If operator commutes with Hamiltonian, why is the variance of the system equal to zero

Question

Say we have an operator $\hat{A}$ and an observable $a$. if $\hat{A}$ commutes with Hamiltonian operator, what is the meaning behind this, or why is the variance of the system equal to 0?

Answer 1

So, you have an operator $\hat{A}$ and I'am representing the corresponding observable as $A$ for consistency. To understand what happens when this operator $\hat{A}$ commutes with the Hamiltonian $\hat{H}$, consider the Heisenberg equation of motion (which is a fundamental equation just as like the Schrodinger's equation):

$$\frac{d\hat{A_H}}{dt}=\frac{1}{i\hbar}[\hat{A},\hat{H}]\tag{1}$$

where $\hat{A_H}$ is the same operator in the Heisenberg picture. In the Heisenberg picture, operators change with time and the state kets are frozen in time. (It doesn't matter what picture you take, the ultimate result, which is of course the expectation value of some observable, is the same).

When $[\hat{A},\hat{H}]=0$, then we get from equation $(1)$

$$\frac{d\hat{A_H}}{dt}=0\implies\hat{A_H}=\text{constant in time}$$

Since the operator that commutes with the Hamiltonian do not change in time, the corresponding observables (or their expectation values) are independent of time. The expectation value of observable $A$ do not vary with time.

Another interesting result is that if an operator commutes with Hamiltonian, then the eigen kets of the operator are also energy eigen kets. In that sense, the energy eigen kets are named stationary kets and they represent stationary states.