0
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 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). $\endgroup$ – Roger Vadim May 12 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 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.