2

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.

Qmechanic
  • 201,751
Victor
  • 107

2 Answers2

2

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.

Qmechanic
  • 201,751
  • Thank you! Can I interpret you answer as: The quantum operator $x(t),p(t)$ satisfy $dx/dt=p/m, dp/dt=-\partial V/ \partial x$ because otherwise they won't be a solution of the Heisenberg equation. However, a solution to the above equations may not satisfy $[x,p]=i\hbar$ because the ordering ambiguity for compositing $x(0),p(0)$ in the solution. – Victor May 21 '23 at 03:52
  • One should first of all pick a representation/realization that respects the CCR. This answer is mainly concerned with what happens to composite operators. – Qmechanic May 21 '23 at 06:31
1

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.

Relativisticcucumber
  • 1,180
Tarik
  • 470
  • Thank you for the detailed question! By "the order ambiguity of operators like p⋅x still remains, even if H is completely separated in x and p", do you mean since the solution $x(t),p(t)$ are functions involving the initial value $x(0), p(0)$, then the order such as the order ambiguity like $x(0)\cdot p(0)$ remains? – Victor May 22 '23 at 01:22
  • Can I ask a follow-up question: Although, the Heisenberg equation and Hamilton equation have the same form $dx/dt=p/m,dp/dt=-\partial V/\partial x$, the forms of the solution seem to be different. The general form of the solution to Heisenberg equation is $x(t)=e^{iH t/\hbar} x(0) e^{-iH t/\hbar}, p(t)=e^{iH t/\hbar} p(0) e^{-iH t/\hbar}$. How can this also serve as the solution for the classical case, in which $\hbar$ doesn't even appear. – Victor May 22 '23 at 01:48
  • You're welcome! About the order ambiguity: That's basically what I meant. The classical expression $x \cdot p$ won't look the same as the quantum mechanic one, even in the Heisenberg picture, since in quantum mechanics, $x \cdot p \neq p \cdot x$. So you'll need some sort of symmetrization (most popularly Weyl) which you don't need in classical mechanics. – Tarik May 22 '23 at 14:03
  • About the follow-up question: That's a good question. I couldn't come up with a proof of why that x(t) is always the same as in classical mechanics and I don't know of any solution to Hamilton's equations that looks similar (though I'm really not an expert on analytical mechanics). I would assume that the solution has to look the same since they both obey the same differential equations which should already be a proof. – Tarik May 22 '23 at 14:13
  • 1
    Since in classical mechanics everything commutes, $x(t) = e^{...} x(0) e^{-...} = x(0)$, so that form of the solution doesn't really work in classical mechanics, but once you've massaged that solution a bit, it should look the same as the classical one. I know that that's really not a satisfactory answer, but it's the only one I could come up with. Btw: There is another form of the general solution which works in both cases: $x(t) = x(0) + t {x,H} + t^2/2 {x, {x, H} } + ...$, for the QM solution you have to replace ${...,...}$ with $\frac{1}{i \hbar} [...,...]$. – Tarik May 22 '23 at 14:14