# Electrons - What is Waving?

If an electron is a wave, what is waving?

So many answers on the internet say "the probability that a particle will be at a particular location"... so... the electron is a physical manifestation of probability? That doesn't sound right.

This page http://mwolff.tripod.com/see.html seems to suggest that the spherical wave pair is what gives the electron the properties of a particle. I could be misinterpreting, but this answer makes more sense to me - if only I understood what was waving!

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Y'know, I'm not convinced that the author of that web page you linked to really knows what he's talking about. Sure, you can represent the wavefunction of an electron in an atom (or any bound spherically symmetric system) in terms of spherical waves, but there's nothing particularly magical about that representation as the web page seems to be getting at. –  David Z Mar 5 '12 at 4:05

In QM, a "wave" isn't what we normally imagine: something that moves up and down and moves in one direction, like water. It's just a function that evolves with time and has a (in general) different value at every point in space. See this applet for some examples of atomic orbitals which are infact electron wavefunctions (the applet actually shows the absolute value squared $|\psi|^2$ of the wavefunction; or the probability density). The wave does not "exist" per se in physical space. It can be drawn (superimposed) on physical space, but that just means that it has a value at every point there.

The wave associated with an electron shows the probability of finding it at a particular point in space. If an electron is moving, it will have a "hump" in its vicinity, which shows it's probability at every point in time. This hump will move just like the electron does. For more info on this (though you may have read stuff like this before), see the "why don't they need to be close" section of this answer. When you observe the electron, you collapse the hump to a peak. This peak is still a wave, just narrowly confined so it looks like a particle.

Your issue is that you're trying to look at the "electron" and "wave" simultaneously. This isn't exactly possible. The wave is the particle. You can look at it as if you exploded the electron into millions of fragments and spread it out over the hump. There is a fraction of an electron at every point. The fraction corresponds to the probability of finding it there. At this point, there is no electron-particle. So there's nothing that's "waving". Of course, we never see a fraction of an electron, so these fellows clump together the minute you try to make an observation.

Edit by OP -- This is the section referenced above that I found most helpful

Quantum mechanics has a nice concept called wave particle duality. Any particle can be expressed as a wave. In fact, both are equivalent. Exactly what sort of wave is this? Its a probability wave. By this, I mean that it tracks probabilities.

I'll give an example. Lets say you have a friend, A. Now at this moment, you don't know where A is. He could be at home or at work. Alternatively, he could be somewhere else, but with lesser probability. So, you draw a 3D graph. The x and y axes correspond to location (So you can draw a map on the x-y plane), and the z axis corresponds to probability. Your graph will be a smooth surface, that looks sort of like sand dunes in a desert. You'll have "humps" or dunes at A's home and at A's workplace, as there's the maximum probability that he's there. You could have smaller humps on other places he frequents. There will be tiny, but finite probabilities, that he's elsewhere (say, a different country). Now, lets say you call him and ask him where he is. He says that he's on his way home from work. So, your graph will be reconfigured, so that it has "ridges" along all the roads he will most probably take. Now, he calls you when he reaches home. Now, since you know exactly where he is, there will be a "peak" with probability 1 at his house (assuming his house is point-size, otherwise ther'll be a tall hump). Five minutes later, you decide to redraw the graph. Now you're almost certain that he's at home, but he may have gone out. He can't go far in 5 minutes, so you draw a hump centered at his house, with slopes outside. As time progresses, this hump will gradually flatten.

So what have I described here? It's a wavefunction, or the "wave" nature of a particle. The wavefunction can reconfigure and also "collapse" to a "peak", depending on what data you receive.

Now, everything has a wavefunction. You, me, a house, and particles. You and me have a very restricted wavefunction (due to tiny wavelength, but let's not go into that), and we rarely (read:never) have to take wave nature into account at normal scales. But, for particles, wave nature becomes an integral part of their behavior. --Manishearth Feb 14, 2012

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The applet displays charge densities (i.e., absolute values squared of the wave functions), not the wave functions themselves. –  Arnold Neumaier Mar 3 '12 at 16:33
Aah good point.. I'll edit that in.. –  Manishearth Mar 3 '12 at 16:45
It is important when answering questions like this to say that two electrons are described by a wave in 6 dimensions, so that the wave is not in physical space. This usually resolves the confusion. –  Ron Maimon Mar 4 '12 at 2:16
@RonMaimon I'll add that, thanks. –  Manishearth Mar 4 '12 at 2:17
"Why we don't need to be close" proved to be exactly what I needed. Thanks for the reference. I hope you don't mind, but I want to append a block quote from that answer to yours for completeness. –  Brien Malone Mar 5 '12 at 3:33

In QM, the "waves" lie on the Hilbert space, that is not our "real life" space (where we live). All we can do with this "wave", or with this wave vector (that could be called $\psi(x)$ or $|\psi \rangle$, depending on your choice of representation), to make prediction about the properties of a particle (here on your real lifes), like a electron in your question, its to calculate its square modulus:

$|\psi(x)|^2 \mathrm{d}x \equiv P(x, x+\mathrm{d}x)$,

where $P(x, x+\mathrm{d}x)$ is the probability to find the electron between $x$ and $x+\mathrm{d}x$, if you want to measure the position of the electron; the rule above its called the Born Rule.

Well, this said, it's time to go to your question: "If an electron is a wave, what is waving?".

First of all, what determines the wave or corpuscular behavior of your subject of study its the type of experiment that you do; but, the outcomes (the results of the measure process) will always be probabilistic. So, in QM, we "accumulate statistics", to make a histogram and compare with the prediction of the Born rule to the system in question.

Therefore, the electron could be a wave (existence of interference of electrons) or a particle (existence of quantas), depending on what you want to know about it. Both behaviors are complementary, that is, with the two descriptions the properties of the electron can be achieved completely, but the two cannot be measured simultaneously.

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According to quantum electrodynamics (QED), which encodes the properties of electrons and photons, electrons are excitations of an electron fiueld in the same way as photons are excitations of the electromagnetic field.

The fields wave, and the electrons (or photons), as far as they can be considered to be particles, are localized wave packets of excitations of these fields.

The particle interpretation is appropriate, however, only to the extent that the so-called geometrical optics approximation is valid. This means, in a particle interpretation, you shouldn't look too closely at the details, as then the particle properties become more and more fuzy and the wave properties become more and more pronounced.

But if you just look at quantum mechanics (QM) rather than QED, your question cannot be answered as the wave function is something unobservable, existing only in an abstract space,

What can be given an interpretation in QM are certain things one can compute from the wave function. The stuff of interest to chemists is the charge distribution, given by $\rho(x)=e|\psi(x)|^2$ for a single electron, by $\rho(x)=\int_{R^3} ~dy~ e|\psi(x,y)|^2$ for a 2-electron system, etc.; here $e$ is the electron charge. For electrons in a molecule, nothing is waving here anymore, as the wave aspect is eliminated by taking the absolute values.

Indeed, if this charge density is concentrated in a tiny region, one sees the particle aspect of electrons; if it is very spread out, one sees the wave aspect, revealed by high frequency oscillatory patterns in the charge density.

This is the chemist's interpretation. See Chapter A6:The structure of physical objects of A theoretical physics FAQ.

Physicists (especially if not well acquainted with the use of charge density information) are often brainwashed by the teaching tradition, and then think and express everything in terms of probabilities, giving QM an unnecessary flair of mystery.

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When the charge is multiplied by the probability density one gets the charge density. This is only true in the limit of large number of particles. Otherwise one cannot interpret that multiplication as a charge density for 1 electron say because it would mean that the electron is an extended object which is not! –  Revo Mar 3 '12 at 20:52
Electrons in an atom are indeed extended objects. Otherwise you'd have to explain the least number of electrons to make the charge distribution describable by my recipe. There is no such least number. –  Arnold Neumaier Mar 3 '12 at 23:05
@Revo: The first paragraph in mat.univie.ac.at/~neum/physfaq/topics/atoms contains the quote of a published statement by a famous physicist (with reference) that explicitly supports my reading of the matter: ''The product of -e and |psi(x)|^2 is usually interpreted as charge density, because the electrons in an atom move so fast that the forces they exert on other charges are essentially equal to the forces caused by a static charge distribution -e|psi(x)|^2.'' –  Arnold Neumaier Mar 3 '12 at 23:11
A single electron is not a smeared object. Let me quote Feynman: "the wave function of an electron in an atom does not, then, describe a smeared-out electron with a smooth charge density. The electron is either here, or there, or somewhere else, but wherever it is, it is a point charge." He goes on further on explaining this which can be found in student.fizika.org/~jsisko/Knjige/Opca%20Fizika/… in section 21-4 (2nd paragraph). –  Revo Mar 4 '12 at 4:30
Feynman just exposes his beliefs. His belief cannot be checked, as observation changes the picture. My exposition is of the chemist's beliefs, who universally treat electrons as smeared objects. Famous physicists share this much better founded belief, as the citation in the previous comments shows. (Jochen Mannhart is one of the 10 winners of the Leibniz prize 2008, dfg.de/aktuelles_presse/preise/leibniz_preis/2008 among others for the achievement that, for the first time, he made pictures of atoms with subatomic resolution possible.) –  Arnold Neumaier Mar 4 '12 at 11:36