How come a photon acts like it has mass in a superconducting field?

Question

I've heard the Higgs mechanism explained as analogous to the reason that a photon acts like it has mass in a superconducting field. However, that's not too helpful if I don't understand the latter. Why does this occur, and how?

Answer 1

A quick answer: "screening" currents in the superconductor are proportional to the vector potential. With an appropriate choice of gauge, the screening current appears as a mass term in the wave equation for the vector potential. From "An Informal Introduction to Gauge Field Theories":

(This excerpt from Google books)

Answer 2

This is a simple way to understand the screening currents in Alfred Centauri's answer. Consider the simplest model of a Superconductor--- the Landau Ginsburg model. Here you have a nonrelativistic scalar field which is both charged and has an expectation value. The situation is described by a Schrodinger field Hamiltonian:

$$ H = \int_{x} \bar{\psi}{(\nabla-qA)^2\over 2m}\psi - \mu\bar\psi\psi d^dx + \int_{x,y} \bar{\psi}(x)\psi(x)\bar\psi(y)\psi(y) V(x-y) d^dx d^dy $$

Where V(x-y) is the interaction potential between the bosons, $\psi(x)$ annihilates a boson at position x, and $\bar{\psi}$ creates a boson. The operator $\bar\psi(x)\psi(x)$ counts the number of particles at x. The total number of particles is conserved by all terms in the Hamiltonian, so the $\mu$ term is really just acting as a chemical potential, minimizing the energy with a $\mu$ is picking out the particle number you are interested in. If you don't want to include the $\mu$ term because it isn't a physical energy, just declare you start out with a certain number of particles in a periodic box.

If you choose a short-range repulsion, like $V(x-y)=\delta(x-y)$ you reproduce the nonrelativistic limit of the Abelian Higgs mechanism, a quartic term and a quadratic term. But whatever form of repulsive V you choose, you get a reasonable limit which is qualitatively the same.

The lowest energy state is the one where $\psi$ has a certain magnitude, let's call it C. This defined by the expected value of the particle number

$$ \bar{\psi}\psi = C^2 $$

So that the number density is the square of C. Now notice that a gauge transformation on $\psi$ and $A$ does the following:

$$ A\rightarrow A+\nabla \theta $$ $$ \psi \rightarrow e^{iq\theta(x)}\psi $$

So that the condensate picks out a preferred phase. If you choose the gauge so that the condensate in the vacuum is real and positive everywhere (you rotate the complex field so that it all points in one direction in the complex plane) the action is gauge-fixed, and the ground state cannot change phase.

To see what this means, consider the same theory but not coupled to electromagnetism--- this is a neutral superfluid. The superfluid phase tells you the wavefunction current, the superflow, and this flow has kinetic energy proprortional to the velocity squared.

$$ E \propto |\nabla \psi|^2 $$

which is the square of the phase-variation in $\psi$ (the amplitude variation has a restoring force, it is gapped). So there are flow modes of arbitrarily low energy, corresponding to arbitraily slow superflows.

But when you add a coupling to the vector potential A, the superflow is no longer visible, because a vacuum field $\phi$ of constant magnitude can be gauge rotated to be constant. So where is the superfluid flow degree of freedom?

It's still there, because now you gauge fixed the vector potential A without a condition on A, but using a condition on $\psi$. You see that making a superflow isn't changing the phase of $\psi$,

$$ \psi \rightarrow e^{i\theta}\psi$$

because you would rotate that away. When you rotate it away, making a superflow adds to A instead

$$ A + {1\over q} \nabla \phi$$

And the energy you add is the kinetic energy of the superflow:

$$ {m\over 2} |\nabla\phi|^2 $$

So that the effective energy of the A modes has an additional contribution:

$$ {m\over 2q} |C|^2 $$

And this is the mass term. The "current proportional to A" is saying that the superflow velocity shows up as a contribution to A, rather than as a velocity, because the gauge invariance mixes them up.

This is the content of the papers that won Landau a Nobel prize. The original papers are a little confusing in their presentation (although the ideas were clear to the authors, of course). The thing wasn't presented fully clearly until Anderson's presentation in the 1960s.

Answer 3

It is only photons in empty space which are necessarily massless. Photons in a waveguide or a plasma have cut-off frequencies $f_C$ and follow the equations of particles with rest mass $m_0=hf_C/c^2$. The group-velocity of the photon is the particle velocity, just as it is when considering an electron or any other particle as made up of a wave-packet.