# Can the thermal state be associated with a single pure state?

I'm trying to understand better the quantum thermal state defined by

$$$$\rho_{0}=\frac{e^{-\hbar\omega_{\mu}}\left|n_{\mu}\right\rangle \left\langle n_{\mu}\right|}{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}}$$$$ More specifically, I'm interested whether or not we could associated to the above density matrix a state ket defined through by $$\rho_{0} =\left|\psi_{0}\right\rangle \left\langle \psi_{0}\right|$$ with perhaps

$$$$\left|\psi_{0}\right\rangle =\sum_{n_{\mu}}\frac{e^{-\frac{\hbar\omega_{\mu}}{2}}}{\sqrt{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}}}\left|n_{\mu}\right\rangle$$$$

I believe this is not the correct answer since if I use this formula it will give rise to terms like $$\left|n_{\mu}\right\rangle \left\langle n_{\mu}+l\right|$$. Any thoughts on that?

Thanks

No. The state $$\rho_0$$ is not a pure state, i.e. it cannot be written in the form $$\rho_0=|\psi_0\rangle\langle\psi_0|$$.
This can be seen by noting that $$\mathrm{trace}(\rho_0^2)<1$$, while for a pure state, the trace would have to be $$1$$.
$$\rho_0$$ can, however, be seen as one half of the "thermofield double" state $$$$\left|\psi_{0}\right\rangle =\sum_{n_{\mu}}\frac{e^{-\frac{\hbar\omega_{\mu}}{2}}}{\sqrt{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}}}\left|n_{\mu}\right\rangle \otimes \left|n_{\mu}\right\rangle \ .$$$$