I have a system of $N$ uncoupled 1D quantum harmonic oscillators, each with its own frequency $\omega_i$. The density of states for a single quantum harmonic oscillator shall be defined as

$$ \rho(E) = \frac{dn}{dE} $$ It is trivial in the 1-D case to obtain $n$ as function of energy by rearranging the eigenenergy equation, $$ E(n) = \hbar \omega(n+1/2) \leftrightarrow n(E) = \frac{E}{\hbar \omega } - \frac{1}{2\hbar \omega} $$ The density for a single oscillator is then $$ \rho(E) = \frac{dn}{dE}=\frac{1}{\hbar \omega} $$ So far so good. My problem is now to derive the density of a system with the eigenenergy defined as $$ E(n_1,n_2,\dots, n_N) = \sum^N_{i=1}\hbar \omega_i(n_i+1/2) $$

Is there a way to obtain/construct the density from the 1D solution for the N-dimensional system ? There seems no straightforward way to define a total $n$ and to apply the derivative. Or is there a way ?


1 Answer 1


The definition of the density-of-states is $$ \rho(E)=\frac{dN(E)}{dE}, $$ where $N(E)$ is the number of states with energies less than $E$.

Note, that as the equation involves a derivative, it is not defined for non-continuous spectra, such as that of a Harmonic oscillator. However, it can be still treated in terms of generalized functions, such as Heaviside step-function, $\theta(x)$,a nd delta-function, $\delta(x)$.

The number of oscillator states with energies less than $E$ is: $$ N(E) =\sum_{n=0}^{+\infty}\theta(E-E_n) = \sum_{n=0}^{+\infty}\theta\left(E-\hbar\omega_0(n+\frac{1}{2})\right) $$ The resulting density-of_states is then $$ \rho(E) =\sum_{n=0}^{+\infty}\delta(E-E_n) = \sum_{n=0}^{+\infty}\delta\left(E-\hbar\omega_0(n+\frac{1}{2})\right). $$

With the correct formulas the generalization to the case of multiple oscillators with frequencies $\omega_k$ is trivial: $$ N(E) =\sum_{n=0}^{+\infty}\sum_k\theta(E-E_n^{(k)}) = \sum_{n=0}^{+\infty}\sum_k\theta\left(E-\hbar\omega_k(n+\frac{1}{2})\right), $$ and similarly for the density-of-states. If the oscillators spectrum is continuous, the summation over $k$ becomes an integral, which is easily evaluated using the properties of the delta-function, resulting in a continuous density-of-states.

Continuous approximation for a harmonic oscillator
If we are interested in energy scales much greater than the oscillator energy level spacing, we can approximate the sum in the oscillator dos by an integral, with $\epsilon=\hbar\omega_0 n$, $d\epsilon=\hbar\omega_0$: $$ \rho(E) =\sum_{n=0}^{+\infty}\delta\left(E-\hbar\omega_0(n+\frac{1}{2})\right)\frac{d\epsilon}{\hbar\omega_0}= \frac{1}{\hbar\omega_0}\int_{0}^{+\infty}\delta\left(E-\epsilon -{\hbar\omega_0}{2})\right)d\epsilon=\frac{1}{\hbar\omega_0}. $$ Note that we are essentially in the classical limit here.

  • $\begingroup$ Is there also an approximation to the final expression with the double sums ? I would like to avoid the "counting" if possible. I assumed there would be some sort of approximate function that can be obtained by integration, as often done in statistical mechanics. $\endgroup$
    – Hans Wurst
    May 12, 2021 at 10:11
  • $\begingroup$ It depends on the nature of your oscillators - if they form a continuous spectrum, you will get a continuous function. Also, if you are interested in energy scales bigger than $\omega_0$, you can approximate the sum by an integral. But these are two different things (two different sums). $\endgroup$ May 12, 2021 at 10:21
  • 1
    $\begingroup$ I guess i should post a second question. My system are quantum oscillators. So your answer without approximations seems appropriate and correct. But my actual problem which caused this question is numerically and i need something that i can evaluate. At the moment i have 66 oscillators and that is not a particular large number, just a test system. Doing the sum explicitly is not practical. But i realize now that this was not part of the question as i asked it. $\endgroup$
    – Hans Wurst
    May 12, 2021 at 10:32

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