What is the analog of the equation $\frac{d^2\vec{r}(t)}{dt^2}=\frac{\vec{r}(t)}{\left|\vec{r}(t)\right|}f(\vec{r}(t))$ in Quantum Mechanics?

Question

In Newtonian Physics the general equation for the acceleration when there is a central force is $$\frac{d^2\vec{r}(t)}{dt^2}=\frac{1}{m}\frac{\vec{r}(t)}{\left|\vec{r}(t)\right|}f(\vec{r}(t))$$ with $m$ being the mass, $t$ being the time, and $\vec{r}(t)$ being the distance vector as a function of time. $$\frac{d^2\vec{r}(t)}{dt^2}$$ is equivalent to the acceleration term, and $f(\vec{r}(t))$ is a function of the distance vector $\vec{r}(t)$.

I was wondering if there is an analog of the equation I gave above in Quantum Mechanics, and if so what equation would be analogous to the equation I gave in Quantum Mechanics.

Answer 1

Newtonian-equations-like description is not possible in quantum mechanics, since the position and the momentum are not simultaneously measurable. However, the quantum mechanics, obviously contains the classical mechanics and the Newton's laws as a limiting case, as is stated by the correspondence principle. The correspondence between the classical and quantum equations-of-motion is express by the Ehrenfest theorem: $$m\frac{d\langle \hat{x}\rangle}{dt}=\langle \hat{p}\rangle, \frac{d\langle \hat{p}\rangle}{dt}=-\langle V'(\hat{x})\rangle.$$

Thus, for the central force we could write: $$\frac{d^2\langle \hat{\vec{r}}\rangle(t)}{dt^2} = \frac{1}{m}\langle \nabla_{\vec{R}}V(R)|_{\vec{R}=\hat{\vec{r}}(t)}\rangle,$$ where $V(R)$ is the potential corresponding to the force $f(R)$: $$\nabla_{\vec{R}}V(R) = -\frac{\vec{R}}{R}f(R).$$ The difficulty in writing literally the relation in the question is notational: since one replaces $\vec{r}$ by the position operator, it is impossible to express the radial unit vector as $\vec{r}/r$ without ambiguity.

A couple of things to keep in mind is:

• the intricacies with defining time derivatives in quantum mechanics
• the fact that $\langle V'(\hat{x})\rangle \neq V'(\langle\hat{x}\rangle)$, which means that the equations are not closed in respect to $\langle\hat{x}\rangle$, which makes them impossible to solve.

Answer 2

The easiest way to connect classical and quantum mechanics is to work with the energy, or Hamiltonian, rather than the equations of motion.

You can write the Hamiltonian for your system as

\begin{equation} H = \frac{1}{2} m \dot{\vec{r}}^2 + V(|\vec{r}|) \end{equation} where, to recover your equation of motion, \begin{equation} \nabla V = -\frac{\vec{r}}{|\vec{r}|}f(|\vec{r}|). \end{equation}

Then, the conceptually simplest way to represent the corresponding quantum mechanical system is write the Schrodinger equation for wavefunction, $\Psi(\vec{r},t)$. Explicitly:

\begin{equation} i \hbar \frac{\partial \Psi }{\partial t} = H \psi = -\frac{\hbar^2}{2m}\nabla^2 \Psi + V(|\vec{r}|) \Psi \end{equation}

Answer 3

Your question is ambiguous and some of the answers given above are extremely misleading.

In quantum theory the motion of measurable quantities is represented by operators called observables. In the Heisenberg picture an observable $\hat{A}(t)$ evolves according to the equation: $$i\hbar\frac{d\hat{A}(t)}{dt}=[\hat{A}(t),H]$$ where $H$ is the Hamiltonian.

One answer above suggested that the Ehrenfest theorem and the Correspondence principle explains how to get classical behaviour from quantum systems. There are many papers explaining that this is false, e.g. - Section III of this paper:

https://arxiv.org/abs/quant-ph/0105127

Rather when information about a quantum system is copied out of it, quantum interference is suppressed: this effect is called decoherence. Large systems like grains of dust and people are monitored on timescales much smaller than their motion and this makes their equations of motion effectively classical on the relevant scales of time and space:

https://arxiv.org/abs/1911.06282

So if you want classical equations of motion, then they are a result of decoherence.