Superfluid Hydrodynamics

The notion of superfluid and Bose-Einstein condensation are intimately related. Superfluids can flow through narrow tubes without dissipating heat due to their zero viscosity. A well-known example of superfluid is low-temperature helium-4 (and helium-3, which is a fermionic superfluid). What characterizes a superfluid is that it can only dissipate heat via creation of elementary excitations, and if the flow velocity is slower than the characteristic speed of sound in the superfluid, those elementary excitations cannot be spontaneously created, thus ensuring a frictionless flow (Landau's criterion). Indeed, dilute BEC gases also have a linear dispersion with non-zero speed of sound at low momentum (Bogoliubov dispersion) and feature superfluidity. But it is important to note some differences between a superfluid and a BEC. An ideal BEC will have a quadratic dispersion (zero speed of sound), and there is no energy gap to protect the ground state from excitations, so an ideal BEC cannot be a superfluid. Also, even for the interacting BEC, the condensate density of BEC is not equivalent to superfluid density, for at zero temperature, by definition the superfluid fraction is 100% and normal fraction is 0%, but the condensate fraction is less than 100%, due to quantum depletion.

Now let's try to obtain an equation that describe the superfluid flow of BEC, which will require two variables: density and velocity. We again emphasize that the density in this case is not equivalent to superfluid density because of quantum depletion. In this article, we try a bit formal approach to arrive at the continuity equation.

The original Hamiltonian is invariant under the U(1) transformation $\psi \to \psi e^{i\theta }$. Indeed, the equation of motion obeys this symmetry, even though the superfluid itself has broken the symmetry by selecting a particular phase. This is characteristic of spontaneously broken symmetry. Noether's theorem (from classical mechanics / field theory) tells us that every continuous symmetry is associated with a conserved current $j^{\mu },(\partial _{\mu }j^{\mu }=0)$, and we will see that that the conserved charge is actually the average number of atoms, $N$. To make use of Noether's theorem, we need an action term that produces the GPE. We can obtain the GPE via the action principle if we suppose the action is given by the functional

$S[\psi ,\psi ^{*}]=\int d^{3}rdt\,\psi ^{*}{\Big (}-i\hbar {\frac {\partial }{\partial t}}{\Big )}\psi +{\frac {\hbar ^{2}}{2m}}\nabla \psi ^{*}\cdot \nabla \psi +V\psi ^{2}+{\frac {U_{0}}{2}}|\psi |^{4}=\int d^{3}rdt\,{\mathcal {L}}[\psi ,\psi ^{*}]$

If the field $\psi$ transforms infinitesimally under symmetry with parameter $\theta$ as $\delta \psi =\theta \Delta \psi$, then the Noether current $j^{\mu },\mu =t,x,y,z$ is given by ${\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\psi )}}\Delta \psi +({\text{similar for other independent field = }}\psi ^{*})$. The result is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^0 = n = |\psi|^2\\ j^{x,y,z} = \vec{j} = - \frac{i\hbar}{2m} (\psi^{*} \nabla \psi - \psi \nabla \psi^{*} )\\ \partial_{\mu} j^{\mu} =0 \Rightarrow \partial_t n + \nabla \cdot \vec{j} = 0 }

By using the fact that the equation of motion for the superfluid obeys U(1) symmetry, we have derived the continuity equation, which implies that the average number of atoms $N=\int d^{3}r\,n$ is conserved. Now consider writing the complex order parameter in terms of its amplitude and phase: $\psi ={\sqrt {n}}\exp(iS)$. You will find that the current density ${\vec {j}}$ can be expressed as proportional to gradient of the phase term:

Failed to parse (syntax error): {\displaystyle \vec{j} = n \frac{\hbar}{m} \nabla S = n \vec{v}_s\\ \vec{v}_s = \frac{\hbar}{m} \nabla S }

Since the curl of a gradient term is zero, the expression for the velocity field shows that the condensate flow is irrotational. If you plug in the phasor expression for the order parameter into the GPE, you obtain a complex equation whose real and imaginary part need to set to zero separately. The imaginary part gives back the continuity equation, and the real part gives

$\hbar {\frac {\partial S}{\partial t}}+{\frac {1}{2}}mv_{s}^{2}+V_{trap}+U_{0}n-{\frac {\hbar ^{2}}{2m{\sqrt {n}}}}\nabla ^{2}{\sqrt {n}}=0$

The last term is known as the quantum pressure term and emphasizes the importance of quantum effects in inhomogeneous condensates. Take the gradient of the above equation to obtain the equation of motion for the velocity field. Now we have a set of coupled equations for two hydrodynamic variables (density and velocity field):

${\begin{array}{rcl}m{\frac {\partial {\vec {v}}_{s}}{\partial t}}&=&-\nabla {\Big (}\mu +{\frac {1}{2}}mv_{s}^{2}{\Big )}\\\mu &=&-{\frac {\hbar ^{2}}{2m}}{\frac {\nabla ^{2}{\sqrt {n}}}{\sqrt {n}}}+V_{trap}+U_{0}n\\{\frac {\partial n}{\partial t}}&=&-\nabla \cdot (n{\vec {v}}_{s})\end{array}}$

Note these resemble the Euler equations from fluid dynamics. If the quantum pressure term were zero, the $\hbar$ dependence disappears (except implicitly in $U_{0}$) and the equations describe the potential flow of a non-viscous gas with pressure $U_{0}n^{2}/2$.

In the Thomas-Fermi approximation (ignoring the quantum pressure term), we can linearize the coupled hydrodynamic equations by expanding the variables around their stationary values

${\begin{array}{rcl}n(r,t)&=&n_{0}(r)+\delta n(r,t)\\{\vec {v}}_{s}(r,t)&=&{\vec {v}}_{s,0}(r)+{\vec {\delta v}}_{s}(r,t)\end{array}}$

If we assume ${\vec {v}}_{s,0}(r)=0$, then we obtain the following equations:

${\begin{array}{rcl}m{\frac {\partial {\vec {\delta v}}_{s}}{\partial t}}&=&-\nabla (U_{0}\delta n)\\{\frac {\partial \delta n}{\partial t}}&=&-\nabla (n_{0}{\vec {\delta v}}_{s})\end{array}}$

We see that these produce a wave equation for the density fluctuation:

$m{\frac {\partial ^{2}\delta n}{\partial t^{2}}}=U_{0}\nabla (n_{0}\nabla (\delta n))$

For homogeneous $n_{0}$, the speed of sound $c$ is ${\sqrt {U_{0}n_{0}/m}}$. A droplet of condensate can have shape resonances, waves, and many other physical behaviors, captured by the solutions to the hydrodynamic equations. For an early discussion of collective excitations in trapped BEC, see PRL 77, 2360 (1996). For the description of the time-of-flight expansion of BEC out of a time-dependent harmonic trap, see PRL 77, 5315 (1996).

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