# Stability of Electron/Nucleus; Heisenberg Unc.

1. Jun 14, 2010

### Enjolras1789

In Six Easy Pieces: Fundamentals of Physics by Feynman, the claim is made that the reason why electrons don't just collide “ontop” of the nucleus is due to the Heisenberg uncertainty principle. This perplexes me greatly; it sounds wrong to me.

The moon and the Earth do not collide because there is a balance between the kinetic energy “pushing” outward and the gravitational attraction “pulling” inward. Although electrons are not classical objects, I don't see why the same analogy does not hold, albeit a need for modification to be framed in terms of the Schrodinger equation. An electron in hydrogen might start off far away, “feels” the Coulombic potential, and its position is altered until a balance is met between the attraction bringing it closer in and the speed directing it away, resulting in some bound state. How does the physical cartoon we have about planets fail to describe electron/nucleus interaction? I don't see that the cartoon is fundamentally different. This is my first question.

The answer to my first question may lie in my 2nd question. I have taken a year of graduate non-relativistic quantum mechanics; perhaps I need more to understand the answer to the question I am getting to. If so, and an answer is only possible in a QFT framework, just let me know.

My 2nd question concerns how the Heisenberg uncertainty principle relates to an increase in energy based on knowing a position. For example, I have heard the statement that one can attribute the notion of zero-point energy in a harmonic oscillator as consequence of specifying its position; that somehow the Heisenberg uncertainty principle also states that if a position is specified, there MUST be an increase in the kinetic energy. I do not understand this at all. My only understanding of the Heisenberg uncertainty principle is that, at certain scales, the meaningfulness of certain conjugate variables diminishes. One cannot talk about the position and momentum simultaneously beyond a certain uncertainty window. That is, in my mind, very different from saying that because we can't know the value of a quantity, it must be big in overall value, whatever it is.

2. Jun 14, 2010

### alxm

This isn't quite correct, strictly speaking. The uncertainty principle is a derived result of the non-commuting relationship between position and momentum.
The same relationship explains why electrons do not remain in close proximity to the nucleus.

You don't have forces. You have potentials, a potential energy associated with the position of the electron, but since the electron does not have a definite position, you don't have forces in the Newtonian sense.
The electron doesn't have a trajectory, so you cannot construct a 'planetary' type model.

It's also empirically wrong, since the single electron of ground-state hydrogen (for instance) has no angular momentum! It would appear not to be 'orbiting' at all!

It doesn't. But confinement to a smaller physical space leads to an increase in energy.
Surely your course must've covered the particle-in-a-box model and arrived at this result?

No the reason is that the particle is not completely stationary. Although you can use the HUP to rationalize why this is so (uncertainty in position would become 0).

3. Jun 14, 2010

### peteratcam

It is worth adding that the cartoon is wrong because the electron is charged and accelerating. Classical electrodynamics would make us expect the electron to radiate its energy away - so classically there would be no stable bound state of orbiting charges.

4. Jun 14, 2010

### Enjolras1789

alxm:

I don't think you answered the relevant question. I never once said anything about forces or trajectories or planetary models literally in what I wrote; you did. The cartoon to which I refer is not the idea of forces or trajectories; I asked if the idea that there is a balance between an attractive potential and kinetic energy.

Feynman acts as though there is a great mystery about electrons and nuclei. What I don't understand is what is so mysterious that once need invoke the Heisenberg uncertainty principle. Why can't one simply explain it as the balancing of two energy types?

As to your next point, you state that knowing a position does not increase the value of the momentum. OK. Can you explain why Feynman says that? I agree it doesn't follow logically, at least to me, but I'd like to think there is a good reason he said. Is there some limiting case in which this is a true statement?

As for your snarky little "Surely your course must have covered...", my course did not say WHY a zero point energy exists. We in no way discussed why the solution to a particle in a bound state has a latent energy due to confinement, merely that it does. If my course covered this, I probably wouldn't be asking.

Regarding harmonic oscillator, you state that the reason a ZPE exists is because the particle is not completely stationary. Stationary meaning what in this sentence? The usual meaning of stationary in QM refers to a stationary state, that is to say, an eigenfunction of the Hamiltonian operator. That can't be what you mean, since we are talking about energy eigenstates. Stationary in what sense then? Meaning it's still moving?

What is a HUP?

5. Jun 14, 2010

### Enjolras1789

peteratcam:

Can you please tell me how the Heisenberg Uncertainty principle does resolve the problem of a charged particle accelerating should radiate photons and eventually decay in orbit? Or direct me to a text which addresses this directly?

6. Jun 14, 2010

### alxm

Okay, well usually a 'planetary model' is constructed from the starting point of centrifugal force and gravitational force cancelling; i.e. from a force point of view. So I assumed that's what you meant. Obviously the quantum-mechanical picture is not. But yes of course there's a "balance" between the two. You only have those two terms in your Hamiltonian.

Because there's no reason to assume the kinetic energy would increase as the occupied space decreases. This is not the case at all classically. Where does the kinetic energy come from? The HUP makes for a simple heuristic argument for why this is the case.

Because he equates the volume of the electron's orbital with the position-uncertainty as part of this heuristic. You seem to be interpreting the statement overly stringently. It's a popular-scientific book, not a rigorous treatment of QM.
Feynman express the same idea to derive the Bohr radius in his "Lectures on Physics" (vol 3 section 2-4) and underlined: "It must not be taken too seriously; the idea is right but the analysis is not very accurate".

No snark was intended. But since you solved that problem, you know that you didn't invoke the uncertainty principle to do so, yet arrived at the result that kinetic energy increases with decreased box size.
So why think the uncertainty principle is directly involved?

The solution to the wave function is a stationary state. Which is not the same as saying the particle is stationary. An electronic orbital is a stationary state, yet the electron certainly moves, albeit not in the classical sense. (It has no trajectory, but it does have kinetic energy and displays the dynamical effects of motion, such as correlation)

Heisenberg's Uncertainty Principle.

7. Jun 14, 2010

### Enjolras1789

alxm:

Thank you for the clarifications. I assumed Feynman's book was "rigorous" in the sense of not ever saying something wrong, just not saying much; he at least claims that to be the case in his introduction.

I am still missing the key to this. If you can explain, I'd appreciate it, or if it's too involved, a specific reference would be great.

You wrote: "Where does the kinetic energy come from? The HUP makes for a simple heuristic argument for why this is the case."

I am confused. You're saying that the HUP describes that the kinetic energy should increase. But then later you mentioned the fact that, in solving for bound states in QM, one does not explicitly use the HUP, and yet bound states do have a higher kinetic energy.

So...the big question for me is...why do bound states have a ZPE / higher kinetic energy than we might "physically" associate? Or is this not known? Is it due to HUP, or not?

I appreciate your time in helping a rookie.

8. Jun 14, 2010

### JK423

Personally i dont understand where the problem is..
When a particle is confined in tiny space, you have a large dispersion of momentum due to HUP. What large dispersion means? It means that if you make measurements on the particle, you will measure momentum in the region p,p+Δp which means that you will measure higher values --> Higher energy.
Please explain to me why this explanation isnt enough for you

As far as the 'why the electron doesnt fall on the proton' question:
The electron wants to minimize its energy.
Due to the attractive coulomb potential it basically wants to go and fall on that proton..
But in order to do that, it must specify its position so much (Δx->0) that due to HUP there will be huge dispersion in energy.
So, the electron in order to minimize its energy t must take account of TWO things
1) The attractive coulomb potential --> Go very close on that proton (REDUCE energy)
2) HUP, Getting on that proton specifies its position enormously--> INCREASE energy
The 1st bound state of the hydrogen atom is a compromise between these two.

9. Jun 15, 2010

### alxm

Well I said it's a 'heuristic argument'. Remember, the HUP is not in itself a fundamental postulate or some-such of QM, although it's often presented as such in the pop-sci accounts, chronological accounts and introductory texts. Once you have the Schrödinger equation etc, the HUP is arrived at as a result of the non-commutation of position and momentum operators, which is the much more fundamental thing here.

But obviously if you want to just give a simplified account, it's easy to teach the simple HUP inequality, and use this to describe the pre-quantum question of "why the electron doesn't fall into the nucleus". It would be more precise to say it's not because of the HUP, but analogous to the HUP. One can do what Feynman did in his textbook and use the HUP in a somewhat hand-waving fashion to get the Bohr radius, but you can also equate the particle-in-the-box result with Rydberg's formula to arrive at the same result. But the only proper derivation of the Bohr radius means solving the S.E. for the hydrogen atom*.

Most fundamentally you could really just look at the postulates and you see the wave function is related to position, it is normalized, and its derivative is related to momentum. That's enough, really, to conclude that momentum will increase if the particle is more sharply located. I'm not sure I can give a deeper reason for the 'why' question than that; if we knew why we needed these postulates they wouldn't be postulates!

Then you can go off and derive the HUP, and also show that $$\Delta x \Delta p \geq \hbar$$ is the limit of this uncertainty. Since it's a nice simple equation that you could explain to a high-school student, it's a good heuristic for qualitatively illustrating this aspect of QM. Obviously a 'deeper' explanation of commutation relations, Hilbert spaces, the Cauchy-Schwarz inequality, etc is out of the question if you want to reach a broader audience.

(* Of course, now the Bohr radius is defined from the solution to the S.E. with a fixed nucleus. The actual radius is slightly different due to reduced-mass and relativistic effects)

Last edited: Jun 15, 2010
10. Jun 15, 2010

### Enjolras1789

alxm:

"Once you have the Schrödinger equation etc, the HUP is arrived at as a result of the non-commutation of position and momentum operators, which is the much more fundamental thing here."

I disagree on this point; the uncertainty principle can be demonstrated without any particular physics. The uncertainty principle exists in Fourier analysis; it is a mathematical notion. Given that we are discussing some sort of wave-based means of describing matter, the HUP *must* be there, or it would be incompatible with the entire notion of waves mathematically.

I think the specific question I am asking has been overlooked. In this thread, I am not asking about the postulates. I am asking something very specific, from before:

"So...the big question for me is...why do bound states have a ZPE / higher kinetic energy than we might "physically" associate? Or is this not known? Is it due to HUP, or not?"

I am not asking for a heuristic argument, unless it also happens to be rigorous. Just because an argument is heuristic does not imply it is not rigorous; rather, physics is based on the heuristic (classical mechanics).

What is the cause/origination of the notion of ZPE? Beyond just "it falls out of the equations that way?" If the answer is simply "I don't know, it just seems to be a result of the Schrodinger equation, and it has no physical explanation," that's fine. But if there is some physical explanation, that would be desirable.

I apologize if you are answering these questions and I am just not seeing it, but I don't think we're communicating on the same frequency thus far.

11. Jun 15, 2010

### Enjolras1789

JK423:

"When a particle is confined in tiny space, you have a large dispersion of momentum due to HUP. What large dispersion means? It means that if you make measurements on the particle, you will measure momentum in the region p,p+Δp which means that you will measure higher values --> Higher energy."

It doesn't logically follow, to me. If there is a definite position, the uncertainty in the momentum is greater. Not being able to say the value of the momentum except within a range IS NOT the same as saying that the value of the momentum is the extreme value of that range...unless there is something I am missing.

It's the part in bold that makes no sense to me.

For example, to me, this is your argument: I know the value of momentum is between 4 and 8....therefore, because I don't know that it isn't 8, it must be 8.

I also don't understand your use of the word "measure" in the statement I underlined and put in bold. Perhaps my confusion about your use of this word is at the heart of my confusion about the entire problem.

My understanding is that HUP is saying that it is impossible to know the value beyond some uncertainty window of one variable if you know it's conjugate exactly. To me, you're saying something totally different: it seems like you're saying that if you were to do another measurement, you would observe a range of values of the momentum, some higher. That makes no sense to me; you can't go do a measurement of the momentum and still know the value of the position. Every measurement changes the state of the system, such that measurement can't be invoked to discuss the values of the momentum, because doing another measurement means you don't know the position anymore.

I solicit your opinions to clarify my confusion.

12. Jun 15, 2010

### Enjolras1789

alxm:

Can you shed insight to how all of this (does / does not) relate to the issue that a classical charged object in a bound state releases photons, thereby eventually decaying its bound state to collapse? Does that argument not apply and one must use a different logic?

13. Jun 15, 2010

### JK423

Let me rephrase.
Suppose that the electron is in a state |Ψ> with momentum dispersion Δp.
Without the dispersion Δp, the electron would have momentum p.
But if we take account of the dispersion, its momentum will range in the region [p,p+Δp]. If Δx is very small then Δp ιs huge due to HUP.
Which means that the range [p,p+Δp] will be vast. So, the *new* mean value <p> will not be equal to p but will be somewhere inside the range [p,p+Δp] taking large values. Ofcourse it wont take the maximum value, that is just an approximation.
Thats my understanding.
Forget the word 'measure'. Just take into account that momentum will fluctuate in higher values than normal increasing the energy.

14. Jun 15, 2010

### Enjolras1789

JK423:

This is the crux of the confusion, for me. You are using the dispersion of the momentum to mean something different than what I *think* it means (but I am a novice).

That there is a dispersion in the momentum to me means "fundamental ignorance of the expectation value." It seems like you have it mean "possibility of the expectation value being higher." Does the distinction I am drawing make sense? One is about ignorance, the other is about shifting the mean of a probability distribution function.

Are they somehow equivalent in a way I don't appreciate? In my mind, my meaning is narrower and does not include yours. Again, this is all the thoughts of a rookie, however.

15. Jun 15, 2010

### ZapperZ

Staff Emeritus

Zz.

16. Jun 15, 2010

### JK423

No thats not correct. Dispersion in momentum means ignorance of the exact momentum value, not its expectation value!
For example, suppose that momentum is in the range [p,p+Δp] due to HUP. What we mean when we say "its in the range..."?
If we have many identical systems in this exact state with such a momentum dispersion, when measure each one of them we will get different values p1,p2,... but all belonging in the range [p,p+Δp]. Foundamental ignorance means that we cant predict what's the value of momentum if i measure it. We just know that it will be larger than p due to dispersion, which means that the energy will be larger as well.

17. Jun 15, 2010

### humanino

Feynman begun his career in physics with Wheeler, who encouraged his study of advanced and retarded wave for electron (self-)interactions without photon field. He describes it nicely in his Nobel lecture. This was the seed of his diagrammatic technics.

The crux of the electron orbit stability as it was formulated historically is precisely the fact that an accelerated charge radiates. So the question Feynman addresses in his book is "why electrons do not radiate and fall into the center ?".

As it turns out, we can derive a detailed balance between the electron energy loss by radiation and energy gain from vacuum fluctuations using consistency conditions for Heisenberg relations. This is a semi-classical treatment and is done in detailed for instance in Milonni's textbook on QED. There is no ambiguity in this treatment and one is lead to the correct Bohr's orbits. He pushes the game much further and computes corrections as well.

In fact, he even had already shown that Heisenberg's relations for an electron in a box will be damped to zero commutator unless one includes fluctuations of the photon field as well.

Those discussions are too lengthy and technical to be pursued in between QM and QED in regular classes. But I am pretty sure Feynman knew of them$^{(1)}$

Last edited: Jun 15, 2010
18. Jun 15, 2010

### Enjolras1789

JK423:

What you're saying makes logical sense. It's just a bit different than what I had understood the words to mean. Yours is logical, just different...unless I am missing something.

You describe momentum here as something that is well-defined, just having a higher range of possible measurable values, thus shifting the average of whatever probability distribution function characterizes momentum, and THAT is the meaning of "uncertainty."

I assume when you say dispersion, you mean the same as uncertainty (I just normally associate "dispersion" as meaning a function which gives the frequency dependence of some property and keep using "uncertainty"; please let me know if that is problematic in some other way).

This is different than what I how I had always thought of it. I don't want to mis-represent Fourier or Griffiths...but my understanding of HUP is more akin to my readings of Fourier analysis and Griffiths QM book. To my understanding, Griffiths described HUP as a reflection of the idea that, below a certain scale of size, both of these variables don't have meaning. That makes physical sense to me; at some scale, you can't know the speed and position of a wave/particle, because a wave's "position" is irrelevant unless the packet is extremely localized, etc. This also makes sense to me in terms of Fourier analysis: anything obeying the wave equation, mathematically, obeys the uncertainty principle. This is a statement that these terms are NOT well-defined beyond a certain realm. This is the definition of uncertainty I had walking into this discussion.

Do you view that the purely mathematical expression of the uncertainty principle from Fourier analysis is logically equivalent, no more no less, than the HUP? Do you agree that the Fourier analysis description of the uncertainty principle is about the inability to ascribe clear meaning to these variables? This is also nicely represented by Griffiths in his QM book, in a less mathematical way.

19. Jun 16, 2010

### DrChinese

Don't forget that when you have 2 entangled particles, the HUP applies to them. How would you expect that to work using your example? As I see this explanation evolving, you would end up predicting that you could observe their non-commuting observables to unlimited precision. But that is not allowed either.

My point being that the analogy has a limit. And that limit is where you start making incorrect predictions.

20. Jun 16, 2010

Dr. Chinese: