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Hot answers tagged bose-einstein-condensate

3

In the Bogoliubov transformation, it is the case that $$u_{-p} = u^*_p,$$ and $$v_{-p} = v^*_p,$$ so your result is correct. In fact, if you move on to basically the next equation, they assume that both $u$ and $v$ are real, in which case $$u_{-p} = u^*_p = u_p,$$ as mentioned by Mark Mitchison in a comment below this answer. And incidentally, yes it is ...

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I will try to address the first point raised by the OP, i.e. the occurrence of spontaneous symmetry breaking in Bose-Einstein condensation. The free boson gas is described by the hamiltonian: $$H_V=\int_V\frac{d^sx}{2m}\big|\nabla\phi(x)\big|^2.$$ The ground state satisfies $H_V\Psi_0 = 0,\ \forall V$ and hence $\nabla \phi(x)\Psi_0=0,\ \forall x.$ ...

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Superfluid helium finds an application as a coolant in superconducting systems (http://link.springer.com/chapter/10.1007%2F3-540-45542-6_4#page-1 )

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I dug around in the literature a bit and found that this formulation (semiclassical Bose-Hubbard plus Langevin-type dissipation) has been studied before. Here is the relevant reference: http://arxiv.org/abs/1304.5071. What you are trying to do is derive their equation (9). You probably missed it because they refer to their model as the discrete nonlinear ...

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