What does $V(r)$ mean in the Schrodinger equation?

Question

The Schrodinger equation:
$$-frac{hbar^2}{2m}nabla^2Psi(r)+V(r)Psi(r)=EPsi(r)$$
$$textit{kinetic energy} + textit{potential energy}=textit{total energy}$$
Is one of my favourite equations, but there's one term I don't understand: the  $V(r)$ term which is supposed to mean potential energy... but what type of potential energy? In the classic 0 potential well example they say that the potential out side the well is infinite but what type of potential energy are they talking about? I googled it and its also called Bohm quantum potential but I really don't get what it means. Anything would be a great help.

Hans Wurst · Accepted Answer

There is not one single special potential function, rather the opposite. The potential function is a placeholder that takes a different functional form depending on what kind of physical situation you want to model. The physics and the system that we want to describe goes into the Schrödinger equation via this potential function.
The only information that the equation  gives you written in this way is the fact that it has to be a function depending only on the position variable. The function $V(x)$ may not depend on derivatives of $x$ for example.
Some basic examples for potentials are the particle in a box potential,
$$
V(x) = cases{ 0,  -L/2 < x < L/2 \infty,  textrm {otherwise} }
$$
With this we can model situations where a particle can move freely in a certain area, but unable to escape.
Another potential would be a harmonic potential,
$$
V(x) = frac{1}{2}momega^2x^2
$$
With this we can model situations where a particle is for example resting in a local minimum that looks like a parabola. This can describe for example molecules in their stable groundstate geometry. Another example that is described by a harmonic potential would be the time dependent amplitudes of the electromagnetic vector potential.
Potential functions are also often so complicated that we are only able to obtain approximate solutions.

Vadim · Answer

The potential energy in Schrödinger equation is the electrostatic one. Here are a few points to note:

Since the Schrödinger equation is used on a microscale, we exclude the "microscopic" kinds of potential energy familiar from Newtonian mechanics, such as, e.g., the "elastic potential energy", which are really result of the electrostatic interaction between many particles.
This leaves us with four fundamental interactions acting on the particle level: electromagnetic, strong, weak and gravitational.
I am not sure whether there is a generally accepted theory of gravity
on quantum level, so I would say that gravitational forces never
appear in the Schrödinger equation.
Weak and strong interactions, in principle, could appear in the Schrödinger equation, but a) they are rarely reducible to a purely potential interaction, b) they are usually treated using more sophisticated mathematical techniques, and c) they are often treated in relativistic limit, wwhere the Schrödinger equattion does not apply.
This leaves us with electromagnetic interactions, i.e., the scalar and the vector potentials. Thus, the potential energy in question is
$$
V(mathbf{r}) = -evarphi(mathbf{r}),
$$
since the particle in question is usually an electron.
This is sufficient for describing the physics of atoms and condensed matter in non-relativistic limit, and taking properly into account the exchange interaction.

JoshuaTS · Answer

It's just a normal potential energy function. It could be gravitational potential energy, electric potential energy, or any other kind of potential energy from classical mechanics. The way that the particle reacts to the potential energy will be different, but the form of $V(x)$ is exactly the same.
For the infinite square well, we are not really concerned about what is causing the potential. It's mostly used as a teaching example, but it could provide a simplistic model for a particle strongly confined to a region by any type of potential energy.

What does $V(r)$ mean in the Schrodinger equation?

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