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?



1 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}

  • $\begingroup$ thanks! But why the inclusion of the tensor product resolve the previous issue here? $\endgroup$
    – sined
    Commented Jan 12, 2021 at 17:00
  • 1
    $\begingroup$ @denis Why not? The two states have nothing to do with each other. $\endgroup$ Commented Jan 12, 2021 at 17:54

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