Albfey said:
Yes, you are right, they don't comute in general. I confused it with another situation, I'm sorry.
I have already read it in a book, but I didn't understand what it means. The most books say that all operators are time-independent in this picture. So, how can they have an explicit dependence on time?I'm sorry for insisting on this question, I am really confused about pictures in quantum mechanics.
To be more specific, use as example a non-relativistic single particle. Then the fundamental set of operators are the position, momentum, and spin operators ##\hat{\vec{x}}##, ##\hat{\vec{p}}##, and ##\hat{\vec{\sigma}}##, all of which by definition do
not explicitly depend on time.
Now you can choose quite arbitrarily the picture of time evolution, i.e., how these fundamental observable operators depend on time by splitting the Hamiltonian ##\hat{H}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{\sigma}},t)## (where the time dependence here means explicit time dependence, e.g., for a charged particle moving in a time-dependent electromagnetic field) in two parts, ##\hat{H}=\hat{H}_0+\hat{H}_1##.
Then the fundamental observable operators are defined to time-evolve with ##\hat{H}_0##, i.e. according to the evolution equation
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{\vec{x}}(t)=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}}(t),\hat{H_0}],$$
and analogously for ##\hat{\vec{p}}## and ##\hat{\vec{\sigma}}##, while the state kets time-evolve according to ##\hat{H}_1##:
$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d}t} |\psi(t) \rangle= \hat{H}_1 |\psi(t) \rangle.$$
This is a general picture of time evolution (also knows as a "
Dirac picture").
The
Schrödinger picture lumps all time dependence to the states, i.e., you have ##\hat{H}_0=0## and ##\hat{H}_1=\hat{H}##, while the
Heisenberg picture lumps all time dependence to the operators representing observables, i.e., ##\hat{H}_0=\hat{H}## and ##\hat{H}_1=0##.
Now for explicitly time dependent observable operators you have
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{A}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{\sigma}},t) = \frac{1}{\hbar} [\hat{A},\hat{H}_0] + \partial_t \hat{A}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{\sigma}},t),$$
where the partial time derivative is due to the explicit time dependence and the commutator due to the time dependence of the fundamental operators.
For a general state, i.e., a statistical operator, on the one hand one has
$$\hat{\rho}=\sum_{j} P_j |\psi_j(t) \rangle \langle \psi_j(t)|,$$
and thus due to the equation of motion for the state kets
$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}= [\hat{H}_1,\hat{\rho}].$$
On the other hand, if read as a function of the fundamental operators (and mabe explicitly on time), it must fulfill
$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}=[\hat{\rho},H_0]+\mathrm{i} \hbar \partial_t \hat{\rho}.$$
Subtracting both equations yields the famous von Neumann equation
$$\mathring{\hat{\rho}}=\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}] + \partial_t \hat{\rho}=0.$$
Indeed
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]+\partial_t \hat{A}$$
defines the operator for the time derivative ##\dot{A}## of the observable ##A##, represented b the operator ##\hat{A}##, i.e., that defines the "covariant time-derivative" in QT, and the von Neumann equation for the stat. op. states that the covariant time-derivative vanishes, as it must be in statistical mechanics.
