I'm reading an article about environmental monitoring and information transfer. Suppose $S$ represents a quantum system and $E$ is the environment. Assume at time $t=0$ there are no correlations between $S$ and $E$: $\rho_{SE}(0)=\rho_{S}(0)\otimes\rho_{E}(0)$, and this composite density operator evolved under the action of $U(t) = e^{-iHt/h}$, where $H$ is the total Hamiltonian. Let $P_\alpha$ be a projective operator on $E$. Then, the probability of obtaining outcome $α$ in this measurement when $S$ is described by the density operator $\rho_s(t)$ is given as

$$ \text{Prob}(\alpha|\rho_s(t))=\text{Tr}_E (P_αρ_E(t)) $$

and the density matrix of $S$ conditioned on the particular outcome $\alpha$ is

$$ \rho_s^{\alpha}(t)= \frac{\text{Tr}_E\{(I\otimes P_\alpha)\rho_{SE}(t)(I\otimes P_\alpha)\}}{\text{Prob}(\alpha|\rho_s(t))} $$

I'm wondering how those two equations coming from? Also, since the indirect measurement aims to yield information about S without performing a projective (and thus destructive) direct measurement on S, why there's $P_\alpha$ in the equation? Thanks!!

  • $\begingroup$ I think there is a typo in the equation for $\text{Prob}(\alpha|\rho_s(t))$: don't you mean $\rho_{SE}(t)$ in place of $\rho_E(t)$? $\endgroup$ Jan 29, 2021 at 7:00
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    $\begingroup$ The two equations - for outcome probability and post-measurement state - are density matrix variants of the equations in this question. They are also equations $(2.159)$ and $(2.160)$ in Mike & Ike with the caveat about subsystems described in this answer. $\endgroup$ Jan 29, 2021 at 7:03
  • $\begingroup$ @Adam Zalcman Thanks for the comment! No it's not a typo. The two equations come from eqn 13&14 in the Quantum Decoherence paper: doi.org/10.1016/j.physrep.2019.10.001 $\endgroup$
    – ZR-
    Jan 30, 2021 at 3:02
  • $\begingroup$ You're right. I'm not sure why I thought otherwise. Perhaps I missed the fact that $P_\alpha$ are projectors on $E$ which is implicit in the use of partial trace and tensor product. I made a small change to your question to clarify this. Anyway, do the pointers to the other question and section 2.4.2 in Mike&Ike help you or is there something else that's not clear? :-) $\endgroup$ Jan 30, 2021 at 4:03
  • $\begingroup$ @Adam Zalcman Thank you so much!! I noticed the analogy, but I'm still wondering why the denominator doesn't have the square root in this case, and what's the meaning of two $I\otimes P_\alpha$. $\endgroup$
    – ZR-
    Jan 30, 2021 at 4:26

1 Answer 1


The two equations are part of the measurement postulate of quantum mechanics which states that probability of the outcome $m$ in a measurement described by operators $M_m$ on a state $\rho$ is

$$ p(m) = \mathrm{tr}(M_m^\dagger M_m \rho)\tag1 $$

(c.f. $(2.159)$ in Nielsen & Chuang) and the post-measurement state is

$$ \frac{M_m\rho M_m^\dagger}{\mathrm{tr}(M_m^\dagger M_m \rho)}\tag2 $$

(c.f. $(2.160)$ in Nielsen & Chuang).

The first equation in the question follows from substitutions

$$ m = \alpha \\ \rho = \rho_E(t) \\ M_m = P_\alpha $$

in $(1)$. The second follows from substitutions

$$ \rho = \rho_{SE}(t) \\ M_m = I\otimes P_\alpha $$

followed by partial trace over the environment on the post-measurement state.


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