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

Physics Asked on November 6, 2021

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

2 Answers

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}

Answered by Andrew on November 6, 2021

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: $$mfrac{dlangle hat{x}rangle}{dt}=langle hat{p}rangle, frac{dlangle hat{p}rangle}{dt}=-langle V'(hat{x})rangle.$$

Thus, for the central force we could write: $$frac{d^2langle hat{vec{r}}rangle(t)}{dt^2} = frac{1}{m}langle nabla_{vec{R}}V(R)|_{vec{R}=hat{vec{r}}(t)}rangle,$$ 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:

Answered by Roger Vadim on November 6, 2021

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