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.

  • $\begingroup$ What is the problem with your formula for the quantum partition function with a multivariable $\epsilon$? $\endgroup$ Commented Apr 17, 2015 at 12:23
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    $\begingroup$ This is essentially identical to this question, for the specific case of two modes. Note that the energies are labelled by two integers, $n_1$ and $n_2$, with energies $\epsilon(n_1,n_2) = \epsilon_{n_1} + \epsilon_{n_2}$. Summing over all values of $n_1$ and $n_2$ gives you the answer. $\endgroup$ Commented Apr 17, 2015 at 12:23
  • $\begingroup$ @BySymmetry only one $n$ not two, $n_1$ and $n_2$ $\endgroup$
    – Trajan
    Commented Apr 17, 2015 at 12:25
  • $\begingroup$ @sandstone that doesn't matter. you just sum over all the eigenstates of $H$ $\endgroup$ Commented Apr 17, 2015 at 12:27
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    $\begingroup$ @sandstone Yes, the oscillators are independent, as you can see from the Hamiltonian and from your expression for the energies. Your attempted solution is just the product of two very similar sums, which you can carry out in exactly the same way as your other question. $\endgroup$ Commented Apr 17, 2015 at 14:50

1 Answer 1


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 $\sum_{n_1}\sum_{n_2}$.


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