Let's say I have the following state:

$$ |\psi\rangle = \sqrt{\frac{2}{3}} |0000\rangle_{a_1b_1a_2b_2} + \sqrt{\frac{1}{6}} \big( |0011\rangle_{a_1b_1a_2b_2} + |1100\rangle_{a_1b_1a_2b_2} \big). $$ I measure the $b_1$ subsystem and find it to be $|0\rangle$. Then the post measurement state is: $$ |\psi'\rangle = \frac{\sqrt{\frac{2}{3}} }{\sqrt{\frac{2}{3} + \frac{1}{6}}}|0000\rangle_{a_1b_1a_2b_2} + \frac{\sqrt{\frac{1}{6}} }{\sqrt{\frac{2}{3} + \frac{1}{6}}} |0011\rangle_{a_1b_1a_2b_2} $$ Now, I want to trace out subsystems $a_1b_1$ and calculate the fidelity of the remaining system $a_2b_2$ with $|\Phi^+\rangle = \frac{1}{\sqrt{2}} (|00\rangle + |11\rangle)$. This value is found to be: $$ |\langle \psi' |\Phi^+\rangle|^2 = .9 $$ Here is my question. If instead of tracing out subsystems $a_1b_1$, I trace out $a_2b_2$ and calculate the remaining subsystem's fidelity with $|\Phi^+\rangle$, would I find $.9$ again?


1 Answer 1


No, you wouldn't find $0.9$ again. To make the partial trace calculation simpler you can note that the state $|\psi'\rangle$ is separable under the bipartition $a_1b_1 | a_2 b_2$, i.e. $|\psi'\rangle = |00\rangle \otimes (\sqrt{a} |00\rangle + \sqrt{1-a} |11\rangle)$. So irrespective of the value of $a$ we have $\operatorname{Tr}_{a_2b_2}[|\psi'\rangle\langle \psi'|] = |00\rangle\langle 00|$.

The fidelity calculation should then be relatively straightforward.


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.