Observables commuting with the Hamiltonian

Question

Why is it that observables which commute with the Hamiltonian are constants of the motion? I can't understand why that should be true.

Answer 1

The time evolution of an operator $O$ in the Heisenberg picture is given by

$$ \dot O(t) = i [H,O(t)] $$ and therefore if the operator commutes with $H$ we get $\dot O(t) = 0$, AKA it is a constant of motion.

Answer 2

You got already two fine answers using the Heisenberg picture. But it may be more intuitive (at least in my opinion) to use the Schrödinger picture for answering your question.

Then you have the time-dependent Schrödinger equation for the time evolution of a state $|\Psi(t)\rangle$ $$\frac{d|\Psi(t)\rangle}{dt}=-\frac{i}{\hbar}H|\Psi(t)\rangle$$ and, equivalently, the adjoint of this equation (using $H^\dagger=H$) $$\frac{d\langle\Psi(t)|}{dt}=\frac{i}{\hbar}\langle\Psi(t)|H$$

Now let's examine how the expectation value $\langle O \rangle$ of an observable $O$ changes over time. $$\begin{align} \frac{d\langle O \rangle}{dt} &=\frac{d\langle\Psi(t)|O|\Psi(t)\rangle}{dt} \\ &=\frac{d\langle\Psi(t)|}{dt}O|\Psi(t)\rangle +\langle\Psi(t)|O\frac{d|\Psi(t)\rangle}{dt} \\ &=\frac{i}{\hbar}\langle\Psi(t)|HO|\Psi(t)\rangle -\frac{i}{\hbar}\langle\Psi(t)|OH|\Psi(t)\rangle \\ &=\frac{i}{\hbar}\langle\Psi(t)| \underbrace{(HO-OH)}_{=0}|\Psi(t)\rangle \\ &=0 \end{align}$$

That means the observable $O$ is a constant of motion. The important step above was the last one, where we used that $O$ commutes with the Hamiltonian $H$.

Answer 3

The Hamiltonian gives you the time evolution operator. So a general observable evolves (in the Heisenberg picture) as $$ O(t) = e^{i H (t-t_0)} O(t_0) e^{- i H (t-t_0)} \,.$$ If $O$ and $H$ commute, you can bring the exponentials together, they cancel out and you find that $O(t)$ is the same at every time $t$.