Fidelity With Bell State Calculation

Question

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?

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.