Weakly interacting Bosons in a 3D harmonic oscillator

[rmiller70015]

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.

 

[Twigg]

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?

 

[rmiller70015]

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?

 

[Twigg]

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?

 

[Twigg]

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.