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When we use the formula to calculate two-qubit entanglement, like these:

$$ C(\rho)=\max \left\{\sqrt{e_{1}}-\sqrt{e_{2}}-\sqrt{e_{3}}-\sqrt{e_{4}}, 0\right\}\tag{18} $$

with the quantities $e_{i}\left(e_{1} \geq e_{2} \geq e_{3} \geq e_{4}\right)$ are the eigenvalues of the operator

$$ R=\rho\left(\sigma^{y} \otimes \sigma^{y}\right) \rho^{*}\left(\sigma^{y} \otimes \sigma^{y}\right),\tag{19} $$

where $\rho^*$ is the complex conjugate of the reduced density matrix $\rho$ given by Eq. (12), and $\sigma^y$ is the Pauli operator.

Why do we use the complex conjugate of the density matrix instead of its complex conjugate transpose?

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    $\begingroup$ If it’s a density matrix, it’s Hermitian, meaning that the complex conjugate transpose is equal to itself. $\endgroup$
    – DaftWullie
    May 24, 2019 at 5:22

1 Answer 1

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I believe the question is:

"why does Eq. 19 use $\rho^*$ instead of $\rho^\dagger$?"

I believe this is because $\rho^* = \rho^\dagger$ for Hermitian matrices such as $\rho$, so it can be written either way.

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    $\begingroup$ I don't think so. If $\rho^* = \rho^\dagger$, the density matrix $\rho$ is not a Hermitian matrice unless it's real symmetric. $\endgroup$
    – karry
    May 26, 2019 at 3:16
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    $\begingroup$ This is wrong. Hermitian means $\rho=\rho^\dagger$, therefore $\rho^*=\rho^T$ for an Hermitian matrix $\endgroup$
    – glS
    May 27, 2019 at 20:43

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