The potential well for a single atom, e.g. enter image description here

is basically a graphical representation of the possible electron positions at various energy levels. The horizontal axis represents distance from the nucleus of the atom. For higher energy levels (increasing $n$), we generally expect an electron within that energy level to be further from the nucleus than an electron at a lower energy level.

The square of the magnitude of the wave function $\psi(x,t)$ (from the Schrödinger equation) gives the probability density associated with finding an electron at a certain position $x$ from the nucleus. This is also sometimes plotted on the potential well (but for the above picture $\psi(x,t)$ is shown because the square of the function should be symmetrical about the central vertical line). We can observe that an electron at a particular energy level (e.g. $n = 1$) can, at one instance, be closer to, or further from, the nucleus than if we were to measure the distance at another instance, and this is based on $\psi(x,t)$ for that particular energy level. An electron at a higher energy level (e.g. $n = 2$) will have an increased "maximum" distance from the nucleus (represented on the potential well diagram), but can sometimes be closer to the nucleus than an electron at a lower energy level, but the probability is again based on the wave functions $\psi(x,t)$ for the two energy levels. More likely than not an electron at a higher energy level will be further away than an electron at a lower energy level.

In reality, the energy levels are split into subshells and we represent 4 different subshells with the letters s, p, d, and f. In a Cartesian coordinate system, we can use 3 different potential well diagrams to represent each spacial dimension. The aomic orbital for any given subshell is a region of space where an electron of that subshell has a 90% chance of occupying at any time. The atomic orbital for the s subshell is spherical (or dome shaped) because the wave function is the same for each spacial dimension, but, for other subshells such as p, the wave function is not the same for every dimension so their atomic orbitals appear to extend in particular directions.

So basically my question is that, based on my explanation, am I right in my understanding, or am I way off? If there is anything that wasn't quite right, please feel free to correct me. I also appreciate additional explanatory comments in general.

Right now, I don't really know much about the Schrödinger equation. As far as I am aware $|\psi(x,t)|^2$ gives the probability density of finding an electron at a certain position, but I don't know how it came about, how to interpret the Schrödinger equation, and how to solve for $\psi(x,t)$. So I plan to do more digging on this. However it would also be great if anyone can provide a dummy explanation to help me understand the Schrödinger equation.

  • $\begingroup$ There's no maximum distance from the nucleus in quantum theory of atoms. It can be anywhere. You need to discard all classical ideas of orbits. $\endgroup$ Sep 22, 2017 at 20:47
  • $\begingroup$ Oh I see, thanks. What do the "walls" represent when we plot a Potential Well? $\endgroup$
    – Royalrange
    Sep 22, 2017 at 20:53
  • $\begingroup$ They're not walls in the quantum world. They're just the value of the potential field at given radii. And that's all they are. As quantum tunneling shows, potentials are not walls. $\endgroup$ Sep 22, 2017 at 21:23
  • $\begingroup$ If an electron were a circular standing wave around the nucleus, the length of the orbit would contain a whole number of wavelengths. This view is incorrect, because electrons are clouds and there are no orbits, but it gives you an intuitive feeling of why the wave mechanics produces discrete levels. $\endgroup$
    – safesphere
    Sep 22, 2017 at 22:00
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    $\begingroup$ Please read our FAQ on writing good question titles. $\endgroup$
    – DanielSank
    Sep 22, 2017 at 22:46

2 Answers 2


I've read your explanation, and I agree with it. As someone who is formally studying physics, I hope I can help.

Think of the electrons creating a cloud. You can't make out exactly where they are, but they are in this region. The cloud becomes thicker the closer to the center you get. The higher energy electron spend more time further away from the nucleus. In this sense, they could be considered further away.

There are two forms of the Schrodinger equation, time dependent and time independent. You wrote |ψ(x,t)|2, which is time dependent. However, orbitals are an example of time independent states. The relation given in the graph you posted is strictly spatial, meaning there is no change with time of this system. Since you said you are new to this, I imagine you are actually studying time independent states. Time dependent states can become much more complicated. I believe I didn't study them until I was in graduate level classes.

ψ represents the particle. In the Copenhagen interpretation, this is the most complete description of a physical system. It describes how a particles wave equation (also called state function) changes with time and space. To find this, you solve the Schrodinger equation. The Schrodinger equation is like the F=ma of quantum mechanics. For the most part, you will just plug in what you need the equation and solving it will yield ψ.

The Schrodinger equation came about when more evidence was being presented about wave-particle duality. Schrodinger thought that if particles could behave as waves, there should be an equation that describes them as such. Using other equations for inspiration, Schrodinger was able to come up with a description of particles using a wave equation. Schrodinger was able to calculate the spectral lines of hydrogen using this equation. The Schrodinger equation isn't the only way to calculate the states of quantum systems, but it is a very useful one.

  • $\begingroup$ "Using other equations for inspiration, Schrodinger was able to come up with a description of particles using a wave equation" - It may be worth mentioning that the Schrodinger equation comes from quantizing the classical Hamilton principle of least action by replacing energy with the energy operator. Therefore the Schrodinger equation is simply the least action principle restated in the language of quantum mechanics. $\endgroup$
    – safesphere
    Sep 22, 2017 at 21:10
  • $\begingroup$ @safesphere I agree with that statement. The user said he was an electrical engineer, so I didn't think that information would really help his understanding of the situation. He did ask for a "dummy explanation". Or perhaps it is helpful to him. $\endgroup$
    – Hugger
    Sep 22, 2017 at 21:16

Actually the plots are rather misleading as they only show the part of the wave functions that are in the classical region, where the solutions are oscillatory in nature. Upon reaching classical turning points of motion the solutions decay exponentially.

Also, the small $r/r_0$ the behaviour of the wave function for the hydrogen atom is $R(r)\sim (r/r_0)^{\ell}$ so the probabilty distribution "close" to the nucleus is largely determined by the angular quantum number $\ell$ rather than the principal quantum number $n$. Classically this is because the effective potential contains a centrifugal part which pushes the particle away from the origin when $\ell\ne 0$. In the Coulomb problem, this part is dominant at small distances. For the hydrogen atom, the tail of the wave functions depends on the energy and is of the type $\sim e^{-r/(nr_0)}$, with $r_0$ the Bohr radius.

Schrodinger was apparently inspired a mixture of optics and classical mechanics. In its simplest incarnation the Hamilton-Jacobi equation of classical mechanics is given by $$ H\left(q,\frac{\partial S}{\partial q}\right)=E $$ Schrodinger replaced the function $S$ by $k\log\psi$ to obtain $$ H\left(q,\frac{k}{\psi}\frac{\partial \psi}{\partial q}\right)=E $$ which, after straightforward manipulations and appropriate identification, lead to variational problem common in optics and mechanics and also used in the old Sommerfeld theory: $$ \delta J=\delta \int d^3 r\,\left((\nabla \psi)^2 -\frac{2m}{k^2}(E-V)\psi^2\right) $$ and (if I did not make any error) to a differential equation $\psi$ that is the time-independent Schrodinger equation. The final key step was to show that the spectrum of hydrogen was recovered by forcing $\psi$ to satisfy some boundary conditions, most notably at $\infty$. Fully time-dependent solutions are of the form $\Psi_n(r,t)=\psi(r)e^{-itE_n/\hbar}$. A good historical account can be found in the book by Max Jammer The conceptual development of quantum mechanics.

The hydrogen atom is an exceptional case in that multiple values of $\ell$ can produce the same energies. This energy degeneracy is actually good as it produced the correct number of states of a given energy, something the "old" Bohr model did not predict correctly. The values of $\ell$ predicted by Schrodinger for each energy level are in agreement with experiment. (The harmonic oscillator in $n>1$ dimensions is another notable potential with the property that energy levels do not depend on $\ell$.) In general however, you can expect the energies to depend on both $n$ and $\ell$.

If one applies Schrodinger's prescription to an atom with $2$ electrons, the solution contains $6$ spatial coordinates ($3$ for each electrons). Thus, $\Psi(\vec r_1,\vec r_2,t)$ cannot be a wave in ordinary $\mathbb{R}^3$ space. Moreover, even for a single electron, $\Psi(\vec r,t)$ will be generally complex. Born proposed that $\vert \Psi\vert^2$ - which is necessarily real and non-negative - be interpreted and treated as a probability density. This removes the two aforementioned problems in understanding "what" is $\psi(r)$ or $\Psi(\vec r,t)$.

Edit: As pointed out by @dmckee there is also the problem that the solutions are not $0$ at the turning points. Finally, for good measure, the number of nodes does not necessarily increase with energy. For hydrogen, the number of nodes is $n-\ell-1$: for a given $n$, the highest allowed $\ell$ is $\ell_{max}=n-1$ so in particular for $\ell_{max}$ the solution has $n-n+1-1=0$ nodes. In fact, for $\ell_{max}$ the probability density has a maximum at the value of $r$ predicted by the Bohr model.

The wavefunctions illustrated here are more appropriate for the 1d infinite well than the hydrogen atom, although the information and the form of the potential strongly suggest a Coulomb-type potential

  • $\begingroup$ "Actually the plots are kinda misleading [...]" They also show the amplitude going to zero at the classical limit of motion which is very wrong. $\endgroup$ Sep 23, 2017 at 16:10
  • $\begingroup$ @dmckee good point and completelly correct. $\endgroup$ Sep 23, 2017 at 16:14

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