I Time dependence of operators in the Schrödinger picture

[The Tortoise-Man]

I found in wikipedia following formula describing the derivative of operator $A_H$ considered in Heisenberg picture, where $A_S$ is it's representation in Schroedinger picture:

$\frac{d}{dt}A_\text{H}(t)=\frac{i}{\hbar}[H_\text{H},A_\text{H}(t)]+\left( \frac{\partial A_\text{S}}{\partial t} \right)_\text{H} .$Note that $A_H$ and $A_S$ are related via $A_{\mathrm{H}}(t) = e^{iHt/\hbar} A_{\mathrm{S}}(t) e^{-iHt/\hbar}$.What I not understand is why $\frac{\partial A_\text{S}}{\partial t}$ isn' t zero ? So far I learned it in Schrödinger picture every operator is time independent, so it' time derivative must be zero, right?Or, how does exacty the the 'sloppy' posed assumption on time independence of operators in Schrödinger picture reads in precise mathematical terms? Should it be that there is no explicit time dependence or total time dependence, ie do operators considered through glasses of Schrödinger picture satisfy $\frac{\partial A_\text{S}}{\partial t}$ or $\frac{d A_\text{S}}{d t}$ ?

 

[renormalize]

So far I learned it in Schrödinger picture every operator is time independent, so it' time derivative must be zero, right?

Not true. For example, you can add a time-dependent potential $V(t)$ to a time-independent Hamiltonian operator $H_{0}$ to get a total Hamiltonian that is time-dependent: https://ocw.mit.edu/courses/5-74-in...c55f9ea1ff1faf17c847bb2f_MIT5_74s09_lec02.pdf

 

[The Tortoise-Man]

So for operators represented in Schrödinger picture time independence is not an obligatory condition, but happens just often in practice? So one may say that the choice to work in Schrödinger picture provides most advantages when the operators one is dealing with are time independent, but it's not required at all? But what one can say that in Schrödinger all operators satisfy $\frac{\partial A_\text{S}}{\partial t} = \frac{d A_\text{S}}{d t}$

 

[vanhees71]

I found in wikipedia following formula describing the derivative of operator $A_H$ considered in Heisenberg picture, where $A_S$ is it's representation in Schroedinger picture:

$\frac{d}{dt}A_\text{H}(t)=\frac{i}{\hbar}[H_\text{H},A_\text{H}(t)]+\left( \frac{\partial A_\text{S}}{\partial t} \right)_\text{H} .$Note that $A_H$ and $A_S$ are related via $A_{\mathrm{H}}(t) = e^{iHt/\hbar} A_{\mathrm{S}}(t) e^{-iHt/\hbar}$.What I not understand is why $\frac{\partial A_\text{S}}{\partial t}$ isn' t zero ? So far I learned it in Schrödinger picture every operator is time independent, so it' time derivative must be zero, right?Or, how does exacty the the 'sloppy' posed assumption on time independence of operators in Schrödinger picture reads in precise mathematical terms? Should it be that there is no explicit time dependence or total time dependence, ie do operators considered through glasses of Schrödinger picture satisfy $\frac{\partial A_\text{S}}{\partial t}$ or $\frac{d A_\text{S}}{d t}$ ?

That's what one should better call "explicit time dependence".

A concrete physical system is described with self-adjoint operators in Hilbert space, representing observables, that are themselves built with a few basic operators with which you can built all other observables of the system. E.g., for a single non-relativsitic particle with zero spin the few basic operators can be chosen as $\hat{\vec{x}}$ and $\hat{\vec{p}}$, which obey the usual commutation relations
$$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \frac{\hbar}{2} \hat{1}.$$
These operators are by definition time-independent in the Schrödinger picture of time evolution.

Nevertheless you can have physical situations, where you need explicitly time-dependent operators that represent observables. This is, e.g., the case if you consider an electron in a time-dependent external electromagnetic field. In the usual semi-classical approximation, where the field is not quantized, then your Hamiltonian reads
$$\hat{H}=\frac{1}{2m} [\hat{\vec{p}}-q \vec{A}(t,\hat{\vec{x}})]^2+q \Phi(t,\hat{\vec{x}}).$$
Here the Hamiltonian is explicitly time dependent in the Schrödinger picture, i.e., $\hat{\vec{x}}$ and $\hat{\vec{p}}$ are time-independent, but $\hat{H}$ has an explicit time dependence through the electromagnetic potentials.

In the Schrödinger picture and arbitrary self-adjoint operator, representing an observable with explicit time dependence you have
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{A}(t,\hat{\vec{x}},\hat{\vec{p}})=\frac{\partial \hat{A}(t,\hat{\vec{x}},\hat{\vec{p}})}{\partial t},$$
and of course the partial derivative here refers only to the explicit time dependence.

You get from the Schrödinger picture to the Heisenberg picture by an explicitly time dependent unitary transformation, which is given in the case of a Hamiltonian that is NOT explicitly time dependent,
$$\hat{A}_{\text{H}}(t,\hat{\vec{x}}_{\text{H}},\hat{\vec{p}}_{\text{H}})=\exp(\mathrm{i} t \hat{H}/\hbar) \hat{A}(t,\hat{\vec{x}},\hat{\vec{p}}) \exp(-\mathrm{i} t \hat{H}/\hbar).$$
From this you get
$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{A}_{\text{H}}(t,\hat{\vec{x}}_{\text{H}},\hat{\vec{p}}_{\text{H}}) = \frac{1}{\mathrm{i} \hbar} [\hat{A}_{\text{H}},\hat{H}] + \partial_t \hat{A}_{\text{H}}.$$
In the time derivative of the Heisenberg picture the commutator with the Hamiltonian takes care of the time dependence of $\hat{\vec{x}}_{\text{H}}$ and $\hat{\vec{p}}_{\text{H}}$ and the partial time derivative again of the explicit time dependence of $\hat{A}$.

 

[PeroK]

So for operators represented in Schrödinger picture time independence is not an obligatory condition, but happens just often in practice?

In general, something being a function of time does not prohibit something being constant in time. That's just a special case.

 

[vanhees71]

As I said in #4, by definition the fundamental operators of the observable-algebra representation in the Schrödinger picture of time evolution are time-independent, and the entire time evolution is in the states (statistical operators/state kets). Consequently, in the Schrödinger picture a possible time-dependence of an observable is entirely an explicit time dependence.