Clarifying Neumaier's Interpretation of Quantum Mechanics: Wave or Particle?

[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..

 

[A. Neumaier]

Is Neumaier's interpretation similar to the statistical interpretation? but what happened to single quantum system?

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.

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?

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.

 

[cube137]

(Please spell my name correctly, and edit in particular the title of your thread!)

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.

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?

 

[A. Neumaier]

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 one detector (the wave function collapses). In your view.. what really happened?

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.

 

[cube137]

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.

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.

 

[A. Neumaier]

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.

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.

 

[naima]

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!

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?

 

[A. Neumaier]

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?

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.

 

[naima]

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

 

[A. Neumaier]

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

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.

 

[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.

 

[A. Neumaier]

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

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.

 

[cube137]

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.

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.

 

[Nugatory]

Supposed the detectors were located 5 miles diameter. If you emit a photon at the center.

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

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.

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.

 

[A. Neumaier]

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.

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.

 

[vanhees71]

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

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

 

[cube137]

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.

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?

 

[A. Neumaier]

Neumaier suggested that wave functions are physical fields

Please re-read more closely what I had stated.
You misunderstand what I said although I had corrected your statement already:

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.

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?

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.

Differences only start to appear when one considers nonclassical light and correlated detections.

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.

What are their objections etc?

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.

 

[A. Neumaier]

I'd like to understand Neumaier programme to make quantum mechanics not weird.

Please don't use any math. Just verbal in the meantime.

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...

 

[vanhees71]

Please re-read more closely what I had stated.
You misunderstand what I said although I had corrected your statement already: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.

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).

 

[A. Neumaier]

A single electron is never observed as a smeared charge distribution as this interpretation would suggest but always as a single "point particle".

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.

M. Herz, F.J. Giessibl and J. Mannhart http://prola.aps.org/pdf/PRB/v68/i4/e045301 Phys. Rev. B 68, 045301 (2003)
write in the introduction:
''quantum mechanics specifies the probability of finding an electron at position x relative to the nucleus. This probability is determined by |psi(x)|^2, where psi(x) is the wave function of the electron given by Schroedinger's equation. The product of -e and |psi(x)|^2 is usually interpreted as charge density, because the electrons in an atom move so fast that the forces they exert on other charges are essentially equal to the forces caused by a static charge distribution
-e|psi(x)|^2.''
One of the authors, Jochen Mannhart, is one of the winners of the German Leibniz prize 2008 and the EPS Europhysics Prize 2014. He obtained the Leibniz prize among others for the achievement that, for the first time, he made pictures of atoms with subatomic resolution possible. The Leibniz prize is the highest German academic prize, endowed with a research grant of up to 2.5 Million Euro for each winner, awarded each year to a few excellent younger scientists from all sciences.

The atomic shape that one can measure is actually a 3-dimensional charge density rho(x) (x in R^3).
''It is important to stress here that, unlike a many-electron wave function, the electron density is an observable; for example the X-ray scattering from an atomic or molecular gas gives rather direct information about the spatial distribution of electrons.'' (This is quoted verbatim from the introduction of the chapter N.H. March, Origins - The Thomas-Fermi theory, pp. 1-77 in: Theory of the Inhomogeneous Electron Gas. Physics of Solids and Liquids, Springer 1983.

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

See also the insight article ''Can we see an atom?''

 

[A. Neumaier]

A single electron is never observed as a smeared charge distribution as this interpretation would suggest but always as a single "point particle".

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.

 

[vanhees71]

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?)

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.

 

[vanhees71]

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.

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!

 

[A. Neumaier]

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.

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.

 

[A. Neumaier]

you'll always detect a single electron at a point (in the sense of the resolution of your position measurement).

I was asking how you distinguish the two possibilities experimentally. What you detect is a macroscopic impact region, which is just the geometric cross section of the cigar or point impacting on the measurement element. From the macroscopic size of the impact region one cannot deduce more than a bound on the diameter of the cross section. Thus the measured impact cannot distinguish between a charge distributions in form of a cigar, a point, or a ball, once their width is below the size of the oil drops or steam bubbles or silver corns, or whatever is used for detection.

That you deduce the presence of a point charge is pure speculation.

Deducing the cigar form is consistent with the experiments, and it is much more in line with what is actually observable once the objects become slightly larger (atom size, but still quantum) where one can probe their surface. Since it allows the extrapolation from bigger objects to smaller objects it is the more natural conclusion.

 

[A. Neumaier]

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!

Schroedinger was right for the single electron, but not for multi-electron states. There the square of the absolute value of the wave function means nothing, but $\rho(x):=\mbox{tr}~\rho j_0(x)$, where $\rho$ is the multi-electronic state and $j_0(x)$ is the time component of the current operator, still describes the measurable charge density. At least all quantum chemists rely on this!

 

[vanhees71]

Schrödinger was not right with his initial interpretation of the wave function. He took it as a classical field and considered the electron as the corresponding extended charge distribution $\propto |\psi(t,\vec{x})|^2$. This does not hold empirically and that's why Born got a Nobel prize for the now accepted probabilistic interpretation. Schrödinger didn't like this at all, and he lamented about having invented wave mechanics later on.

 

[A. Neumaier]

Born got a Nobel prize for the now accepted probabilistic interpretation

... of the multiparticle situation, where Schroedinger didn't assert anything since it was a function $\psi(x_1,\ldots,x_N)$ of too many coordinates.

But in the single particle case, the two descriptions are experimentally compatible with each other if one invokes ergodicity (as Mannhart does). Even in the case of electrons in a well-collimated beam!

 

[vanhees71]

How can the two descriptions be experimentally compatible? Take a single electron prepared with pretty well determined momentum. Then the spatial spread of its wave function will be large. Put a high-position resolving detector in its way. According to Schrödinger each single (!) electron should give a wide-spread continuous distribution, while we always find a point-like track, and only an ensemble of such prepared electrons give the distribution function. That's a qualitative difference between Schrödinger's original classical-wave and Born's probabilistic interpretation of the single-particle wave function.

I don't know, how Schrödinger interpreted multi-particle wave functions either. If needed, I can look for the original papers, which should be online (I think it was published in Zeitschrift für Physik, which is available online completely).

 

[A. Neumaier]

I don't know, how Schrödinger interpreted multi-particle wave functions either.

He didn't. I think he just remarked that his 1-particle recipe no longer worked.

According to Schrödinger each single (!) electron should give a wide-spread continuous distribution, while we always find a point-like track

Why should it? In the situation you describe, the charge distribution of the single electron before the first ionization is not measurable at all!

The attempt to measure it heavily affects the state of the electron. As discussed in another thread in the context of Mott's analysis, the very first ionization effected by the single electron changes its state to one with a tiny cross section (reasonably well-defined transversal position) and a reasonably well-determined longitudinal momentum only, with a cigar-shaped charge distribution, whose track is measured!

The only situations where the charge distribution of a single electron can be measured are those where the measurement doesn't influence this distribution significantly, so that repeated measurements can be made. This is the case

• (i) for an electron in a well-collimated beam, and
• (ii) for an electron in a temporally stable stationary state.

In the first case, the charge distribution is cigar shaped and the experimental tracks confirm this. In the second case, the electron must be in the ground state, the charge distribution is shaped according to an equipotential surface of the static electric potential it experiences, and atomic resolution microscopy confirms this, too.

 

[vanhees71]

Sure, and that's why Schrödinger's original interpretation doesn't make physical sense!

Case (i) measures not a single-electron position but the position of many electrons. Very often you can consider the electrons in such a "macroscopic beam" as independent particles (which, however, is no longer the case with modern accelerators of high-intensity beams; the space-charge effects are a complicated problem there!) and thus take it as (approximate) ensemble. So you measure (approximately) a single-particle distribution function through ensemble averaging.

Case (ii) describes what's measured with a tunnel microscope!

 

[A. Neumaier]

Sure, and that's why Schrödinger's original interpretation doesn't make physical sense!

Of course it does. In all cases where one can measure the charge distribution of a single electron it conforms to his interpretation. Moreover, his interpretation is just a special case of the interpretation of the expectation of the charge density operator as the observable charge density, which is something that is verifiable both macroscopically through the classical limit of electrodynamics and microscopically through a tunneling microscope.

Thus it is safe to assume it also in cases where one cannot measure it. After all, we also assume that the standard properties of macroscopic bodies exist and behave according to the laws of physics as found where we can measure it - even in cases such as the deep interior of the Earth or the sun where we cannot measure anything (except indirectly by assuming the laws)!

 

[A. Neumaier]

Case (i) measures not a single-electron position but the position of many electrons.

?

A single track measures a single electron, as you had asserted before:

each single (!) electron [...] we always find a point-like track

 

[vanhees71]

If we have a single electron and measure its position we see a pointlike trace on a screen (or nowadays a pixel in a digital device) but not a spread charge distribution. Only an ensemble of electrons, each leaving such a pointlike trace, measures the spread position probability distribution. If you were right, and Schrödinger's interpretation were correct, we'd not have all these discussions on interpretation at all, because then quantum theory would be simply a classical field theory as classical electrodynamics, but that's not what's observed in nature! I don't know, if you've ever seen this in a lab. As physicists we have had to do such experiments in the mandatory labs, and that's why it's clear to us!

BTW: I tried to find a paper by Schrödinger on many-electron atoms. So far to no success. He only discusses the single-particle wave function and clearly rejects the probabilistic interpretation (in the last section of the book containing his "Four lectures on wave mechanics").

 

[A. Neumaier]

clearly rejects the probabilistic interpretation (in the last section of the book containing his "Four lectures on wave mechanics").

Can you please quote his rejection statement and his reasons for it?

 

[vanhees71]

Schrödinger, "Four Lectures on Wave Mechanics" (1928)

An obvious statistical interpretation of the $\psi$-function
has been put forward, viz. that it does not relate to a
single system at all but to an assemblage of systems,
$\psi \bar{\psi}$ determining the fraction of the systems which happen
to be in a definite configuration. This view is a little
unsatisfactory, since it offers no explanation whatever
why the Quantities $a_{ki}$ yield all the information which they
do yield*. In connexion with the statistical interpretation
it has been said that to any physical quantity which would
have a definite physical meaning and be in principle (prin-
cipiell) measurable according to the classical picture of the
atom, there belong definite proper values (just as e.g.
the proper values $E_k$ belong to the energy); and it has
been said that the result of measuring such a quantity
will always be one or the other of these proper values,
but never anything intermediate. It seems to me that
this 1st statement contains a rather vague conception, namely
that of measuring a quantity (e.g. energy or moment of
momentum), which relates to the classical picture of the
atom, i.e. to an obviously wrong one. Is it not rather bold
to interpret measurements according to a picture which
we know to be wrong? May they not have quite another
meaning according to the picture which will finally be
forced upon our mind? For example: let a beam of
electronic rays pass through a layer of mercury vapour, and
measure the deflection of the beam in an electric and in a
magnetic field before and after the beam has traversed
the vapour. According to the older conceptions this is
interpreted as a measurement of differences of energy-
levels in the mercury atom. The wave-picture furnishes
another interpretation, namely, that the frequency of
part of the electronic waves has been diminished by an
amount equal to the difference of two proper frequencies
of the mercury. Is it quite certain that these two
interpretations do not interfere with one another, and that the
old one can be maintained together with the new one?
Is it quite certain that the conception of energy,
indispensable as it is in macroscopic phenomena, has any
other meaning in micro-mechanical phenomena than the
number of vibrations in h seconds?

 

[A. Neumaier]

If we have a single electron and measure its position we see a pointlike trace on a screen (or nowadays a pixel in a digital device) but not a spread charge distribution.

As mentioned before in post #31, this is because what is measured is the track after the first ionization, where the single electron has a cigar-shaped charge distribution and moves along the cigar's long direction. Nobody expects to measure a spread charge distribution in this case.

 

[vanhees71]

Exactly! So why do you claim the correctness of Schrödinger's original interpretation of the single-particle wavefunction, which clearly is a misconception.

 

[A. Neumaier]

Exactly! So why do you claim the correctness of Schrödinger's original interpretation of the single-particle wavefunction, which clearly is a misconception.

Because in all cases where one can measure the charge distribution of a single electron it conforms to his interpretation. Before the first ionization, we cannot measure the charge distribution of the electron. So we have either deny that it has a charge distribution (which is the take of the Copenhagen interpretation), or to find a definition of the charge distribution that makes sense both in the observable and in the unobservable case (which, for a single electron, is Schroedinger's interpretation). Both points of view completely agree on the measurable part, hence are equally valid.

But the Copenhagen view has a very strange side effect: The mean of something nonexistent exists!?? How can this be? The ensemble interpretation inherits this strange property, though in a less outspoken way.

 

[vanhees71]

No they do not agree on the measurable part, because the one states that we measure a point-like location and the other states that we measure a smeared distribution. Everybody who has ever made an experiment with a weak radioactive source finds points on the scintillation screen. No single event gives a smeared distribution. QT is about what's measurable and observable not about theoretical constructs.

Scanning a probe with a tunnel microscope measures ensembles and not locations of a single particle! That's why it can resolve the distributions.

The ensemble interpretation doesn't claim that something nonexistent exists. Also the no-nonsense flavor of Copenhagen (Bohr himself included; Heisenberg has to be taken with a grain of salt ;-)) hasn't ever claimed such a thing. The only, in my opinion correct, claim is that an observable has a predetermined value if and only if the state of the system says so, i.e., if the probability to find this value is 1. That's a tautology, if you ask me!

Take the example of an electron in a hydrogen atom in its ground state. It has no definite position but a probability distribution of positions. Whenever measured you find a point-like particle (for which you usually have to destroy the atom, i.e., you have to kick out the electron, if you want a sufficient spatial resolution). If you want to measure the probability distribution you need an ensemble of hydrogen atoms. You can also use the same atom many times, you only have to make sure that you always prepare it in its ground state again before you measure the electron's position again.

 

[A. Neumaier]

Take the example of an electron in a hydrogen atom in its ground state. It has no definite position but a probability distribution of positions.

What does it mean for a single electron to have a probability distribution of positions? The latter is a property of an ensemble, not of a single system!

QT is about what's measurable and observable not about theoretical constructs.

Then what is measurable and observable about the probability distribution of positions of a single electron?

 

[A. Neumaier]

Everybody who has ever made an experiment with a weak radioactive source finds points on the scintillation screen. No single event gives a smeared distribution.

Again this is since, by Mott's argument, the first ionization changes the state of the electron to one with a point-like geometric cross section, which is then measured, and of course leads to a point-like flash. The measurement (irreversible amplification) is not one of the unperturbed electron in its spherical state emanating from the source but one of the perturbed state caused by the interaction leading to the first ionization.

 

[A. Neumaier]

Take the example of an electron in a hydrogen atom in its ground state. It has no definite position but a probability distribution of positions. Whenever measured you find a point-like particle (for which you usually have to destroy the atom, i.e., you have to kick out the electron, if you want a sufficient spatial resolution). [...] you need an ensemble of hydrogen atoms.

Don't you agree that destroying the atom definitely changes the state of the electron?

Consider the following analogous situation: To measure the chemical structure of a complex molecule you have to destroy sufficiently many copies of it, split them into pieces that can be analyzed by a mass spectrometer, and afterwards find out how overlapping fragments can be combined to a whole. At the end, if you are able to reconstruct a unique chemical formula, your measurement was successful.

What you try to argue is like saying that in the analogous situation that whenever you measure the complex molecule you always find fragments, and conclude that the complex molecule must be an ensemble of fragments with the remarkable property that they seem to come from a single molecule that is never observed in the measurement process. Whereas the natural interpretation is that there is a single molecule, which changes its state during the measurement, so that only its fragments are measured.

In the same vein, rather than arguing that when analysing an atom one always measures rays impinging on a screen that seem to come from a spherical charge distribution which is only visible in the ensemble, it is far more natural to assume that the spherical charge distribution exists but is destroyed by the measurement process, so that the latter has to be repeated often enough to recover the original, spherical structure.

The example discussed in posts #47 and #49 of the thread Are tracks in collision experiments proof of particles? suggest the same.

 

[vanhees71]

What does it mean for a single electron to have a probability distribution of positions? The latter is a property of an ensemble, not of a single system!

Then what is measurable and observable about the probability distribution of positions of a single electron?

As you say: You prepare an ensemble of independently prepared single-electrons and measure their positions, leading to the probability distribution by the usual statistical evaluation procedures. The position of the single electron is not determined when prepared in the hydrogen-atom ground state, but you know the probability distribution by solving the time-independent Schrödinger equation. You can verify the predicted probability distribution only on an ensemble. This is the standard point of view of the ensemble interpretation.

 

[vanhees71]

Again this is since, by Mott's argument, the first ionization changes the state of the electron to one with a point-like geometric cross section, which is then measured, and of course leads to a point-like flash. The measurement (irreversible amplification) is not one of the unperturbed electron in its spherical state emanating from the source but one of the perturbed state caused by the interaction leading to the first ionization.

That's what makes such a scientillator a good position-measurement device. It's one way to define the position observable operationally, and without such a definition the idea of observables is empty.