In my statistical mechanics class, my professors did a problem in which he calculated the density of states, however I am having trouble justifying his approach. I did the problem beforehand in an entirely different way, and got the same answer, but his method is much quicker and I'd like to understand why it works.

Consider a gas of $N$ identical spin-$0$ bosons confined by an isotropic three-dimensional potential.The energy levels in this potential are $ε_n= \dfrac{n}{\hbar \omega}$ with $n$ a nonnegative integer and $ω$ the classical oscillation frequency. The degeneracy of level $n$ is $g_n = \dfrac{(n+1)(n+2)}2$.

Find a formula for the density of states, $D(ε)$ for an atom confined in this potential. Assume that $n\gg1$ and recall that the density of states $D(ε)$ is defined by the fact that $D(ε)dε$ is the number of orbitals of energy between $ε$ and $ε+dε$

My method:

The total number of energy levels below $ε_n$ will simply be:

$$N(ε_n) = \sum_{k = 0}^n g_k$$

Because $n \gg 1$, we can approximate this sum as an integral and $g_n$ as $\frac{n^2}2$

$$N(ε_n) \approx \int_0^n \frac{x^2}2 dx = \frac{n^3}6 = \frac{ε_n^3}{6 \hbar^3 \omega^3}$$ Taking the logarithm of both sides:

$$\log(N(ε_n)) = 3 \log(ε_n) - \log(2 \hbar^3 \omega^3)$$ Differentiating:

$$\frac{dN}{N} = 3 \frac{dε}{ε} \implies D(ε) = \frac{dN}{dε} = \frac{3N}{ε} = \frac{ε^2}{2 \hbar^3 \omega^3}$$

His method:

$n >> 1 \implies g_n \approx \frac{n^2}2$ is the degeneracy of the energy level $ε_n$. The spacing between energy levels is $ε_{n+1} - ε_n = \hbar \omega$

Apply $D(ε) = \frac{dN}{dε}$ in a discrete setting:

$$D(ε) = \frac{n^2/2}{\hbar \omega} = \frac{\epsilon^2}{2 \hbar^3 \omega^3}$$

I can understand why $dε$ can be said to be $ε_{n+1} - ε_n$, but why does $dN = g_n$ here? It seems more reasonable, by this argument, that we would want $g_{n+1} - g_n$, which does not appear to yield the correct answer.

I believe it boils down to some confusion on what "the number of orbitals of energy between $ε$ and $ε + dε$" actually means. It's not an easy task to visualize when you are given discrete variables (despite them behaving somewhat continuously for $n\gg1$)

  • $\begingroup$ In the main Problem shouldn't the energy be given by $\epsilon_n=n\hbar\omega$? $\endgroup$ Commented Mar 12, 2021 at 4:24

1 Answer 1


From first formula of your method for $n\gg 1$ we get $$dN = N(\varepsilon_{n+1}) - N(\varepsilon_n) = g_{n+1} \approx g_n$$.


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