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?



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
    Jan 12 at 17:00
  • 1
    $\begingroup$ @denis Why not? The two states have nothing to do with each other. $\endgroup$ Jan 12 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.