Weakly interacting Bosons in a 3D harmonic oscillator

AI Thread Summary
The discussion revolves around the calculation of the chemical potential for weakly interacting bosons in a 3D harmonic oscillator. The initial approach involved ignoring the kinetic energy term and deriving the density function, leading to a formula for the chemical potential. Concerns were raised about the professor's assertion that the chemical potential should scale as N^(2/5), prompting a review of the integration limits for normalization. It was noted that the density function could become negative when the potential exceeds the chemical potential, indicating a need to set the density to zero beyond a certain radius. The conversation highlights the importance of considering the kinetic energy term and the correct limits of integration in the Thomas-Fermi approximation.
rmiller70015
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Homework Statement
N identical, weakly interacting bosons are trapped in a 3-dimensional harmonic oscillator. Use the Gross-Pitaevskii equation and N = large to find the chemical potential as a function of the number of bosons.
Relevant Equations
##V(r) = \frac{1}{2}m\omega ^2 r^2##
GP: ##[-\frac{\bar{h}^2 \nabla ^2}{2m} - \mu + V(r) + U|\Psi (r)|^2]\Psi (r) = 0###
1. Since N is large, ignore the kinetic energy term.
##[-\mu + V(r) + U|\Psi (r)|^2]\Psi (r) = 0##

2. Solve for the density ##|\Psi (r)|^2##
##|\Psi (r)|^2 = \frac{\mu - V(r)}{U}##

3. Integrate density times volume to get number of bosons
##\int|\Psi (r)|^2 d\tau = \int \frac{\mu - V(r)}{U}d\tau###
## = \frac{4\pi}{U} \int_0^r \mu \rho ^2 - \frac{1}{2}m\omega ^2 \rho ^4 d\rho## where ##\rho## is a dummy variable for integration
## N = \frac{4\pi}{U}( \frac{1}{3}\mu r^3 - \frac{1}{10}V(r)r^3 )##4. Solve for ##\mu##
##\mu = \frac{3NU}{4\pi r^3} - \frac{3}{10}V(r)##

The problem is that my professor said that chemical potential should go like ##N^\frac{2}{5}## or something like that. So I am concerned that I didn't do something correctly. She also recalls things from memory incorrectly a lot of the time so I may actually be correct. I would just like a second opinion.
 
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rmiller70015 said:
The problem is that my professor said that chemical potential should go like ##N^{\frac{2}{5}}## or something like that.
It does.

Notice that your chemical potential is a function of r. You're missing a step. Think about the limits of integration in the normalization condition. Where should you stop integrating the density?
 
Twigg said:
It does.

Notice that your chemical potential is a function of r. You're missing a step. Think about the limits of integration in the normalization condition. Where should you stop integrating the density?
I found a paper that does this in 1-dimensions and I can kind of expand that to 3-dimensions, but they integrate between ##\pm \sqrt{\mu}##. Is this because at ##\sqrt{\mu}## you have a density that drops below the level where you can still be in the Thomas-Fermi regime and the kinetic energy term is no longer negligible?
 
rmiller70015 said:
Is this because at you have a density that drops below the level where you can still be in the Thomas-Fermi regime and the kinetic energy term is no longer negligible?
You're on the right track, but no.

The density you got was $$n(r) = \frac{\mu - V(r)}{U} = \frac{\mu}{U} - \left( \frac{\frac{1}{2}m\omega^2}{U} \right) r^2 $$ Try plotting this density vs r for ##\frac{\mu}{U} = 1## and ##\left( \frac{\frac{1}{2}m\omega^2}{U} \right) = 2## (I made up random numbers, but you'll see what I mean pretty quickly.) Notice anything funky?
 
I think the OP is gone, but here's the solution for anyone browsing this thread.

If you look at the density obtained from the Thomas-Fermi approximation, it eventually goes negative when ##V(r) > \mu##. The missing step was to set the density to 0 for all ##r > R## wgere ##R## is the radius of the atom cloud obtained by solving ##V(R) = \mu##.

In reality, these corners are smoothed out by the kinetic energy Hamiltonian as the density approaches 0, so there are no cusps. But for high average density, these corners are small in extent.
 
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