Is the evolution of a density matrix linear with respect to density matrices?

Question

The evolution of a density matrix in quantum mechanics is given by $$i\hbar\dot\rho=[H,\rho]=H \rho-\rho H.$$ If it is linear, can be written the rhs as $A \rho$ for a linear operator $A$?

Answer 1

Short answers : yes and yes.

The equation of motion for the density matrix, namely von Neumann's equation $\dot{\rho} = -\frac{i}{\hbar}[H,\rho]$, is linear with respect to $\rho$, as expected, since the commutator is bilinear. We can check it again, in order to convince ourselves : $$ \begin{array}{rcl} \partial_t(\lambda_1\rho_1+\lambda_2\rho_2) &=& -\frac{i}{\hbar}[H,\lambda_1\rho_1+\lambda_2\rho_2] \\ &=& -\frac{i}{\hbar}\left(H(\lambda_1\rho_1+\lambda_2\rho_2) - (\lambda_1\rho_1+\lambda_2\rho_2)H\right) \\ &=& -\frac{i}{\hbar}\left(\lambda_1(H\rho_1-\rho_1H) + \lambda_2(H\rho_2-\rho_2H)\right) \\ &=& -\frac{i}{\hbar}\left(\lambda_1[H,\rho_1] + \lambda_2[H,\rho_2]\right) \\ &=& \lambda_1\dot{\rho}_1 + \lambda_2\dot{\rho}_2 \end{array} $$ As it is linear, it can be rewritten as $\dot{\rho} = A\rho$, with $A = -\frac{i}{\hbar}[H,\,\cdot\,]$; however, since $A$ is a linear operator acting itself on a vector space made of linear operators, it is called a superoperator, as mentioned by Tobias Fünke in the comment section, in order to emphasize that $A$ and $\rho$ do not live in the same spaces.