# Common quantum question

1. May 21, 2007

### muh1616715

I'm sure this has been asked before, but I've been searching for a clear answer as to why the Uncertainty Principle is a representation of physical reality and not just a result of our imperfect measuring techniques. Is an electron still considered to be a discrete particle orbiting the nucleus, or is the distribution of probability wave an accurate representation of how the atom physically exists? I understand that the act of measuring an electron's momentum will decrease the accuracy to which we can measure its position (and vice versa), but how does the electron cloud physically exist at any moment in time? With the electrons as discrete particles, or as a distribution of probability?

Also, I'm pretty sure that protons and neutrons are actual spherical objects. Is the same true for electrons, or is this an unanswered question?

2. May 22, 2007

### Anonym

Perhaps the wrong wording is confusing you. The Heisenberg “Uncertainty Principle” is the Heisenberg Dispersion Relation and has nothing to do neither with the measurements nor with the accuracy of the measurements.

The dispersions are eigenvalues of the self adjoint operators (observables). They clearly indicate that the physical object under consideration is field and provide the parameters for description of the corresponding wave packets. For example, in case when the physical system (physical reality) contains the single (discrete,n=1) electron moving freely (v<<c) or bounded by external spherically symmetric EM field, it is spherical object. Indeed, the waveforms are entirely different in each case.

How does the electron cloud physically exist at any moment in time? That QM is all about.

The same true for free protons and neutrons. If v~c they more similar to pita (discus).

Regards, Dany.

Last edited: May 22, 2007
3. May 22, 2007

### Meir Achuz

The HUP is just the Fourier transform relation that (Delta k)(Delta x)>1/2 for any wave packet. This holds even for a classical light wave of finite extent.
The connection of the HUP with measurement problems was just bad advice given to Heisenberg by Bohr.

4. May 22, 2007

### Meir Achuz

The proton and neutron are not spherical. The two u quarks in the proton are further apart than the u-d quark distance.
The electron is believed to be a point particle, so its shape is not defined.

5. May 22, 2007

The HUP isn't really the representation, the wave and states of the particles are more like it, but of course the HUP governs these.

It isn't just a result of imperfect measurement. As it turns out, the probability functions are the only way to describe what is happening, and they describe it perfectly, as far as we can tell. Since Newtons laws of motion perfectly describe our classical world, we accept them as factual, and the same goes for QM.

Electrons haven't been considered actual orbiting objects for 100 years, even before QM proper.

Well, that's the mystery. QM doesn't make the distinction. It has a model that describes and predicts the electron in it's various possible situations, and calculates it as a wave or particle according to context. The function is the electron. You must first accept this. The reality of the particle in QM is the function. It's the only way it all works and makes sense. It's bizarre, but there you go. You can only convince yourself that it is true by reading and understanding the actual experiments yourself.

It's not true of protons, neutrons, or even atoms, though by the time you start talking about atoms, the wave function fringes are so small that it begins to behave more predominately spherical. On the other hand, all particles have spherical probability domains, so the distinction is not a useful one.

6. May 22, 2007

I think it might be OK to ask some basic questions in this thread...

What do you mean? If the HUP defines the relative resolution of the probability components, how isn't it part of the accuracy of measurement?

Could you state this in more conceptual terms, i.e. how an "observable" translates into the above?

Could you explain more about "field" vs physical object? Are you talking about a mathematical, vector, or EM field?

I'm having trouble looking up pita in this context, i.e. I keep getting "pita bread". Pointers?

7. May 22, 2007

### Dr.Calpol3

electrons have always been in orbit around the nucleous which is made up of protons and neutrons!

8. May 22, 2007

### Anonym

The inequality has only qualitative meaning. To make it the quantitative statement you should consider the specific realization (R. Jackiw, JMP, 9, 339 (1968)).

I have impression that it was much more than advice (W. Heisenberg as well as E. Schrödinger were ill after discussions with N.Bohr).

At high energy region. At low energy it is coherent wave packet.

Regards, Dany.

9. May 22, 2007

### Anonym

HDR is the terminology used originally by W.Heisenberg. It expresses the experimental evidence that QM system is described by the dynamical variables which are not commuting in general (matrices). The measurement instruments are the macroscopic objects and obey the laws of the classical physics. The accuracy of measurement is well defined notion within the classical physics (statistical sampling, etc.).

Delta A=<A^2>-<A>^2 for every A self-ajoint operator.

OP asks about the single electron “orbiting the nucleus”. It described by the Dirac field or by Pauli-Schrödinger field in the non-relativistic limit (v<<c). I add the free electron consideration (the coherent wave packet) for completeness.

Your questions make me feel that you arrived from Alpha Centauri half hour ago.

Good morning!

Regards, Dany.

10. May 22, 2007

### muh1616715

Ok, now I'm more confused than when I asked the question. Can someone explain this in understandable terms without using wave function math (and ideally with proper grammar)?

11. May 22, 2007

### Mentz114

No. This well covered in the posts above. The HUP follows from the wave-particle duality. It can't be explained in 'billiard ball' terms.

Completely answered by the posts above. You have to stop thinking classically, or none of this will ever make sense.

12. May 23, 2007

### Anonym

You apparently discuss physics using English. I discuss physics apparently using English.

When you write:” Is an electron still considered to be a discrete particle orbiting the nucleus, or is the distribution of probability wave an accurate representation of how the atom physically exists?”, it perhaps sounds for you meaningful in English, but it is nonsense physically, since you use undefined notions (for you) and it were already demonstrated that they are not adequate within the Quantum World.

When I write:” the single (discrete, n=1) electron, I mean that in the discussed problem the operator of number of particles is observable (self-adjoint which commute with Hamiltonian and has eigenvalue equal to one).

Physics is a science and not a philosophy, journalism or Hollywood movie. It based on the results of the measurements and uses the mathematics for its processing. In addition, the relevant math is simple, deep and much more beautiful than any science fiction book or movie.

Regards, Dany.

13. May 23, 2007

### XVX

muh,

Since QM has wave solutions, like analyzing any properties of a wave, there are relations. The Uncertainty Principle is taking those universal properties of waves and applying it to matter and energy. It is not a result of imperfection in equipment or interaction of particles, is has to do with that your experimenting on something that has wave properties.

Based on my QM interpretation:
And as far as what the electron looks like is entirely dependent upon what interpretation you have. I believe in matter waves. Just like if I took a wave from the ocean and changed its potential by putting it into my pool, it would change its form to the new conditions. A free electron with a Gaussian for a minimum wave packet, which appears localized, would change its form when encountering a proton. An electron in hydrogen is a trapped wave. The electron's matter wave has formed to the new potential and it no longer appears like a Gaussian wave that is localized.

If you could actually measure the position of the electron as it flies around the proton, you would force the electron to localize into a Gaussian, somewhere within its probability cloud and then it would smear out again according to the Schrodinger equation. Having a Gaussian wave packet for the electron as the initial state for the hydrogen equation is do-able. But the actual act of measuring, causing the wave packet to become Gaussian or a delta function is what’s controversial.

Overall, there is no true consensus on a QM interpretation. The Copenhagen Interpretation is the most popular and in that case, the electron is a point particle with properties that can be described by a mathematical wave equation.

14. May 23, 2007

### Meir Achuz

The wave packet describes the position of a particle, not the particle.
A "point particle" is usually defined as one whose form factor is a constant:
F(q^2)=1.

15. May 23, 2007

### Anonym

How you define and calculate form factor in case of the Schrödinger (Die Naturwissenschaften, 14,664,(1926)) coherent state?

Regards, Dany.

16. May 23, 2007

### lightarrow

Can you explain me this better? You mean "coherence length" greater than source-detector distance or what exactly?
(spassibo).

17. May 23, 2007

### muh1616715

Thanks, your response is the only one I've really understood so far. Are you saying that we don't actually know how the electron physically exists, and that using wave functions is the only way to accurately describe how it behaves at this point?

18. May 23, 2007

### muh1616715

I'm sure the physical state of the electron can be explained in words without using equations. I'm not asking for an analogy, but a straightfoward answer to a straightforward question: How does an electron exist physically? A) as a point particle with wave properties, B) just as a wave function, C) we don't know, D) something else.

Also, doesn't all matter have wave properties? Don't people have wave properties? Why can macroscopic objects be thougt of as actual objects, while electrons can only be accurately represented by wave functions? Is it because of their small size, or is it because they HAVE NO actual size? And if the latter is true, how is that possible?

19. May 23, 2007

### Mentz114

In different circumstances it can look like A) or B). Evidence from scattering experiments indicate that the electron has spatial extent. But it looks smaller when probed with higher energy particles. Evidence from diffraction experiments shows it is also a wave.

I don't have a view about what it 'really' is because it is not relevant to making calculations.

20. May 23, 2007

### muh1616715

What about protons and neutrons, and other subatomic particles? Do any of these have defined sizes (I know they have defined masses) the way macroscopic objects do? Is there some fundamental difference between matter at the subatomic level and matter at larger scales? Since everything is made of this subatomic matter, why should macroscopic objects have defined sizes and shapes, but not the individual particles they are made of? I'm just trying to clear up whether there IS such a distinction, or if we simply don't understand what is going on at these small scales yet.

21. May 24, 2007

### Mentz114

Protons and neutrons display similar characteristics to the electron when probed.
Not sure what you mean by this. You can use quantum mechanics to solve classical problems so the difference is not so fundemental.

It's like Dolby noise reduction - adding up the random uncertainties in position (say) of zillions of atoms reduces the randomness to give more precise boundaries and positions. Macroscopic bodies still obey quantum rules, so to speak.

22. May 24, 2007

I'll have to take this as a friendly jocular way of saying that I'm way behind in my understand of QM. This is of course evident. If you would forgive this, you could help my learning process.

Your original answer is cryptic for someone asking a basic question. It doesn't illuminate the distinction you are making. But, I've done some more reading, and I think I understand.

First of all, are you saying that "HUP" is obsolete in light of "HDR"?

A macroscopic instrument set, with its own intrinsic precision and accuracy, which measures the position and momentum of a quantum object will yield measurements which will have to be understood in terms of the dispersion relation between position and momentum, meaning that the two measurements cannot simultaneously be known to be 100% certain (not "accurate", right?) representations of the object.

You can have an instrument set with 99% accuracy, so you know that you have position and momentum measurements with an apparatus error of 1%. With respect to HDR, you can put this error in a box and incorporate it some other time.

So, by now you've written down some numbers from your measurements, and you know that there is an uncertainty relation between them. Again, sorry if this is a basic question...

I assume that

$$\Delta x_i \Delta p_i >= h/2$$

means that you have to apply this to two data sets, and never to single measurements?

This is why you said, or rather implied, that "accuracy" has no representation in the HDR formulas?

Ok, I see now. I still have the words "Dirac equation", probability amplitudes, and "Schrödinger equation" ringing in my ears, and "wavefunction" instead of "field".

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Last edited: May 24, 2007
23. May 24, 2007

Yeh, kinda. Macro matter sizes are dependent on electron/valence shell interations (binding/ repulsing) of their component atoms.

Sub-atomic sizes are dependent on other things, like the Pauli exclusion principle, which says that no two particles can occupy the same state at the same time.

As to how a state's diameter is defined, I hope somebody else will answer. I'm guessing it has to do probability distribution over an area, but even so, what would the area be. We'll away somebody who knows.

See above. It's also a question of aggregate diameter, and averaging. In macro objects, you *think* that an object has a specific border, but if you were to try to find that border with an electron microscope, you would find a huge jagged terrain with no specific defining line. What is the height of the grass on your lawn? Overall you could say, 2 inches, but as you can see, that's an average.

24. May 24, 2007

### Anonym

For hep, see Meir Achuz, post #14 and my question #15. Follow it, I expect receiving very interesting answer.

Low energy. I consider single electron set-up only (A. Tonomura et al., 1989). Detailed explanations you may find in M.P. Silverman “More Than One Mystery”, Ch.1, Springer-Verlag, 1994.

I guess you mean longitudinal coherence length (You still worried about whether the size is a matter?). The simple answer is no. M.P. Silverman: ”In order for wave packets to overlap and interfere, their optical path length difference between source and detector must not be much in excess of longitudinal coherence length.”

But in this experiment the longitudinal coherence length is ~10^(-4)cm, which may be considered macroscopic. Now, your question looks to me very interesting, but I don’t know how to arrange the measurement. What happens when electron still did not finish penetration through beam splitter but already arrived to the detector? I have no doubt that in usual case beam splitter is not a measurement device and no collapse have place caused by it, but in your case something interesting may happens. What? I have no idea (Pojaluysta).

Regards, Dany.

25. May 24, 2007

### Anonym

Sure.

No. It is perversion of Heisenberg original explanation. Let me translate V.A.Fock that may be not available for you:” Most easily accepted the mathematical formalism of QM, and easier when it more and more complicated. When one wrote the relation between coordinate and momentum in the form of CR, i.e. as the connection between rather complicated mathematical notions, they did not meet objections. However, when W. Heisenberg presented the deduced from them but much more simple relation
Delta(x)*Delta(p)>=h/2

Then storm of objections rose”. Today the situation is not essentially better.

No. One data set contained about 100 000 outcomes. I use the QM property that the identical particles are indistinguishable and apply standard repeatability requirement. The same picture I will obtain in any other laboratory (inertial frame) as required from the measurements to present the objective reality. If the physical object is extended (like your face) you never will get the picture using one pixel.

Let me explain my attitude. Suppose you came to restaurant. What you prefer to eat? Originally prepared delicatessen or second hand vomiting? There are very good textbooks on QM. But in addition I send someone asking basic questions to obtain first hand answers from E. Schrödinger, W. Heisenberg, P.A.M. Dirac, V.A. Fock, A. Einstein, C. N. Yang and R. Feynman. That someone suddenly will find that to read and understand paper by A. Einstein is much easier than Moishe Zuchter. After that I suggest to study D.Hilbert, E. Cartan, H. Weyl, E. Wigner and J. von Neumann (for the theorists only). If that someone will survive, then I will start to explain him my story.

Regards, Dany.