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If I want to measure an observable $A$ but the measurement apparatus has $(1-p)$ probability of measuring the observable $B$ and probability $p$ that a measurement of $A$ would be done. So how can I write the post-measurement state if the initial state is $\rho$? I was thinking in write the projection operator $P_{ij}=p|a_i\rangle\langle a_i|+(1-p)|b_j\rangle\langle b_j|$ but I don't know if it's correct.

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Let's assume that that both observables $A$ and $B$ have projectors $P^{A/B}_i$. I'm going to assume that the set of outcomes for the two is the same (and therefore, implicitly, that I don't know whether $A$ or $B$ has been performed if I get a particular result, $i$).

Now, if I actually measured $A$, and got result $i$, my final state would be $$ \rho_A=\frac{P^A_i\rho P^A_i}{\text{Tr}(\rho P^A_i)}. $$ Similarly, if you measured $B$, the outcome would be $$ \rho_B=\frac{P^B_i\rho P^B_i}{\text{Tr}(\rho P^B_i)}. $$

So, if you have a measurement that is $A$ with probability $p$ and $B$ with probability $1-p$ (and you do not know which), your outcome will be $$ \rho'=(1-p)\rho_A+p\rho_B. $$ There is not a simple projector that you can use to describe this (if you need such a description, you probably have to introduce an ancilla qubit). Note, in particular, that your proposed $P_{ij}$ is not a projector: it does not satisfy $P_{ij}^2=P_{ij}$.

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  • $\begingroup$ In this case, can we interpret $\dfrac{P_i^A}{\text{Tr}(\rho P_i^A)}$ like a Kraus Operator? $\endgroup$
    – username9
    Oct 20, 2022 at 3:32
  • $\begingroup$ Well, the problem is having the Kraus operator depend on the state it's acting on. (Also you'd need to include a factor of $\sqrt{p}$ and the denominator would have a square root on it). $\endgroup$
    – DaftWullie
    Oct 20, 2022 at 6:39
  • $\begingroup$ Sure, but there's no form to get rid of this factor in the equation because the probability of obtaining outcome $i$ is $\text{Tr}(\rho P_i^A)$, isn't it? $\endgroup$
    – username9
    Oct 21, 2022 at 20:12
  • $\begingroup$ Yes, so I was trying to imply that you probably shouldn't call it a Krauss operator. $\endgroup$
    – DaftWullie
    Oct 24, 2022 at 8:32

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