When studying gauge theory, I often see the statement that gauge invariance does not allow the Lagrangian of the theory to contain terms that are quadratic in the gauge field. For example, to quote Folland's book on quantum field theory (page 295):

...gauge invariance forbids the inclusion of terms that are quadratic in the fields $A_\mu$ themselves (without derivatives), such as the Lorentz-invariant $\langle A_\mu | A^\mu \rangle$; hence the quanta of the fields $A_\mu$ are massless (like photons).

How does the absence of quadratic terms imply the quanta must be massless?

  • $\begingroup$ Confidently saying much of anything about the quanta (including the fact that their mass is tied to the presence of quadratic terms) requires the coupling to be small. $\endgroup$ Commented Mar 25, 2023 at 15:59

3 Answers 3


I will expand a bit on one of the answers to be clearer why a quadratic term usually leads to the 'mass' term. Because the coefficient in front of the $A_\mu A^\mu$ term has dimension $2$ doesn't by itself imply that it's the mass, but rather that it's some energy scale, squared. Now, the question is, how do we determine if that energy scale is indeed the mass? For that, we need to first define mass: mass is the threshold energy required to produce a massive particle at rest, that is, where the 3-momentum $\vec{p}$ (in some frame) is zero. The mass can then be read off from the pole of the propagator in a QFT because the propagator is just another way of writing the dispersion relation of a particle, but in momentum space. The propagator comes from the quadratic part of the Lagrangian. In a free theory, if there is no term quadratic in $A_\mu$, you don't get a constant term in the propagator which could play the role of the mass of the particle. Finally, even if you include quantum corrections to the photon propagator (self energy), it still remains massless because of Ward identity/gauge invariance (for instance, see the following questions: What prevents photons from getting mass from higher order Feynman diagrams, Is there a rigorous proof that photons do not acquire mass through renormalization?).

  • $\begingroup$ Nice compelement, thanks:) $\endgroup$
    – schris38
    Commented Mar 24, 2023 at 17:04

This is because if you have a term in the Lagrangian that is proportional to $A_{\mu}A^{\mu}$, with some coefficient in front of it, then its coefficient is going to automaticlly be considered the mass of the gauge field $A_{\mu}$ (squared).

The term proportional to $A_{\mu}A^{\mu}$ breaks gauge invariance because if you apply a gauge transformation, i.e. $$A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\lambda$$ then the Lagrangian peaks up some extra terms and, therefore, can not be considered invariant under the abovementioned transformation. The physics, then, or equivalently the equations of motion, change $$\delta\mathcal{L}\ne0$$

I hope this helps

Edit#1: It has occurred to me that your question is "why is the coefficient of the quadratic term in the Lagrangian considered the square of the mass of the gauge field?" The answer to this comes from dimensional analysis. Each of the fields $A_{\mu}$ has units of mass. Therefore $A_{\mu}A^{\mu}$ has units of mass squared. The Lagrangian (density) has units of mass to the fourth power. So, the coefficient accompanying $A_{\mu}A^{\mu}$ has units of mass squared.

Edit#2: I will rephrase because @Avantgarde's comment has indeed a point. The fact that we usually (and sometimes naively) assume that every factor, accompanying the quadratic terms in the Lagrangian, is indeed a mass term is wrong (despite the fact that it is some sort of mass, since it has units of mass). The correct phrasing would be: if I want to appoint a mass to the gauge field, then the $\underline{\text{only}}$ way to do it is by adding a quadratic term with some coefficient to the Lagrangian. That coefficient is the square of the gauge field's mass...

  • 3
    $\begingroup$ This was exactly what I was looking for, thank you! $\endgroup$
    – CBBAM
    Commented Mar 24, 2023 at 8:18
  • 4
    $\begingroup$ Just because the coefficient has dimensions (also, not 'units') of mass (or energy) squared doesn't imply that it represents the (square of the) mass itself. It just means that the coupling is dimensionful. Lagrangians have several different kinds of couplings. How do you know that a coupling is indeed the mass? In fact, you can rewrite perturbation theory and treat the quadratic, non-derivative part of the Lagrangian as an interaction term. It turns out that you can absorb it into the propagator. $\endgroup$
    – Avantgarde
    Commented Mar 24, 2023 at 12:07

If by mass we mean the coefficient in front of the quadratic terms, then the answer is straightforward... (although one might ask whether such a situation could be also treated as a case of infinite mass.)

See also:
Are photons inside the media massive? If yes, why there is no Meissner effect?
Do quasiphotons have mass?
Photon effective mass in plasma


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