I'm looking at some examples, but I cannot get the expected result when it comes down to making the measurement on the following state where we measure the first qubit which is the ancilla state.

Here is $|\psi\rangle = \frac{1}{2} |{0}\rangle \otimes (|\psi_a\rangle |\psi_b\rangle + |\psi_b\rangle |\psi_a\rangle) + \frac{1}{2} |{1}\rangle \otimes (|\psi_a\rangle |\psi_b\rangle - |\psi_b\rangle |\psi_a\rangle)$

My calculation suggests that the probability of the ancilla being in the state $|0\rangle$ is:

$p_{0} =\frac{1}{2} + \frac{1}{2}|\langle\psi_a | \psi_b \rangle|^2 $

However the text suggests that there is a minus in place of the plus. I'm not sure if I'm doing anything wrong here or this is a typo.

  • $\begingroup$ Does the text say that $p_{0} =\frac{1}{2} - \frac{1}{2}|\langle\psi_a | \psi_b \rangle|^2 $ or $p_{1} =\frac{1}{2} - \frac{1}{2}|\langle\psi_a | \psi_b \rangle|^2 $ $\endgroup$ Commented Feb 19, 2020 at 21:27
  • $\begingroup$ $p_{0}$ it says. I can see a minus coming in if its $p_{1}$. $\endgroup$
    – disruptive
    Commented Feb 20, 2020 at 7:21

1 Answer 1


Your calculation is correct, given the state and conclusion that you have written. The book may be unclear, or have a typo.


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.