The polite fiction of symmetry breaking

The concept of spontaneous symmetry breaking (SSB) is well-studied and used in condensed matter. Most prototypically, the first textbook example is that of a weakly interacting Bose gas, where the global $U(1)$ symmetry is broken to give a Bose condensed phase, with massless Goldstone bosons as the residual evidence of that continous symmetry. This is usually described as $\langle \phi \rangle$ being an order parameter, and becoming non-zero, where $\phi$ is the annihilation operator for the a particle. The intuition is clear: given a potential like $V = |\phi|^4 - |\phi|^2$ there is a degenerate ring of non-zero $\phi$ which has the lowest energy, and by picking a particular phase we can minimise the energy.

The story becomes a bit messier upon closer inspection. The overall resolution is actually not that complex, but it seems to be mostly folklore, in that no-one has written it down.

To begin with, notice that the Hamiltonian $H$ commutes with the number operator $N = \phi^\dagger \phi$ — we do not, after all, randomly create or destroy rubidium atoms in a trap (losses are accounted for in a density matrix formalism, something we will come back to). Thus, all energy eigenstates, and particularly the ground state, will be an eigenstate of $N$. Thus by elementary arguments, $\langle \phi \rangle=0$, always. Fundamentally, by picking a definite phase for $\phi$, we introduce uncertainty in $N$.

There are some conventional ways around this. One is to say that what we really care about is off-diagonal, long-range order (ODLRO). That is, $\langle \phi^\dagger (r) \phi(0) \rangle$ as $r \rightarrow \infty$. This certainly is completely well-defined... except for real systems. The first experimental systems were a few hundred to thousand atoms, and very small. It is not clear that infinity range is applicable, and since experimentally they did in fact condense, nor necessary. In addition, ODLRO is usually quite hard to really compute.

The other standard method is to pass to a grand canonical ensemble, and argue that in the thermodynamic limit this will produce the same answers for means of macroscopic variables. This looks like it lets us have fluctuations in $N$. This shares the same problem as the previous method, but allows us to use easier mathematics. On the other hand, it's just wrong: the grand canonical ensemble is defined by $H - \mu N$, which again has eigenstates labelled by $N$ — so $\langle r \rangle$ is still zero! Indeed, what we end up with is a density matrix, with eigenvalues defined by $\exp(H - \mu N)$ in the $N$-basis! Even there, the expectation value of $\phi$ is still zero.

Antony Leggett, having thought about this quite hard, comes at the problem from an experimental side — for him the important thing is to make contact with the available experiments, which is quite reasonable. He works without 2nd quantisation, and considers the single-particle density matrix; he shows that in the normal phase, this has eigenvalues of the same order, whereas in the broken symmetry phase, one eigenvalue is near unity, with other eigenvalues of order $1/N$ or smaller. From a theoretical point of view however, this seems to throw out so much of the elegance in SSB as a generic mechanism of describing critical behaviour.

A truly mathematically correct justification is to add a term $-\epsilon \phi$ to $H$, which explicitly breaks the symmetry. Then we carefully take the limit of $N \rightarrow \infty$ first, before $\epsilon \rightarrow 0$. Analysis of the Hilbert space shows that it breaks into separate super-selection sectors, indexed by the phase. Indeed, for things like ferromagnetic transitions, one can reasonably declare that it would be unphysical for there not to be some stray fields in an experiment which would provide a small but definite symmetry breaking potential. Unfortunately for Bose condensation, there is no way to introduce a non-Hermitian term such as $\phi$ into $H$ — physical loss of atoms from the trap needs to be accounted for by a density matrix again, and the loss term only appears in the Lindbladian, not the Hamiltonian. Indeed, even for the ferromagnetic transition, one could argue that this simply transfers the problem — the universe as a whole is presumably symmetric under $SO(3)$, so the symmetry breaking term which is classical is just an approximation to a properly invariant quantum interaction term with a quantum environment.

So far, the analysis is standard, and may be found in Leggett's book Quantum Liquids. What follows is the folklore.

The final analysis of the above is actually pretty close to the mark. The missing ingredient is decoherence. A useful view point is that the symmetry breaking system is acting as an amplifier — it turns a very small asymmetric potential into a macroscopically observable one, and fluctuations in the environment certainly counts. More precisely, the ground state of the total system with its environment is a symmetric combination of states with entanglement between the large coherent fluctuation in the system and the much smaller fluctuation in the enviroment. Upon taking the thermodynamic limit and tracing out the environment, we are left with a density matrix of states equivalent to the superselection sectors description above, justifying its use.