Neumaier quantum 101

1. Feb 20, 2016

cube137

I'd like to understand Neumaier programme to make quantum mechanics not weird. But his math is incredibly dense so I can't understand the thread "Quantum Mechanics is not weird.. unless presented at such"...

1. is Neumaire interpretation similar to the statistical interpretation? but what happened to single quantum system?

2. does he treat wave function as just information of the system and not real.. if so.. between emission and detection in the double slit experiment.. what is the photon or electron doing? does it have a trajectory.. is it wave or particle?

just simple questions first.. he or others can answer in verbal.. thanks..

Last edited by a moderator: Feb 20, 2016
2. Feb 20, 2016

A. Neumaier

It is similar but not identical, since it ascribes some uncertain properties to each single system. Namely a real quantity (Hermitian operator) $A$ has approximately the value $\bar A =\langle A\rangle$ to an accuracy of approximately $\sigma(A)=\sqrt{\langle(A-\bar A)^2\rangle}$. Thus if $\sigma(A)\ll |\bar A|$, the quantity is well-determined (though only with limited accuracy). Macroscopic quantities typically have this property! This is fully consistent with a statistical description, and matches exactly the way how in statistical physics, classical thermodynamics appears as a thermodynamical limit of quantum mechanics in equilibrium. This is why I call it the thermal interpretation of quantum mechanics. For more details, see my thermal interpretation FAQ.

Everything has wave nature; particles are localized concentrations of mass (except for photons) and energy. Wave functions have no meaning, except as idealizations for systems with very few and only discrete degrees of freedom. The state is instead described by a density matrix (or, more technically, a monotone linear operator). The state is real and objective, not just information.

Last edited by a moderator: Feb 20, 2016
3. Feb 20, 2016

cube137

I can't understand what you were saying in the first paragraph. Please don't use any math. Just verbal in the meantime.

For example you have detectors in a circle and a photon was emitted from the center. Does the photon travel with a trajectory to a particular detector or does it travel as wave.. if wave.. that is the meaning it travels as wave function.. upon detection.. it only appears at once detector (the wave function collapses). In your view.. what really happened?

4. Feb 20, 2016

A. Neumaier

It travels as a wave, as one knows since Huygens 1690. It doesn't appear on the detector; it disappears there, and the transmission of energy leaves random spots at a rate determined by the impinging energy density.

5. Feb 20, 2016

cube137

To make quantum mechanics not weird.. you treat the wave function as classical wave and the position of the particles as energy density impinging on the detectors. This is so far from standard view and I think there may be problems to it. Hope others can point out what may be possible problems with this view.

6. Feb 20, 2016

A. Neumaier

No; wave functions don't appear at all in my interpretation, except in very special circumstances.

I treat the field operator expectations as classical waves, with quantum corrections coming from quantum field theory.

This has a good classical limit, hence (unlike decoherence) fully accounts for a macroscopic classical world.

Because of the quantum corrections built into quantum field theory, it also completely accounts for all nonclassical behavior.

Last edited: Feb 20, 2016
7. Feb 21, 2016

naima

You give definitions, axioms for QM. When you talk about a single system or a macroscopic system, is it something inside this language or is it in our usual language?

8. Feb 21, 2016

A. Neumaier

In the usual language. My axioms are essentially those in the entry ''Postulates for the formal core of quantum mechanics '' of Chapter A1: Fundamental concepts in quantum mechanics of my theoretical physics FAQ, together with the interpretation rules mentioned above. it is the language used in statistical mechanics.

9. Feb 21, 2016

naima

As you defined quantities, ensembles and so on, could you define occupation numbers and coherent states or do you drop that things?

10. Feb 21, 2016

A. Neumaier

Coherent states are pure states, defined as eigenvectors of the corresponding annihilation operators, made out of the free field operators in the usual way by Fourier transform. No reference to particles is needed.

I never use occupation numbers - they can be dropped without loss of substance and without loss of calculational efficiency.

11. Feb 21, 2016

naima

I found no annihilation operator in your arxiv paper and the name of Fock is not cited. Are they somewhere else?
It is not a criticism just a question.

12. Feb 21, 2016

A. Neumaier

The arxiv paper is very old, and only has rudiments (not even the name) of my interpretation, which grew over time and still improves the more i think about it. Chapter 10 of my online book is the best source in publication quality, but it also says very little about the quantum field aspects. For the latter see the thread ''Are tracks in collision experiments proof of particles?''. It will take some time before something along this line will be ready for publication.

There is nothing nonstandard in my view of coherent states. How it can be constructed from annihilators is stated in many place, e.g. in wikipedia. Fock space can be constructed in several ways without using particles. One is through the representation theory of the Heisenberg groups; another is through coherent states (Bargmann construction). This is also standard. The only truly nonstandard step in my thermal interpretation is spelled out in post #2. Together with removing references to the particle picture from the fundamental level (where it can be avoided completely without loss of substance) everything follows naturally, though it took men many years to see how much - essentially everything! - can indeed be reframed in this way.

Last edited: Feb 22, 2016
13. Feb 21, 2016

cube137

Supposed the detectors were located 5 miles diameter. If you emit a photon at the center. You are saying the wave can spread literally and one detector would be hit depending on the energy density? But the physical wave can dilute while spreading.. while in the mainstream view the omnipotent wave function spreads and doesn't dilute and upon reaching the detectors.. one of the detector is hit and the wave function physically collapses (and the photon reappear) so it won't hit other detectors. This makes better sense.

14. Feb 21, 2016

Staff: Mentor

This particular formulation of the problem bothered Einstein too.... However...

If thinking in terms of "physical waves" that spread and dilute and collapse works for you, then you are free to think about it in those terms. But the only things that you can take to the bank are: one detector will trigger; and quantum mechanics correctly predicts the probability of any giiven detector triggering.

15. Feb 22, 2016

A. Neumaier

In your situation, my view is identical to the mainstream view. The wave function dilutes like the physical wave. This dilution has a very natural explanation:

The intensity of classical light (equivalently the probability of detecting a particular photon) from a distant star decays like the inverse squared distance. To get a significant detection probability at a particular position you need to emit sufficiently strong light (equivalently sufficiently many photons) so that enough light (equivalently enough photons; just one in case of an ideal detector) falls on your detector.

The quantum mechanics of the detector alone predicts, and experiments confirm, that both for classical electromagnetic waves and for quantum photons, the detection rate is exactly the same. Differences only start to appear when one considers nonclassical light and correlated detections.

16. Feb 22, 2016

vanhees71

This is like demanding to explain you something but forbidding to talk at all!

17. Feb 22, 2016

cube137

Can someone suggest an experiment to resolve it? Neumaier suggested that wave functions are physical field and the particle is a result of the energy density appearing just in one part resulting in what you called "particle". Why didn't Schroedinger think of this or the other QM pioneers? What are their objections etc?

18. Feb 23, 2016

A. Neumaier

Schroedinger interpreted in 1926 the squared absolute value of the single-particle wave function as the (scaled) charge density of an electron, in agreement with the thermal interpretation.
This is why Schroedinger's interpretation didn't extend to multi-particle states. In this case his interpretation must be modified: In the general case, the charge density is the expectation of the charge operator in the given state, in agreement with the thermal interpretation. This is the way the charge distribution of molecules is today determined in computational quantum chemistry by packages such as GAUSSIAN.
Quantum field theory is a more fundamental theory than multiparticle quantum mechanics. The latter is obtained in some approximation. My thermal interpretation is based on quantum field theory and fully consistent with its predictions. Therefore also with quantum mechanics to the extent that the latter is a valid approximation to quantum field theory.

19. Feb 23, 2016

A. Neumaier

It is very difficult if not impossible to understand quantum mechanics and quantum field theory without using formulas. One can get only an extremely superficial picture of it and is prone to forming lots of misunderstandings. Already the concept of a wave function is a mathematical concept. So is the concept of entanglement, the notion of entropy, the Schroedinger equation, and in fact the concept of everything people are talking about when discussing quantum mechanics informally....

20. Feb 23, 2016

vanhees71

Schrödinger's initial interperation does not hold against experiment. A single electron is never observed as a smeared charge distribution as this interpretation would suggest but always as a single "point particle". That's why finally Born's probability interpretation became the standard interpretation for the physics content of the quantum state (or the wave function for non-relativistic particles, which is just representing a pure state in the position representation of the corresponding state vector).

21. Feb 23, 2016

A. Neumaier

Never is far too strong! Your statement holds at best for free electrons in a high energy beam. But most electrons are bound - then the point particle picture gives a very poor description of the electron's properties.

In a classical external Coulomb field (approximately realized for an electron in a hydrogen atom), a single low energy electron is never observed as a point particle but as a charge distribution, which has in the ground state the form of a fuzzy sphere whose radius is approximately given by 1.5 times the Bohr radius. This is the special case of the charge distribution of an atom or molecule, as used in quantum chemistry and as measured in subatomic resolution microscopes.

(Taken from the introduction to my theoretical Physics FAQ entry How do atoms and molecules look like?)

Last edited by a moderator: May 7, 2017
22. Feb 23, 2016

A. Neumaier

Even for the electrons in an electron beam you cannot decide this experimentally.

How would you distinguish experimentally a fast flying point particle with charge $e$ from a narrow cigar-shaped charge distribution (integrating to a charge of $e$) with the long side in the direction of flight? The latter is the charge distribution of an electron in a Gaussian coherent state with fairly sharp well-determined momentum and well-determined traversal position but therefore poorly determined position in longitudinal direction.

Last edited: Feb 23, 2016
23. Feb 23, 2016

vanhees71

It should be clear from the abstract of the above cited paper that here you measure averages, i.e., you measure the probability distribution. A tunnel microscope always does so, and that's its purpose. It's not the detection of the position of a single electron.

24. Feb 23, 2016

vanhees71

No matter, how poorly determined the position of the single electron might be, you'll always detect a single electron at a point (in the sense of the resolution of your position measurement). You get the broad distribution of the longitudinal position only when measuring an ensemble of such prepared electrons. That's at the heart of the probabilitistic interpretation (Born's rule) of the quantum states. If Schrödinger were right with his interpretation, we'd never have had debates about the meaning of quantum theory, and electrons were simply described by classical complex valued fields!

25. Feb 23, 2016

A. Neumaier

They don't claim that it is ''the detection of the position of a single electron''; neither did I. But they do claim that they measure the stationary charge density, resp. electron density of the electrons in a stationary state of an atom or molecule.