# Three-part question on HUP

1. Jun 8, 2005

### El Hombre Invisible

Apologies to those I've already driven mad with questions on HUP and particle-antiparticle annihilation, but here's three more.

I was reading Feynman's Lectures wherein he explains why electrons do not combine with protons in hydrogen atoms. This was the same answer I got to my question of why quarks of opposite charges in nuclei have separation despite having both electromagnetic and strong attractions in an earlier thread. HUP states that as the electron nears the proton, its position becomes more certain, so the uncertainty of its momentum greatly increases, therefore its kinetic energy is great enough to escape the proton's pull.

Q1. Why does greater uncertainty in momentum necessarily lead to greater kinetic energy? Surely if the momentum is uncertain, the kinetic energy mall fall into a range of values, some of which may be small (we do not know if it is great or small)?

Q2. Furthermore, why does the certainty in position increase as the electron nears the proton? If there is uncertainty in the position of the proton itself, then surely the electron's proximity to it sheds no more light on the electron's position? If protons and electrons could combine (as in neutron stars?), they could do so with high uncertainty as to exactly where they do so. (This, then, is a question of why the HUP forbids the event to happen at all rather than why we cannot observe it happening.)

Q3. Why does this not forbid positrons from annihilating with electrons? As the positron approaches the electron, surely the same increase in certainty as to the position of the positron is gained as the electron as it nears the proton. Therefore the same increase in UNcertainty of its momentum will ensure high enough kinetic energies to avoid annihilation?

Simply being reassured that this is fully explained will be partially satisfactory as I will (hopefully) hit this later on in my studies. If the answer can be explained to me now, conceptually, that is a huge bonus. I am probably not at a high enough mathematical level to understand the mechanics of the explanation, or even the question, but maybe there's a way to explain it at a high conceptual level. Also please correct any misintepretation of Feynman's lecture I have made - I tend to read him in short bursts late at night when I'm not at my best.

Thanks...

El Hombre

2. Jun 8, 2005

### Meir Achuz

Your use of the HUP confuses the issue.
All your questions are answered clearly using the Schrodinger equation and the electron's wave function.
The HUP is like Galileo's version of mechanics, and the Schrodinger equation like Newton's.

3. Jun 8, 2005

### El Hombre Invisible

I don't understand. It's not my use of HUP - it's Feynman's! The outline given was exactly what I read last night. Are you basically saying Schrodinger supersedes HUP?

4. Jun 8, 2005

### pmb_phy

Kinetic energy is meaningless in QM.
I don't believe that's true in general. It depends on the particular quantum state.
Hold on there bucko! Where did the positrons come from? You were speaking about an atom with protons and electrons. A proton does not anihilate an electron. Particles only anihilate their anti-particles.

Pete

5. Jun 9, 2005

### El Hombre Invisible

Stick with momentum then, though Feynman does say momentum, thus kinetic energy.

The picture put forward by Feynman is one of an electron in a nucleus. I thought that would not correspond with any quantum state, since the base state of an atomic electron would require some distance between that electron and the nucleus? I know this is abstract, but can we put Schrodinger aside (imagine an electron CAN be in the nucleus) and focus on HUP (why it won't stay there). While this may seem an odd thing to do, and even useless to yourselves, this isolation will aid my understanding of HUP. Thanks.

The positrons came from my own question about them. I did not suggest protons and electrons annihilate, in fact you can ignore annihilation altogether. This isn't a question about annihilation, but proximity. If an electron's proximity to a proton ensures it has enough momentum to escape that proton's attraction, how come a positron's proximity to an electron does not ensure enough momentum to escape the electron's attraction, or vice versa.

But it's okay - I think I've figured it out. Ignoring Schrodinger altogether, the HUP does not disallow an electron from 'sitting on top of' a proton, it just states that if it were to do so, it would have enough escape velocity. Likewise a positron may 'sit on top of' an electron, but the question of momentum becomes irrelevant because the pair will annihilate before either can escape from each other. Schrodinger forbids the former (electrons and protons) but not the latter (electrons and positrons). This, as I see it, is an issue that can be considered separately (at least within the context of my question).

This answers my third question, and you have answered my second. This just leaves the question of why high uncertainty in momentum necessitates high momentum. I don't understand the link. Can this be explained in similarly simple terms?

Again, please correct any misunderstandings (preferably in a nice way).

Thanks,

Bucko.

6. Jun 9, 2005

### Meir Achuz

"I don't understand. It's not my use of HUP - it's Feynman's! The outline given was exactly what I read last night."
Feynman also picked locks. He did like to write for effect.

"Are you basically saying Schrodinger supersedes HUP?"
Yes.

7. Jun 9, 2005

### El Hombre Invisible

Ha ha! Okay, point taken.

8. Jun 10, 2005

### El Hombre Invisible

On the question of why high uncertainty in momentum necessitates high momentum itself, while I do not yet have an answer on this, is this explanation for why an electron cannot stay in the atomic nucleus rather redundant, even putting Schrodinger aside? Whatever increase in momentum an electron acquires as it is pulled toward the nucleus will be the exact momentum required to allow it to escape it again and return to the same distance from it on the other side (in an atom at rest). Why do we need Heisenberg to tell us why an electron cannot stay in a nucleus?

9. Jun 12, 2005

### Allday

Perhaps

El Hombre,

It sounds like you are thinking of both the proton and the electron as point particles with some exact location and the uncertainty in their positions and momentum as our inability to measure their positions and momentum. This is a confusing model for this situation. I think the problem is best thought about in terms of waves which is what the schroedinger equation describes the evolution of. The HUP then just becomes a statement relating two parameters of the wave that represents the electron. The proton can be approximated as a point particle more accurately than the elecron because it is on the order of a 1000 times heavier. This makes the approximation where the proton is a point mass at rest a natural one. This is why people speak of the electron approaching the proton instead of the proton approaching the electron.

One other thing. The HUP while being a cornerstone of QM is (as far as I've seen) always applied with a factor of 2*pi fuzziness. ie it does not give exact results but rather upper limits and things like that. On the other hand, if you have some initial conditions (states or wavefunctions) for your electron and proton you can work out the exact time evolution of the statse from the Schroedinger Eq. and at any point calculate the probability density. Take the fourier transform of your state and you can calculate a momentum probability density. At any time you do this the product will satisfy the HUP.

hope that helped a little

10. Jun 13, 2005

### El Hombre Invisible

Allday, thanks for explanation. That helped a lot. You have certainly cleared up one of my questions and, as usual, when the answer is presented I see the stupidity in the question. I have read, though only in vague terms, that the 'wavelength' of a particle is dependant on energy - that a proton would have a smaller wavelength than an electron. How far off the mark is this summary to myself: as a proton is heavier than an electron, the uncertainty in its position/momentum is less than that of an electron. As an electron approaches a proton, therefore, the uncertainty in position of the electron converges with that of a proton, and so the uncertainty in momentum increases accordingly. I'm only after a vague, high level, approximate understanding of this. I will study it in more depth in a year or two. I just need to get past the explanation repeated in my original post.

I was aware that the HUP only provides an upper limit on measurement. However, I am still unclear on why a high uncertainty in momentum necessitates a high momentum itself. Why does the probability density of an electron in close proximity to a proton exclude lower momenta?

11. Jun 14, 2005

### Allday

Hi Hombre,

Every particle from quarks on up to baseballs has an associated De Broglie wavelength or just wavelength. To calculate it (without worrying about relativity) simply use the formula below

$$\lambda = h/p = h/mv$$

where h = 6.625*10^-34 Js is Plank's constant. Now momentum appears downstairs and can be associated with an energy in our case, therefore more energy = smaller wavelength = point particle better approximation = higher frequency

Now for the second part. The HUP is a powerfull thing, but you just cant use it to predict a particles motion very cleanly. Think of a one dimensional case to simplify things. Now imagine the "position" of the particle as a gaussian (bell shaped curve) centered at x=0. There is a corresponding gaussian curve for the momentum, except it is centered at some momentum p. The uncertainty in the position or momentum of the particle is the width of the respective curve. The particle can take on any value of position and momentum that lies under the curves and satisfies the HUP. As you squeeze the positon gaussian thereby eliminating uncertainty in position you have to widen the momentum gaussian. This doesn't mean that the particle necessarily has a higher momentum, but the chances are greater and if you took a million identically prepared systems and made measurements you'd find more of them with high momenta then if you hadn't restricted the postion.

Last edited: Jun 14, 2005
12. Jun 15, 2005

### Tom Mattson

Staff Emeritus
That's not true. The nonrelativistic operator for KE is $T=-\frac{\hbar^2}{2m}\nabla^2$. Any nonrelativistic momentum eigenstate is also a KE eigenstate, with eigenvalue $\frac{p^2}{2m}$. And there are corresponding relativistic versions as well.

13. Jun 15, 2005

### redwraith94

To the original op: Maybe what happens is that the electron accelerates towards the proton. Since it is able to smack the surface of the proton without any annhiliation the electron bounces off of it, and then because the electron is being pulled back because of coloumbic forces, it starts orbiting the proton.

14. Jun 15, 2005

### RandallB

I've seen how Galileo was an early version of Relativity Theory in his mechanical solution to cannon ball trajectory. Vertical and Horizontal movement independent of each other and relative to initial Vertical and Horizontal speeds vs.Time.
But I don’t see how his work is analogous to Hinesburg Uncertainty Principal – do you have a reference on the concept.

ON HOW Electron should hit Proton:
Fact is this is a true paradox that’s been around for over a century.
I don’t think it’s fair to claim that any theory – QM, Field, Feynman, Schrodinger, etc. has SOLVED or explained this paradox.
At best they can claim to have a model that gives good reliable predictions of what observations, i.e. test results, we will see from looking at what goes on between the Electron and Proton interactions.

SO El Hombre don’t expect “that this is fully explained ………later on in my studies”. To fully understand and explain the paradox. - The best you can expect is a wink and a “because I said so” just see the formula already given.
Hey it works for parents - & it’s the best current theory(s) can offer.
Same deal applies for quarks vs. quarks and for how the Strong Force even works and why.
There are many other true paradoxes that unlike the “Twins” are just not explained yet.

So there is still work to do, - but then that’s why we learn right, to help look over the hill and maybe someday see and explain whats not yet known.

Randall B

15. Jun 16, 2005

### Meir Achuz

"The HUP is like Galileo's version of mechanics".
I meant that rhetorically. That is, Galileo gave a description of some principles of mechanics, but Newton solved the whole (classical) subject.
In the same way, the HUP can't be used to solve specific problems.
It can only point out some difficulties.
Galileo was uncertain about some things, but never knew Heisenberg.

16. Jun 16, 2005

### masudr

The HUP is a point about pairs of non-commuting operators, and an inequality that the product of the expectation values of the observables represented by those operators. This is not "superseded" by the Schrodinger formalism.

17. Jun 17, 2005

### El Hombre Invisible

I'm not sure what you're saying here. I'm aware of the de Broglie wavelength, but not how the smaller wavelength of the proton can effect the position certainty of the electron as the distance between them decreases. Sorry if I'm being thick. I really haven't done a great deal of QM, but I have difficulty reading things I can't explain and moving on.

This is exactly what I wanted to know - and exactly what I had thought. A higher momentum will be more probable, but smaller momenta are not forbidden.

So, to summarise:

1. The HUP alone does not explain why an electron will not be FOUND in a nucleus, but why it will not STAY there.
2. Pending allday's reply, the certainty in position of an electron is not necessarily going to increase just because of its closer proximity to a proton.
3. Lower uncertainties in position lead to higher uncertainties in momenta, meaning higher momenta are more likely than lower momenta, but does not mean that the lower momenta are forbidden.
4. My question about pair annihilation was stupid, but kind of based on Feynman's bad wording relating to 1. above.

I'm almost there! Thanks everyone, particularly allday, for all your help. I'm most grateful.

Redwraith - this is similar to my question about why you can't use classical EM to explain why electrons do not stay in nuclei. However, I do not think an electron will smack into a proton and bounce back, as the EM force between them is attractive. As the electron and proton are different flavors, I don't know of any reason why they cannot occupy the same point in space time, therefore the electron would continue past the nucleus until the EM force overcomes its momentum once more. Again, I will bow to superior knowledge on this one.

18. Jun 17, 2005

### El Hombre Invisible

On the increase in certainty of position of the electron as it nears the proton, would this have any link to the number of positions of the electron possible decreasing as the distance decreases?

19. Jun 20, 2005

### Allday

Hi Hombre,

To answer the questions about wavelengths would take a little more time than I have, but heres a rough start. It turns out that if you try to come up with a wavefunction that represents a proton and satisfies the HUP while minimizing both uncertainties what you get is called a wave packet. Its a sinusoidal wave with an envelope that causes it to be localized. Another way to say that is, its a superposition of many different waves that have different momenta. As you localize the particle more and more, ie shrink the envelope, you introduce more and more waves into the superposition (more and more momenta increasing the uncertainty) Now a smaller envelope means that the state can be better approximated by a point particle.

Google wave packets if youd like more

20. Jun 20, 2005

### El Hombre Invisible

Hi Allday

I'll take a look, though my question was more about why the uncertainty in position of the electron decreases as it nears the proton, rather than about the certainty in position of either one of them alone. However, your post gave me a decent layman's grasp of something I had no real knowledge of til now, so for that - I thank you! You've been really helpful.