# Why do we use complex-conjugate instead of complex-conjugate-transpose when calculating the concurrence?

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?

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

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