# Mass gap for photons

I am puzzled by the answers to the question:

What is a mass gap?

There, Ron Maimon's answer gives a clear-cut definition, which I suppose applies to any quantum field theory with Hamiltonian $H$, that the theory has a mass gap if there is a positive constant $A$ such that $$\langle \psi| H |\psi \rangle\geq \langle 0 |H | 0 \rangle +A$$ for all nonzero (normalized) $\psi$.

But then, Arnold Neumaier says

QED has no mass gap, as observable photons are massless states.

I would quite appreciate a brief explanation of this statement. The definition is concerned with the minimum possible energy for non-zero states. So I don't see why the photons having zero mass would imply the absence of a mass gap.

• to : "for all nonzero (normalized) ψ" : must be "for all normalized ψ that are orthogonal to the ground state" Commented May 18, 2013 at 18:17
• Commented Oct 20, 2016 at 18:21

Because you can prepare a state with an arbitrarily long wavelength, hence arbitrarily low energy, photon. That's essentially the definition of a massless particle. If you put in an IR regulator, by putting the system in a box for example, a gap appears since there is now a largest possible wavelength. This can be mimicked by giving the photon a small mass. However, in the limit where the IR regulator disappears so goes the mass gap and photon mass.

• But the universe is compact, is it not? Commented May 18, 2013 at 1:16
• @MinhyongKim Is it? That's an unsettled cosmological question. If you are referring to the Hubble volume fair enough... but now we're not talking about vanilla QED, which exists (platonically) in Minkowski space. In a version of QED on some compact space there will be a gap in the spectrum corresponding to the lowest mode of the photon. Commented May 18, 2013 at 1:36
• That seems to make sense. Could I trouble you to explain the phrase 'mimicked by giving the photon a small mass?' I presume this has to do with some effective action describing the regulated theory? Commented May 18, 2013 at 7:16
• Regarding the universe, am I correct that the standard view is still that there is a finite radius? In this case, I don't see how it can be non-compact, unless the metric is incomplete. Perhaps that's the point you're making: the completeness is itself conjectural. In any case, it seems the wavelengths must be bounded. Commented May 18, 2013 at 12:49
• I see I should correct myself on one point. It seems the universe is conjectured to be complete except for the interior of black holes.(Cosmic censorship) So at the classical level, maybe the universe is supposed to be non-compact. Commented May 19, 2013 at 9:25

The point is that if you have particles of zero mass, there are states with an arbitrary positive mass. The reason is that n-particle states made up of particles with momentum $p_1,...,p_n$ the total momentum is $p=p_1+...+p_n$, which is a state of positive mass $m=\sqrt{p^2}$. If all of the $p_k$ come from a photon, it is a simple mathematical exercise to see that $m$ can take any positive value. Thus the mass spectrum has no gap.

This is purely a terminology issue. In relativistic quantum field theory, mass is considered to be a form of energy (by $E = m c^2$). In nonrelativistic quantum mechanics, one should indeed talk about the "energy gap," not the mass gap. People still sometime call it a "mass gap" by analogy with the relativistic case, but this is just a slightly sloppy word choice.

• As jjcale points out, your "for all nonzero (normalized) $| \psi \rangle$" should be "for all normalized states $| \psi \rangle$ orthogonal to the ground state $| 0 \rangle$." Despite the potentially misleading notation, the ground state $| 0 \rangle$ is not the zero vector, which is usually denoted 0 to avoid confusion even though it's a ket.
• When deciding whether a theory is massive or massless, one usually (though not always) considers the theory defined on (infinite) Minkowski space - because as you say, when defined on a spatially compact spacetime any theory is trivially gapped, with a minimum energy on the order of $\hbar c / R$ (where $R$ is the radius of the universe).