# Why quantum map must be hermitian?

Quantum maps transform a density matrix into another one, Assume we are in the Hilbert space :$$H_A$$ the quantum map on the density matrix $$\rho_A$$ living in $$H_A$$ is : $$\mathcal{L}_A$$

Why $$\mathcal{L}_A$$ must conserve the hermiticity : $$\mathcal{L}_A(\rho_A)^{\dagger}=\mathcal{L}_A(\rho_A)$$ = $$\mathcal{L}_A(\rho_A)^{\dagger}=\mathcal{L}_A(\rho_A)$$ ?

• Your last equality seems duplicated. I'm not sure if the answer "Density matrices are Hermitian and therefore the map must conserve hermiticity to be a valid quantum map" is what you're looking for? – user1936752 Jan 23 at 14:03

## 1 Answer

A sensible density matrix needs to be hermitian, i.e. $$\rho^{\dagger} = \rho$$. This is because the diagonal elements of the density matrix can be interpreted as probabilities in any basis and therefore they should be real. This is guaranteed for hermitian matrices. Furthermore, they should sum to $$1$$, which is why we also require $$\textrm{tr}\ \rho = 1$$.

A map $$\mathcal{L}$$ from one valid density matrix to another one must preserve these properties, i.e. it must satisfy (among other things) $$\mathcal{L}(\rho)^{\dagger} = \mathcal{L}(\rho)$$ if $$\rho^{\dagger} = \rho$$ and $$\textrm{tr}\ \mathcal{L}(\rho) = 1$$ if $$\textrm{tr}\ \rho = 1$$. Otherwise $$\mathcal{L}(\rho)$$ would not be a proper density matrix.