1
$\begingroup$

I have earlier posted this question in Physics StackExchange but I feel that it is more relevant here. The question is about a contour integral and I have written most of the equations needed. Please feel free to ask for any clarification.

I have the thermal partition function and the density of states for the 3D simple harmonic oscillator, which is 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 can 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 are 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.

$\endgroup$
  • $\begingroup$ could you please outline your calculations a bit along with the answer you expect? Would help speed up the checking of your work. You are correct that this is the right site for a question like this, and I am more than happy to take a look at it if you show proper effort :) $\endgroup$ – Brevan Ellefsen Jan 3 at 18:03
  • 1
    $\begingroup$ If $E$ is just an independent variable, then $\mathcal L_{E \to \beta}[E^2/(2 \omega^3) - 1/(8 \omega)]$ is $1/(\omega \beta)^3 - 1/(8 \omega \beta)$. $\mathcal L[\rho]$ just happens to coincide with the principal part of the Laurent series of $Z(\beta)$ at zero. $\endgroup$ – Maxim Jan 3 at 19:43
  • $\begingroup$ @BrevanEllefsen The partition function $Z(\beta)$ has poles at $\beta = (2 \pi i) n$ (where $n$ is all integers). Now, I am getting the residue of all these poles to be the same equal to $\rho(E)$ given above. Hence, if we sum up these residues we get the answer to be infinity. I expect the answer of the inverse Laplace transform to be $\rho(E)$. $\endgroup$ – rahuldan Jan 3 at 19:45
  • $\begingroup$ @Maxim Can you explain what you are implying from your statement? $\endgroup$ – rahuldan Jan 4 at 5:59
  • $\begingroup$ See also physics.stackexchange.com/questions/451541/… and the links in my answer to this question. $\endgroup$ – user575517 Jan 4 at 8:00
1
$\begingroup$

The way it's written, $\rho(E)$ is just a polynomial, the Laplace transform of which is just a sum of negative powers of $\beta$. What is meant here is $$\rho(E) = \frac 1 {2 \omega} \left( \frac E \omega - \frac 1 2 \right) \left( \frac E \omega + \frac 1 2 \right) \sum_{n \geq 0} \delta {\left( \frac E \omega - \left( n + \frac 3 2 \right) \right)} = \\ \sum_{n \geq 0} \frac {(n + 1) (n + 2)} 2 \,\delta {\left( E - \omega \left( n + \frac 3 2 \right) \right)}, \\ \mathcal L[\rho] = \sum_{n \geq 0} \frac {(n + 1) (n + 2)} 2 e^{-\omega (n + 3/2) \beta} = \frac {\operatorname{csch}^3 \frac {\omega \beta} 2} 8.$$ It's related to compositions of integers because when we convolve a Dirac comb with itself twice, we get a sum of the form $\sum_{i, j, k} \delta(t - i - j - k)$, and $(n - 1) (n - 2)/2$ is the number of ways to write $n$ as the sum of an ordered triple of positive integers.

$\endgroup$
  • $\begingroup$ Thanks a lot. This clarified my doubt. $\endgroup$ – rahuldan Jan 4 at 17:27

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.