Why Must the Density Operator Be Hermitian? Exploring Its Physical Significance

[KFC]

Hi there,
In all text of QM I have, they tells that the density operator is hermitian. But without considering the math, from the physics base, why density operator must be hermitian? What's the physical significane of the eigenvalue of density matrix?

Thanks

 

[Demystifier]

Eigenvalues of the density matrix are probabilities. They clearly must be real, which is why the matrix must be hermitian. (In addition, they also must be non-negative and not larger than 1.)

 

[KFC]

Eigenvalues of the density matrix are probabilities. They clearly must be real, which is why the matrix must be hermitian. (In addition, they also must be non-negative and not larger than 1.)

Thanks. So before diagonalization, the diagonal elements of density matrix don't tell anything , right?

 

[KFC]

Well, I understand now how the diagonalized density matrix works. But for the following case how do I use it? For example, I have a two-level system without interaction, and hamiltonian gives

$$H|\Psi_n\rangle = \hbar\omega_n |\Psi_n\rangle$$

Now I consider the Heisenberg equation of the density operator

$$\dot{\rho} = -\frac{i}{\hbar}[H,\rho]$$

to findout the element of $$\dot{\rho}_{nm}$$, I apply an eigenstate of H on both side

$$\dot{\rho}|\Psi_i\rangle = -\frac{i}{\hbar}(H\rho - \rho H)|\Psi_i\rangle = -\frac{i}{\hbar}(H-\omega_i)\rho|\Psi_i\rangle$$

So, what is $$\rho|\Psi_i\rangle$$? In the text, it gives

$$\dot{\rho}_{nm} = -i\omega_{nm}\rho_{nm}$$ ?

 

[Jano L.]

Eigenvalues of the density matrix are probabilities. They clearly must be real, which is why the matrix must be hermitian. (In addition, they also must be non-negative and not larger than 1.)

Real eigenvalues does not imply hermiticity - for example, look at "[URL Hermiticity of the density matrix follows from its definition, which is

$$\rho := \sum_k p_k |\psi_k\rangle\langle\psi_k|$$

where $$p_k$$ is the probability that system will be found in the state $$\psi_k$$ (states $$\psi_k$$ can be arbitrary possible states of the system).