Partition function problem

[merkamerka]

I need some help with this problem:

Consider a diatomic molecule closed in a cubic container of volume $V$ which hamiltonian is:
$$H=\frac{p_1^2}{2m}+\frac{p_2^2}{2m}+\frac{K}{2}| \vec r_2 - \vec r_1|^2$$
where $\vec r_1, \vec r_2$ are the positions of the two atoms.

a) Find the partition function in the limit $V \longrightarrow \infty$.
b) Find (also for $V \longrightarrow \infty$ and without using the equipartition theorem) the mean value of $|\vec r_2 - \vec r_1|^2$.

In particular I have some problems evaluating the integral

$$\frac{1}{N!h^{3N}}\int e^{- \beta H(\vec q, \vec p)} d\vec q \ d\vec p$$

Thanks!

 

[lippyka]

a) The partition function in the limit V→∞ is given by\begin{equation}Z = \frac{1}{N! h^{3N}} \int_{-\infty}^{\infty} e^{-\beta H(\vec q, \vec p)} d\vec q \ d\vec p \end{equation}where N is the number of atoms, h is Planck's constant, and β is the inverse temperature. For a diatomic molecule, N=2. Using the Hamiltonian given in the problem, we obtain\begin{align}Z &= \frac{1}{2h^6} \int_{-\infty}^{\infty} e^{-\beta \left(\frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \frac{K}{2}|\vec r_2 - \vec r_1|^2 \right)} d\vec q \ d\vec p \\&= \frac{1}{2h^6} \int_{-\infty}^{\infty} e^{-\beta \left(\frac{p_1^2}{2m} + \frac{p_2^2}{2m} \right)} \int_{-\infty}^{\infty}e^{-\beta\frac{K}{2}|\vec r_2 - \vec r_1|^2} d\vec r_1 \ d\vec r_2 \ d\vec p \\&= \frac{1}{2h^6} \int_{-\infty}^{\infty} e^{-\beta \left(\frac{p_1^2}{2m} + \frac{p_2^2}{2m} \right)} \int_{-\infty}^{\infty}e^{-\beta\frac{K}{2}r_1^2} \int_{-\infty}^{\infty}e^{-\beta\frac{K}{2}r_2^2} d\vec r_1 \ d\vec r_2 \ d\vec p \\&= \left