# Is there a spontaneous $U(1)$ symmetry breaking in atomic BECs?

In the theory of Bose-Einstein condensation, one way to define the order parameter is by using the concept of spontaneous symmetry breaking. One says that, below the critical temperature, the condensate aquires a well-defined phase by spontaneously breaking a U(1) symmetry. This is analogous to the technique used to define the classical electric field, $E_{\text{class}}(r,t) = \langle\hat{E}(r,t)\rangle$, where $\langle\hat{E}(r,t)\rangle$ is the quantum mechanical electric field operator in terms of the standard creation and annihilation operators. So, in a similar fashion, we say that $$\Psi(r,t) = \langle\hat{\Psi}(r,t)\rangle$$ where $\hat{\Psi}(r,t)$ is the Bose field operator.

I see two problems with this approach however. One is that while superposition of states corresponding to different photon numers can exist in nature, the same cannot be said about atoms however, as one cannot create or destroy them!

The second point is more technical, and presented in these notes, on page 87.

if $\langle\hat{\psi}\rangle(t=0)\neq 0$ the state of the system necessarily involves a coherent superposition of states with different total number of particles; such a state cannot be stationary (as states with different number of particles have also different energies) and it experiences a phase collapse $\langle\hat{\psi}\rangle(t)\rightarrow 0$ making the description of the evolution of the system more involved.

So, my question is, is it correct to think of BEC as a $U(1)$ symmetry breaking transition? In particular, is this the only way of explaing things like the zero sound mode (a Goldstone mode due to this symmetry breaking). Or can one avoid this concept altogether?

• (1/2) I'm not sure the second point is necessarily true. If one has a BEC in thermal equilibrium with a thermal reservoir, so that atoms can freely leave the condensate and re-enter, then the number really should be able to fluctuate without any energy cost, don't you think? Apr 5 '16 at 23:38
• (2/2) However, Leggett has made essentially the same critisisms of the SSB paradigm as applied to superfluids, and if you are interested in this subject I strongly recommend that you pick up a copy of his book Quantum Liquids and take a look at section 2.2 (it looks like you can see the relevant part in Google Books). Apr 5 '16 at 23:42
• Thanks! Indeed, it seems Leggett argues that the total number of atoms in the system + rezervoir must still be conserved in that case. Anyway, this book is from 2006, and I was wondering if the dispute has been settled meanwhile. Apr 6 '16 at 8:57

## 1 Answer

One can avoid the concept of symmetry breaking in this context, to avoid "non-conservation of the particle number".

People have devised way to do that, see for example http://arxiv.org/pdf/cond-mat/0105058v1.pdf. However, all these approaches gives the same results than standard Bogoliubov-like methods in the thermodynamic limit. This is not too surprising, given that in the thermodynamic limit, the fluctuations of the total number of particles vanishes (much in the same way that the equivalence of ensemble insure that canonical and grand-canonical ensembles give the same results). These approaches only make things more complicated for the fun of it (or consistency, depending on your community...).

• I'm no expert, but I think some of the cold atom experiments are not in the thermodynamic limit. Apr 6 '16 at 17:32
• Let's say that 10^5 atoms is far from the few-body limit, and that the precision of the experiment is not good enough to see 1/N corrections... No experiment is ever in the thermodynamic limit, yet we clearly observe second order phase transition...
– Adam
Apr 6 '16 at 21:10
• @Adam Would you please describe quickly the other method you mention ? Especially because the article you refer to is more than 100 pages long ... Thank you in advance. Feb 2 '17 at 18:12
• @FraSchelle: If I remember correctly, one does not use the standard creation operators as Bogoliubov, but instead quadratic operators $a_0 a^\dagger$ (which thus conserve the density). You then assume that the state $0$ is the most populated, go on with that, and have all factors needed to conserve the density. At the end of the day, you use the fact that $N\gg 1$, and you recover the standard Bogoliubov results...
– Adam
Feb 2 '17 at 21:16