# How to write post-measurement states when the measurement apparatus measures one of two observables?

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.

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 $$P_{ij}^2=P_{ij}$$.
• In this case, can we interpret $\dfrac{P_i^A}{\text{Tr}(\rho P_i^A)}$ like a Kraus Operator? Commented Oct 20, 2022 at 3:32
• 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). Commented Oct 20, 2022 at 6:39
• 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? Commented Oct 21, 2022 at 20:12