[Physics] $V(x)$ in Schrödinger’s equation

potential energyquantum mechanicsschroedinger equation

In the time-independent Schrödinger equation it is stated that

$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$$

And it is common to give $V(x)$ some standard "forms": the infinite well, the finite square well, etc.

The problem is that I'm finding it pretty difficult to get what $V(x)$ actually is. I've read in books that it is the potential energy, but…

As far as I know, the potential energy depends on the properties of the particle itself (its mass, its charge), but, for instance, the finite square well is defined as

$$V(x) =
\left\{
\begin{array}{ll}
-V_0 & \mbox{if } -a<x <a \\
0 & \mbox{if } |x| >a
\end{array}
\right.$$
where $V_0$ is a positive constant. In this case, $V_0$ seems to be fixed, but wouldn't it depend on the properties of the particle?

Apart from that, I don't know how to understand what these wells actually are. Why is a particle uncapable of, for example, escaping from an infinite well? A concrete and intuitive example of this would be appreciated.

Best Answer

As far as I know, the potential energy depends on the properties of the particle itself (its mass, its charge)

No, not really. Think of a classical mass-spring system. The potential energy is $$ V(x)=\frac{1}{2}kx^2 $$ which is independent of the properties of the mass. In some other cases, such as a charged point particle, the potential energy could be $$ V(x)=\frac{e^2}{r} $$ and in this case it does depend on the charge of the particle.

Apart from that, I don't know how to understand what these wells actually are. Why is a particle incapable of, for example, escaping from an infinite well? A concrete and intuitive example of this would be appreciated.

Wells don't exist in Nature. They are used to mathematically model cases where the actual potential is unknown or where it's too complex to be analytically solvable.

But particles can escape from wells: the probability of scape is given by the transmission coefficient: $$ T\sim \text{csch}\left[a\sqrt{2m(E-V_0)}\right] $$ which is in general non-zero. Only in the case of infinite $V_0$ we get $T=0$, but then again this is not physical: infinities don't exist in Nature.

As an example where finite wells are used, you can find a lot of information online. Google nuclear shell model and you'll find that, as the potential energy of nucleons is unknown, people use the finite well approximation to get a qualitative description of the nucleus of atoms.