I've read chapter about superfluids of Anthony Zee's QFT in nutshell and it is still not clear, why the energy spectrum of massless field only is taken into account to describe superfluid? In that chapter Zee imposes two fields instead of one complex field, h (stands for "higgs" boson) and $$\theta(x)$$ as goldstone boson. To show, that the energy spectrum of excitations in He is proportional to momentum, Zee integrates over h (the path integral) and shows, that only massless field is left. The question is why we can leave only goldstone boson to check the energy spectrum of excitations, without taking into account massive field h ?

thank you Anton


First of all, I remember AZee is discussing about global U(1) in the case of superfluid. By $U(1)$ we mean that that both Hamiltonian and Lagrangian are invariant under $\phi\rightarrow e^{i\theta}\phi$ transformation.

I think he mentioned that what he is doing in that section is called "integrate out" high energy fluctuations (fields.)

The goal here is---to study excitations around (classical) ground state (with AZee's notations $\phi=\sqrt{\rho}e^{i\theta}$), $$ \phi_0=\sqrt{\rho_0}e^{i\theta_0}, $$ where $\phi_0$ and $\theta_0$ correspond to the ground state.


  1. Note that this classical ground state is asymmetric because it dose not obey the same symmetry, $U(1)$ of the Lagrangian.

  2. If one write down a self-consistent bosonic theoy in a not hand-wavy way, he will find that tuning the chemical potential (which is correspond to $\rho$ in above) can obtain the transition between different phases. And for the same reason, for some range of chemical potential, one can have symmetry-breaking phase---superfluid phase in our case. Otherwise, one would get a trivial phase with symmetric ground state with no degeneracy.

Based my own experience of using AZee's book as textbook in QFT course, I prefer to derive everything in detail, independently, with my notations. Thus I'll slight change notations from here.

To be more quantitative, we will apply the language of Lagrangian for non-relativistic non-interacting bosons: $$ \mathcal{L}=i \Phi(x)^\dagger\dot{\Phi}(x)-\frac{1}{2m}\nabla\Phi(x)^\dagger\nabla\Phi(x)+\mu\Phi(x)^\dagger\Phi(x), $$ where $\mu=ng$ is the chemical potential and $n$ is the particle density of the ground state. We can further include interaction term which triggers the symmetry breaking $$ \mathcal{L}=i \Phi(x)^\dagger\dot{\Phi}(x)-\frac{1}{2m}\nabla\Phi(x)^\dagger\nabla\Phi(x)+\mu\Phi(x)^\dagger\Phi(x)-\frac{g}{2}\big(\Psi(x)^\dagger\Phi(x)\big)^2\nonumber\\ =i\Phi(x)^\dagger\dot{\Phi}(x)-\frac{1}{2m}\nabla\Phi(x)^\dagger\nabla\Phi(x)-\frac{g}{2}\big(n-\Phi(x)^\dagger\Phi(x)\big)^2+\text{constant}. $$ To see the Goldstone modes, we can use polar coordinates to decoupled the amplitude and phase fields, $$\Phi(x)\equiv\sqrt{\rho(x)} e^{i\theta(x)}$$. Then we have $$ \mathcal{L}=\frac{i}{2}\dot{\rho}-\rho\dot{\theta}-\frac{1}{2m}\big[\frac{1}{4\rho}(\nabla\rho)^2+\rho(\nabla\theta)^2\big]-\frac{g}{2}(n-\rho)^2, $$ where $\frac{i}{2}\dot{\rho}$ can be ignored since it's total derivative. In order to find the minimum of this Lagrangian, we can take derivative with respect to $\rho$, set it to zero and get $\rho_{min}=0$ as minimum This vacuum value has degeneracy under global $U(1)$ symmetry which will be broken if we pick out a particular vacuum---$\theta_0$. Fixing the phase of the ground state as $\theta_0$ is the same at every point in space and can take any value.

By Goldstone's theorem, breaking a continuous symmetry, global (or local with extra subtleties), will lead to emergence of Goldstone bosons. To find the Goldstone modes, we will first isolate out the excitations and integrate it out in path integral. If we observe the potential carefully, we will see the pattern that changing $\rho$ will cost more energy that changing $\theta$, therefore $\rho$ corresponds to excitations. Thus, we decompose the excited field as $$ \sqrt{\rho}=\sqrt{n}+v, $$ where $\sqrt{n}$ corresponds to ground state value ($\sqrt{\rho_0}$) and $v$ corresponds the excitations. The excitation $v$ can be interpreted as climbing the wall of Mexican hat slightly from the ground state. Thus, we can expand the Lagrangian in terms of $v$, $$ \mathcal{L}=-2\sqrt{n}\dot{\theta}v-2\mu v^2-\frac{1}{2m}(\nabla v)^2-\frac{n}{2m}(\nabla\theta)^2+\cdots, $$ and put it into path integral formalism to integrate out the fast excitation part.

In path integral, we have the functional integral $$ \int\mathcal{D}\phi e^{\frac{i}{2}\int\phi A\phi+i\int J\phi}=e^{-i\frac{1}{2}\int J(iA)^{-1}J}, $$ which help us to transform to propagator form $$ -\frac{1}{2}\phi(\partial^2+m^2)\phi+J\phi\rightarrow\frac{1}{2} J(x)\frac{1}{\partial^2+m^2}J(y). $$ Thus, we have $$ -2\mu v^2-\frac{1}{2m}(\nabla v)^2-2\sqrt{n}\dot{\theta}v\rightarrow -h\big[2\mu-\frac{1}{2m}(\nabla v)^2\big]h-2\sqrt{n}\dot{\theta}v \rightarrow \sqrt{n}\dot{\theta}\frac{1}{-\frac{1}{2m}\nabla^2+2\mu} \sqrt{n}\dot{\theta}, $$ where $J\equiv\sqrt{n}\dot{\theta}$ and $\phi\equiv-\sqrt{2}h$. Thus, our Lagrangian becomes $$ \mathcal{L}=\sqrt{n}\dot{\theta}\frac{1}{-\frac{1}{2m}\nabla^2+2\mu} \sqrt{n}\dot{\theta}-\frac{n}{2m}(\nabla\theta)^2+\cdots, $$ which corresponds to non-excited part---$2\mu$ term dominates in low-energy limit. We arrive the global-$U(1)$-invariant Lagrangian that describes the massless Goldstone modes $$ \mathcal{L}=\frac{1}{2g}\dot{\theta}^2-\frac{n}{2m}(\nabla\theta)^2, $$ whose excitations tend to approach zero as $\mathbf{p}\rightarrow 0$ (gapless as the standard dispersion of Nambu-Goldstone modes.) This leads to the same low energy dispersion relation as previous $$ E_{\mathbf{p}}=(\frac{\mu}{m})^{1/2}|\mathbf{p}|. $$ Note that we don't even need Bogoliubov transformation to get this gapless mode. This gapless mode is accordingly the Nambu-Goldstone boson associated with SSB of global $U(1)$ symmetry.

In short,

To study the low-energy excitations in the symmetry-breaking phase, one normally derived an effective theory at low energy region. This kind of effective theory can be derived by integrating out the fast (high energy) fluctuations.

The steps are roughly stated in above.

The question is why we can leave only goldstone boson to check the energy spectrum of excitations, without taking into account massive field h ?

After go through those calculations, I think your doubt is demonstrated in the part we take low-energy limit, in which the kinetic part of the propagator $\nabla^2$ is ignored.

  • $\begingroup$ Thank you for reply, So is it correct, that mass is taken into account in the effective lagrangian, but it behaves as massless field as we neglect nabla^2 ? $\endgroup$ – Anton Jun 7 '17 at 13:26

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