How to properly get low energy effective field theory of superfluid?

Question

I am following chapter 3 of X. G. Wen's book "Quantum Field Theory of Many-Body Systems". The following action for a weakly interacting Bose gas is derived: $$S[\varphi,\varphi^*] = \int dt \ d^D \textbf{r} \bigg[\frac{1}{2}i(\varphi^*\partial_t \varphi -\varphi \partial_t \varphi^*)-\frac{1}{2m}\nabla \varphi^* \nabla \varphi+\mu |\varphi|^2 -\frac{V_0}{2}|\varphi|^4\bigg]$$

The equations of motion can easily be found to be \begin{align} &\bigg(-i\partial_t - \frac{1}{2m}\nabla^2-\mu + V_0|\varphi|^2\bigg)\varphi=0\\ &\bigg(i\partial_t - \frac{1}{2m}\nabla^2-\mu + V_0|\varphi|^2\bigg)\varphi^*=0 \end{align}

Now if $\mu>0$ we are in the superfluid phase so the ground state is $|\varphi_0| = \sqrt{\frac{\mu}{V_0}} \equiv \sqrt{\rho_0}$. To get the low energy excitations I consider small fluctuations from this ground state by setting $\varphi = \varphi_0 + \delta \varphi$. After linearising the equations of motion I get that \begin{align} &\bigg(-i\partial_t - \frac{1}{2m}\nabla^2+\mu\bigg)\delta \varphi=-\mu \delta \varphi^*\\ &\bigg(i\partial_t - \frac{1}{2m}\nabla^2+\mu \bigg)\delta \varphi^*=-\mu \delta \varphi \end{align} which can be combined to give \begin{equation} \bigg|-i\partial_t - \frac{1}{2m}\nabla^2+\mu\bigg|^2 \delta \varphi = \mu^2 \delta \varphi \end{equation} The resulting dispersion relation is \begin{equation} \omega = \frac{1}{2m}k^2 \ \text{ and } \ \omega = \bigg|\frac{1}{2m}k^2-2\mu\bigg| \end{equation}

However the second of these looks very odd to me.

Furthermore, I was expecting to get a gapless, linearly propagating Goldstone mode $\omega \propto k$ (due to the spontaneously broken U(1) symmetry). Indeed section 3.3.3 in Wen's book explains how to get these modes by introducing the fields $\theta$ and $\delta \rho$ (the heavy field) as \begin{equation} \varphi = \sqrt{\rho_0+\delta \rho}e^{i\theta} \end{equation} and integrating $\delta \rho$ out to get a low energy effective field theory for $\theta$.

So what excitations has my method produced instead? Have I found some sort of higher-energy excitations, perhaps due to density fluctuations? To me it seems like setting $\rho = \rho_0+\delta \rho$ corresponds to taking $\theta=0$ and $\delta \phi \sim \delta \rho$, so that my method found the dispersion for the heavy $\delta \rho$ fluctuations.

Answer 1

$$ {\cal L} = \varphi^\dagger(i\partial_t +\frac 1{2m} \nabla^2 +\mu)\varphi -\frac \lambda 2 (\varphi^\dagger\varphi)^2. $$ Here $\mu$ is the chemical potential for the bosons.

The potential part of this expression is $$ V(\varphi)= \frac\lambda 2 (\varphi^\dagger\varphi)^2- \mu \varphi^\dagger\varphi. $$ This has its minimum at $\varphi^\dagger\varphi=\mu/\lambda$, so the possible stationary solutions have $$ \langle \varphi\rangle = \varphi_c= e^{i\theta} \sqrt{\frac \mu\lambda}. $$ Any choice of $\theta$ gives a stationary solution, and is a candidate classical ground state.

Let us look for small oscillations about one of these stationary points. Without loss of generality we may take $\theta=0$. We set $\varphi = \varphi_c+ \eta$ so $$ V(\varphi)=\frac \lambda 2\left( \varphi^\dagger\varphi -\varphi^\dagger_c\varphi_c\right)^2- \frac {\mu^2}{2\lambda}\nonumber\\ = \frac \lambda 2\left( (\varphi^\dagger_c+ \eta^\dagger)(\varphi_c+ \eta) -\varphi^\dagger_c\varphi_c\right)^2- \frac {\mu^2}{2\lambda}\nonumber\\ =const.+\mu \eta^\dagger\eta+\frac \mu 2 \eta\eta+ \frac \mu 2 \eta^\dagger\eta^\dagger + O(\eta^3). $$ Remembering that $\mu = \lambda \rho_o$, where $\rho_0$ is the density of particles in the condensate, we see that each of these terms corresponds to the interaction between the $\eta$ field and the condensate.

Keeping only the quadratic terms gives the linearized equations of motion $$ i\partial_t \eta = -\frac 1{2m}\nabla^2 \eta +\mu \eta + \mu \eta^\dagger\nonumber\\ -i\partial_t \eta^\dagger = -\frac 1{2m}\nabla^2 \eta^\dagger +\mu \eta^\dagger + \mu \eta. $$

If we look for plane-wave solutions of the form $$ \eta= a e^{ikx-i\omega t} + b^\dagger e^{-ikx+i\omega t}, $$ we find that $$ \left[ \matrix{ \frac 1{2m} k^2-\omega +\mu & \mu\cr \mu & \frac 1{2m} k^2+\omega +\mu\cr}\right] \left [\matrix{ a\cr b\cr}\right]=0. $$

Setting the determinant of this to zero yields $$ \omega^2 = ( \frac 1{2m} k^2 +\mu)^2 - \mu^2 $$ so $$ \omega \approx \sqrt{\frac \mu m} k + O(k^2). $$ Recalling that $\mu = \lambda |\varphi_c|^2 = \lambda\rho_0$ shows that this is exactly the expected sound-wave dispersion curve