1
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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 $ $\endgroup$
    – Victory Omole
    Feb 19, 2020 at 21:27
  • $\begingroup$ $p_{0}$ it says. I can see a minus coming in if its $p_{1}$. $\endgroup$
    – disruptive
    Feb 20, 2020 at 7:21

1 Answer 1

Reset to default
1
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

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

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