# Bose condensation in a harmonic potential

1. Feb 9, 2015

### CAF123

1. The problem statement, all variables and given/known data

Consider a gas of N weakly interacting bosons trapped in a 3d harmonic potential $V = \frac{1}{2}mw^2 (x^2 + y^2 + z^2)$. The single particle quantum states have energies $\epsilon = \hbar w (n_x + n_y + n_z + 3/2)$.

Calculate the total number of quantum states with energies less than $\epsilon$ and from this deduce that the density of states $g(\epsilon)$ is $$g(\epsilon) \approx \frac{\epsilon^2}{2 (\hbar w)^3} \,\,\,\text{for large}\,\,\,\,\epsilon$$
Hint: $\epsilon$ is a function of $\mathbf n$ rather than just the magnitude $n$. A surface of constant energy is a plane in $n$ space and the number of states with energy less than $\epsilon$ is given by the volume of a tetrahedron.
2. Relevant equations

$V = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon$

3. The attempt at a solution

A plane intersecting the $n_x, n_y$ and $n_z$ axes form a tetrahedron with the sides the $n_x - n_y$ plane, $n_y - n_z$ plane and so on. The volume of such a tetrahedron is $V = n_x n_y n_z/6$. Therefore per the hint, $$\frac{n_x n_y n_z}{6} = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon = G(\epsilon_{max}) - G(0)$$ I am just a bit unsure of how to progress. Many thanks!

2. Feb 9, 2015

### TSny

nx, ny, and nz are variables. What are the values of these variables in your expression for V?

3. Feb 10, 2015

### CAF123

Hi TSny,
I could reexpress any one of $n_x, n_y$ or $n_z$ using the equation for the energy $\epsilon = \hbar w(n_x + n_y + n_z + 3/2)$ to reduce the number of variables by one. Is it what you meant? The plane in n-space can be written as $n_x + n_y + n_z = C$ for C a constant given by rearranging the above equation. Thanks.

4. Feb 10, 2015

### TSny

The volume of the tetrahedron in n-space should depend only on C. Can you express V in terms of C?

5. Feb 10, 2015

### CAF123

Yes sorry, I was confusing my own notation. The volume is $C^3/6$ and so we have the integral $$\frac{C^3}{6} = \int_0^{\epsilon_{max}} g(\epsilon) d \epsilon = G(\epsilon_{\max}) - G(0) = \frac{1}{6} \left(\frac{\epsilon_{max}}{\hbar w} - \frac{3}{2}\right)^3$$ Then $$\frac{d}{d\epsilon} \left(G(\epsilon) -G(0)\right) = \frac{d}{d\epsilon} \frac{1}{6} \left(\frac{\epsilon}{\hbar w} - \frac{3}{2}\right)^3 \approx \frac{\epsilon^2}{2 (\hbar w)^3}$$ Is this along the right lines? I feel like I have sort of fudged the result in getting $g(\epsilon)$, in particular how I can neglect further terms in the expansion of the square at the end and in going from $G(\epsilon_{max}) - G(0)$ to $d/d\epsilon (G(\epsilon) - G(0))$.

6. Feb 10, 2015

### TSny

That looks good.

$\epsilon$ is assumed "large". Presumably, that means relative to $\hbar \omega$. So, you can neglect the 3/2 relative to $\epsilon/(\hbar \omega)$.

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