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Every textbook of statistical mechanics will derive the Bose-Einstein condensation(BEC) of free boson. I know how to derive in this way. It gives the heat capacity of free boson is $C\propto T^{3/2}$. But for $He^4$ that is interacting boson we know $C\propto T^3$. So to explain this, every condensed matter textbook will use 2nd quantization to rewrite the orginal many-body hamitonian in field formalism. For interacting boson, the Hamitonian of field is $$H=\int d^3 x \frac{1}{2m} |\nabla\phi|^2 + U |\phi(x)|^2 + g |\phi(x)|^4$$ For Helium 4, $U<0$ and $g>0$, so field configuration of minimum energy is $\phi = const.\neq0$, thus there is a spontaneous symmetry breaking of $U(1)$ , and due to Goldstone theorem it gives a massless excitation which explains the $C\propto T^3$.

It's all reasonable above. However how to use 2nd quantization back to explain BEC of noninteracting boson?

My questions:

  1. For free boson, the field Hamitonian is now $$H=\int d^3 x \frac{1}{2m} |\nabla\phi|^2 $$ The saddle point of field configuration is now $\phi(x)=const.$ for $\forall const. \in \mathbb{C}$, not necessary to be nonzero. How to explain the macroscopic occupation number of ground state?

  2. How to use Hamitonian of field to explain the BEC of noninteracting boson in potential well, like harmonic potential well? In this case, $$H=\int d^3 x \frac{1}{2m} |\nabla\phi|^2 + \frac{1}{2} x^2 |\phi(x)|^2$$ Certainly the minimum energy configuration is $\phi(x)=0$.

  3. Certainly the above two cases must be able to have BEC. Then for free boson and noninteracting boson in potential well, does BEC spontaneously breaking the $U(1)$ symmetry? If yes, there must exist massless excitation due to Goldstone theorem, so why the heat capacity of free boson is not propotional to $T^3$? If no, it will contradict with Landau's paradigm of phase transition that SSB results in the 2nd order phase transition. How to explain? Related to my another question.

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  • $\begingroup$ The condensation phenomena appears when considering the equilibrium physics associated to a given Hamiltonian. Writing the Hamiltonian in first or second quantization should not make any difference when constructing the equilibrium scenario. As for the Goldstone theorem thing, are you sure it is the reason for the $C \propto T^3$ ? It could also be related to the dispersion relation which differs for interacting bosons. $\endgroup$ – Tony May 19 '17 at 14:52
  • $\begingroup$ @Tony $C\propto T^3$ is always related to the dispersion relation of $\epsilon(p)\propto p$ in 3 dimension. That's a massless excitation. $\endgroup$ – user153663 May 19 '17 at 14:55
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    $\begingroup$ Ok, $ϵ(p) \propto p$ at low $p$ is only true for interacting bosons indeed. But $ϵ(p) \propto p^2$ seems massless to me. $\endgroup$ – Tony May 19 '17 at 14:56
  • $\begingroup$ @Tony $\epsilon (p) \propto p^2$ is a massive excitation. The coefficient of this radio is related to mass. $\endgroup$ – user153663 May 19 '17 at 14:59
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    $\begingroup$ No, it is massless, $ϵ(p) \propto p^2 + m$ is massive. $\endgroup$ – Tony May 19 '17 at 15:01
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As was stated in the comments, the fact that $\phi = 0$ minimizes the classical action is not sufficient to establish absence of SSB. In fact, even calculating the expectation value $\langle \phi(x) \rangle$ is not sufficient; in fact, this expectation value is always zero when calculated with respect to the grand-canonical ensemble (it's a fun exercise to prove this). Rather, the correct way to diagnose spontaneous symmetry-breaking is through the long-rangedness of the two-point correlation $$ \lim_{|x - y| \to \infty} \langle \phi^{\dagger}(x) \phi(y) \rangle$$ When there is SSB, this goes to a nonzero value, otherwise it goes to zero. In fact, by expressing $\phi(x)$ as the Fourier transform of the fermion annihilation operator $a_k$ in momentum space, one finds that this limit is actually equal to the macroscopic occupation of the $k=0$ state, which is non-zero even for free bosons.

As for the question of the Goldstone bosons, a non-interacting BEC does have gapless excitations; you just add one more boson in a state other than the $k=0$ state. But since they are non-relativistic particles, the dispersion relation is $E \sim k^2$. This correctly gives a $\sim T^{3/2}$ specific heat. To get a $\sim T^3$ specific heat you would need a linear dispersion relation $E \sim |k|$, which is what happens in the interacting case. In non-relativistic systems, Goldstone's theorem doesn't necessarily require that the Goldstone bosons have linear dispersion; another example of a case where you get a quadratic dispersion is the spin-wave excitations of ferromagnets.

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