Electrons orbiting the nucleus: angular momentum

[barnflakes]

If an electron does not orbit the nucleus in the classical sense, then how can we define an angular momentum operator that is analogous to the classical angular momentum?

$$\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$$

This angular momentum depends on both the position of the electron, and the momentum of it. If we know the position of the electron, then we have infinite uncertainty of its momentum and vice versa.

The point being that the angular momentum in the quantum sense depends on the position and momentum of an electron, and in the quantum sense, the position of an electron is actually in some probability cloud distributed around the nucleus.

The problem is that we only know this because we have assumed the above form for the angular momentum operator and solved the Schrodinger equation accordingly.

How we can justify defining the angular momentum of the electron as $$\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$$ in the first place?

 

[mfb]

If we know the position of the electron, then we have infinite uncertainty of its momentum and vice versa.

We don't know either very precise, but we know a particular relation of them (the angular momentum and one if its components) very precise.

 

[barnflakes]

The point still stands: if electrons do not "orbit" a nucleus at all, how is it meaningful to define a angular momentum in this way?

 

[leright]

First, something doesn't not need to orbit something else to have an angular momentum. Second, the quantum mechanical angular momentum operator is simply defined that way. Sorry if my explanation is vague. Maybe someone can provide more insight.

 

[strangerep]

The point still stands: if electrons do not "orbit" a nucleus at all, how is it meaningful to define a angular momentum in this way?

OK, do you understand that, in QM, one represents physical dynamical quantities by hermitian operators acting on a Hilbert space of wavefunctions? (Orbital) angular momentum corresponds to one such dynamical quantity. "Quantization" of a classical system essentially boils down to finding a representation of the algebra of classical dynamical quantities as operators on a Hilbert space. The familiar $\hat L$ that you wrote above becomes such an operator by this procedure.

I'm not sure what level of elaboration is needed here, but since you categorized the thread as "I", perhaps you've already studied some QM? If so, from which textbook(s)?

 

[barnflakes]

OK, do you understand that, in QM, one represents physical dynamical quantities by hermitian operators acting on a Hilbert space of wavefunctions? (Orbital) angular momentum corresponds to one such dynamical quantity. "Quantization" of a classical system essentially boils down to finding a representation of the algebra of classical dynamical quantities as operators on a Hilbert space. The familiar $\hat L$ that you wrote above becomes such an operator by this procedure.

I'm not sure what level of elaboration is needed here, but since you categorized the thread as "I", perhaps you've already studied some QM? If so, from which textbook(s)?

I have a PhD in physics. I also got the top grades in my quantum mechanics exams during undergrad.

I have no problem solving the mathematical problems of QM. My question is of a conceptual nature. Unfortunately, I am used to people responding to question like this with mathematical statements like yours: "one represents physical dynamical quantities by hermitian operators acting on a Hilbert space of wavefunctions?" that are completely meaningless at getting to the bottom of a conceptual understanding of the problem. How does that statement answer my question in any way? It is, as far as I'm concerned, irrelevant to the question at hand.

As far as I can tell, the answer you're trying to give is along the lines of "we just try to find a quantum version of the classical angular momentum, which means that we invent some new mathematical machinery that deals with the problem in a "quantum" way".

The wavefunction is simply a way of describing the properties of the system. A hermitian operator acts on that wave function and spits out an eigenvalue along with the original wavefunction. I understand fully that the angular momentum operator has these properties and acts in this way.

That doesn't answer my original question. My question is, what are r and p in the context of a quantum system.

How does having a position and a momentum even make sense for a particle that doesn't orbit? Or is it in fact that electrons DO orbit an atom?

Let me try one more attempt at making this crystal clear:

We agree that an electron does not orbit an atomic nucleus. So what is the position r of the electron, and what is the momentum p, of the electron if this is true? It can clearly not be its position in the orbit or its momentum at a particular position of that orbit.

However, we must also agree that an electron *does* orbit a nucleus, or at least moves around the nucleus, otherwise it would be impossible to create the correct probability density functions of its position: if in one measurement you measure the electron on one side of the nucleus, and in another measurement the other, and so on, the implication is that an electron must in general be moving.

 

[Nugatory]

If an electron does not orbit the nucleus in the classical sense, then how can we define an angular momentum operator that is analogous to the classical angular momentum?

You started this thread by defining classical angular momentum as $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$.

However, an alternative definition is that angular momentum is the conserved quantity associated with rotational symmetry, so that $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$ is merely a convenient formula for calculating it in classical systems that are are easily described in terms of $\hat{\mathbf{r}}$ and $\hat{\mathbf{v}}$. Although [understatement]somewhat less intuitive[/understatement], this approach has the nice property that it is equivalent to the classical formula when we have a classical orbit but also works when we don't.

Ballentine ("Quantum Mechanics: A Modern Development", and check out Strangerep's signature - he's posted in this thread already) devotes several chapters to this approach. It's also worth googling for "Noether's theorem".

 

[strangerep]

[...] That doesn't answer my original question.

Sorry if my attempted answer annoyed you.

 

[barnflakes]

Sorry if my attempted answer annoyed you.

I'm sorry if I sounded curt, it's just that I get extremely tired of people trying to explain quantum mechanics in terms of mathematical jargon. I am trying to build a more conceptual understanding of QM, based on visual representations of what's happening in my head: "hermitian operator acting on a wavefunction that lives in a hilbert space" does nothing to develop a mental representation of what's going on. I encourage everyone to go through these processes because visual representations last a lot longer than the equations, which you'll soon forget after your degree.

 

[strangerep]

Are you able to apply the concepts of symmetry operations (such as translation, rotation, etc) to whatever visual representation(s) you currently imagine?

And how about distributions of intrinsic properties of a field over a spatial region?

 

[Nugatory]

We agree that an electron does not orbit an atomic nucleus. So what is the position r of the electron, and what is the momentum p, of the electron if this is true?

As far as the mathematical formalism of QM is concerned, neither question has an answer because neither concept even exists in the formalism. All the formalism will do is tell you what the probability of getting particular measurement results. You may find this situation to be completely exasperating, and if so you're in good company - Einstein didn't like it either (check out the EPR paper, published in 1935). But exasperating or not, it's the best model that we have. (I should note that there are alternatives such as GHZ and Bohmian mechanics - they are unsatisfactory for various other reasons that are off-topic in this thread).

However, we must also agree that an electron *does* orbit a nucleus, or at least moves around the nucleus, otherwise it would be impossible to create the correct probability density functions of its position: if in one measurement you measure the electron on one side of the nucleus, and in another measurement the other, and so on, the implication is that an electron must in general be moving.

If the electron does have a position even when it's not measured, then this argument would be ironclad But, no matter how natural that assumption feels, it's still an assumption... and if it turns out not to be a valid assumption then the argument doesn't work.

Although there are no practical experiments that will test whether this assumption is valid for position, it is possible to experimentally test it for other properties such as particle spin and photon polarization - google for "Bell's theorem". The experimental results are inconsistent with the intuitively satisfying way of attributing definite values to unmeasured observables that you're assuming in the argument above.

 

[vanhees71]

If an electron does not orbit the nucleus in the classical sense, then how can we define an angular momentum operator that is analogous to the classical angular momentum?

$$\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$$

This angular momentum depends on both the position of the electron, and the momentum of it. If we know the position of the electron, then we have infinite uncertainty of its momentum and vice versa.

The point being that the angular momentum in the quantum sense depends on the position and momentum of an electron, and in the quantum sense, the position of an electron is actually in some probability cloud distributed around the nucleus.

The problem is that we only know this because we have assumed the above form for the angular momentum operator and solved the Schrodinger equation accordingly.

How we can justify defining the angular momentum of the electron as $$\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$$ in the first place?

The justification for the operator algebra comes from symmetry considerations. In classical physics we have a spacetime model (here the Galilei-Newton spacetime of non-relativstic physics), which has symmetries. In this case we have symmetries under space-time translations (homogeneity of space and time), rotations (isotropy of space), and Galilei boosts (principle of inertia). These symmetries are generated in the sense of Lie algebras, realized by the Poisson brackets and represented by canonical transformations (local symplectomorphisms) by phase-space distribution functions and define the general conserved quantities of the system (energy, momentum, angular momentum, center-of-mass velocity) via Noether's theorems.

This structure is inherited by quantum theory. Instead of Poisson brackets you have commutators realizing the Lie algebra. The continuous symmetries can be represented by unitary transformations through the operator exponential. The operators
$$\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}$$
indeed obey the Lie algebra of the rotation group SO(3) and thus it's justified to identify them with the component of orbital angular momentum.

However, this is too naive, because in quantum theory a (pure) state is represented not by the Hilbert space vectors but by rays in Hilbert space. Thus more representations are allowed, and this is very lucky for non-relativistic quantum theory, because unitary representations of the classical Galileo group do not lead to a meaningful quantum dynamics, but you have to extend the group to its quantum version, which is a central extension of the Galileo group (with mass as a non-trivial central charge) and its universal covering group, which substitutes the SO(3) rotation group by its covering group SU(2). The latter extends the allowed representations of angular momentum to half-integer representations and extends the naive orbital angular momentum by the notion of spin, i.e., the realization of the rotation group for particle states which represent a particle at rest ($\vec{p}=0$). Only for a particle with spin 1 (e.g., pions) the total angular momentum is identical with the orbital angular momentum.

 

[barnflakes]

As far as the mathematical formalism of QM is concerned, neither question has an answer because neither concept even exists in the formalism. All the formalism will do is tell you what the probability of getting particular measurement results. You may find this situation to be completely exasperating, and if so you're in good company - Einstein didn't like it either (check out the EPR paper, published in 1935). But exasperating or not, it's the best model that we have. (I should note that there are alternatives such as GHZ and Bohmian mechanics - they are unsatisfactory for various other reasons that are off-topic in this thread).

If the electron does have a position even when it's not measured, then this argument would be ironclad But, no matter how natural that assumption feels, it's still an assumption... and if it turns out not to be a valid assumption then the argument doesn't work.

Although there are no practical experiments that will test whether this assumption is valid for position, it is possible to experimentally test it for other properties such as particle spin and photon polarization - google for "Bell's theorem". The experimental results are inconsistent with the intuitively satisfying way of attributing definite values to unmeasured observables that you're assuming in the argument above.

This is an interesting discussion. It is amazing how one can go through an undergraduate education, get the top grades, and not have comprehended some of the more fundamental principles of QM. I remember studying and deriving Bell's equations (since those are the things that you get asked about in homework questions and on exams) with no idea that that was the implication.

Let's take a step back then:

Q1) Do electrons move around a nucleus at all? I.e. Can we say anything about the motion of electrons around a nucleus from the wavefunction of the hydrogen atom?

Or, can we simply say, "well, if you measure the position of the particle, you'll find it at x with probability p" and similarly with momentum: this has always been my understanding of QM. Therefore the only conclusion from this is that we cannot say anything about the motion of the electron around the nucleus.

In that case, what is an intuitive understanding of the angular momentum of a wave function? I'm frankly not buying this symmetry argument. Yes, what you all say here is true, but I am not convinced that's how the angular momentum operator was justified when it was developed in the 20's/30's, and therefore there must be a more intuitive explanation to it and what it represents.

 

[dextercioby]

The answer to these questions lies with complicated mathematics. There's no way to rephrase it, so that a HS student would "get" it. You have a PhD, you got high grades in QM, you should know the complicated answer. We're just telling you that there's no simple one. That's it. Sorry if you felt ridiculed.

 

[blue_leaf77]

In that case, what is an intuitive understanding of the angular momentum of a wave function?

By the way, do you have an intuitive explanation of the spin of an electron in such a way that you can explain it to layman without using math at all or at least minimal math? Or is it also something you already planned to ask sometimes in the future? What about the result of the double slit experiment, any intuitive, math-free explanations?
Every physical quantities in quantum mechanics is associated with an operator, so are position, momentum, angular momentum, etc. That's, for now, the most intuitive explanation of angular momentum to me - a physical quantity being represented as an operator. It may sound ignorant, but if someone says why one can associate the angular momentum with an operator $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ and not another formula? The answer is because it works. As far as I can remember, physical quantities being represented as operators in QM is a hypothesis, in other words, it's still indeed open for revisions. But why hasn't there been any revisions? Again, because it works. Why hasn't there been any accepted rejection to the hypothesis that the upper limit of velocity is the speed of light? Because observations have yet to prove it wrong.
And lastly, that's how we developed QM - it's counterintuitive but it works, so people use it.

 

[barnflakes]

OK, I have finally figured out how to ask this question so that I stop getting these kind of tautological responses.

Angular momentum can only exist if something is rotating.

In the hydrogen atom solution of Schrodinger's equation, what is rotating? If it's not the electron around the nucleus, what is it?

 

[barnflakes]

By the way, do you have an intuitive explanation of the spin of an electron in such a way that you can explain it to layman without using math at all or at least minimal math? Or is it also something you already planned to ask sometimes in the future? What about the result of the double slit experiment, any intuitive, math-free explanations?
Every physical quantities in quantum mechanics is associated with an operator, so are position, momentum, angular momentum, etc. That's, for now, the most intuitive explanation of angular momentum to me - a physical quantity being represented as an operator. It may sound ignorant, but if someone says why one can associate the angular momentum with an operator $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ and not another formula? The answer is because it works. As far as I can remember, physical quantities being represented as operators in QM is a hypothesis, in other words, it's still indeed open for revisions. But why hasn't there been any revisions? Again, because it works. Why hasn't there been any accepted rejection to the hypothesis that the upper limit of velocity is the speed of light? Because observations have yet to prove it wrong.
And lastly, that's how we developed QM - it's counterintuitive but it works, so people use it.

Spin is a fundamental property of an elementary particle, just like mass or charge.

Double-slit: quantum particles have both wave and particle like nature. A single quantum particle can pass through two-slits at once, just as a wave can. However, when we measure the position of the particle on the detector, it registers at a single point (particle nature). If we make enough of these measurements of single hits on the detector and look at the pattern, it forms a diffraction pattern, just like a wave would.

Although the idea of an electron behaving as a wave seems counter-intuitive, it is not counter-intuitive if you never considered it to be a particle in the first place i.e. you always considered it having both wave-like and particle-like nature.

Either way, it answers the question nicely: a quantum particle has both wave-like and particle-like properties.

In my question:

An angular momentum only exists if something is rotating. What is rotating when we consider the physics of an electron and a proton interacting via the Coulomb potential? i.e. the wavefunction of the hydrogen atom.

Something is rotating (otherwise "angular momentum" is meaningless). What is it?

 

[mfb]

Angular momentum can only exist if something is rotating.

That is not true. You don't need rotation for a non-zero r x p.

Spin is a fundamental property of an elementary particle, just like mass or charge.

That is not really an explanation of spin. You just said "it is a property", but not what the property does.

quantum particles have both wave and particle like nature.

That is at best misleading.

 

[blue_leaf77]

In the hydrogen atom solution of Schrodinger's equation, what is rotating? If it's not the electron around the nucleus, what is it?

If you consider the nucleus to be of finite mass, the Schrodinger equation can be separated into the equation for the center of mass and that of the relative position. The angular momentum appearing in such a Hamiltonian is the relative angular momentum observed by either particle.

Spin is a fundamental property of an elementary particle, just like mass or charge.

Sounds like you simply recite the popular definition about the spin of a particle, the thing I would also do if I were to be asked the same question. But why don't you argue, why are we justified to define spin as it is, why electrons have spin 1/2 etc etc, just like you did with

How we can justify defining the angular momentum of the electron as

^L=^r×^pL^=r^×p^​

\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}} in the first place?

?

Angular momentum can only exist if something is rotating.

Just like mfb said, you don't need something rotating to define angular momentum, it's also true in classical mechanics. In particular, whether or not there is non-zero angular momentum depends on the chosen frame of reference. Imagine a particle is moving straight along the x axis, and you stand on a point on this axis - you will observe zero angular momentum for this particle. But as soon as you walk away from the x axis, the angular momentum is not zero. In quantum mechanics, there's no stopping you calculating the angular momentum of a free electron.

Although the idea of an electron behaving as a wave seems counter-intuitive, it is not counter-intuitive if you never considered it to be a particle in the first place i.e. you always considered it having both wave-like and particle-like nature.

That's it! Intuition is subject to experience. One says that the notion of a particle being a wave is counterintuitive because he/she is used to think that a particle takes the form of a minute entity. I bet that's what all physicists used to think in their childhood before being introduced to quantum mechanics.

 

[PeroK]

OK, I have finally figured out how to ask this question so that I stop getting these kind of tautological responses.

Angular momentum can only exist if something is rotating.

In the hydrogen atom solution of Schrodinger's equation, what is rotating? If it's not the electron around the nucleus, what is it?

I think you've been deflected from your original question which was to try to gain a more intuitive view of what is going on at the quantum level and to relate it by analogy to the macro-world.

I've got 3-4 pages of my own notes and calculations on the hydrogen atom to try to get a picture of what is happening. The one line summary I have is that an electron simply does not "move" in a way that is directly analogous to motion in the macro-world.

I don't claim any deep insight here (I found a long exchange on Physics Stack Exchange - which I can't find again now - on this subject) but it seems to me you can only push macro-world analogies so far in QM.

I would say that nothing "rotates" at the quantum level in a way that is directly analogous to a planet spinning on axis and orbiting the Sun, hence an intuitive picture of the hydrogen atom can only take you so far.

 

[barnflakes]

Just like mfb said, you don't need something rotating to define angular momentum, it's also true in classical mechanics. In particular, whether or not there is non-zero angular momentum depends on the chosen frame of reference. Imagine a particle is moving straight along the x axis, and you stand on a point on this axis - you will observe zero angular momentum for this particle. But as soon as you walk away from the x axis, the angular momentum is not zero. In quantum mechanics, there's no stopping you calculating the angular momentum of a free electron.

That's a good point and something I hadn't thought about.

That is not true. You don't need rotation for a non-zero r x p.That is not really an explanation of spin. You just said "it is a property", but not what the property does.That is at best misleading.

What is mass? What is charge? At some point you boil everything down into things that are "irreducible", at least as far as physics knows. That is clearly not the case for angular momentum, which is defined in terms of other quantities.

Good points about the reference frame argument. My feeling is that for the hydrogen atom solution of the Schrodinger equation, one couldn't consider the reference frame of the proton and the electron separately. Do you agree?

 

[blue_leaf77]

My feeling is that for the hydrogen atom solution of the Schrodinger equation, one couldn't consider the reference frame of the proton and the electron separately. Do you agree?

It's not like you can't, but if you write the Schroedinger equation in terms of $\mathbf{r}_e$ and $\mathbf{r}_p$, it is not separable, therefore you can't solve it analytically. But of course, a computer program can help.

 

[mfb]

Good points about the reference frame argument. My feeling is that for the hydrogen atom solution of the Schrodinger equation, one couldn't consider the reference frame of the proton and the electron separately. Do you agree?

There is not even such a reference frame. You can consider both particles separately in a different reference frame, but that would make things more complicated than necessary.

 

[strangerep]

OK, I have finally figured out how to ask this question so that I stop getting these kind of tautological responses

@barnflakes, can we please take deep breath and try again? Like dextercioby in post #14, I too was surprised and concerned that a physics PhD needed to ask this level of question. But I do not mean this in any derogatory way. Indeed, there were various things that were inadequately explained to me during my formal education, so I have considerable sympathy with your situation. That's also why I asked the questions I did in my post #10 (which you have not answered, afaict). I was trying to find out what you are comfortable with -- so I could try to answer in those terms.

Angular momentum can only exist if something is rotating

This is not correct. There is a distinction between orbital angular momentum (which we have been discussing in this thread), and intrinsic angular momentum (also known as "spin"). The latter is not associated with rotational motion. If you have access to Misner, Thorne & Wheeler's textbook on Gravitation, there's a more extensive explanation in Box 5.6 (from memory).

Now, going back to something you said earlier,

Or, can we simply say, "well, if you measure the position of the particle, you'll find it at x with probability p" and similarly with momentum: this has always been my understanding of QM. Therefore the only conclusion from this is that we cannot say anything about the motion of the electron around the nucleus.

Your 1st sentence is correct, though incomplete since we can also find higher moments of the probability distribution in principle (i.e., variance, etc). But your 2nd sentence does not follow since we can say something about it, albeit only in probabilistic terms. This is more than "nothing".

Regarding (orbital) angular momentum, the usual states for the H-atom are eigenstates of orbital angular momentum, so the probability distribution for that observable reduces to a delta function: if the orbital angular momentum were measured multiple times for identically prepared H-atoms, we'd get the same result every time. In this sense, the orbital angular momentum is deterministic. This, even though position and ordinary momentum are non-deterministic for that state.

 

[barnflakes]

So let me summarise what I think I've learned from this thread:

Noether's theorem tells us that invariance with respect to certain physical transformations leads to a conservation law. Invariance with respect to time leads to conservation of energy, invariance with respect to translation leads to conservation of momentum, and invariance of rotation leads to conservation of angular momentum.

Therefore, we conclude that conservation of angular momentum is simply a physical property that all systems must obey, whether or not that system is actually rotating in your reference frame or not (the fact that a free particle moving in a straight line also has angular momentum depending on your reference frame shows this).

Therefore, we should not think of the fact that the hydrogen atom has orbital angular momentum as meaning that an electron is orbiting around an atom, or indeed than anything is "orbiting".

I would be interested to hear your opinions on the correctness of my conclusion from what I've read in this thread.

Best wishes, barnflakes

 

[mfb]

Therefore, we conclude that conservation of angular momentum is simply a physical property that all systems must obey, whether or not that system is actually rotating in your reference frame or not (the fact that a free particle moving in a straight line also has angular momentum depending on your reference frame shows this).

Right.

Therefore, we should not think of the fact that the hydrogen atom has orbital angular momentum as meaning that an electron is orbiting around an atom, or indeed than anything is "orbiting".

It is right, but it does not follow from the previous statements.

 

[vanhees71]

OK, I have finally figured out how to ask this question so that I stop getting these kind of tautological responses.

Angular momentum can only exist if something is rotating.

In the hydrogen atom solution of Schrodinger's equation, what is rotating? If it's not the electron around the nucleus, what is it?

The entire atom is "rotating". Take the non-relativistic hydrogen atom whose energy eigenstates can be calculated exactly. You start separating the problem into center-of-mass motion and relative motion of electron and proton. Then you go into the rest frame of the atom (total momentum 0) and look only at the relative-motion part. It's a good exercise to think about what this wave function means in terms of proton and electron degrees of freedom. The answer is found in the marvelous AJP article

http://arxiv.org/abs/quant-ph/9709052

 

[barnflakes]

A quote from Griffiths "Introduction to Quantum Mechanics" page 171:

"In addition to orbital angular momentum, associated (in the case of hydrogen) with the motion of the electron around the nucleus (as described by the spherical harmonics), the electron carries another form of angular momentum, which has nothing to do with motion in space..."

So it seems one of best selling QM books claims that orbital angular momentum is indeed related to the motion of the electron around the nucleus, contrary to everything that has been said in this thread. This is part of the reason that trying to understand these concepts is so difficult: conflicting information.

 

[PeroK]

A quote from Griffiths "Introduction to Quantum Mechanics" page 171:

"In addition to orbital angular momentum, associated (in the case of hydrogen) with the motion of the electron around the nucleus (as described by the spherical harmonics), the electron carries another form of angular momentum, which has nothing to do with motion in space..."

So it seems one of best selling QM books claims that orbital angular momentum is indeed related to the motion of the electron around the nucleus, contrary to everything that has been said in this thread.

It's not contrary. It is motion at the quantum level. It's relating this too closely to real-world motion that is the problem.

The mathematics of QM is different but the same terms are used.

Elsewhere in his book, Griffiths is at pains to stress that classics analogies only go so far.

Put simply if it was motion in the classical sense then you'd simultaneously need the position and velocity of the particle.

 

[vanhees71]

Well, particularly this bestselling book seems sometimes a bit sloppy. Why it sells so good, is not clear to me...

If you have an atom in an energy eigenstate there's nothing moving by definition, because the energy eigenstates are stationary states. Quantum theory tells you that you don't need anything moving to have angular momentum (spin).

 

[barnflakes]

Well, particularly this bestselling book seems sometimes a bit sloppy. Why it sells so good, is not clear to me...

If you have an atom in an energy eigenstate there's nothing moving by definition, because the energy eigenstates are stationary states. Quantum theory tells you that you don't need anything moving to have angular momentum (spin).

Yes but I'm referring specifically to orbital angular momentum here.

 

[vanhees71]

Have you looked at the suggested paper on the hydrogen atom in #31? If not, do so! It should really answer your question.

 

[PeroK]

Spin is another good example. Everyone calls it spin, but it's clearly not spin in the classical sense.

As soon as you use classics terms in QM you are open to argument and misunderstanding. The only way to avoid this is to retreat into the formalism of the mathematics, and that is partly why some people may prefer to stick to explaining things in terms of operators on a Hilbert space!

 

[barnflakes]

Have you looked at the suggested paper on the hydrogen atom in #31? If not, do so! It should really answer your question.

Yes. It answered nothing. Perhaps you can refer to a specific part of the paper that apparently answers my question?

 

[barnflakes]

Spin is another good example. Everyone calls it spin, but it's clearly not spin in the classical sense.

As soon as you use classics terms in QM you are open to argument and misunderstanding. The only way to avoid this is to retreat into the formalism of the mathematics, and that is partly why some people may prefer to stick to explaining things in terms of operators on a Hilbert space!

It is nothing to do with classical or quantum intuition. It is a simple physical question: does an electron move around a nucleus? I do not mean in the sense of an electron orbiting in the sense of the Earth orbiting the sun. I simply mean: does the electron move around a nucleus at all?