# Helium-4 superfluidity and gauge symmetry breaking

Is there an accessible account of superfluidity in Helium-4 as a manifestation of "global gauge symmetry" breaking?

And what is meant by "global gauge symmetry"? I was taught that gauge symmetries were by definition local. Is it just a different terminology in condensed matter?

It's just a global phase symmetry that's being broken--- they mean the wavefunction of the He, the condensate wavefunction, defines a definite notion of phase at every point, and this is a breaking of phase symmetry. The word "gauge" is being misused here, the phase symmetry of the field is not a gauge symmetry in this case (although it would be if the He wasn't neutral).

The precise statement is that the He is described by a Schrodinger field $\Psi(x)$, a (nonrelativistic) bosonic complex quantum field $\Psi$, or alternatively two real fields, the real and imaginary part, with a Lagrangian:

$$S = \int i \Psi^* i{\partial \over\partial t} \Psi - \Psi {\nabla^2\over 2m} \psi d^3x dt - \int dx dy \Psi^*(x)\Psi(x) V(x-y)\Psi^*(y)\Psi(y) d^3x d^3y dt$$

Where $V(x-y)$ is the pair-potential for He atoms. This quantum field Lagrangian gives the many-particle bosonic Schrodinger equation.

The Lagrangian is phase invariant, corresponding by Noether's theorem to the conservation of particle number. The particle-number current is what is called the "probability current" in elementary quantum mechanics books (this is a misnomer: quantum probabilities are global notions. The phase symmetry of the quantum field version of the SE explains why you have a local current for the probability density--- in the quantum field context, the probability $\Psi^*\Psi$ is the particle number operator, and particle number is locally conserved).

You can multiply $\Psi$ by a phase and nothing happens to the Lagrangian. But in a dense condensed state, where the He atoms are superfluid, there is an expectation value for $\Psi$ in this state.

$$\Psi(x) = \psi(x)$$

Where $\psi$ is the superfluid condensate. This $\psi$ has a definite phase which breaks the phase-invariance. Since "phase invariance" is the "gauge invariance" for a charged field, people call this (inappropriately) breaking global gauge invariance, which sounds like an oxymoron.

The Schrodinger equation in this classical context is sometimes called the Gross-Pitaevski equation.

There is a minor paradox associated with a definite phase--- the particle number must be indefinite for the phase to make sense! This is resolved either by thinking of this as Yang's "off-diagonal long range order" in the density matrix formulation (although I still have to find an example where this is not the same as saying a quantum field has an expectation value! I am sure such examples exist, I just haven't seen one), or just by imagining you have a superposition of different numbers of He atoms in your container (this can happen in an open system).

• Regarding the phase (warning: no relevance to the OP) --- it's not quite that simple. In an open system, the density matrix has eigenvectors which correspond to states of definite number; the statistical fluctuations in number is purely classical (as it must be since number is a conserved quantity of $H$ and thus $H - \mu N$). ODLRO is good for theoretical purposes, but real condensates are not that big --- until quite recently, a few thousand atoms was an achievement! The expectation value has its own problems in that for finite systems it is zero, and (cont.) – genneth May 17 '12 at 5:45
• (cont.) the square of it is non-zero always. Tony Leggett has put forwards the suggestion of a large (order unity) separation between eigenvalues of the single particle density matrix, which clearly reproduces the thermodynamic limit, but is also practically quite well-defined for experimental purposes. He punts on the question of how symmetry breaking ever really happens, however. – genneth May 17 '12 at 5:48
• The best reference I can find (aside from Leggett's book) is springerlink.com/content/g1rjw320120755k4 which is unfortunately Springer so depends on your institution's access. – genneth May 17 '12 at 6:04
• @genneth: These things are directly analogous to any other SSB, so for example a solid crystallizing in a liquid can't pick a location, but it does, and a Higgs field can't pick a direction in finite volume (but it does), these things don't bother me. The thing that bothers me (perhaps this should be a question) is when is ODLRO ever something other than a classical field (composite or elementary) with an expectation value? It's a more general notion, so there should be an example, but I don't know any. – Ron Maimon May 17 '12 at 6:17