Aether existance in alternate universe made of Bose-Einstein condensate

Question

I came across an interesting question which was shown to me by my professor, it is as follows:

Investigation of an alternative Universe:

This Universe contains three spatial and one time dimensions, and space can be described as a large cuboid. The lengths of the sides of the cuboid are denoted by $L_{x},L_{y}$ and $L_{z}$, its volume is $V$ . The universe is prescribed cyclic boundary conditions. The alternative Universe is filled with a Bose-Einstein condensate which consists of ultra-cold atoms. The condensate is characterized by a complex wavefunction $\psi(r, t)$. The dynamics of the wavefunction is described by the Gross-Pitaevskii equation

$$\boxed{\mathrm i\hbar\frac{\partial{\psi(r,t)}}{\partial{t}}=\left(-~\frac{\hbar^2}{2m}\Delta-\mu\right)\psi(r,t)+g\psi^{*}(r,t)\psi(r,t)\psi(r,t)}$$

Where $m$ is the mass of the ultracold atoms, $\mu$ is the chemical potential and g is a constant which characterizes the atomic collisions. The number of condensed atoms is denoted by $N_{c}$. The local density of the condensate is given by the absolute value of the wavefunction squared. Let $c =\sqrt{\frac{gN_{c}}{mV}}$ denote a parameter which has the dimension of velocity and let $\xi =\frac{\hbar}{mc}$ denote another parameter, which is small, and which has the dimension of length.

The alternative Universe is habited by intelligent creatures and the creatures live in two colonies, whose distance is large. The habitants of the two colonies communicate by creating small wavelike perturbations in the otherwise homogeneous condensate. Due to the limitations of the moderately advanced technology of the creatures, they can only create waves in the condensate, whose wavelength is much larger than the length $\xi$.

In one of the colonies, the leading physicists of the creatures are arguing about the fundamental nature of the physical laws of their world. While Alice says that the aether should exist, Bob argues that the aether hypothesis is not needed, because their world is fundamentally Lorentz invariant.

Who is right, Alice or Bob?

To decide who is right in the controversy, write up the equation of the small perturbations of the complex wavefunction, and then make a real wave equation from this complex equation, which will contain derivatives higher order than the previous complex one. In this equation, the parameters $c$ and $\xi,$ which were introduced above, will appear. Analyze this equation from the perspective of the debate.

How do I go about solving this question?

My attempt:

I have found the solution to the Gross-Pitaevskii equation and have hence found the local density of the condensate. Bose-Einstein statistics tells us that the occupation of the ground state is unbounded as the energy of the system decreases. For dilute systems this can be achieved by reducing the temperature of a system below some critical $ T$, given by $Tc$, such that the particles want to be in the lowest state. For a free gas of non-interacting particles in 3D, with particle mass m, this critical temperature is given by : $$\boxed{T_{c}=\frac{2(\pi)\hbar^{2}}{mk_{B}}\left(\frac{n(r,t)}{\zeta\left(\frac{3}{2}\right)}\right)^{\frac{2}{3}}}$$ Where $\zeta$ is the Riemann Zeta function and $\zeta\approx2.612$ Below this temperature we would expect the onset of Bose-Einstein condensation. A quantum particle is smeared out over a region of space as a wavepacket. When the temperature of a system is decreased, these wavepackets (whose typical size is given by the de Broglie wavelength) increase in length and begin to overlap, behaving as one giant matter wave. The thermal de Broglie wavelength can be derived by equating the quantum mechanical kinetic energy of free particles, $E_{K} = \pi k_{B}T$, with kinetic energy in terms of momentum, $E_\textrm{kinetic} = \frac{p^2}{2m}$ and using the standard de Broglie wavelength $\lambda_{dB} = \frac{2\pi}{p}$ to obtain $$\boxed{λ_{dB} =\sqrt{\frac{2(\pi)\hbar^2}{mk_{B}T}}}$$

where we note $λ_{dB} ∝\frac{1}{T}$ so it increases in length as $T$ decreases. Note that using the former relation in terms of $λ$ we can write $nλ^3 ∝ nT^{\frac{-3}{2}} ≈ 2.612$.

At temperatures T much smaller than the critical temperature Tc, the BEC is well described by the macroscopic wave function $\psi=\psi(r,t)$ whose evolution is governed by a self-consistent, mean field nonlinear Schrödinger equation (NLSE) known as the Gross-Pitaevskii equation. If a harmonic trap potential is considered, the single particle equation becomes:

$$\boxed{i\hbar\frac{\partial{\psi(r,t)}}{\partial{t}}=-\frac{\hbar^2}{2m}\Delta(\psi)+V(r)\psi+NU_{0}\mid{\psi^2}\mid\psi}\tag{1}$$

Where $r=(x,y,z)^{T}$ is the spatial-coordinate vector, $U_{0}=\frac{4(\pi)(\hbar^2)a}{m}$ with a, the s-wave scattering; and $$V(r)=\frac{m}{2}(\omega_{x}^{2}x^2+\omega_{y}y^2+\omega_{z}z^2)$$is the harmonic trap potential.

Multiplying $(1)$ by $(\frac{1}{m\omega_{x}^2 a_{0}^{\frac{1}{2}}})$ and normalising the function I got:

$$\boxed{i\frac{\partial{\psi(r,t)}}{\partial{t}}=-\frac{1}{2}\Delta(\psi(r,t))+V(r)\psi+\beta\mid{\psi^2}\mid\psi(r,t)}\tag{2}$$

$\beta=\frac{4(\pi)aN}{a_{0}}$. Here positive/negative β corresponds to the defocusing/focusing NLSE, respectively.

Hence, $(2)$ is the local density, I.e., $\frac{N_{c}}{V}=\frac{N_{c}}{L_{x}L_{y}L_{z}}$

Now, I am aware of the dark solition (oscillating dark solutions-arxiv), am I to use that along with introducing perturbations? How do I write down an equation for the small perturbations in terms of a complex wavefunction and then take a real value of it? Any help will be deeply valued and appreciated.

Answer 1

I believe this question is a reference to the superfluid vacuum theory.

First and foremost, I believe the Gross-Pitaevskii equation in the problem statement is incorrect. It seems to have mixed up the time dependent and independent case. There should not be a $-\mu$ term on the right hand side.

Before I move on to my answer, I want to also point out that in the equation that was presented in the problem, there is no external potential acting on the condensate, due to the absence of the $V(r)\psi$ term. Considering this is a model of the universe, I would say it is a reasonable assumption. Therefore your attempt with the quadratic potential does not seem to apply to Alice and Bob's universe.

To the zeroth order, the (mean field) wave function is $\psi_0 = \sqrt{n}e^{-i\mu t/\hbar}$, where $\mu$ is the chemical potential, and $n$ is the particle density,or $N_c/V$, in the notation of this problem. Now to the first order, assume that the wave function of the universe is $\psi(\mathbf r, t) = \psi_0 + \delta\psi(\mathbf r, t)$. If we plug this into the GP equation, and keep only terms that are linear in $\delta\psi$, we get $$ i\hbar\partial_t \delta\psi = -\frac{\hbar^2}{2m}\nabla^2 \delta\psi + g (2n\delta\psi + \psi_0^{2}\delta\psi^*) \>. $$ Here is when the famous Bogoliubov approximation comes in. We assume that $\delta\psi(\mathbf r, t) = e^{-i\mu t/\hbar} (u(\mathbf r) e^{-i\omega t} + v^{*}(\mathbf r) e^{i \omega t})$. This essentially means that we assume the perturbations $\delta\psi$ are waves with frequency $\omega$ superimposed on the $e^{-i\mu t/\hbar}$ background, and we use $u$ and $v$ to express their amplitude. Plugging these into the above equation, and matching the Fourier components for $\omega$ and $-\omega$ respectively, we get the following equations, \begin{align} \left(-\frac{\hbar^2}{2m}\nabla^2 + 2gn - \mu - \hbar\omega \right)u & = gn v\\ \left(-\frac{\hbar^2}{2m}\nabla^2 + 2gn - \mu + \hbar\omega \right)v & = gn u \end{align} If we further assume $u$ and $v$ are plane waves, that is, $u = u_0 e^{i\mathbf p\cdot \mathbf r/\hbar}$, then we can obtain the dispersion relation for these waves, $$ (\hbar\omega)^2 = \left(\frac{p^2}{2m} + 2gn -\mu \right)^2 - (gn)^2 \>. $$ Finally for a free interacting BEC, $\mu = gn$, so the dispersion relation further simplifies to $$ \hbar\omega = \sqrt{\frac{p^2}{2m}\left(\frac{p^2}{2m}+ 2gn\right)} \>. $$

Now since the people inhabiting the BEC universe can only send waves with wavelengths much longer than $\xi$, the energy of the waves will be much less than $mc^2 = gn$, which means that the above dispersion relation can be reduced to $\hbar\omega = pc$. This is a linear, and more importantly, Lorentz invariant dispersion relation, so the scientific community in the BEC universe will likely agree with Bob. However, if the BEC universe people invent particle accelerators or some other high energy apparatus that helps them reach into energies much higher than $mc^2$, the above dispersion relation will reduce to the free particle form (in the normal universe), $E = p^2/2m$, and the existence of the BEC aether will be revealed! Alice wins in the end!