4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The potential energy term is just whatever it would be if you only knew about classical mechanics. If you're studying 2 charged particles, you get the usual Coulomb term. If you're studying a particle that classically can't escape because it would have negative KE, the potential is as per the classical model that gives that conclusion. The TISE shows that quantising with that potential actually allows a positive transition probability (unless the "barrier" is infinite, an unphysical model used when there's negligible chance of tunneling out, as in the "particle in a box"). $\endgroup$ – J.G. Apr 13 '16 at 17:31
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$V(x)$ is a potential energy function for the system of a particle or particles interacting with a set of constraints. These constraints can be thought of as fields which produce a force on the particle(s) of interest.

In the infinite square well (ISW), we examine a particle which has no interaction at all until it gets to some impenetrable constaint, i.e., and infinitely large barrier. Classically thinking, the particle experiences an infinitely large force for an infinitesimally small time, so that the resulting impulse is finite with the particle reversing direction. The ISW is a simply approximation to very strongly bound systems with short-range restoring forces. A carbon dioxide molecule might be approximated by the ISW problem, as well as other linear molecules.

Even in the finite square well, while a particle may have a small possibility of moving "into" the wall, it will eventually bounce back. For the particle to escape, energy must be added to the system from outside the analyzed system. It will not spontaneously "pop" out of the system.

For the electron in the hydrogen atom, $V(r)$ is the electrical potential energy of the system of the electron and proton due to the Coulomb attraction of the proton on the electron.

Radioactive decay involves the interaction of coulomb and nuclear systems which result in the transformation of mass-energy into kinetic energy. The mass of the system after the decay will be less than the mass of the system before the decay. The system thus has more kinetic + potential and can either escape the well completely or has a higher probability to tunnel through the constraint.

$\endgroup$
1
$\begingroup$

The potential energy definitely depends on the properties of the particle. For example, if you were using electric fields to contain the particle, the potential energy would depend on the charge of the particle; if you were using gravity to contain the particle, the potential energy would depend on the mass of the particle.

When there is just one particle in your system, this is still easy to write down. You can just write $V(r)=\frac{eq}{r}$, or $V(r)=\frac{GmM}{r}$, or whatever. When there are multiple particles, you have to keep track of each one; for example, if you had two particles with different charges, the total potential energy would be

$$ V(r_1,r_2) = \frac{e_1q}{r_1}+\frac{e_2q}{r_2} $$

In both cases, you make explicit reference to the properties of the particle when writing down Schrodinger's equation.

$\endgroup$
1
$\begingroup$

The potential $V(x)$ used in Schrodinger's equation is used to either model the motion of particles in real systems (in which case the parameters of the potential depend on the particle properties, from Coulomb forces where the property of interest is electric charge and spin or nucleon-nucleon interactions where isospin and spin is important) through to systems which are introduced merely because they admit analytical solutions (which may approximate real systems) and are useful to introduce concepts of quantum mechanics.

$\endgroup$

protected by Qmechanic Apr 13 '16 at 19:43

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.