# Is the superfluid state a coherent state?

In the normal to a superfluid phase transition, U(1) symmetry related to particle number conservation is spontaneously broken which seems to imply that the superfluid state is a state in which there is no definite number of particles? This property is shared by that of a coherent state or any arbitrary superposition of number operator eigenstates.

Is there any property of the superfluid state (the condensate wavefunction) that is not shared by a coherent state?

• "which seems to imply that the superfluid state is a state in which there is no definite number of particles" - why would it imply that? The breaking of a symmetry means that the operator for the conserved charge does not exist (this is, in some sense, much "worse" than simply its value being undefined on the SSB state), see this answer by David Bar Moshe Commented Apr 2, 2018 at 19:38
• @ACuriousMind U(1) symmetry is related to particle number conservation. See here hitoshi.berkeley.edu/misc/CERN.pdf Are you referring to Fabri-Picasso theorem?
– SRS
Commented Apr 2, 2018 at 19:40
• That's not what my comment asks about, and yes, David Bar Moshe's answer explicitly refers to the Fabri-Picasso theorem. Sure, the broken symmetry is related to particle number conservation. But why would its breaking "imply that the superfluid state is a state in which there is no definite number of particles"? I don't see that implication, and in fact the answer I linked there implies something different - that the number operator is ill-defined to begin with in the Hilbert space of the broken theory. Commented Apr 2, 2018 at 19:46
• I get your point but there can be caveats to it in a non-relativistic theory. @ACuriousMind
– SRS
Commented Apr 2, 2018 at 19:49

The ground state of a superfluid can be indeed (very precisely) approximated by a coherent state. More accurately by a squeezed coherent state. Please see Zhang equation (72):

$$|\{z_k \beta_k\}\rangle = \prod_k \exp\{ z_k a^{\dagger}_k - \bar{z}_k a_k\} \exp\{ \beta_k a^{\dagger}_ka^{\dagger}_{-k} - \bar{\beta}_k a_ka_{-k}\} |0\rangle$$

Where, $|0\rangle$ is the unbroken vacuum and $z_k$, and $\beta_k$ are parameters dependent on the details of the underlying many-body Hamiltonian.

This type of ground states is characteristic to collective ground states of strongly interacting systems. The squeezing is obtained due to the Bogoliubov transformation required to diagonalize the Hamiltonian in the large $N$ limit. (squeezing means "flattening" the circualr uncertainty region of an oscillator into an ellipse).

A quite transparent derivation of this type of ground states is given, for example, by: Solomon, Feng and Penna.

• Is it not in violation of Fabri-Picasso theorem that ACuriousMind was alluding to? If not why? @DavidBarMoshe
– SRS
Commented Apr 3, 2018 at 15:25
• This ground state is still a many-body ground state with $N$ (the condensate number of particles) very large but not infinite. I think that when the thermodynamic limit is taken, the number operator will cease to be well defined. Commented Apr 3, 2018 at 15:44
• But the coherent states don't have a definite number of particles. @DavidBarMoshe
– SRS
Commented Apr 5, 2018 at 16:00
• The above state does not have a definite number of paricles. $N$ means the expectation value of the number operator. However, the distribution is very narrow around $N$, having a width of something like$\frac{1}{\sqrt{N}}$, which is very small in the large $N$ . Commented Apr 8, 2018 at 6:18