In the theory of superfluidity in weakly interacting Bose gases, one finds that in the symmetric phase the exctitations have the dispersion relation

$\omega = \frac{k^2}{2m}-\mu$

with gap $\Delta=-\mu>0$.

In the symmetry-broken phase the low energy excitations have energy

$\omega = c|\mathbf{k}|$

As far as I understood, according to Landau's criterion for superfluidity ($v_c=min_k \frac{\omega_k}{k}$), in order for a critical velocity $>0$ to exist, the spectrum must either be gapped (as in the case of superconductivity) or linear.

So why is the symmetric phase not superfluid? Where am I going wrong?


1 Answer 1


The Landau criterion is not in itself a criterion for superfluidity, but a criterion for the breakdown of superfluidity. Indeed, if applied to insulators or ordinary fluids, it would tell you that all of these are also superfluid...

What the Landau criterion tells you is the velocity of the superfluid flow at which excitations are created from the superfluid component (note that any finite velocity can create excitation from the normal component).

The correct definition of the superfluidity is in terms of the response of the system to a (Galilean) boost, which defines the superfluid density as the difference of the longitudinal and transverse current-current correlation function.


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