Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering if anyone here has calculated before the partition function for the Jaynes-Cummings Hamiltonian: $H_{JC}=\omega_0 (a^\dagger a + \sigma^+ \sigma^-)+g (a \sigma^+ +a^\dagger \sigma^-)$ (Here I have assumed resonance and $a,a^\dagger$ are bosonic operators and $\sigma^-,\sigma^+$ are ladder operators for a spin one half particle) For the Hamiltonian above the energy of the ground state is zero and corresponds to 0 excitations in the harmonic oscillator and the spin being down. The excited eigenenergies come in pairs and are given by: $\omega_{n\pm}=n \omega_0 \pm \sqrt{n}g$. I am interested in knowing the partition function: $\mathcal{Z}=\text{tr}( \exp(-\beta H_{JC} ))=1+2\sum_{n=1}^\infty\exp(-\beta \omega_0 n )\cosh(\beta g \sqrt{n}) $ I tried Mathematica to get an analytic expression for the above sum and it did not work. Any thoughts on whether the summation can be expressed in terms of some special function or how to calculate it numerically in an efficient and reliable way?

share|cite|improve this question

it is not exact and is impossible to compute exactly i recommend to use the Euler method to approximate your series by an integral plus some extra corrections , this Euler-Maclaurin summation converges fast to the exact solution with only a few terms.

share|cite|improve this answer
Thanks a lot Jose Javier, I will just use what you just wrote :) – Nicolás Quesada Jul 26 '12 at 1:53
Also do you know any reference related to this specific problem? – Nicolás Quesada Jul 26 '12 at 2:32
no, unfortunately i know no reference to this problem :/ or how the partition funciton is obtained. – Jose Javier Garcia Jul 26 '12 at 8:42

The function $\cosh(x)$ expands in even powers as $\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}$ and so $\cosh(a\sqrt{x})$ is actually a nice function.

The sum over $\exp(-\beta \omega_0 n )$ has a similar expansion in terms o Bernulli numbers - that's really the source of the MacLaurin formula as well.


$\sum_{n=1}^\infty\exp(-\beta \omega_0 n )\cosh(\beta g \sqrt{n})$

$=\cosh\left(\beta g \sqrt{-\frac{1}{\omega_0}\frac{\partial}{\partial\beta}}\right)\dfrac{1}{\exp({\beta\omega_0})-1}.$

$=\sum_{j=0}^\infty\dfrac{B_j\,\omega_0^{j-1}}{j!}\sum_{k=0}^\infty\dfrac{(-1/\omega)^k(\beta g)^{2k}}{(2k)!}\dfrac{\partial^k\beta^{j-1}}{\partial\beta^k}.$

Apart from the $j=0$ term, the sum is truncated whenever $k$ is bigger than $j$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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