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In Bose-Einstein condensate (BEC), people often say there is a well defined macroscopic phase. What exactly the macroscopic phase is? (a phase factor $\mathrm{e} ^{i\phi}$ in a many-body wavefuction?) Is the macroscopic phase the same as the coherent phase?

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There are different ways to define this phase. In mean-field (low temperature, weak interaction regime), the many-body wave function

$\psi(x_1, x_2,...)=\prod_i \Phi(x_i)$

where $\Phi(x)$ is sometimes called the macroscopic wavefunction (because all the bosons are in the same state described by $\Phi$). In the simplest case (homogeneous system), one can show that (by finding the solution of the time-independent Gross-Pitaevski equation (wiki))

$\Phi(x)=|\Phi_0| e^{i\phi}$,

where $|\Phi_0|^2$ gives the density of the condensate (and in this regime, a good approximation of the density of particles), whereas $\phi$ is a phase which is arbitrary (in the sense that all value are allowed, and that it is not given by the parameters of the problem). This is the macroscopic phase.

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  • $\begingroup$ Thanks! If all the bosons are in the same state described by $\Phi$, the the many-body wave function should be the product of identical single particle states: $\psi=(\Phi)^N$, wherer N is the total number of bosons. Actually the $\Phi$ in your answer is the single-site many-body wave function which is equivalent to the coherent state in second quantization, and it is a superposition of different atom number states. $\endgroup$ – Timothy Nov 21 '13 at 16:09
  • $\begingroup$ @Jeremy: I'm not sure what you mean by single site in this context. But yes, it is indeed a superposition of states with different atom number, which allow a well defined phase (phase a number operators are canonical conjugate). $\endgroup$ – Adam Nov 21 '13 at 16:23
  • $\begingroup$ :You are right. I am talking the BEC in context of cold atoms in an optical lattice. Each boson condenses to the zero momentum state, so the ground state can be described by $(b_{k=0}^{\dagger})^{N}|0\rangle$. In each lattice site, the many-body wave function is a coherent state, so the ground state can be described by a product of single-site states over the entire lattice $\prod_{i}|\Phi\rangle_{i}$ $\endgroup$ – Timothy Nov 21 '13 at 16:36
  • $\begingroup$ @Jeremy: Agreed. Just keep in mind that this is correct only in the limit $t/U\gg1$ (in standard notation of the Bose-Hubbard model). $\endgroup$ – Adam Nov 21 '13 at 16:40

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