6
$\begingroup$

The normal to the superfluid transition of liquid helium breaks is a U(1) global symmetry. Since it is a continuous, global symmetry (unlike superconductivity which is a gauge theory), I expect that there be Goldstone bosons in the theory like phonons or magnons which result from the spontaneous breakdown of translational invariance in crystals and rotational symmetry in ferromagnets, respectively. However, while reading online, and textbooks on statistical mechanics, I sparsely encounter Goldstone bosons in the context of superfluid transition.

Can someone suggest a reference (a condensed matter physics reference, in particular) which mentions about Goldstone modes in superfluids? If my guess is incorrect (i.e., there are no such modes) do correct me. If the presence of such modes is debated and controversial in the condensed matter community, and therefore not a standard textbook material, also let me know.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Actually, A.Zee (QFT in a nutshell, chapter V.1 superfluids) derives from a non-relativistic scalar boson field with short-range repulsion the existence of gap-less modes which he identifies with a Nambu-goldstone boson. $\endgroup$
    – Frederic Thomas
    Commented Dec 6, 2017 at 21:45

1 Answer 1

Reset to default
6
$\begingroup$

The Goldstone boson for a superfluid is the phonon.

See Wikipedia:

A version of Goldstone's theorem also applies to nonrelativistic theories (and also relativistic theories with spontaneously broken spacetime symmetries, such as Lorentz symmetry or conformal symmetry, rotational, or translational invariance).

It essentially states that, for each spontaneously broken symmetry, there corresponds some quasiparticle with no energy gap—the nonrelativistic version of the mass gap. [...] However, two different spontaneously broken generators may now give rise to the same Nambu–Goldstone boson. For example, in a superfluid, both the U(1) particle number symmetry and Galilean symmetry are spontaneously broken. However, the phonon is the Goldstone boson for both.

And also this article:

An example of spontaneous symmetry breaking is the breaking of the global U(1)-symmetry in $^4$He and the appearance of superfluidity below a certain critical temperature. Associated with the breaking of the symmetry, there is a nonzero value of an order parameter and a condensate of particles in the zero-momentum state. When a global continuous symmetry is broken, the Goldstone theorem states that there is a gapless excitation for each generator that does not leave the ground state invariant. In the case of nonrelativistic Bose gases, one identifies the Goldstone mode with the phonons.

Other relevant sources:

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is useful. I thought phonons exist only in solids. Can you suggest a textbook which discusses how such modes appear? @valerio92 $\endgroup$
    – SRS
    Commented Jul 18, 2017 at 14:39
  • 1
    $\begingroup$ @SRS Not that I can think of right now, sorry. I think that Huang (Statistical Mechanics) briefly mentions that, but it is not a very thorough discussion...If I can think of something I will update the answer. $\endgroup$
    – valerio
    Commented Jul 18, 2017 at 14:44
  • 1
    $\begingroup$ Any book discussing interacting Bose systems should discuss this. Off the top of my head, one example is Xiao-Gang Wen's quantum many-body book. $\endgroup$
    – Rococo
    Commented Jul 18, 2017 at 16:20
  • 1
    $\begingroup$ @SRS Actually I took a look at Huang and the discussion is better than I remembered. I also included a useful document that I found on the arxiv. $\endgroup$
    – valerio
    Commented Jul 19, 2017 at 7:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.