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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.

Community
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Minhyong Kim
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3 Answers3

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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.

Michael
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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.

Arnold Neumaier
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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.

A few other stray comments:

  • 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).
  • Whether or not our universe is spatially finite or infinite is a completely open observational question. Observationally, it appears to be extreme close to being perfectly flat - right on the border between having a gigantic radius and an infinite radius. The observable universe is certainly finite in radius, but homogeneity of the large-scale metric suggests that the universe extends far, far beyond the observable horizon ("horizon" in the colloquial sense, not the general relativity sense).
  • Even if the real universe does turn out to be spatially finite, the resulting mass gap would be so tiny that the photon would still be massless for all intents and purposes - its "mass" would be dozens of orders of magnitude below the mass of every other elementary particle.
tparker
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