# How can I understand these two equations about the indirect measurement?

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!!

• 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 $\rho_E(t)$? – Adam Zalcman Jan 29 at 7:00
• 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. – Adam Zalcman Jan 29 at 7:03
• @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 – ZR- Jan 30 at 3:02
• 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? :-) – Adam Zalcman Jan 30 at 4:03
• @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 $I\otimes P_\alpha$. – ZR- Jan 30 at 4:26

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.