Given two mixed states $\rho$ and a $\sigma$, does it make sense to say that the probability of $\rho$ being in the state $\sigma$ is given by $Tr(\rho \sigma)$?

It seems to me that the answer must be no (e.g., take $\sigma = \rho$ and this is not necessarily one as I naively expect it to be), but then, what is the physical meaning of $Tr(\rho \sigma)$ in this context?

  • $\begingroup$ I'm not sure it makes sense. Given $\rho$ the (mixed) state is uniquely determined; if $\sigma \neq \rho$ the states are different and there's probability $0$ to find the system in $\sigma$. It makes sense if $\sigma$ describes a pure state, in which case $Tr(\rho |\sigma><\sigma|)$ answers the questions "what's the probability of finding the system in the pure state $\sigma$?" $\endgroup$ – tbt Feb 16 '18 at 20:25
  • $\begingroup$ Yes, I know that when $\sigma$ is pure this is the case, my question is when both $\sigma$ and $\rho$ are mixed. My mistake to not be clear about it. I'll edit it. $\endgroup$ – Gabriel Cozzella Feb 16 '18 at 20:27

It seems to me that it makes sense. Let start with pure states: $$\rho=|\psi\rangle\langle\psi|$$ $$\sigma=|\phi\rangle\langle\phi|$$ Then $${\rm Tr\ \!}\rho\sigma ={\rm Tr\ \!}|\psi\rangle\langle\psi|\phi\rangle\langle\phi| =|\langle\psi|\phi\rangle|^2$$ which is the probability to observe the system in the state $|\phi\rangle$ during a quantum measurement if it was prepared in the state $|\psi\rangle$. The formula are easily extended to mixed states $$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$$ $$\sigma=\sum_i q_i|\phi_i\rangle\langle\phi_i|$$ Then $${\rm Tr\ \!}\rho\sigma =\sum_{i,j} p_iq_j|\langle\psi_i|\phi_j\rangle|^2$$ where $p_ip_j$ is the probability that system 1 be in the state $|\psi_i\rangle$ and system 2 in $|\phi_j\rangle$. ${\rm Tr\ \!}\rho\sigma$ is the probability that the two systems be observed in the same quantum state during a quantum measurement.


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