# Ideal Bose gas and Goldstone modes

In an ideal Bose gas there is a symmetry breaking phase transition, namely Bose-Einstein condensation. In a weakly interacting Bose gas or in helium-4 there is a longitudinal phonon because of the symmetry breaking, which leads to a linear dispersion relation for small energies and momenta.

I would expect to have something similar also in the ideal Bose gas case, instead there is no phonon-like dispersion relation. Where is the Goldstone mode corresponding to the symmetry breaking?

From another point of view, it would be strange to have a collective excitation if particles don't interact with each other. Does this have anything to do with the absence of a Goldstone mode? What other hypothesis (that is not satisfied in the ideal Bose gas) is needed in order to have a Goldstone mode derived from a symmetry breaking?

• What makes you think that there is a broken symmetry when a non-interacting gas condenses?
– Buzz
Jun 14, 2020 at 19:16
• Because for $T \to 0$ all the particles condense in the $k=0$ state. Hence, I'd say that the symmetry of the phase is broken, since the condensate acquires a well-defined phase. Is this wrong? Jun 14, 2020 at 19:42
• Related: see comments in physics.stackexchange.com/q/768837/226902 "Are there phonons in air?" Jun 19, 2023 at 18:55

The action in Fourier space for the field $$\psi_{k,n}$$ reads $$S[\bar{\psi},\psi] = - \beta \bar{\psi}_{0,0} \mu \psi_{0,0} + \sum_{k,n} \bar{\psi}_{k,n} \left(-i\omega_n + \xi_k \right) \psi_{k,n}$$ where I have separated the $$k=0,n=0$$ term from the rest of the sum. Here $$\omega_n$$ are Matsubara frequencies, $$\xi_k = \varepsilon_k - \mu$$, with $$\varepsilon_k = k^2/2m$$ the energy corresponding to momentum $$k$$ and $$\mu$$ the chemical potential. The equation for $$\psi_{0,0}$$ is simply $$-\beta \mu \psi_{0,0} = 0$$, and it has the trivial solution $$\psi_{0,0}=0$$ as long as $$\mu \neq 0$$. As you may know, above the critical temperature we have $$\mu<0$$, so the solution is trivial. However below the critical temperature we have $$\mu=0$$ for consistency (see the book) and the equation has no solutions. We conclude that below the critical temperature, the action is unbounded with respect to the variable $$\psi_{0,0}$$. In other words there is no minimum around which we can expand, no ground state and no Goldstone mode. How can we determine the value of $$\psi_{0,0}$$ then? Well it is determined by imposing that the total number of particle is fixed to the value $$N$$, so $$\psi_{0,0}$$ is not even a dynamical variable, but is just a parameter of the theory. In other words, since $$\mu$$ is constrained to $$0$$ for consistency, you need a new parameter to set the correct number of particles, and this is precisely $$\psi_{0,0}$$.
The picture above is clearly not satisfactory, because we would like to treat $$\psi_{0,0}$$ as a dynamical variable, find the minimum of $$S$$ with respect to it and so on. If we introduce a weak interaction of strength $$g/V$$, with $$V$$ the volume of the system as follows: $$T S[\bar{\psi}_0,\psi_0] = - \bar{\psi}_{0} \mu \psi_0 + \frac{g}{V} |\psi_0|^2,$$ then the pathology is solved for all $$g > 0$$, no matter how small it is. Notice the cheaper notation $$\psi_0 = \psi_{0,0}$$. If you compute the minimum of the action, you find two solutions: $$\psi_0 = 0$$ and $$|\psi_0| = \sqrt{\mu V/g}$$. The latter makes sense when $$\mu>0$$, which is now possible below the critical temperature due to the interaction (again see the book for the details) and it turns out to be the minimum.