I have the thermal partition function and the density of states for the 3D simple harmonic oscillator, which are given below $$ Z(\beta) = \frac { 1 } { \left( 2 \sinh \left( \frac { \beta \omega } { 2 } \right) \right) ^ { 3 } } $$ and $$ \rho ( E ) = \frac { \left( \frac { E } { \omega } - \frac { 1 } { 2 } \right) \left( \frac { E } { \omega } + \frac { 1 } { 2 } \right) } { 2 \omega } $$ where $\beta$ is the inverse temperature and $\omega$ is the natural frequency of the SHO.

Now, the partition function can also be written as the following integral $$ Z(\beta) = \int_{0}^{\infty} dE \: \rho(E) e^{-\beta E} $$ which is a Laplace transform, hence we can calculate the asymptotic of the density of states by inverting this Laplace transform. The inverse Laplace transform can be done using the Bromwich contour integral which is $$ \rho(E) = \frac{1}{2\pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} d\beta \: Z(\beta) e^{\beta E} $$ where $\gamma$ is greater than the real part of all the singularities of $Z(\beta)$. Now, I tried doing this integral using the residue theorem by closing the contour on the left half of the complex plane. But somehow I am getting the correct density of states using the residue of just one pole, whereas there is an infinite number of poles on the imaginary axis due to the periodicity of the hyperbolic sine function (on the imaginary axis).

Can someone explain where I am wrong in this process?

P.S. We also know that $n = \frac{E}{\omega} - \frac{3}{2}$ is a positive integer.

  • $\begingroup$ See also math.stackexchange.com/questions/3060442/… $\endgroup$ – LonelyProf Jan 4 at 8:00
  • $\begingroup$ As I mentioned in my answer, your first expression for $\rho(E)$ is not the density of states for this system, it is just a (smooth) approximation valid at large $E$. The links in my answer explain where this kind of approximation comes from. Also, when discussing an integral over energy $E$, it is not possible to say (as you do in your "PS") that it only takes discrete values, unless you write it in terms of Dirac delta functions or similar. $\endgroup$ – LonelyProf Jan 4 at 8:22

I am by no means an expert in this field, but it seemed an interesting question, so I did some searching around. Since nobody else has answered, here goes.

This approach goes back a long way, and has a lot of relevance to molecular systems with far more general quantum states (combinations of anharmonic vibrations, rotational levels and so on) and applications to the theory of unimolecular reactions. I believe that the first paper dealing with your specific system of multiple harmonic oscillators is by E Thiele J Chem Phys, 39, 3258 (1963). It tackles the problem in exactly the way you describe. The relevant quotation from that paper, which answers your question, is

Here again, the singularities in $\Phi/\beta$ are all poles located on the imaginary axis. Due to the term $e^{\beta\epsilon}$ in the integrand the residues at all poles except for the pole at the origin will contribute oscillating terms to the expression for $W(\epsilon)$. We may, therefore, reasonably expect to find a good smooth approximation to $W(\epsilon)$ by considering only the contribution from the residue at $\beta=0$.

(Obviously your expression for the density of states is only an asymptotic smooth approximation, presumably valid at high energy, since the actual energy spectrum of a finite system of quantum oscillators is discrete. So this is presumably what you are after).

There's plenty of literature on this sort of thing, including an old review by Wendell Forst Chemical Reviews, 71, 339 (1971), and papers citing this review, which you may be able to view online (unfortunately I haven't found any open access versions of these papers).

I hope this is helpful: I doubt that I can say any more since, as stated, I'm not really an expert. Maybe others will give more detail, if you need it.

  • $\begingroup$ Thanks a lot for these comments. $\endgroup$ – rahuldan Jan 3 at 19:49

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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