I'm wondering how one precisely defines a superfluid in terms of the effective field theory description. In Nicolis's paper http://arxiv.org/abs/1108.2513 there seems to be an extremely simple characterization of the superfluid: it is just a scalar field $\psi$ with a nonlinearly realized $U(1)$ shift symmetry, $\psi\mapsto \psi + a$.

Does this mean I can call any theory with such a shift symmetry a relativistic superfluid? E.g., the simplest action you could write for $\psi$ is just

$$S=\int d^4x \sqrt{-g}\partial_a\psi \partial^a\psi$$

which is just a massless free scalar field. Is it correct to regard this as a superfluid? Or is there some other feature that is needed as well? For definitiveness, I'm just concerned with zero temperature superfluids, so I'm not worried about including the second fluid that is needed for vortices and the like.

  • 2
    $\begingroup$ This question does not make much sense. What is a "nonlinearly realised U(1) shift symmetry" ? I note your example is not even non-linear... A superfluid is a fluid realising a spontaneously breaking of the U(1) rotation symmetry, that's in principle all. From this you infer the Goldstone mode (second sound phenomena), the Josephson-like physics (coherent tunnelling), vortex quantisation (quantisation of defect) and phase-slip due to vortex accumulation, ... Actually, you do not need a "second-fluid" for vortex, so the last part of your question does not make much sense neither. $\endgroup$
    – FraSchelle
    Oct 23, 2014 at 20:28
  • $\begingroup$ @FraSchelle Nonlinearly realizing a $U(1)$ symmetry is exactly what I wrote: $\psi\mapsto \psi+a$, as opposed to a linear realization $\phi\mapsto e^{ia} \phi$. This is standard stuff in spontaneous symmetry breaking. A shift is a nonlinear realization of a symmetry in the sense that $\psi$ is not living in a representation of $U(1)$, in which the group can only act by matrix multiplication. And if all I have is the the phase field $\psi$, there is no way to get a vortex: a scalar cannot give rise to a flow that is twisting. Perhaps look at the reference I posted for clarification. $\endgroup$
    – asperanz
    Oct 24, 2014 at 20:13
  • 1
    $\begingroup$ Sorry, but it seems to me that neither you nor the paper you're referring to define what they mean by $\psi$. If it's the phase of a complex wave-function, like $\Psi=\Psi_{0}e^{i\psi}$ then I would agree with you, otherwise your question makes absolute no sense, and my first comment still applies: your action is trivial, and can not give any phase transition, nor symmetry breaking, until $g$ is itself $\psi$-dependent of course, and then reproduce the Ginzburg-Landau functional. Of course your action can still be used as the low-energy effective theory for a boson gas, though. $\endgroup$
    – FraSchelle
    Oct 28, 2014 at 18:26
  • $\begingroup$ Since a superfluid is a free gas of bosons in the low-energy sector, you can apply your contentless action. It nevertheless describes nothing else than the second-sound as far as I see, except it does this in an explicit covariant way, but this also is highly artificial in your notations, since multiplying an Laplacian by the space-time volume is trivial, isn't it ? Perhaps the paper cited by Nicolis would help you: arxiv.org/abs/hep-ph/0204199 to clarify your question. Thanks in advance for doing so. $\endgroup$
    – FraSchelle
    Oct 28, 2014 at 18:30

1 Answer 1


Physicists aren't really big on rigorously defining broadly used terms for complicated physical phenomena, so I'm sure that for any putative definition of a superfluid there's some physicist who would disagree with it. That having been said, I'd say that the field-theoretic definition of a superfluid is just any scalar field theory with a spontaneously broken global $\mathrm{U}(1)$ symmetry, e.g. an action of the form $S = \int d^d x\ \left[ \partial_\mu \phi^\dagger \partial^\mu \phi - \mu\, \phi^\dagger \phi - \lambda (\phi^\dagger \phi)^2 \right]$, where $\mu$ is negative.

  • $\begingroup$ Good answer. Apart from Nicolis and Son papers, I also found this one very useful and rigorous: arxiv.org/abs/1906.03140. $\endgroup$
    – Quillo
    Mar 24 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.