The wavefunctions for atomic orbitals have always been described to me one of two ways:

  1. As a "smeared out" electron standing wave with integer number circumference of de Broglie wavelengths

  2. As a "probability cloud" wherein a point particle electron is likely to be observed.

While those 2 interpretations have never seemed compatible to me, one question seems incompatible with BOTH of these interpretations:

If an electron does not orbit classically, i.e. is not an accelerating point charge in a circular orbit, how does it create a magnetic field of magnitude equivalent to a current carrying loop of orbital circumference?

In other words, why does an electron that is not moving classically appear to produce a magnetic field one would expect from classical movement?

I'm having difficulty understanding how either a "smeared out" or "probability distribution" interpretation of electron orbitals under QM accounts for the non-spin related magnetic moment of an electron, just as if that electron were a point charge in a circular orbit, which QM emphatically refutes?


4 Answers 4


It's neither. Those descriptions are just mental imagery for beginners. You need not waste any time on them. All of reality obeys the laws of quantum mechanics. What we call classical physics is just an approximation of quantum mechanics under certain circumstances.

This is important: classical physics derives from QM and NOT the other way round.

You have to stop thinking in terms of shiny classical billiard balls jumping around in some weird way and you have to start thinking about a complex field like object, that, when looked at from 30,000 feet, can produce distributions of expectation values that may look like particle tracks or electromagnetic waves.

For low energies the interactions of the different components of this field, i.e. those that stand for electrically charged fermionic states ("electrons") and those of uncharged bosonic states ("photons") can, on average, reproduce the usual phenomenology for electrons interacting with electromagnetic fields like Lorentz forces.

That is only a small fraction of what the QFT model can reproduce, though. It can also calculate magnetic moments with high precision, explain the hyperfine structure of atomic transition, explain matter and anti-matter and describe its production in e.g. pair-production processes and then some. Taken together, the most complete model that we have, describes almost every aspect of elementary particle physics and motivates the behavior of a wide variety of nuclear phenomena.

  • $\begingroup$ Thanks for the detailed response - I thought I was getting away from the particle view by embracing the idea of a standing electron wave, but I hit a mental brick wall when I try to envision how a non-local object produces magnetic moment that seems to be best described by a discrete object moving classically in a circular path. It seems such an object should radiate EM radiation which is exactly why the Bohr model was abandoned - in other words I'd gladly embrace a different way of thinking about electrons than point particles but the (non-spin) magnetic moment calculation seems to demand it $\endgroup$
    – JPattarini
    Dec 29, 2014 at 14:32
  • 2
    $\begingroup$ @JamesPattarini: The standing wave model is not quite sufficient to make the theory self-consistent. For one thing it does not cover all the properties of a quantum field. There is really no way getting around the definition of something fundamentally new (i.e. the quantum field), which in one approximation (large numbers of particles) has wave-like and i another (weak measurements) has particle track like properties. The good news is that once we understood that, the early ontological questions fell away completely. The bad news is that this new object is mathematically hard to deal with. $\endgroup$
    – CuriousOne
    Dec 29, 2014 at 15:53
  • $\begingroup$ I think my hangup is on the "realness" of the wavefunction, since it's been beaten into me that it's a mathematical tool and not something real in and of itself, if picturing the 3D Schrodinger solutions for orbitals is not "really" what's going on then what CAN I picture in my mind's eye, or put another way: considering how perfectly the orbital picture describes physical atomic bonding behavior, why don't we consider the orbitals "real?" I imagine it's because treating the wavefunction as real in other scenarios seems nonsensical? $\endgroup$
    – JPattarini
    Jan 5, 2015 at 16:14

Actually, in a sense, there does exist something which moves. It's not something which can be measured to move, but rather part of quantum mechanical description, evolution of which much resembles classical electron motion: the phase of wavefunction.

See this java demonstration of hydrogen atom orbital. If you select "Complex Orbitals (phys.)" and then choose a state with $m\ne0$, then you'll see how the colours "rotate". This shows that the phase of the wavefunction does rotate around the $z$ axis. In a sense, this can be said to create the orbital magnetic moment.

But please note, I'll repeat: this is not an observable rotation by itself! Moreover, the angular velocity of phase is actually not fixed - it can be shifted by arbitrary constant (provided it's the same for all the states). But it does show how the wavefunction evolves in time. If you try to measure anything, all you can get is the magnitude squared of the wavefunction (for a large enough number of identical experiments). Still, the phase does play a role in creation of superpositions of states and finally in allowing the probability densities change in time, while for stationary states the probability densities are constant in time, as you already know.

EDIT in response to comment:

First, the wavefunction is not real in the sense $\psi\not\in\mathbb R$. It's complex, while all the directly measurable quantities must be real.

Second, despite in eigenstate the atom is stationary, we can make linear combinations of eigenstates with the same direction of magnetic moment such, that the result will be a wave packet measurably rotating in a definite direction.

For a simpler example of translational motion instead of rotational, consider a plain wave. Here's how it looks (blue real part, purple imaginary, yellow square of magnitude):

enter image description here

Now let's make up a Gaussian wave packet, like in this demonstration. It'll have the same peak state, but additionally some more states with higher and lower phase velocities (with smaller amplitudes). Here's its wave function (blue real part, purple imaginary):

enter image description here

Now this packet has measurable motion. Here's its square of magnitude, which is the probability density by Born rule:

enter image description here

You see that for a particle eigenstate, which is a plane wave, you can't measure its motion. But OTOH, it's the limit of infinitely delocalized wave packet, and for a localized wave packet you can measure its motion (in probabilistic sense, of course).

The same is for rotational motion, it just has some more technical features, which aren't relevant for understanding here.

  • $\begingroup$ I feel like I'm 90% there but hanging up on this: If the wavefunction is not something physical in and of itself and merely the mathematical tool we use to describe the system we're looking at, I have trouble seeing how the change of phase in the WF can result in the physical property of magnetic moment - in other words the WF describing electron orbitals certainly seems VERY REAL and produces measurable consequences in atomic bonding behavior, magnetic moment, etc however we interpret it as non-real. Why? $\endgroup$
    – JPattarini
    Jan 5, 2015 at 16:06
  • $\begingroup$ @JamesPattarini I've updated my answer to address your comment. $\endgroup$
    – Ruslan
    Jan 5, 2015 at 17:26
  • $\begingroup$ @JamesPattarini: The wavefunction (in at least some sense) is real, as it can alter the results of measurements. However, if the wavefunction is multiplied by $e^{ix}$, where x is some real, the probabilities of the outcomes for every measurement will be the same. There exists a good analogy with symmetries in classical mechanics: the following two sentences describe exactly equivalent two physical systems. 1) Particle A moves with velocity $1m/s$ and B with $2m/s$ to the left. 2) Particle A moves with velocity $5m/s$ and B with $6m/s$ to the left. $\endgroup$
    – kristjan
    Jan 5, 2015 at 17:53
  • $\begingroup$ @Ruslan Serious thanks for the detailed explanation, and the visuals help with the conceptualization. If you have any insight on physics.stackexchange.com/questions/156746/… it would be greatly appreciated. $\endgroup$
    – JPattarini
    Jan 8, 2015 at 17:11

Although the standing wave is indeed a stationary state, don't make the mistake of thinking this means the electron is not moving. As you say, we know the electron is moving because there is a very real electric current and magnetic moment. So the electron is moving and its wavefunction is not moving.

The confusion comes from our everyday experience, where inertia and dynamics always come together in one package, which we call "motion". But with quantum mechanics we must separate these ideas. What I mean is, a quantum wave can contain motion (inertia) without itself being in motion (dynamic).

[By the way, of the two viewpoints you note, the probability cloud viewpoint is a much weaker view. I mean, from the wave function you can calculate the probability cloud, but the reverse is not possible.]

  • $\begingroup$ Many thanks for the quick answer - for clarification, when you say "the electron is moving, but the wavefunction is not moving" how best to picture what the electron is "actually" DOING in this case? I ask because the whole issue with the Bohr model of imagining a tiny electron ball wizzing around a nucleus was abandoned because such an electron should radiate EM radiation - QM fixed that (I thought) by doing away with the idea of classical electron motion. So saying that the electron IS moving leaves me asking "in what manner should I conceive of the electron moving?" $\endgroup$
    – JPattarini
    Dec 29, 2014 at 13:54
  • $\begingroup$ I find it hard to visualize, too. But really you have to abandon any hope that your visual picture will involve a particle in a well-defined place. I like to think of the electron standing wave as a smeared electron where it is simultaneously on all sides of the nucleus. It doesn't radiate since the charge and current patterns are steady. But even this view is limited. Another way to look at it is that the electron is a point particle but it is superposed on all sides of the nucleus---each superposed copy does radiate but the radiation is destructively interfered in the superposition. $\endgroup$
    – Nanite
    Dec 31, 2014 at 15:27

The standing wave describes the relative electron-nucleus motion. The electron itself moves since the atomic center of mass (a plane wave) involves the electron coordinates too.


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