Time derivative in the Heisenberg picture?

[pellman]

On the Wikipedia page for http://en.wikipedia.org/wiki/Heisenberg_picture#Mathematical_details" we find this relation

$$\frac{d}{dt}A(t)=\frac{i}{\hbar}[H,A(t)]+\left(\frac{\partial A}{\partial t}\right)$$

I don't understand what the distinction between

$$\frac{d}{dt}A(t)$$ and $$\left(\frac{\partial A}{\partial t}\right)$$

is supposed to be. That is, what is the difference between the meaning of these two expressions?

For regular old c-number functions, the difference between total and partial derivatives is something like

$$\frac{df}{dt}=\frac{\partial f}{\partial u}\frac{du}{dt}+\frac{\partial f}{\partial t}$$.

where f = f(u,t). If f doesn't depend on other variables, then $$\frac{df}{dt}=\frac{\partial f}{\partial t}$$.

 

[jambaugh]

Recall that in the Heisenberg Picture we express the time evolution of the observables for a dynamically evolving system by time varying the operators, leaving the Hilbert space vectors (e.g. wave-functions) fixed at their initial values.

So for a fixed observable, there will be a time dependence relating to the evolution of the system being observed. That is the component of the total time derivative expressed by the commutator.

But in the general setting we can also consider a time varying observable i.e. imagine a polarizer which is rotating over time. This explicit time evolution of the observational device is expressed by an explicit time dependence of the operator and that's the partial derivative w.r.t. t component.

Again this is somewhat confusing since in Physics we too often express both function and functional value (variable) with the same symbol.

Imagine the observable A(t) in the Heisenberg picture is expressed as a function 'a' of the initial operator (the operator corresponding to measuring the system at time t=0) and also a function of t directly indicating a time evolution of the measuring device:
$$A(t) = a(t,A_0)$$.

Then
$$\frac{dA}{dt} = \frac{i}{\hbar} [H,A(t)] + \frac{\partial a}{\partial t}$$

[edit]
It may be helpful to think of the A as the correspondent of the value of a classical observable. Then consider say a position as seen by a moving observer relative to a given frame. There are two components, the motion of the particle (in the given frame) and the motion of the observer. With frame transformation x' = x+ X(t) we have:
dx'/dt = dx/dt + dX/dt which becomes in QM [H,x] + dX/dt.

 

[pellman]

That was very helpful, jambaugh. thank you!