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I was studying constants of motion in quantum mechanics, and at first, I don't understand the condition to be a constant of motion. Generally, the temporal variation of an operator $A$ is given by the quantum Liouville equation:

$$\dfrac{dA}{dt}=\frac{i}{\hbar}[H,A]+\dfrac{\partial A}{\partial t}.$$

For time-independent operators, $[H,A]=0$ implies $A$ is a constant of motion. This equation is equivalent to the classical one if we replace $[H,A]/(i\hbar)$ with $\{H,A\}$. However, in classical mechanics, there are functions that explicitly depends on time, such as $x'=x-vt$, which have a null total derivative with respect to time because the Poisson bracket compensates for the partial derivative.

I wonder what happens in the quantum equivalent of this problem. An operator that commutes with the Hamiltonian, but the total derivative is not null. So, in this case, it will be not a constant of motion? I could not think in an example.

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If I understand you correctly, you want to compute $$\frac{d}{dt}\underbrace{\left(X(t)-\frac{P(t)}{m} t\right)}_{X^\prime(t)}$$ using the Heisenberg equation of motion $$\frac{d X^\prime(t)}{dt} =\frac{i}{\hbar} \left[H(t), X^\prime(t) \right] + \frac{\partial X^\prime(t)}{\partial t} $$ with the Hamilton operator$$H(t)= \frac{P(t)^2}{2m}.$$ The commutator term gives $$\frac{i}{\hbar} \left[ \frac{P(t)^2}{2m}, X(t)- \frac{P(t)}{m} t\right]=\frac{i}{2m \hbar}\underbrace{\left[P(t)^2, X(t) \right]}_{-2i \hbar P(t)}=\frac{P(t)}{m}. $$ The partial derivative in the Heisenberg equation of motion refers to the time dependence not contained in $X(t)$ or $P(t)$, such that $$\frac{\partial X^\prime(t)}{\partial t}= -\frac{P(t)}{m}. $$ Adding both contributions, one finds $$\frac{d X^\prime(t)}{dt}= 0,$$ in accordance with the explicit solutions of the Heisenberg equations of motion for $X(t)$ and $P(t)$ given by $$X(t)=\frac{P(0)}{m} t +X(0), \qquad P(t) =P(0).$$

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  • $\begingroup$ Thank you for the answer. Certainly, the example you gave (analogous to what I suggested) shows that $dX'/dt \neq 0$. I imagine this is the general case, but I can't think of any theorem that proves this holds for all cases, so I was looking for a counterexample. $\endgroup$ Apr 24 at 1:03
  • $\begingroup$ @QuantumBrachistochrone I am somewhat puzzled by your comment. In fact I have shown that $dX^\prime/dt =0$ whereas $\partial X^\prime / \partial t \ne 0$ and $[X,H]\ne 0$. As the last line shows, $X^\prime(t)$ is nothing else than $X(0)$ being, of course, time-independent. It is still not clear to me what you are after. Let me stress again that there is no difference between QM and classical mechanics in this respect. $\endgroup$
    – Hyperon
    Apr 24 at 2:36
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Take any time-independent operator $X$ which commutes with the Hamiltonian and multiply it by an arbitrary real-valued function of time : $$Y(t) = f(t)X$$ Then $Y(t)$ commutes with the Hamiltonian, but is not a constant of motion.

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  • $\begingroup$ That's not the case, i want an example of a operator that depends explicitly on time, is a constant of motion and comuutes with the Hamiltonian, as exists in classical physics. $\endgroup$ Apr 23 at 22:57
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    $\begingroup$ @QuantumBrachistochrone In classical physics, you have $dA/dt= \{ A,H\}+\partial A / \partial t$. You cannot have $dA/dt=0$, $\{A,H\}=0$ and $\partial A / \partial t \ne 0$ at the same time. In this respect, the situation in QM is exactly the same. $\endgroup$
    – Hyperon
    Apr 23 at 23:28

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