Deep in the superfluid phase, the superfluid order parameter $\phi$ can be decomposed into the amplitude (density) mode $\rho$ and the phase mode $\theta$ as $$\phi=\sqrt{\rho} e^{\mathrm{i}\theta}.$$ It is believed that the density fluctuations are gapped as they correspond to "climbing up" the potential $V(\phi)$. What remains gapless at low energy are the phase fluctuations, or the Goldstone modes, described by $$\mathcal{L}[\theta]=\frac{1}{2g}\big((\partial_{t}\theta)^2-(\partial_{\boldsymbol{x}}\theta)^2\big).$$ However, there is an emergent density operator $\rho=-\partial_t\theta$ and current operator $\boldsymbol{j}=\partial_{\boldsymbol{x}}\theta$ in terms of the phase field $\theta$, such that the continuity equation $\partial_{t}\rho+\partial_{\boldsymbol{x}}\boldsymbol{j}=0$ is satisfied on-shell. This means that the density fluctuation is actually gapless, as seen from the density-density correlation function in the momentum-frequency space: $$\langle\rho_{k}\rho_{-k}\rangle=-\omega^2\langle\theta_{k}\theta_{-k}\rangle=\frac{\omega^2}{\omega^2-\boldsymbol{k}^2}.$$ If the density fluctuation in the superfluid is indeed gapless, how can we ignore them and claim $\mathcal{L}[\theta]$ alone as the effective description of the superfluid dynamics at low energy? But on the other hand, the classical picture of "claiming up" the Maxican-hat potential $V(\phi)$ does imply that the density fluctuation should be gapped. How to reconcile this contradiction?

  • $\begingroup$ If we start from the action of amplitude mode and phase mode of the order parameter, i.e. $S[\rho,\phi]$, then calculate the amplitude correlation function, shouldn't we get a different result? In other words, I think the emergent density operator you defined here should be different from the amplitude mode of order parameter. $\endgroup$ – Chuan Chen Oct 23 '17 at 5:49
  • $\begingroup$ @ChuanChen Yes, I believe if we work with $S[\rho,\theta]$, we could get a gapped correlation function for the amplitude mode. If the emergent density operator is really different from the amplitude mode, how should we understand that there are actually two different kinds of density fluctuations in the superfluid? $\endgroup$ – Everett You Oct 23 '17 at 14:58

The density mode in a superfluid is indeed gapless (in fact, the density mode in a normal phase is gapless, too. This mode is called sound).

The confusion arises because in order to arrive at the effective lagrangian for $\theta$, you have to integrate out the amplitude mode. As a result, the $\theta$ parameter in the effective lagrangian couples to the density.

Postscript: The amplitude is not directly related to the superfluid density $\rho_s$. The superfluid density is defined by $$ \vec\pi = \rho_s v_s + \rho_n v_n $$ where $\vec\pi$ is the momentum density and $v_s= i\hbar\nabla\theta/m$ is the superfluid velocity. This means that $\rho_s$ governs the response in momentum to gradients of the phase. Experimentally, $\rho_s$ is extracted by measuring the velocities of first and second sound.

  • $\begingroup$ The superfluid density = (the amplitude of the condensate wave function)$^2$. Is that right? $\endgroup$ – Everett You Oct 22 '17 at 2:46
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    $\begingroup$ Superfluid density is a hydrodynamic quantity, that cannot be directly extracted from the order parameter. The amplitude of the order parameter is more directly related to the condensate fraction. $\endgroup$ – Thomas Oct 22 '17 at 3:55
  • $\begingroup$ @Thomas Can you give more details on how to get the superfluid density? $\endgroup$ – Chuan Chen Oct 24 '17 at 3:46
  • $\begingroup$ @ChuanChen added a short postscript $\endgroup$ – Thomas Oct 24 '17 at 22:30
  • $\begingroup$ @EverettYou I strongly disagree with this answer : $\rho$ is definitely not the square of the condensate wave-function $\phi=\langle\psi\rangle$, it is indeed the density. In 2D, at finite temperature, $\phi=0$ due to fluctuations even below the superfluid transition, but $\rho=\langle\psi^*\psi\rangle$ is definitely finite. Furthermore, at zero-temperature, one shows that $\rho=\rho_S>|\phi|^2$. $\endgroup$ – Adam Oct 25 '17 at 18:23

Superfluids are Galilean invariant therefore it is a good idea to start from a Galilean invariant model when trying to understand their dynamics. For example the Gross-Pitaevski (GP) model that comes from the action integral $$ S[\phi, \phi^\dagger]= \int d^3x dt\left\{ \phi^\dagger (i \partial_t + \frac 1{2m} \nabla^2) \phi + \mu\phi^\dagger \phi -\frac 12 \lambda (\phi^\dagger\phi)^2\right\}. $$ is Galilean invariant.

This action contains a Mexican hat potential $$ V(\phi)= \frac 12 \lambda (\phi^\dagger\phi)^2-\mu \phi^\dagger\phi $$ which is minimized at $\phi^\dagger\phi = \mu/\lambda$. The possible stationary solutions are therefore $$ \langle\phi \rangle= \phi_c= e^{i\theta} \sqrt{\frac \mu \lambda} $$ In the GP model the $\rho$ in $\phi= \sqrt{\rho} e^{i\theta}$ really is the particle density because this model applies at very low temperature when essentially every particle is in the condensate. So the equilibrium particle density is $\rho= \mu/\lambda$.

If we look for small oscillations $\phi=\phi_c+\eta$ then $$ V(\phi+\eta) \sim const+ \mu \eta^\dagger \eta +\frac 12\mu (\eta^2+(\eta^\dagger)^2)+O(\eta^3) $$ and the linearized equations of motion become $$ i\partial_t \eta = -\frac1{2m} \nabla^2 \eta +\mu\eta +\mu \eta^\dagger,\\ -i\partial_t \eta = -\frac1{2m} \nabla^2 \eta^\dagger +\mu\eta +\mu \eta^\dagger. $$ If we seek a solution $$ \eta= a e^{ikx-i\omega t} +b^\dagger e^{-ikx+i\omega t} $$ we find that $(a,b)^T$ must obey $$ \left[\begin{matrix} k^2/2m =\omega +\mu & \mu\cr \mu & k^2/2m +\omega+\mu\end{matrix}\right] \left[\begin{matrix} a \cr b\end{matrix}\right]=0, $$
so the allowed frequencies are given by $$ \omega^2 =(k^2/2m +\mu)^2 -\mu^2. $$ At small $k$ this becomes becomes $\omega^2=c^2k^2$ with $ c^2 =\lambda \rho_0 /m$. These modes are the gapless sound waves. During the motion the tip of the $\phi$ vector describes an ellipse about the equilibrium $\phi_c$. These sound modes are therefore a combination of Goldstone-like motion along the bottom of the Mexican hat potential well and an out of phase radial ``Higgs- like'' radial oscillation. There are no separate circumferential "Goldstone" and radial "Higgs" modes in the non-relativistic bose-condensed superfluid. The coupled modes are sound waves with a density fluctuation $\rho_0\to \rho_0+\delta \rho$ and a simultaneous (in-phase) back-and-forth velocity given by $v=\nabla\theta$.

We had two equations for $\eta$ that were first order in time. We can, if we like, eliminate $\rho$ to get a second order wave equation involving only $\theta$, or eliminate $\theta$ to get a wave equation involving only $\rho$ --- but we won't get a second-order-in-time equation involving both variables. Beware, however, the linearized wave equations are not Galilean invariant. Futher, focusing only on the equations of motion risks discarding the $i\rho_0 \partial_t\theta$ in the action integral on the grounds that it is total derivative. This topological winding-number term is essential for vortex dynamics where it is responsible for the Magnus effect. In its absence (as asked recently on this site) a propeller would not work in a superfluid.

  • $\begingroup$ Thanks for the detailed answer. There is indeed one issue that I should clarify about my answer. However, I would like to point out that GP is a very specific model. It assumes that the superfluid density is equal to the density, and the condensate fraction is one. The effective lagrangian $L[\theta]$, on the other hand, is quite general. $\endgroup$ – Thomas Oct 25 '17 at 20:45
  • $\begingroup$ @Thomas Agreed, in general--- but Everett You's $L(\theta)$ is not Galilean invariant. There is a nice general discussion of the consquences of tgalilean invariance and effective fluid actions in: M. Greiter, F. Wilczek, and E. Witten, Hydrodynamic Relations in Superconductivity, Mod.Phys.Lett. B3 (1989) 903. $\endgroup$ – mike stone Oct 25 '17 at 21:03

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