6
$\begingroup$

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?

$\endgroup$
1
  • 3
    $\begingroup$ If it’s a density matrix, it’s Hermitian, meaning that the complex conjugate transpose is equal to itself. $\endgroup$
    – DaftWullie
    Commented May 24, 2019 at 5:22

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ 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
    Commented May 26, 2019 at 3:16
  • 4
    $\begingroup$ This is wrong. Hermitian means $\rho=\rho^\dagger$, therefore $\rho^*=\rho^T$ for an Hermitian matrix $\endgroup$
    – glS
    Commented May 27, 2019 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.