4

I have a pretty good intuitive understanding of quantum mechanics. But one thing that I don't really intuitively understand is why electrons end up in bound states.

An electron might have some positional uncertainty, but no matter what that position is, it still experiences a force towards the nucleus. I would have expected, given my understanding of QM, that each individual possible position of the electron gets individually pulled toward the nucleus, and you end up with a coherent superposition of all of the possibilities. But, as the electron gets closer it also gets pulled with more strength, so I would have thought that essentially all the possibilities get pushed infinitely close to the nucleus. Yes, then that implies that there is maxium uncertainty in the momentum distribution, but how exactly does this uncertainty counteract the strong pull of the nucleus? I guess a bound state is just an equilibrium of the electron having its position wavefunction pulled to the center but having a component that is trying to escape due to propagation of the uncertainty in momentum?

Steven Sagona
  • 2,222
  • "that implies that there is maxium uncertainty in the momentum distribution" — what exactly implies this? The fact that you chose position representation doesn't make the final state have maximum uncertainty. You still integrate over all the position states to get a bound state, and this makes uncertainty finite. Also, if you try to watch evolution of one single position state from this superposition, you'll find that it expands in all directions really quickly (and also scatters on the nucleus), and you can't easily interpret motion of this wave packet any more. – Ruslan Mar 16 '22 at 16:58
  • @Ruslan, if an electron position wavefunction gets confined infinitely small, shouldn't its momentum wavefunction get infinitely large? I'm asking if it's the fact that momentum increases as its position gets more confined that's preventing it from becoming infinitely small. – Steven Sagona Mar 16 '22 at 19:05
  • Indeed if you know the position the standart deviation of momentum becomes Infinity – Marc Barceló Mar 16 '22 at 19:35
  • The fact that expected value of momentum magnitude increases leads to spatial spreading of initially localized wave packet as time goes on. Existence of this effect doesn't depend on any external potential like that of the nucleus: it happens even with a free electron. The potential just makes it possible to stop this expansion at some distance by reflecting the expanding wave back. This gives us the localized bound states. – Ruslan Mar 16 '22 at 20:15
  • 2
    Apart from the rather unhelpful attempt at "intuition", what in this question is actually about quantum mechanics? What is the difference between this and asking what keeps the Earth from falling into the Sun? Do you think the Earth needs "uncertainty in momentum" to keep "escaping the Sun"? – ACuriousMind Mar 16 '22 at 21:01
  • @ACuriousMind, "What is the difference between this and asking what keeps the Earth from falling into the Sun?" The difference is that the earth's position (approximately) doesnt need to be represented by a wavefunction of probability amplitudes. The question is (assuming you didn't read it), is what is it that is keeping the position wavepacket from condensing into a delta function when it's being pulled by a potential? – Steven Sagona Mar 16 '22 at 21:43
  • I would argue that if you want to keep this intuitive image of "each individual possible position of the electron gets individually pulled toward the nucleus" you also have to account for every possible velocity associated to that position. An object that starts with zero velocity may fall directly into the sun but if it is moving it might orbit or even just pass by. – Mauricio Mar 18 '22 at 16:28

5 Answers5

3

I think your confusion really comes from a misunderstanding at the classical level – that attractive forces work like vacuum cleaners.

A vacuum cleaner creates a wind with a certain velocity that tends to make objects move with that velocity. It's an Aristotelian force, more or less. Electromagnetism and gravity cause acceleration, not velocity. An accelerated object picks up speed as it gets closer to the source, overshoots it, recedes while slowing down, and reverses direction, and the process starts over. In other words, it orbits. If it weren't for the second law of thermodynamics, it would keep orbiting forever.

A hydrogen atom in the ground state is a quantum version of that process. The electron is in a superposition of approaching and receding from the nucleus (and orbiting circularly, etc.), and no direction dominates over any other. There are no dissipative processes to break the time symmetry because the system is already in the lowest energy state that's compatible with the uncertainty principle.

benrg
  • 26,103
  • Thanks for the answer. I think you're correct that I didn't think very much about the classical case where accelleration should be considered. – Steven Sagona Mar 16 '22 at 22:18
  • Though an important part of the question is only mentioned briefly here: "...because the system is already in the lowest energy state that's compatible with the uncertainty principle." This is an important point I'm trying to gain a better understanding of. Somehow if a force tries to confine a quantum particle, its almost as if there's a "counter force" due to uncertainty principle that defines the size of the bound state. – Steven Sagona Mar 16 '22 at 22:19
2

The basic response here must be that your "intuition" already fails at the classical level - if you take a classical probability density in phase space (i.e. classical statistical mechanics) and let it orbit around a star, it doesn't get "pulled into" the star, either - just every dot in that "probability cloud" will follow the orbit it would if it was a definite state. "Forces pull things straight into the sun" isn't how forces work classically, either, and it is unclear why it would be "intuitive" to suggest that they do in quantum mechanics.

The longer response (and the rest of this answer) is that none of this classical thinking (or "intuition") should be applied to quantum mechanics in the first place if you actually want to understand quantum mechanics.

"An electron might have some positional uncertainty, but no matter what that position is, it still experiences a force towards the nucleus."

No, it doesn't. Classical intuition does not apply to quantum mechanics, and there are no "forces" that "pull" particles here - there are just solutions to the Schrödinger equation.

"I would have expected, given my understanding of QM, that each individual possible position of the electron gets individually pulled toward the nucleus, and you end up with a coherent superposition of all of the possibilities."

Your understanding of quantum mechanics is wrong. Quantum mechanics doesn't work like that. Superposing all classical states ("positions") and then applying classical thinking like forces to each of these states isn't quantum mechanics - that's just classical statistical mechanics, i.e. doing classical mechanics for "clouds" (= probability densities) of particles in phase space.

There is a limit in which quantum mechanics works like that, and it is precisely the classical limit. Quantum mechanics itself doesn't function like that. In particular, a stable state in quantum mechanics is simply an eigenstate of the Hamiltonian, i.e. a state of definite energy. Doesn't matter if you think there's "forces" on such a state, if it has a definite energy, it's not going to do anything. Classical mechanics doesn't work like that at all - all the planets in their orbits have constant energy and yet their position is constantly changing, there isn't any "steady" state for a planet except after having fallen into the sun - but in quantum mechanics there is.

Note further that, due to the boundedness of the hydrogen Hamiltonian, there is an actual ground state of minimal energy among these steady states, and it is not a state that corresponds to "the particle just sits in the nucleus" (which would classically be the state of minimal energy where the classical force "wants" to get things to). That is, even if you started with a wavefunction localized tightly around the nucleus, the time-dependent Schrödinger equation will cause it to evolve into a different, less localized wavefunction. The idea that there is some sort of "force" that "pulls the wavefunction inwards" is simply not applicable here - the very same Hamiltonian that classically produces that inward force leads to this rich and very different picture in quantum mechanics!

ACuriousMind
  • 124,833
  • I am trying to gain a mechanistic understanding of what is happening to the wavefunction when it experiences a force. The use of "intuitive" is common in physics to trying to gain such a mechanical understanding. I am fully aware of the differential equations that could be used to solve for such a state, and want to gain an understanding of why the wavepacket does not compress to a dirac delta. – Steven Sagona Mar 16 '22 at 22:09
  • @StevenSagona There is no such "mechanistic understanding". How would your approach work even for a free particle? If you start with a localized wavefunction for a free particle, it will spread over time, with no "force" at all acting on it. – ACuriousMind Mar 16 '22 at 22:10
  • Sounds like to me from your answer that you just don't like the language that I'm using to express my problem, and are picking at the language rather than focusing on the actual question. I do not think it is that uncommon to ask what happens to different parts of a wavepacket when it is in a superposition - and I certainly don't think that asking these kinds of questions merits such obvious eyerolling. – Steven Sagona Mar 16 '22 at 22:10
  • "There is no such "mechanistic understanding". How would your approach work even for a free particle? If you start with a localized wavefunction for a free particle, it will spread over time, with no "force" at all acting on it." - I think this is a perfectly fine response (maybe focus on that in your answer instead). So then my question is basically if the bound state is essentially an equilibrium with the free evolution (the wavepacket essentially wanting to expand) and a "force" (potential V(x)) that is trying to compress it. Maybe even thinking about it like that isn't simple enough. – Steven Sagona Mar 16 '22 at 22:15
  • 1
    @StevenSagona The issue might be trying to think of a physical force acting on a wave function, which is a mathematical object we can use to determine probabilities. The wave function is not a collection of particles at different positions. – BioPhysicist Mar 17 '22 at 00:23
1

You are supposing that the electron is at all positions possible, but the electron in QM is a wave that isn't ruled by the classical equations such as Coulomb's Law. Indeed you find the wave equation of the electron (from which you get probability density) from the Hamiltonian (energy) using Schrodinger's equation. This energy will include the potential energy $-\frac{Ke^2}{r}$. You can not suppose the electron is a dispersed zone that follows classical equations. So what this "attraction" really does is keep the probability density close to the nucleus.

So, you can not use the concept of pulling and forces in QM as the concepts of acceleration and force are not applicable in this theory. In QM you can only think of potentials that condition the wave equation (probability density).

Marc Barceló
  • 847
  • I am in agreement that it makes sense why the wavefunction probability is pulled by the nucleus. My issue is that there needs to be essentially a "restorative force" to the cloud itself, otherwise the cloud will get pulled to the center like a dirac delta. My question is if it's simply its momentum being less well defined that's causing this sort of effective force. – Steven Sagona Mar 16 '22 at 18:58
  • If It helps you could suppose that the "cloud"'s centripetal force will contrarest but It is not completely correct. The concept of force in QM doesn't exist so there isn't any pulling. – Marc Barceló Mar 16 '22 at 19:10
  • switch force with "attraction" then, as you described in your answer. – Steven Sagona Mar 16 '22 at 21:46
  • also, what do you mean by "contrarest"? – Steven Sagona Mar 16 '22 at 21:46
  • There is still no attraction. There are only potentials. By contrarest I mean balance. I am sorry, English is not my native tongue. – Marc Barceló Mar 16 '22 at 22:12
1

Quantum mechanics was invented because of three contradictions of classical mechanics and thermodynamics.

The black body radiation, the photoelectric effect and the spectra of atoms cannot be calculated by using classical mechanics and classical electrodynamics.

Your:

I guess a bound state is just an equilibrium of the electron having its position wavefunction pulled to the center but having a component that is trying to escape due to propagation of the uncertainty in momentum?

italics mine.

In quantum mechanics there is no push and pull to overcome, which is the classical mechanics expectation. It is not uncertainty that builds quantum mechanics, but quantum mechanics that , by its probabilistic nature, leads to uncertainty relations.

In the question Why don't electrons crash into the nuclei they "orbit"? I give an answer.

The classical potential between an electron and a proton, enters the quantum mechanical equation to get the solutions $Ψ$ where $Ψ^*Ψ$ is the probability of finding the electron in a bound state around the proton, superseding the Bohr model which solved the problem ad hoc with quantization of angular momentum.

We now have the theory of quantum mechanics and at a higher level quantum field theory.

anna v
  • 233,453
  • I'm not looking for a classical understanding of electrons orbiting atoms. I am looking for a physical explanation for why the electron cloud doesn't end up getting pulled up infinitely close to the nucleus. There is a force pulling the electron cloud to the center, so what is stopping the cloud from getting infinitely small? For example, is it momentum uncertainty due to its position being too well defined? – Steven Sagona Mar 16 '22 at 19:00
  • I am trying to explain that there is no force pulling the electron cloud to the center ( electron cloud the probable locations of the electron taking as center the proton) , this is a classical intuition. There are no forces in an atom analogous to the classical forces in planetary systems. In scattering of particles , yes, there is the dp/dt transfered at the vertices of feynman diagrams, but in bound systems no. – anna v Mar 16 '22 at 19:13
  • For example, in an S state of the hydrogen atom there is a probability of the electron to be in the proton, but no force can stick it there, The hydrogen atom is stable. – anna v Mar 16 '22 at 19:17
0

I guess a bound state is just an equilibrium of the electron having its position wavefunction pulled to the center but having a component that is trying to escape due to propagation of the uncertainty in momentum?

And from the answer

< "An electron might have some positional uncertainty, but no matter what that position is, it still experiences a force towards the nucleus."

No, it doesn't. Classical intuition does not apply to quantum mechanics, and there are no "forces" that "pull" particles here - there are just solutions to the Schrödinger equation.

I think - and this is my private opinion - that the interaction of the electron with the nucleus is quite accessible to a model conception.

For this one must first realize that the approach to the nucleus is connected with a release of energy. This energy does not come from a kinetic energy of the electron, because also from a rest position the electron is attracted by the nucleus and loses energy in the form of photons.

Where does the energy release come from then? This is a second aspect of the consideration of the nature of the electron and the proton. Nobody has ever measured the charges of these particles in the bound state and - again my modeled conception - the emission of the photons at the mutual approach comes from their electric field. The common field becomes weaker.

Last point of the model. At certain distances from the nucleus this energy conversion of the field into radiation comes to a standstill, a further energy conversion cannot take place from the inner structure of the particles and their electric field.

To give this model a realistic basis, the electric field as well as the photon would have to be granted an inner structure. EM fields, however, are regarded as structureless, which is a peculiar view, because up to now progress in physics was actually always achieved by the description of structures in the end. Why should a field have no structure? Only because we do not think about it?

HolgerFiedler
  • 10,334