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?
2 Answers
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.
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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.
Hence,
$\sum_{n=1}^\infty\exp(-\beta \omega_0 n )\cosh(\beta g \sqrt{n})$
$=\cosh\left(\beta g \sqrt{-\frac{1}{\omega_0}\frac{\partial}{\partial\beta}}\right)\dfrac{1}{\exp({\beta\omega_0})-1}.$
$=\sum_{j=0}^\infty\dfrac{B_j\,\omega_0^{j-1}}{j!}\sum_{k=0}^\infty\dfrac{(-1/\omega)^k(\beta g)^{2k}}{(2k)!}\dfrac{\partial^k\beta^{j-1}}{\partial\beta^k}.$
Apart from the $j=0$ term, the sum is truncated whenever $k$ is bigger than $j$.
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