Consider a system of two harmonic oscillators with different frequencies $\omega_1,\omega_2$ and masses $m_1,m_2$ so the hamiltonian is

$$\mathcal{H}(p_1,q_1;p_2,q_2)=\sum_{i=1}^2 \left[\frac{p_i^2}{2m_i}+\frac{m_i\omega_i^2q_i^2}{2} \right]$$by quantising each oscillator separately one can show that the energy levels of the system are

$$\varepsilon(n_1,n_2)=\sum_{i=1}^2 \hbar \omega_i\left(n_i+\frac{1}{2} \right)$$

where $n_1,n_2$ are integers (independent quantum numbers).Calculate the (quantum) partition function.

I have the formula from my notes for the quantum canonical ensemble partition function as

$$Z(T)=\sum_n e^{-\beta\epsilon_n}$$

where $\beta=1/K_BT$. However this formula does seem to work for multivariable $\varepsilon$.

So in the (non-quantum) section of my notes it states that the partition function is

$$Z=\sum_{\Gamma}e^{-\beta\mathcal{H}(\Gamma)}$$. Can I just lift this to the quantum case?

If so I think I can try

$$Z=\sum_{\Gamma}e^{-\beta\mathcal{H}(\Gamma)}=\sum_{n_1,n_2=0}^{\infty}e^{-\beta \sum_{i=1}^2 \left[\frac{p_i^2}{2m_i}+\frac{m_i\omega_i^2q_i^2}{2} \right]}$$

but doesnt seem very tractable. Why doesnt this method work as I cannot see the error in my line of thinking.

EDIT: This is not the same question as Calculating quantum partition functions because I am looking for a reason why my line of thinking does $\underline{not}$ work.

## Best Answer

The partition function $\mathcal Z$ is the sum over all those these exponential indexed by the state index. In German it's called "$\mathcal Z$ustandssumme", literally "state sum".

You could device a scheme of indexing the possible overall energies so that you could write it as $\sum_n$, i.e. finding an injection ${\mathbb N}\to{\mathbb N}^2$, but if the thing converges nicely, the result is the same as if you write $\sum_{n_1}\sum_{n_2}$.