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So I have the exercise to apply a Cz gate to the following 2 Qubit state

$|a\rangle \otimes |b\rangle = (a_0 |0\rangle + a_1 |1\rangle) \otimes (b_0 |0\rangle + b_1 |1\rangle)\\\\$

Afterwards, I should find the reduced density matrix for both the qubits and show that they are the same for both. So i got the following reduced density matrix.

\begin{pmatrix} \mid b_0 \mid^2 & b_0\overline{b_1}(\mid a_0\mid^2 - \mid a_1\mid^2) \\ b_1\overline{b_0}(\mid a_0\mid^2 - \mid a_1\mid^2)& \mid b_1 \mid^2 \end{pmatrix}

for the Subsystem of my second qubit and for the subsystem for the first

\begin{pmatrix} \mid a_0 \mid^2 & a_1\overline{a_0}(\mid b_0\mid^2 - \mid b_1\mid^2) \\ a_0\overline{a_1}(\mid b_0\mid^2 - \mid b_1\mid^2)& \mid a_1 \mid^2 \end{pmatrix}

But now I cant figure out how the eigenvalues of these matrices are the same

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1 Answer 1

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A relatively quick way is to take the determinant of both matrices. If the same, that proves the products of their eigenvalues are the same. But since their traces are both 1, the sums of the eigenvalues are the same as well. That must mean the eigenvalues are the same.

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  • $\begingroup$ Thanks @DaftWullie for your quick answer. But I'm still confused, because i don't see how these two matrices have the same determinant. Or do I just have some mistakes in my density matrices? $\endgroup$
    – Ruebli
    Oct 19, 2023 at 16:05
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    $\begingroup$ The density matrices look fine to me, and they both have a determinant $4|a_0a_1b_0b_1|^2$. $\endgroup$
    – DaftWullie
    Oct 20, 2023 at 6:22

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