Consider the following passage, via this image:

5.3.1 Density of states

Almost all of the spin-polarized fermionic atoms that have been cooled to ultralow temperatures have been trapped by magnetic fields or focused laser beams. The confining potentials are generally 3D harmonic traps. So let's consider this case in more detail. You might be interested to note that Fermi's original paper on fermionic particles considered this case, not the 3D box case above. As we saw previously, ignoring the zero-point energy in each dimension the eigenvalues (accessible energy states) are given by $\epsilon(n_x, n_y, n_z)=n_x\hbar\omega_x + n_y\hbar\omega_y + n_z\hbar\omega_z$. In order to evaluate the various integrals, we first need to obtain the density of states per unit energy. A rough way to do this is to simply set $k_i=n_i$, so that $$\epsilon^2 = k_x^2(\hbar\omega_x)^2 + k_y^2(\hbar\omega_y)^2 + k_z(\hbar\omega_z)^2 \equiv k^2(\hbar \overline\omega)^2,$$ where $\overline \omega = (\omega_x\omega_y\omega_z)^{1/3}$ is the mean frequency, and $dk_i/\epsilon_i=1/\hbar\overline \omega$. Because $k_i=n_i$ now rather than $k_i=\pi n_i/L$, th 3D density of states is given by $$g(\epsilon) = \frac{k^2}{2} \frac{dk}{d\epsilon} = \frac{\epsilon^2}{2(\hbar\overline\omega)^3}.$$

for the first displayed equation,

shouldn't be $\epsilon^2 =\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 + 2\epsilon_{n_x}\epsilon_{n_y} + 2 \epsilon_{n_x}\epsilon_{n_z} + 2\epsilon_{n_y}\epsilon_{n_z}$..?

if I assume $\omega_i=\omega$ for $i=x,y,z$


$\epsilon_{n_x}=\hbar \omega n_x $

$\epsilon_{n_y}=\hbar \omega n_y $

$\epsilon_{n_z}=\hbar \omega n_z $

$\epsilon_{n_x,n_y,n_z}=\hbar \omega(n_x +n_y +n_z)$

let $\vec{k}=(k_x,k_y,k_z)$ where $k_i=n_i$

$$\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 = \hbar^2 \omega^2 (k_x^2 + k_y^2 +k_z^2 ) = \hbar^2 \omega^2 k^2 \not=\epsilon^2~?$$

And for second displayed equation, why it's not $$\frac{\pi k^2}{2} = \frac{1}{8}4\pi k^2~?$$

  • $\begingroup$ You should provide the source of that passage, for completeness. $\endgroup$ Mar 7, 2016 at 2:28

3 Answers 3


Absorbing the irrelevant ħω constants into the normalization of the suitable quantities, for the 3D isotropic oscillator, $\epsilon=n+3/2$, while for each n the degeneracy is $(n+1)(n+2)/2$; (see SE ). Scoping the power behavior of a large quasi-continuous n, leads you to the answer.

The number of states then goes like $N\propto n^3 \propto \epsilon^3$, and hence the density of states like $dN/d\epsilon\propto \epsilon^2$.


The author is working by analogy with the 3D box case. The 3D box worked out easily because $\epsilon\propto k^2$, so that the number of states up to each energy could use the volume of an ellipsoid, which is well known. He/she makes the volume of an ellipsoid work in a "rough way" for the harmonic oscillator. He/she puts a state where each 3D box state is (with $k_i=n_i$), and calls this $\epsilon^2$ to get the energies right along each axis.

Your suggestion of $4\pi k^2/8$ was correct for the 3D box, but here he/she is using the 3D box result and correcting using $k_i=\pi n_i/L \to k_i=n_i$.

One way of calculating the 3D density of states directly is to use integral 4.634 of Gradshteyn and Ryzhik (7th). I can supply more details if interested. Also see http://arxiv.org/abs/cond-mat/9608032.


we only need to separate the energies $$\epsilon_i = \hbar \omega_i n_i$$ then $$\epsilon^2 =\epsilon_x^2 + \epsilon_y^2 + \epsilon_z^2 = \left(\hbar \omega_x n_x\right)^2 + \left(\hbar \omega_y n_y\right)^2 + \left(\hbar \omega_z n_z\right)^2$$. we can make it better into

$$\epsilon^2 = \hbar^2 \left(\omega_x^2 n_x^2 + \omega_y^2n_y^2+\omega_z^2n_z^2\right)$$

yeph. same with the answer


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