# Fidelity With Bell State Calculation

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?

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 $$\operatorname{Tr}_{a_2b_2}[|\psi'\rangle\langle \psi'|] = |00\rangle\langle 00|$$.