- #1

sergiokapone

- 302

- 17

\begin{equation}

-\frac{\hbar^2}{2m} \partial_i^2\psi = i\hbar\partial_t \psi.

\end{equation}

Let us be interested in the motion of a free particle in quantum mechanics. We say ok, we have a solution to the Schrödinger equation for a free particle (1) with momentum ##\vec{p}## and energy ##E = \frac{p^2}{2m}##

\begin{equation}

\psi = Ae^{-\frac i\hbar(Et - p_ix_i)}.

\end{equation}

But this solution is generally of little use, since it does not carry any useful information at all, since it gives the same probability of detecting a particle anywhere.

Second example. If we consider the problem of the state (I do not say motion) of a particle in a central field, then we must solve the Schrödinger equation:

\begin{equation}

-\frac{\hbar^2}{2m} \partial_i^2\psi - \frac{e}{r}\psi = i\hbar\partial_t \psi.

\end{equation}

When we want to define a state with a well-defined energy (stationary states), we easily separate the phase factor ##e^{\frac{i}{\hbar} E t}##

\begin{equation}

\partial_i^2\chi - \frac{2m}{\hbar}\left(E + \frac{e}{r}\right)\chi = 0.

\end{equation}

Let me remind you that for the ground state we have the solution:

\begin{equation}

\psi = \sqrt\frac{1}{\pi}\frac{1}{a^{3/2}}e^{r/a}e^{-\frac{i}{\hbar} E t},

\end{equation}

where ##a## is the Bohr radius, ## E = - \frac {me ^ 2} {2 \hbar ^ 2} ##.

Again, did we get a motion? Of course not, we can only find out where the electron can be with one or another probability. Yes, the electron does not move, since it is in a stationary state. And naturally, the paradox with the emission of an electron is solved in quantum mechanics right away - since the electron does not move, it means that it does not emit.

In order for there to be movement, there must be a trajectory. It is obvious that classical trajectories are an emergent phenomenon of quantum mechanics. We need to understand how quantum mechanics, with its wave functions, manifests in the classics what we call motion and trajectory.