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I have solved this question by this, but can't able to get my mistake in it.

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  • $\begingroup$ Please do not post screenshots and photos, use LaTeX. $\endgroup$ – Martin Vesely Feb 6 at 8:22
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To start with, you only know that $e^{i\theta A/2}=\cos\frac{\theta}{2} I+i\sin\frac{\theta}{2} A$ if $A^2=I$. Since your $A^2\neq I$, that doesn't work. (It does, however, work on the central $2\times 2$ block.) So, in your final matrix, you should end up with the top-left and bottom-right matrix elements being 1.

Of course, that does not sort everything out. I don't see another problem jumping out at me....

In fact, I'm pretty confident that what you've been asked to prove is not correct. I've convinced myself of this by working backwards -- supplying $|00\rangle$ to the output of both. $U^\dagger(\theta)|00\rangle=|00\rangle$ because $|00\rangle$ is an eigenstate of $XY-YX$. For the circuit, however, cNOT takes $|00\rangle$ to $|00\rangle$, and then the single-qubit unitaries convert it to $|1\rangle(R_Y(-\theta)|0\rangle)$, which is certainly not $|00\rangle$!

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