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Helium-4 superfluidity and gauge symmetry breaking

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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?

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user Jay Bigman
asked May 16, 2012 in Theoretical Physics by Jay Bigman (30 points) [ no revision ]
See: physics.stackexchange.com/questions/13870/… for what a gauge symmetry really means.

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user genneth

1 Answer

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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).

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user Ron Maimon
answered May 17, 2012 by Ron Maimon (7,535 points) [ no revision ]
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.)

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user genneth
(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.

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user genneth
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.

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user genneth
@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.

This post imported from StackExchange Physics at 2014-04-11 15:49 (UCT), posted by SE-user Ron Maimon

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