Differential of Quantum mean value or expectation value How to take differential of Quantum mean value over hermitian operator (mean or expectation value)?
$$d\langle \hat A\rangle$$
remark:
or time evolution of mean value over operator
$$\frac {d\langle \hat A\rangle}{dt}$$

what is the problem here?
  ok let me talk a little more special in three steps.
  1. In classical phyaics differential of momentum is:
  $$dp=dmv=mdv$$
  2. In relativistic physics differential of a momentum is:
  $$dp=dmv=mdv + vdm$$
  3. how about differential of relativistic quantum mean value over momentum operator?
  $$d\langle \hat P\rangle=?$$

I register ,here is my account:
mare
but i dont have access to my questions!.
 A: In fact, you missed another type of force:
In Lagrangian Mechanics there is a scalar potential field V in
which the gradient of  V is the force:
$$F=-\nabla V$$
and this is exactly what we dealing with in QM.
$$\frac {d \langle \hat P\rangle}{dt}=\langle \hat F\rangle=\langle -\nabla V\rangle$$
A: I assume that you only work in the 'wavefunction' approximation then $\langle \hat A \rangle = \langle \Psi | \hat A | \Psi \rangle$
The differential or time derivative follow in the usual way. For instance
$$\frac {d \langle \hat A \rangle}{dt}
=
\frac {d \langle \Psi |}{dt} \hat A | \Psi \rangle
+
\langle \Psi | \frac{\partial \hat{A}}{\partial t} | \Psi \rangle
+
\langle \Psi | \hat A \frac {d | \Psi \rangle}{dt}$$
Using the Schrodinger equation for the 'wavefunction' and working a bit one finally arrives to
$$\frac {d \langle \hat A \rangle}{dt}
=
\frac{1}{i\hbar} \langle [\hat{A},\hat{H}] \rangle
+
\left\langle \frac{\partial \hat{A}}{\partial t} \right\rangle$$
A result known as Eherenfest theorem. The differential can be obtained by multiplying everything by $dt$. The momentum operator $\hat{P}$ does not depend explicitly on time and the last term vanishes
$$\frac {d \langle \hat P \rangle}{dt}
=
\frac{1}{i\hbar} \langle [\hat{P},\hat{H}] \rangle$$
