quantum field theory corresponding situation appears when the photon vacuum polarization amplitude has the pole at zero momentum.

There are few models in which such situation appears. The first one is the model with the Higgs mechanism (often called "spontaneous breaking" of the gauge symmetry), where the particles spectrum doesn't form the representation of the gauge group below some scale $\Lambda$. Instead of $A_{\mu}$ we're dealing with $V_{\mu}$; people say that $A_{\mu}$ eats the "Goldstone boson" $\varphi$. But note that in fact $V_{\mu}$ is not the "photon"; it is the combination of the photon and longitudinal polarization.

The prototype of this idea is the classical theory of EM field interacting with the plasma (this was pioneered by Anderson).

The second example is strong interaction regime bozonization. It happens in Schwinger's massless 2D QED. In this case, two massless fermions form the massless bound state shifting the photon propagator pole. One can't write down the local term generating the photon mass without introducing the new field representing this bound state.

  • $\begingroup$ Why is it often said in introductory texts that gauge symmetry protects the photon (and other particles, without the Higgs mechanism) from acquiring mass then? Is this simply an abuse of terminology from the fact that (before any spontaneous symmetry breaking) they are in gauge groups with SU(2) and U(1) symmetry? $\endgroup$
    – user35305
    Jan 12, 2017 at 8:19
  • $\begingroup$ @user35305 : in the way of redundancy. Really, the Ward identity by forbidding the mass forbids also the generating of the longitudinal component of 4-potential $A_{\mu}$, which is nothing but the gauge redundancy. $\endgroup$
    – Name YYY
    Jan 12, 2017 at 17:45
  • $\begingroup$ In what sense does the Ward identity forbid mass terms for the photon? I read the Wikipedia article on it and found it a bit vague - it only mentions that it prevents the generation of a longitudinal component of the 4-potential $A_{\mu}$. $\endgroup$
    – user35305
    Jan 19, 2017 at 10:54

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