# Is the electron actually a standing wave?

I studied in my physics class that de Broglie proposed that electrons are actually standing waves and that is the reason why their energy levels are quantised.

But I studied that the wave function of an electron is what we call the atomic orbital and wave functions of electrons come in various shapes depending on the energy levels (i.e. the eigenfunctions for the corresponding eigenvalues).

For instance, some wave functions are spherical in shape (s-orbitals), some are dumbbell-shaped (p-orbitals), etc. But if the wave functions of the electrons are of these shapes then how can the electron be a standing wave?

For e.g., if the wave function of an electron is spherically symmetric (s-orbital) that means that there is 99% probability of finding the electron in that spherical region. But if the electron can be anywhere in that 3-dimensional space then how can it behave like a standing wave as proposed by de Broglie?

This is because if the electron were a standing wave it would be a standing wave in its orbit and it will be a 2-dimensional 'thing'. But on the other hand we are also saying that the wave function of the electron is spherically symmetric and thus can be anywhere in the 3-dimensional space.

How can an electron be both a standing wave and have its wave function spherically symmetric? Is the electron even a standing wave? I am so confused. Can someone please provide the explanation?

• Beware that there are already related questions & answers. If you won't get a response dig more in this site. Possibile starting point physics.stackexchange.com/q/638189 May 22, 2021 at 13:01
• For instance, some wave functions are spherical in shape (s-orbitals), some are dumbbell-shaped (p-orbitals), etc. But if the wave functions of the electrons are of these shapes then how can the electron be a standing wave? Standing waves can take all kind of shapes. See e.g. the solutions of the Classic wave equation $\psi_t=c^2\nabla^2 \psi$ (plus boundary conditions).
– Gert
May 22, 2021 at 13:16
• Related/possible duplicates: physics.stackexchange.com/q/137207/50583, physics.stackexchange.com/q/196002/50583 and their linked questions May 22, 2021 at 13:25
• My understanding is that a "standing wave" doesn't have to be 2D. The functions involved describing the bounded electron are spherical harmonics, which are 3D standing waves. The 3D model is a more accurate model than the 2D model, because it handles all three dimensions and we live in a 3D world. The 2D model (developed by Bohr, de Broglie, etc) is just a crude approximation that was used when people didn't have a full understanding of quantum mechanics. May 22, 2021 at 13:32

The Bohr model is a primitive first-exploration of quantization model, . The de Broglie standing wave interpretation of the Bohr model is not the wave function calculated with the Schrodinger equation and the Coulomb potential. The wave in the wavefunction is a probability wave.

A probability means that many measurements should be taken of the electron to define its position in (x,y,z) at time t, i.e.to define the orbitals, here for hydrogen.

The electron itself is a point particle according to the standard model, not a wave. Depending on the boundary conditions a wave function can be calculated, as happens with hydrogen. If you look at the hydrogen orbitals (calculated locations from the wavefunctions where electrons can be bound) you will see that they are not all spherically symmetric.

• What I've never quite understood is that if you take the particle in a $1D$ box with zero potential (e.g.) the particle forms standing waves. The unit of measurement of the wave function $\psi$ is $\mathrm{m^{-1/2}}$ but the UoM of a wave would be $\mathrm{m}$?
– Gert
May 22, 2021 at 15:28
• @Gert The wavefunction in any quantum mechanical solution is $Ψ$, no units, and $Ψ^*Ψ$ is the probability for the reaction to happen, which is a number from zero to 1, so I do not understand where you find the units you are talking about. Are you mixing the de Broglie wave , a handwaving wave, with the solution of the QM equation, the wavefunction? May 22, 2021 at 15:39
• Wave function: $\psi_n(x) = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x$ from chem.libretexts.org/Bookshelves/… with $L$ in $\mathrm{m}$
– Gert
May 22, 2021 at 15:44
• @annav Your comment about units is incorrect. For a one-dimensional wave function $\psi(x)$, the normalized total probability must be $1 = \int_{-\infty}^{+\infty} \psi^*\psi\,\mathrm dx$. This integral is dimensionally inconsistent unless $\psi$ has dimension $(\text{length})^{-1/2}$. A probability density over an interval that’s not a one-dimensional length will have some different unit, but in general they’re not dimensionless.
– rob
May 22, 2021 at 15:46
• @rob thanks, one tends to see the wavefunction without any units, as in the hydrogen link I give, But I can see that one has to define the "length" on which the measurement of the probability is taken May 22, 2021 at 16:26

We often think of a standing wave as a one-dimensional function along a line - for example, a vibrating string or a sound wave in an organ pipe - or around a circle. But there are also two-dimensional standing wave functions that live on the surface of a sphere - they are called spherical harmonics.

The shapes shown for atomic orbitals for a hydrogen atom, for example (which are actually depicting the spatial distribution of the amplitude of an electron’s wave function) are based on these spherical harmonic functions.

• The 3D electron standing waves are not restricted to the surface of a sphere, but their resonant states might still be referred to as spherical harmonics. May 22, 2021 at 13:28
• I think this is a simple and clear answer but perhaps you could stress more that the standing wave of concern is the orbital intended as solution of the Schrödinger eq and not the orbital intended as the volume in which the probability of finding e is somewhat close to 1. It seems to me a point which wasn't clear to OP when s/he mixes the two saying about "99%.....“.Anyway plus 1. May 23, 2021 at 8:08

An electron is a (3D) standing wave only when is is bounded (as by the electric potential well of a nucleus). There is no rule that says a 3D standing wave must be spherical.

You are confused and I understand, because there is this wiki article that definitely says yes:

The electrons do not orbit the nucleus in the manner of a planet orbiting the sun, but instead exist as standing waves.

https://en.wikipedia.org/wiki/Atomic_orbital#Electron_properties

Though, as you can see from other answers, the electron cannot always be represented as a standing wave because:

Quantum objects are not waves. Quantum obejcts are not classical point-like particles. They are quantum objects, which may show wave-like and particle-like properties. You may represent a quantum state by its "probability wave" or wavefunction, whose square gives the probability density to find the object "as a particle" at certain locations. It is not a wave in the classical sense that anything physical would be oscillating here, and the Schrödinger equation does not always look like a wave equation. Electrons may be in more than one orbital at once, due to the general possibility of superposition of quantum states. But since the orbital are the solutions to the time-independent Schrödinger equation, being in one - and only one - orbital is the only stable state for an electron, while all other states will be changed by time evolution. The orbitals don't "interfere" because, well, they aren't actual waves.

Electron as a standing wave and its stability

1. an electron is a quantum object, it shows in certain experiments wave like properties, and particle like properties in others, but it is neither, in reality it is best described by QM.

2. the electron is not a wave in the classical sense, nothing is oscillating, and the Schrodinger equation cannot always be expressed as a wave equation.

3. the electron might be in superposition (more then one orbital), but being in only one is a stable state and all other states will be changed by time evolution, and the orbitals do not interfere, because they aren't waves.

So the answer to your question is that although for simplicity it might be advantageous to depict the electron in certain cases as standing waves, in reality the universe is fundamentally quantum mechanical and the atoms (and electrons) are modeled using QM resources (including the Schrodinger equation) which cannot always be expressed as wave equations.