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

Question

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

\begin{equation} \rho_{0}=\frac{e^{-\hbar\omega_{\mu}}\left|n_{\mu}\right\rangle \left\langle n_{\mu}\right|}{\sum_{n_{\mu}}e^{-\hbar\omega_{\mu}}} \end{equation} 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

\begin{equation} \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 \end{equation}

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

Answer 1

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 \begin{equation} \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 \ . \end{equation}