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The Schrodinger equation: $$-\frac{\hbar^2}{2m}\nabla^2\Psi(r)+V(r)\Psi(r)=E\Psi(r)$$ $$\textit{kinetic energy} + \textit{potential energy}=\textit{total energy}$$ 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.

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  • $\begingroup$ Hi and welcome to physics.SE! Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – ACuriousMind
    Sep 25, 2020 at 14:24
  • $\begingroup$ What do you mean by "type" of potential energy here and who calls this "bhom quantum potential" (what's "bhom"?)? $\endgroup$
    – ACuriousMind
    Sep 25, 2020 at 14:26
  • $\begingroup$ Should be Bohm quantum potential. Fixed. $\endgroup$
    – gandalf61
    Sep 25, 2020 at 14:35

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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.

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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, $$ V(x) = \frac{1}{2}m\omega^2x^2 $$

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.

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    $\begingroup$ It's always intrigued me that quantum physics is so radically different from classical physics yet the same classical potential functions can be used in the Schrodinger equation. Is there an intuitive explanation for that? $\endgroup$ Sep 25, 2020 at 23:30
  • $\begingroup$ The laws of interaction between physical entities are independent of the scale of the problem (at least for everything above the Planck scale). That is an explanation why the potentials, which are more or less a more tractable mathematical form of those laws, work in both cases. $\endgroup$
    – Hans Wurst
    Sep 26, 2020 at 9:40
  • $\begingroup$ @HansWurst Can we also say that it is a natural consequence of quantising the classical Hamiltonian? $\endgroup$ Jun 11, 2021 at 5:33
  • $\begingroup$ @IndischerPhysiker My familiarity with quantization procedures starting from classical equations is limited so I'm not able to give a direct "yes or no" answer to that. $\endgroup$
    – Hans Wurst
    Jun 11, 2021 at 21:51
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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 $$ V(\mathbf{r}) = -e\varphi(\mathbf{r}), $$ 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.
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    $\begingroup$ "The potential energy ... is the electrostatic one": no, it's not. It can be virtually anything you want (e.g. square potential wells). Whether it corresponds to a physically realizable situation is another matter. But even unphysical potentials can be rough approximations to physical situations. The point is that in most cases, you impose a potential (with no indication of how it is produced) and you solve the Schrödinger equation for that given potential. In particular, the electron (or whatever) whose behavior you are trying to predict, is assumed to NOT affect the potential. $\endgroup$
    – NickD
    Sep 25, 2020 at 15:11
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    $\begingroup$ @NickD it is not about the shape of the potential, but about its nature. $\endgroup$ Sep 25, 2020 at 16:38
  • $\begingroup$ This is overly bold in ruling things out. For instance, it makes complete sense to talk about scattering an aplha particle off a nucleus (Rutherford experiment) and to model the interaction via scattering off an infinite spherical well where the potential involved is clearly a result of the strong force. It's not difficult to come up with other non-EM examples. $\endgroup$
    – jacob1729
    Sep 27, 2020 at 20:01
  • $\begingroup$ @jacob1729 I didn't rule this out in my answer. I even covered weak and strong interactions in some details. $\endgroup$ Sep 27, 2020 at 20:27
  • $\begingroup$ My objection to the discussion of weak/strong forces is that it certainly reads like you are dismissing them as unlikely. It's also true that gravitational potentials can appear in the SE, see Collela R, Overhauser A W and Werner S A 1975 Phys. Rev. Lett. 34 1472. $\endgroup$
    – jacob1729
    Sep 27, 2020 at 21:59

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