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When working with Bose-Einstein condensates trapped in potentials, how can one tell what the density of state of a system of identical bosons given the Hamiltonian, $H$? (I have been told that it is possible.)

Suppose the Hamiltonian is some 2D harmonic oscillator -- so $$H=p^2/2m+(1/2)(a^2x^2+b^2y^2) \quad ?$$

I think there is some general formula, something like $$\rho(E)=[gV_dS_dp^{d-1}/(2\pi\hbar)^{d}] (dp/dE) \quad ,$$ where $d$ is the dimension of the space we are working in, $g=2s+1$ where $s$ is the spin of the particles and $V_d$ is the $d-$dimensional "volume", so for a fixed volume box, this is the volume of the box.

But what is $V_d$ in this case? And is $p$ simply $$p^2=2m[E-(1/2)(a^2x^2+b^2y^2)] \quad ?$$

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1 Answer 1

This is definitely possible. A good resource is this statistical physics script. The density of states for an electron gas is calculated explicitly there, check out eqn. (5.21) and following. Then, some BEC calculations are made explicitly, starting on page 119. V is the (physical) volume of the system you look at.

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Thanks, Sebastian! :) About to check it out! –  Henrietta James Mar 26 '13 at 17:31
    
not sure if this is what you are looking for, depending on your previous exposure to statistical physics (or depending on whether I actually got your question ?! ;-)) –  lomppi Mar 26 '13 at 17:33
    
it seems the link is broken –  Michael Brown May 28 '13 at 7:15

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