When do Hamilton's equation and Heisenberg equation give different solutions?

Question

I'm self-studying QM and become fascinated about Heisenberg picture. I have a question about the relationship between Heisenberg picture and classical mechanics. Consider a simple form of Halmitonian: \begin{equation*} H=\frac{p^2}{2m}+V(x), \end{equation*} where $V(x)$ is a sufficiently smooth function (or even polynomial if that makes things easier).

The Hamilton's equation \begin{equation*} \frac{dx}{dt}=\{x,H \}; \quad \frac{dp}{dt}=\{p,H\} \end{equation*} and the Heisenberg equation (of course here $x$ and $p$ are understood as operators) \begin{equation*} \frac{dx}{dt}=\frac{1}{i\hbar}[x,H]; \quad \frac{dp}{dt}=\frac{1}{i\hbar}[p,H] \end{equation*} reduce to the same form \begin{equation*} \frac{dx}{dt}=\frac{p}{m}; \quad \frac{dp}{dt}=-\frac{\partial V}{\partial x}. \end{equation*} The difference is in QM we have $[x,p]=i\hbar$.

Griffith's book discusses the solution of the Heisenberg equation on free particle and harmonic oscillator. The solutions in both cases are exactly the same as their classical counterpart. The former is \begin{equation*} x(t)=x(0)+p(0)t; \quad p(t)=p(0) \end{equation*} and the latter is \begin{equation*} x(t)=x(0)\cos(\omega t)+\frac{p(0)}{m\omega} \sin(\omega t); \\ p(t)=p(0)\cos(\omega t)-m\omega x_0 \sin(\omega t). \end{equation*}

My question is: is it generally true that the Hamilton's equation in the classical mechanics and Heisenberg equation in QM always give the same form of solutions (assume $H=p^2/(2m)+V(x))$? Here I'm only interested in the apparent similarity of the solutions, and I understand despite their similar looking, their physical meanings are completely different (for example, free particles have a nontrivial dispersion relation, or QHO has discrete energy levels). I'm just curious why $[x,p]=i\hbar$ does not affect the form of the solutions.

My guess is the solutions only have the same form for these two simple cases. They are special because the right hand side is linear with respect to $x$ and $p$. I appreciate if someone can explain the general case.

Answer 1

It should be said that even if the fundamental phase space variables $(q,p)$ happens to satisfy the same time-evolution equations at the classical and the quantum level, e.g. if the Hamiltonian is separable $H(q,p)=\frac{p^2}{2m}+V(q)$, there still remain the operator ordering ambiguity for composite operators $f(q,p)$: the algebra of composite operators may receive quantum corrections, cf. e.g. my Phys.SE answer here.

Answer 2

It is possible to prove $[p, f(x)] = \frac {\hbar}{i} \frac{d}{dx} f(x)$ for any analytical f(x). That's actually not that difficult, you can prove it for example by induction:

$$[p, x^0] = 0 = \frac {\hbar}{i} \frac{d}{dx} (x^0)$$ $$[p, x^{n+1}] = x^n [p,x] + [p, x^n] x $$ $$= - i \hbar x^n + \frac {\hbar}{i} n \cdot x^{n-1} x $$ $$= \frac {\hbar}{i} (n+1) x^n$$ $\rightarrow$ the statement is true for all $x^n$ $\rightarrow$ since the commutator is linear, the statement is true for all analytical functions. That proves that this is true in all pictures (and not only in the Schrödinger picture where $p = \frac{\hbar}{i} \frac{d}{dx}$ in position-space is the definition of p).

Hence, if $H = \frac{p^2}{2 m} + V(x)$ with analytical V, Heisenberg's equations give: $$\frac{dp}{dt} = \frac{p}{m}$$ $$\frac{dp}{dt} = \frac{1}{i \hbar} \frac{\hbar}{i} \frac{dV}{dx} = - \frac{dV}{dx}.$$ So they are exactly the same as the classical equations of motion.

Knowing about Newton, everybody will probably agree that the equations above are actually the classical equations. However, for completeness, here is a proof that they are the same as Hamilton's equations:

Since the proof of $[p, f(x)] = \frac {\hbar}{i} \frac{d}{dx} f(x)$ only uses the commutation relations and the algebra of Poisson brackets is the same as the algebra of commutators, $\{p, f(x)\} = -$ $\frac {d}{dx} f(x)$ is also true for every analytical f. So, a Hamiltonian $H = \frac{p^2}{2 m} + V(x)$ with analytical V will always give the following Hamilton equations: $$\frac{dx}{dt} = \frac{p}{m}$$ and $$\frac{dp}{dt} = - \frac{dV}{dx}$$

Therefore, as long as $H = \frac{p^2}{2 m} + V(x)$ with analytical V, Heisenberg's equations should give the same result as Hamilton's equations. But I should note that I've never read that anywhere, those are just my own thoughts, so I could be mistaken.

Edit: By now, I've actually found it in the book "Heisenberg's QM" by Razavy (though only the result without proof).

Edit: But, as Qmechanic pointed out, the order ambiguity of operators like $p \cdot x$ still remains, even if H is completely seperated in x and p.