# Statistical operator

1. Mar 18, 2012

### matematikuvol

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

If I have this statistical operator I get
$$i\hbar\frac{d\hat{\rho}}{dt}=0$$
So this is integral of motion and
$$[\hat{H},\hat{\rho}]=0$$

2. Mar 18, 2012

### fzero

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

3. Mar 18, 2012

### matematikuvol

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

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

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

Last edited: Mar 18, 2012
4. Mar 18, 2012

### 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.