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?
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.
