Constants of motion in quantum mechanics What is the meaning of a constant of motion in quantum mechanics (an observable-operator that commutes with the Hamiltonian) in contrary with classical mechanics?
 A: Any self-adjoint operator $A$ on some Hilbert space $H$ generates a one-parameter subgroup of the unitary group $U(H)$ by
$$t\mapsto U(t):= e^{itA}.$$
Here we shall think that $A$ is the Hamiltonian of some mechanical system, and that $U(t)$ is the flow it generates. In the Heisenberg picture, the time evolution of an observable $F(t)$ is given by
$$F(t)\mapsto U(t)^*F(t)U(t).$$
Hence $F(t)$ is a conserved quantity along the dynamics generated by the Hamiltonian $A$ if the total time derivative of $F$ is zero (this can be done on an analytic vector for the whole operator you're taking the derivative of, but I'll skip on this detail as no physicist really cares). Carrying out the computation (using Stone's theorem) one gets
$$\frac{\text d}{\text dt}F(t) = U(t)^*\left[i[F(t),A]+\frac{\partial F}{\partial t}(t)\right]U(t)$$
which can only be zero for any $t$ if the term inside the square brackets vanishes exactly at any time, i.e.
$$i[F(t),A] + \frac{\partial F}{\partial t}(t) = 0,\qquad\forall t\in\mathbb R.$$
Remark For physical reasons you might want to operate the substitution $A\mapsto A/\hbar$ so that $A$ can have the dimensions of an energy.
A: Is the same idea of classical mechanics, but now this quantities can be undetermined if you apply some measurement of the complementary of this quantity.
e.g.
The total angular momentum is a constant of motion $\vec{L}$. You can measure some component $L_i$ of this angular momentum a lot of times and this yields to the same result (conservation), but if you measure a different component $L_j$ after measured $L_i$ and return to measure the first component $L_i$, you may don't get the same result. This is because, despite of the fact that each component of the angular momentum is a constant of motion $[L_i,H]=[L_j,H]=0$, the components are the complementar one to another $[L_i,L_j]\neq 0$.
A: According to the Wikipedia article "Constant of Motion"

A quantity $A$ is conserved if it is not explicitly time-dependent and
  if its Poisson bracket with the Hamiltonian is zero

That is to say, if both
$$\frac{\partial A}{\partial t} = 0 $$
$$\{A, H\} = 0 $$
then
$$\frac{dA}{dt} = \frac{\partial A}{\partial t} + \{A, H\} = 0, \, \Rightarrow A \mathrm{\;is\; a\; constant}$$
For an operator $\hat A$ corresponding to the classical observable $A$, if the operator is not explicitly time-dependent, and the commutator with the Hamiltonian is zero, the expectation of the operator $\hat A$ is a constant.
That is to say, if both
$$\frac{\partial \hat A}{\partial t} = 0 $$
$$[\hat H, \hat A] = 0 $$
then
$$\frac{d\hat A}{dt} = \frac{\partial \hat A}{\partial t} + \frac{i}{\hbar}[\hat H, \hat A] = 0, \, \Rightarrow \langle \hat A\rangle \mathrm{\;is\; a\; constant}$$
