# Formally from $Z=\mathrm{tr} \,e^{-\beta H}$ to $Z=\sum_{n_1, n_2} e^{-\beta (E_1 + E_2)}$ for two non-interacting quantum harmonic oscillators

Suppose we have a Hamiltonian of two noninteracting quantum harmonic oscillators. Then the Hamiltonian can be written $$H = H_1 \otimes I_2 + I_1 \otimes H_2$$. When I start with $$Z=\mathrm{tr}\, e^{-\beta H}$$ I know that end result for the partition function is $$Z=\sum_{n_1, n_2} e^{-\beta (E_1 + E_2)}$$, however, how does one formally, i.e., taking all the tensor products into account arrive at this result?

Let $$\{|1, n_1\rangle\}$$ be an orthonormal basis for the Hilbert space $$\mathscr H_1$$ of oscillator 1, and let $$\{|2, n_2\rangle\}$$ be an orthonormal basis for the Hilbert space $$\mathscr H_2$$ of oscillator 2, then the set of all tensor products $$\{|1, n_1\rangle\otimes |2, n_2\rangle\}$$ is an orthonormal basis for the Hilbert space $$\mathscr H_1\otimes \mathscr H_2$$. Let us abbreviate: \begin{align} |n_1, n_2\rangle \equiv |1, n_1\rangle\otimes |2, n_2\rangle, \end{align} Notice that for this non interacting Hamiltonian, one has \begin{align} e^{-\beta H} &= e^{-\beta(H_1 \otimes I_2 + I_1 \otimes H_2)} \\ &= e^{-\beta(H_1\otimes I_2)}e^{-\beta(I_1\otimes H_2)} \qquad (\text{since the summands in the exponential commute})\\ &= (e^{-\beta H_1}\otimes I_2) (I_1 \otimes e^{-\beta H_2}) \qquad (\text{use the power series for the exponential}).\\ &= e^{-\beta H_1}\otimes e^{-\beta H_2} \end{align} Thus we have \begin{align} \mathrm{tr} e^{-\beta H} &= \sum_{n_1, n_2}\langle n_1, n_2 |e^{-\beta H_1}\otimes e^{-\beta H_2}|n_1, n_2\rangle \\ &= \sum_{n_1, n_2}\langle n_1, n_2 |e^{-\beta E_{n_1}} e^{-\beta E_{n_2}}|n_1, n_2\rangle \\ &= \sum_{n_1, n_2}e^{-\beta E_{n_1}} e^{-\beta E_{n_2}}\langle n_1, n_2 |n_1, n_2\rangle \\ &= \sum_{n_1, n_2}e^{-\beta (E_{n_1} + E_{n_2})} \\ \end{align} as desired.