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.


1 Answer 1


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 $P_{ij}^2=P_{ij}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.