Calculating Statistical Operator $\hat{\rho}$

[matematikuvol]

$$\hat{\rho} = \begin{bmatrix}<br /> \frac{1}{3} & 0 & 0 \\[0.3em]<br /> 0 & \frac{1}{3} & 0 \\[0.3em]<br /> 0 & 0 & \frac{1}{3} <br /> \end{bmatrix}$$

If I have this statistical operator I get
$$i\hbar\frac{d\hat{\rho}}{dt}=0$$
So this is integral of motion and
$$[\hat{H},\hat{\rho}]=0$$
Is this correct? Tnx for your answer.

 

[fzero]

Your operator is a constant multiple of the identity operator, so it commutes with any operator, in particular the Hamiltonian.

 

[matematikuvol]

Of course. My example wasn't so good. Suppose I have matrix

$$\hat{\rho} = \begin{bmatrix}<br /> \frac{1}{3} & 5 & 6 \\[0.3em]<br /> 5 & \frac{1}{3} & 6 \\[0.3em]<br /> 5 & 6 & \frac{1}{3} <br /> \end{bmatrix}$$

Why now $$\hat{\rho}$$ always commute with Hamiltonian?

 

[kith]

What you have written is the state ρ in a fixed base for a given time t₀.

Compare this with a ket vector, which can be represented as a column vector (1 2 3 ...)^T for any fixed time t₀.

In both cases, the states are initial values for the dynamical equations. Not their solutions for all times t.