Two different to do Bogobliubov transformation but sees to be contradictary

Question

When try to do bogoliubov transformation on a weak-interaction cold atoms with uniform velocity $vec{v}$, I used two different approaches, giving two different results.
The Hamiltonian is
$$hat{H}=sum_{r}Psi^{dagger}(r)frac{nabla^{2}}{2m}Psi(r)+frac{g}{2}Psi^{dagger}(r)Psi^{dagger}(r)Psi(r)Psi(r)tag{1}$$
The first approach is a modification of the method given in  Chapeter 8, Bose-Einstein Condensation in Dilute Gases, written by Pethick.
Think
$$Psi(r)=sqrt{n_{0}}e^{imvr}+hat{a}(r)tag{2}$$
$sqrt{n_{0}}e^{imvr}$ is due to clearly most atoms condense in state whose momenta is $mvec{v}$, $hat{a}(r)$ denotes the fluctuation.
Insert (2) into (1), keep up to second order of fluctuation, then do Fourier transform of $hat{a}_{r}$, and use the method mentioned in Some questions about Bogoliubov transformation to do Bogoliubov transformation.
It gives the following spectrum
$$E_{k}=kcdot v+sqrt{frac{k^{2}+(mv)^{2}}{2m}left(frac{k^{2}+(mv)^{2}}{2m}+2gn_{0}right)}tag{3}$$.
The second is in the paper Nondissipative drag of superflow in a two-component Bose gas published by Fil in Phys.Rev.A in 2005. This approach writes
$$Psi_{r}=e^{ipsi+ihat{psi}}sqrt{n_{0}+hat{n}}tag{4}$$
with $hat{psi}$ fluctuations of phase and $hat{n}$ fluctuation of density. Inserting (4) to (1), and keep up to terms quadratic to $hat{n}$ and $hat{psi}$
The spectrum is given as
$$epsilon_{k}=kcdot v+sqrt{frac{k^{2}}{2m}left(frac{k^{2}}{2m}+2gn_{0}right)}$$.
There does not exsit $k_{0}$, such that $E_{k+k_{0}}=epsilon_{k}$. So to me these two seems not to be the same.
I've noticed one strange thing: Terms linear in $hat{n}_{r}$ are neglected in the second approach. In my eyes, $hat{n}(r)=hat{a}^{+}hat{a}$, these terms are terms quadratic in fluctuation in first approach. And terms quadratic in $hat{n}$, which is of order $4$ of $hat{a}_{r}$, is kept in the second approach.
I wonder what's the problem. Is it just because the second approximation is of higher precision than the first? Or is there physically wrong in the first approximation?

mike stone · Answer

If the theory is Galilean invariant then I believe that the energy/momentum relation for phonons in a moving  fluid is $E(k)=E_0({bf k})+{bf v}cdot {bf k}$ exactly. (This is just the Doppler effect with $E=hbar omega$). Here $E_0({bf k})$ is the dispersion relation in the stationary fluid.  I am therefore sceptical of equation (3).
To see that adding ${bf v}cdot{bf k}$ is the required transformation first imagine creating an excitation (phonon) in a material at rest by inelastic scattering some particle (say a neutron) off it. In the neutron starts with energy $frac 12 mv^2$ and momentum $m{bf v}$ and ends with energy $ frac 12 mv'^2$ and momentum $m{bf v}'$ then it has created a phonon with energy
$$
E= frac 12 mv^2-frac 12 mv'^2
$$
and momentum
$$ 
{bf k}= m{bf v}-m{bf v}'
$$
By lots of such scattering experiments one can measure $E({bf k})$.
Now look at exactly the same scattering process from a reference frame in which the material is moving at speed ${bf V}$.
The neutron still has speed ${bf v}$ with respect to the material, but has speed ${bf v}+{bf V}$ with respect to the observer.
The energy transfer from the view point of the stationary observer is now
$$
E'= frac 12  m|{bf v}+{bf V}|^2- frac 12  m|{bf v}'+{bf V}|^2
=E+ m({bf v}-{bf v}')cdot {bf V}
$$
and the momentum transfer is
$$
{bf k}'= m({bf v}+{bf V})- m({bf v}'+{bf V})
= m({bf v}-{bf v}')={bf k}
$$
We have created the same phonon, with the same momentum, but the stationary observer deduces that the  energy momentum relation of the phonon in the moving material is
$$
E_{rm moving} ({bf k})= E_{rm rest}({bf k})+{bf V}cdot {bf k}.
$$
If the material is a moving superfluid and the  phonons in the superfluid are in thermal equilibrium with the stationary walls, then it costs more energy for a phonon to have momentum in the direction of the flow. There will therefore be fewer phonons going with the flow and more going against it. The momentum density of the moving fluid is therefore less than $rho{bf V}$ by the momentum of the phonon gas. The phonon gas momentum density  is proprtional to ${bf V}$ and can be  parametrized  as by $-rho_n {bf V} $ for some temperature dependent number $rho_n$. The net momentum density, which for a single component fluid is the same as the mass flow, is therefore
$$
{bf j}= rho {bf V} -rho_n {bf V}
= rho_s {bf V}.
$$
In the two fluid model of Tisza, London and Landau, the quantity  $rho_s= rho-rho_n$ is called the superfluid density and $rho_n$ is the normal fluid density.
If the phonons are in equilibroum with a frame moving at speed ${bf V}_n$ the phonon momenum is $-rho_n({bf V}-{bf V}_n)$ and the flow equation becomes
$$
{bf j}= rho {bf V} -rho_n( {bf V}- {bf V}_n)
= rho_s {bf V}+ rho_n {bf V}_n.
$$
NOTE ADDED: I asked one of my colleagues, who is an expert, about this and he said that perhaps the first method    keeps the chemical potential fixed rather than the particle density.  If you accelerate a fluid while keeping the chemical potential fixed (by letting it flow through a venturi say) then, by Bernoulli, the pressure drops and so does the density. As a consequence  the moving fluid is not just a moving copy of the fluid at rest, and  creating it is not just a galilean transformation.

Two different to do Bogobliubov transformation but sees to be contradictary

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