QM and electron orbits

Laura wrote:
"Andrew Usher" wrote in message
om...
This message is a continuation of the discussion in the thread
'Neutrino mass'.

It is more like a reiteration of your position, already stated in that
thread.


I admit to not being formally educated in QM.

Neither am I.
But I try not to criticise things I don't understand.

I am nevertheless trying
to criticise a belief normally taught in such education.

If you're referring to the idea of the electron being "smeared" across the
orbital, then it is you who has misunderstood.
"In a general paper on quantum mechanics, Schroedinger discusses and rejects
the interpretation that a single quantum is somehow phyiscally "spread out"
or "blurred" among the different parts of a superposition ."
That is what is being taught.

To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.

Although I
don't understand the math involved in the conventional approach, I
believe that I can understand the basics in terms of logic.

The false idea is that the Bohr-Sommerfeld orbits are an incorrect and
obsolete model of the atom. As we know, the idea of fixed orbits is
not exactly correct, but that does not make it useless.

It is very useful for chemistry and nuclear physics, but it is a model
and not meant to be taken as a true picture of the atom.

As valuable and useful as the QM model then, it would seem. No ?

The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct description. This
leads to the false belief that an electron's position is smeared out
over the orbital,

That's *your* false belief. The electron isn't "smeared out over the
orbital" (just as Schroedinger's cat isn't alive AND dead). It just doesn't
have a position until one has been measured (just as you don't know if the
cat is alive or dead until you open the chamber), and then that position is
one that is affected by the act of measuring, and is therefore not a true
representation of where the electron really was, if indeed it had a position
at all.
That's why such an innate position is done away with entirely; it is
meaningless since it can't be measured without interfering with it.

and that the probability function is independent of
earlier observations. From the uncertainty principle (which states
that particles occupy h^3 in phase space), this can only be strictly
true for the 1s orbital. For all higher n, the relative uncertainty
becomes smaller, and the classical orbit becomes an increasingly
better approximation.

What difference would it make?
If none, then what is the value of your version?


This explains the solar system, for example, where the quantum numbers
are very, very large and thus quantum effects are unobservable.

Which is exactly why QM doesn't explain it.

The solar system obeys the same physical laws as the atom.

Because it looks like the classic atomic model?
You're not the first to see the similarity.
What happens inside an atom is very different from a solar system.

Is it really? A matter of interpretation possibly.

Or do you believe that the sun has a positive charge and the planets
a negative charge?

It is well known, it seems to me, that all matter making up the Sun
and planets is made up of charged particles (electrons, quarks up and
quarks down, the latter two being the charged components of the
nucleons) and that electrostatic interaction is additive and
universally obeys the inverse square law.

A very simple demonstration of this can be made when using only
the invariant masses of the charged components of nucleons instead
of their usual measured effective masses, which, if the real masses
of the constituting quarks really are in the observed range, can
only be mostly made up of relativistic inertia induced by the near
light velocities that the quarks must have to maintain the structures.

Besides, a clean set of calculated invariant masses that falls right
into the experimentally estimated range can easily be obtained from
the Coulomb inverse square law:

d = down quark
u = up quark
e = electron
Q = unit charge
k = Coulomb constant (1/(4 pi epsilon_0))
T = 1/nu_0 = 1 /6.57968391E15 Hz = 1.519829851E-16 sec
alpha_0 = Bohr radius
alpha = fine structure constant
c = speed of light

Calculating the invariant masses of stable elementary particles

m_i[d,u,e] = (k/alpha_0)(3Q / (n alpha c))^2 (n=1,2,3)

Summing up the invariant masses of the proton charged components:

m_ip = 2 m_iu + m_id

Calculating the G applicable to the invariant mass of the proton
components as a central body according to Kepler's third law

G_p = (4 pi^2 alpha_0^3) / (m_ip T^2) = 2.059446471E31 N . m^2 . kg^-2

Calculating the force at the Bohr radius from the gravitational
equation using the G applicable to the hydrogen atom:

F = (G_p m_ip m_ie) / alpha_0^2 = 8.238721758E-8 N

Which is the exact same value given by the standard Coulomb
equation using charges instead of masses:

F = (k Q^2) / alpha_0^2 = 8.238721806E-8 N

So, wouldn't it look like the same physical laws could be applying
in the atom and in the solar system ?

Some will say what is nu_0 and what has it got to do with this.

First, converting from a time basis to a distance basis, the
dimensions of the Newton (N) resolve as:

kg . m . s^-2 = J . m^-1

So, force (F) at the Bohr radius is 8.238721806E-8 J . m^-1

The energy at the Bohr radius can thus be calculated directly as
folows:

E at Bohr radius = F . alpha_0 = 4.359743805E-18 J (27.21138344 eV)

h = Planck's constant

nu_0 = E / h = 6.579683917E15 Hz

Which is the frequency of the energy induced at the Bohr radius.

Why is nu_0 significant in this context:

It is the number of times per second that the electron is deemed to
circle the proton on the Bohr orbit (hydrogen ground state) in the
Bohr model according to the de Broglie hypothesis, which allowed
calculating the traditional velocity assigned to the electron on
that orbit (his 1924 thesis).

v_0 = h / (m_e Lambda_0) = 2187691.253 m / sec

A velocity that can be confirmed from a second source here by obtaining
that traditional velocity from the length of the ground orbit multiplied
by the number of times that it theoretically circles the nucleus per
seconds according to de Broglie and the frequency of the energy induced
at the Bohr radius:

v_0 = (2 pi alpha_0) * nu_0 = 2187691.252 m / sec

So, this all seems to me a simple matter of viewpoint.

As you see, one can cause math to say or "prove" anything that
one cares to believe.

Who is right, who is wrong ?

The future will tell.

André Michaud

"Andr? Michaud" wrote in message
om...
Laura wrote:
"Andrew Usher" wrote in message
om...
This message is a continuation of the discussion in the thread
'Neutrino mass'.

It is more like a reiteration of your position, already stated in that
thread.


I admit to not being formally educated in QM.

Neither am I.
But I try not to criticise things I don't understand.

I am nevertheless trying
to criticise a belief normally taught in such education.

If you're referring to the idea of the electron being "smeared" across
the
orbital, then it is you who has misunderstood.
"In a general paper on quantum mechanics, Schroedinger discusses and
rejects
the interpretation that a single quantum is somehow phyiscally "spread
out"
or "blurred" among the different parts of a superposition ."
That is what is being taught.

To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.

Stationary states aren't subject to uncertainty. The
parameters of an electron in a stationary state can be
measured with precision. Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

[Old Man]

Although I
don't understand the math involved in the conventional approach, I
believe that I can understand the basics in terms of logic.

The false idea is that the Bohr-Sommerfeld orbits are an incorrect and
obsolete model of the atom. As we know, the idea of fixed orbits is
not exactly correct, but that does not make it useless.

It is very useful for chemistry and nuclear physics, but it is a model
and not meant to be taken as a true picture of the atom.

As valuable and useful as the QM model then, it would seem. No ?

The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct description. This
leads to the false belief that an electron's position is smeared out
over the orbital,

That's *your* false belief. The electron isn't "smeared out over the
orbital" (just as Schroedinger's cat isn't alive AND dead). It just
doesn't
have a position until one has been measured (just as you don't know if
the
cat is alive or dead until you open the chamber), and then that position
is
one that is affected by the act of measuring, and is therefore not a
true
representation of where the electron really was, if indeed it had a
position
at all.
That's why such an innate position is done away with entirely; it is
meaningless since it can't be measured without interfering with it.

and that the probability function is independent of
earlier observations. From the uncertainty principle (which states
that particles occupy h^3 in phase space), this can only be strictly
true for the 1s orbital. For all higher n, the relative uncertainty
becomes smaller, and the classical orbit becomes an increasingly
better approximation.

What difference would it make?
If none, then what is the value of your version?


This explains the solar system, for example, where the quantum numbers
are very, very large and thus quantum effects are unobservable.

Which is exactly why QM doesn't explain it.

The solar system obeys the same physical laws as the atom.

Because it looks like the classic atomic model?
You're not the first to see the similarity.
What happens inside an atom is very different from a solar system.

Is it really? A matter of interpretation possibly.

Or do you believe that the sun has a positive charge and the planets
a negative charge?

It is well known, it seems to me, that all matter making up the Sun
and planets is made up of charged particles (electrons, quarks up and
quarks down, the latter two being the charged components of the
nucleons) and that electrostatic interaction is additive and
universally obeys the inverse square law.

A very simple demonstration of this can be made when using only
the invariant masses of the charged components of nucleons instead
of their usual measured effective masses, which, if the real masses
of the constituting quarks really are in the observed range, can
only be mostly made up of relativistic inertia induced by the near
light velocities that the quarks must have to maintain the structures.

Besides, a clean set of calculated invariant masses that falls right
into the experimentally estimated range can easily be obtained from
the Coulomb inverse square law:

d = down quark
u = up quark
e = electron
Q = unit charge
k = Coulomb constant (1/(4 pi epsilon_0))
T = 1/nu_0 = 1 /6.57968391E15 Hz = 1.519829851E-16 sec
alpha_0 = Bohr radius
alpha = fine structure constant
c = speed of light

Calculating the invariant masses of stable elementary particles

m_i[d,u,e] = (k/alpha_0)(3Q / (n alpha c))^2 (n=1,2,3)

Summing up the invariant masses of the proton charged components:

m_ip = 2 m_iu + m_id

Calculating the G applicable to the invariant mass of the proton
components as a central body according to Kepler's third law

G_p = (4 pi^2 alpha_0^3) / (m_ip T^2) = 2.059446471E31 N . m^2 . kg^-2

Calculating the force at the Bohr radius from the gravitational
equation using the G applicable to the hydrogen atom:

F = (G_p m_ip m_ie) / alpha_0^2 = 8.238721758E-8 N

Which is the exact same value given by the standard Coulomb
equation using charges instead of masses:

F = (k Q^2) / alpha_0^2 = 8.238721806E-8 N

So, wouldn't it look like the same physical laws could be applying
in the atom and in the solar system ?

Some will say what is nu_0 and what has it got to do with this.

First, converting from a time basis to a distance basis, the
dimensions of the Newton (N) resolve as:

kg . m . s^-2 = J . m^-1

So, force (F) at the Bohr radius is 8.238721806E-8 J . m^-1

The energy at the Bohr radius can thus be calculated directly as
folows:

E at Bohr radius = F . alpha_0 = 4.359743805E-18 J (27.21138344 eV)

h = Planck's constant

nu_0 = E / h = 6.579683917E15 Hz

Which is the frequency of the energy induced at the Bohr radius.

Why is nu_0 significant in this context:

It is the number of times per second that the electron is deemed to
circle the proton on the Bohr orbit (hydrogen ground state) in the
Bohr model according to the de Broglie hypothesis, which allowed
calculating the traditional velocity assigned to the electron on
that orbit (his 1924 thesis).

v_0 = h / (m_e Lambda_0) = 2187691.253 m / sec

A velocity that can be confirmed from a second source here by obtaining
that traditional velocity from the length of the ground orbit multiplied
by the number of times that it theoretically circles the nucleus per
seconds according to de Broglie and the frequency of the energy induced
at the Bohr radius:

v_0 = (2 pi alpha_0) * nu_0 = 2187691.252 m / sec

So, this all seems to me a simple matter of viewpoint.

As you see, one can cause math to say or "prove" anything that
one cares to believe.

Who is right, who is wrong ?

The future will tell.

André Michaud

Old Man wrote:
"Andr? Michaud" wrote in message
om...

[snip]


To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.


Stationary states aren't subject to uncertainty.

Wrong. Why do you think so???


The
parameters of an electron in a stationary state can be
measured with precision.

Only the energy can (in principle) be measured with precision in
stationary states. Both position and momentum are "uncertain".


Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

It's not clear to me what you mean by "measure the distribution of
degenerate states".

Additionally, AFAIK, "degenerate states" means only that the energies
are equal. The states can have different angular momenta. E.g. the
states |200 and |210 of the hydrogen atom (using a notation |nlm for
the states here) are degenerate, although the l is different.



[snip rest]

Bye,
Bjoern

"Bjoern Feuerbacher" wrote in message
...
Old Man wrote:
"Andr? Michaud" wrote in message
om...

[snip]


To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.


Stationary states aren't subject to uncertainty.

Wrong. Why do you think so???

In what way does Bjoern think that the quantum numbers
of energy, total angular momentum, and parity that together
uniquely define an atomic stationary state carry intrinsic
uncertainty ?

The
parameters of an electron in a stationary state can be
measured with precision.

Only the energy can (in principle) be measured with precision in
stationary states. Both position and momentum are "uncertain".

The quantum numbers of energy, total angular momentum,
and parity together uniquely define a stationary state and are
not subject to inherent uncertainty.

Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

It's not clear to me what you mean by "measure the distribution of
degenerate states".

Additionally, AFAIK, "degenerate states" means only that the energies
are equal. The states can have different angular momenta. E.g. the
states |200 and |210 of the hydrogen atom (using a notation |nlm for
the states here) are degenerate, although the l is different.

The states mentioned are degenerate only in the sense
that the spin orbit interaction has been neglected. Taken
into account, only states of differing z-components of
total angular momentum, J_z, are degenerate in energy.

Even if they are "accidentally" degenerate in energy, a
superposition of atomic stationary states cannot include
states of differing total angular momentum, J, because
they are orthogonal. Without an external force, a transition
between them is impossible. The wave functions don't
overlap.

This is also true for Bjoern's example wherein J = L. The
transition, |200 = |210 is impossible without the
application of an external force. The electron can't exist
in both orbitals at once. They're orthogonal.

[Old Man]



[snip rest]

Bye,
Bjoern

Old Man wrote:
"Bjoern Feuerbacher" wrote in message
...

Old Man wrote:

"Andr? Michaud" wrote in message
e.com...


[snip]



To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.


Stationary states aren't subject to uncertainty.

Wrong. Why do you think so???


In what way does Bjoern think that the quantum numbers
of energy, total angular momentum, and parity that together
uniquely define an atomic stationary state carry intrinsic
uncertainty ?

That *some* parameters of stationary states are *not*
uncertain does in no way imply that "stationary states aren't
subject to uncertainty"!!!


The
parameters of an electron in a stationary state can be
measured with precision.

Only the energy can (in principle) be measured with precision in
stationary states. Both position and momentum are "uncertain".


The quantum numbers of energy, total angular momentum,
and parity together uniquely define a stationary state and are
not subject to inherent uncertainty.

Right. But you were not talking merely about these quantum numbers.
You said simply:
"Stationary states aren't subject to uncertainty."
And that's wrong, plain and simple.


Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

It's not clear to me what you mean by "measure the distribution of
degenerate states".

Additionally, AFAIK, "degenerate states" means only that the energies
are equal. The states can have different angular momenta. E.g. the
states |200 and |210 of the hydrogen atom (using a notation |nlm for
the states here) are degenerate, although the l is different.


The states mentioned are degenerate only in the sense
that the spin orbit interaction has been neglected. Taken
into account, only states of differing z-components of
total angular momentum, J_z, are degenerate in energy.

Right. But that doesn't change the original argument that
"degenerate" means only "have the same energy", not also "have the
same angular momentum".


Even if they are "accidentally" degenerate in energy, a
superposition of atomic stationary states cannot include
states of differing total angular momentum, J, because
they are orthogonal.

So what??? There is no problem with forming a superposition
of mutually orthogonal states! E.g. the spinors (1,0)
and (0,1) (spin in +z or -z direction) are also orthogonal - but
nevertheless the electron can have a spinor 1/sqrt(2) (1,1)
(spin in +x direction), which is obviously a superposition
of (1,0) and (0,1).

And what about the hybrid orbitals of carbon?


Without an external force, a transition
between them is impossible.

A transition between p and s orbitals is impossible???
Ever heard of the selection rule that the angular momentum
has to change by 1 in dipole radiation???


The wave functions don't overlap.

Yes, that's clear. But why on earth should that rule out
a transition between them?


This is also true for Bjoern's example wherein J = L. The
transition, |200 = |210 is impossible without the
application of an external force.

Ever heard of spontaneous emission? The same selection rule
applies here than for stimulated emission.


The electron can't exist in both orbitals at once. They're orthogonal.

The first sentence does in no way follow from the second.


Bye,
Bjoern

"Bjoern Feuerbacher" wrote in message
...
Old Man wrote:
"Bjoern Feuerbacher" wrote in
message
...

Old Man wrote:

"Andr? Michaud" wrote in message
e.com...


[snip]



To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.


Stationary states aren't subject to uncertainty.

Wrong. Why do you think so???


In what way does Bjoern think that the quantum numbers
of energy, total angular momentum, and parity that together
uniquely define an atomic stationary state carry intrinsic
uncertainty ?

That *some* parameters of stationary states are *not*
uncertain does in no way imply that "stationary states aren't
subject to uncertainty"!!!

A stationary state is uniquely defined by it's quantum
numbers. There's no ambiguity. They form a complete
set. There isn't any more information to be had. That
set is a complete description of the unperturbed wave
function, and the wave function is a complete description
of Nature.

The
parameters of an electron in a stationary state can be
measured with precision.

Only the energy can (in principle) be measured with precision in
stationary states. Both position and momentum are "uncertain".


The quantum numbers of energy, total angular momentum,
and parity together uniquely define a stationary state and are
not subject to inherent uncertainty.

Right. But you were not talking merely about these
quantum numbers. You said simply:
"Stationary states aren't subject to uncertainty."
And that's wrong, plain and simple.

Old Man is satisfied with Bjoern's agreement that the
complete set of quantum numbers of a stationary state
can be known and that they aren't subject to inherent
uncertainty. We also probably agree that the wave
function provides a complete description of Nature.

There may be some disagreement in what follows:

Old Man says that there exists no intrinsic uncertainty
in the superposition of orthogonal stationary states that
make-up the unperturbed wave function. Any uncertainty
in state is purely statistical and not inherent.

Atomic excited states with finite lifetimes aren't
orthogonal to the states to which they would decay.
That is, for operator O, O|E_i isn't orthogonal to
E_f|. If the transition probability, E_f|O|E_i ,
contains Planck's constant, h, then the transition is
subject to inherent uncertainty. Clearly, the lifetime
of the excited state, tau, is related to the energy
width, delta_E, of the state by delta_E ~ hbar / tau
which demonstrates intrinsic uncertainty for the process.

Do we agree ? [Old Man]

Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

It's not clear to me what you mean by "measure the distribution of
degenerate states".

Additionally, AFAIK, "degenerate states" means only that the energies
are equal. The states can have different angular momenta. E.g. the
states |200 and |210 of the hydrogen atom (using a notation |nlm for
the states here) are degenerate, although the l is different.


The states mentioned are degenerate only in the sense
that the spin orbit interaction has been neglected. Taken
into account, only states of differing z-components of
total angular momentum, J_z, are degenerate in energy.

Right. But that doesn't change the original argument that
"degenerate" means only "have the same energy", not also "have the
same angular momentum".


Even if they are "accidentally" degenerate in energy, a
superposition of atomic stationary states cannot include
states of differing total angular momentum, J, because
they are orthogonal.

So what??? There is no problem with forming a superposition
of mutually orthogonal states! E.g. the spinors (1,0)
and (0,1) (spin in +z or -z direction) are also orthogonal - but
nevertheless the electron can have a spinor 1/sqrt(2) (1,1)
(spin in +x direction), which is obviously a superposition
of (1,0) and (0,1).

And what about the hybrid orbitals of carbon?


Without an external force, a transition
between them is impossible.

A transition between p and s orbitals is impossible???
Ever heard of the selection rule that the angular momentum
has to change by 1 in dipole radiation???


The wave functions don't overlap.

Yes, that's clear. But why on earth should that rule out
a transition between them?


This is also true for Bjoern's example wherein J = L. The
transition, |200 = |210 is impossible without the
application of an external force.

Ever heard of spontaneous emission? The same selection rule
applies here than for stimulated emission.


The electron can't exist in both orbitals at once. They're orthogonal.

The first sentence does in no way follow from the second.


Bye,
Bjoern

Bjoern Feuerbacher wrote in message ...
Old Man wrote:


The quantum numbers of energy, total angular momentum,
and parity together uniquely define a stationary state and are
not subject to inherent uncertainty.

Right. But you were not talking merely about these quantum numbers.
You said simply:
"Stationary states aren't subject to uncertainty."
And that's wrong, plain and simple.



This is helpful in my trying to learn what is and is not certain. It
sounds like the certainty here is due only to definition, which of
course can never be uncertain. Am I correct in guessing that the
unobserved system is in a superposition of stationary states?

Old Man wrote:

To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.

Are you saying that the collapse of the wave function describes a physical
event involving one electron?

-- Charlie Springer

"Old Man" wrote in message ...
"Andr? Michaud" wrote in message
om...
Laura wrote:
"Andrew Usher" wrote in message
om...
This message is a continuation of the discussion in the thread
'Neutrino mass'.

It is more like a reiteration of your position, already stated in that
thread.


I admit to not being formally educated in QM.

Neither am I.
But I try not to criticise things I don't understand.

I am nevertheless trying
to criticise a belief normally taught in such education.

If you're referring to the idea of the electron being "smeared" across
the orbital, then it is you who has misunderstood.
"In a general paper on quantum mechanics, Schroedinger discusses and
rejects the interpretation that a single quantum is somehow phyiscally
"spread out" or "blurred" among the different parts of a superposition ."
That is what is being taught.

To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.

Stationary states aren't subject to uncertainty. The
parameters of an electron in a stationary state can be
measured with precision. Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

[Old Man]

Strict copenhagen interpretation says that the uncertainty principle
always applies in atoms.

But common sense and, as you say, observation of identically
prepared systems say that you are right.

It is quite unfortunate that the idiotic attitude, initiated by
Heisenberg, immediately followed by Bohr (unfortunately letting
go of his very promising model) according to which no further
progress could be made in understanding the foundations simply
because they considered that if more could not be understood then
meant that more could never again be understood in the future has
stalled progress for so long.

No one seems to realize that this attitude has tainted science
all through history.

I have no doubt that common sense will eventully prevail again
and that more precise mathematical tools will eventually be
developped to deal with the foundation.

But since peer pressure from the followers of the omnipresent
Copenhagen school supporters is such a hindrance, pressure cannot
be successfully exerted from inside the community to initiate
the trend.

I do what I can to help reintroduce both into academia from the
outside, and I hope that many others do the same.

The sooner the academic thinking engine is restarted, the quicker
progress will resume.

Regards

André Michaud

"Andr? Michaud" wrote in message
om...
"Old Man" wrote in message
...
"Andr? Michaud" wrote in message
om...
Laura wrote:
"Andrew Usher" wrote in message
om...
This message is a continuation of the discussion in the thread
'Neutrino mass'.

It is more like a reiteration of your position, already stated in
that
thread.


I admit to not being formally educated in QM.

Neither am I.
But I try not to criticise things I don't understand.

I am nevertheless trying
to criticise a belief normally taught in such education.

If you're referring to the idea of the electron being "smeared"
across
the orbital, then it is you who has misunderstood.
"In a general paper on quantum mechanics, Schroedinger discusses and
rejects the interpretation that a single quantum is somehow
phyiscally
"spread out" or "blurred" among the different parts of a
superposition ."
That is what is being taught.

To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.

Stationary states aren't subject to uncertainty. The
parameters of an electron in a stationary state can be
measured with precision. Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.

[Old Man]

Strict copenhagen interpretation says that the uncertainty principle
always applies in atoms. ....

No it doesn't. The HUP applies to certain pairs of
canonically conjugate variables. It doesn't apply to the
quantum numbers that uniquely define atomic stationary
states. transitions between these states are subject to
strict causality.

There is an inherent uncertainty between degenerate states,
that is, between states that are slightly non-orthogonal, as in
radioactive nuclei. In those cases, the wave function is a
superposition of several states.

{snip verbose extension of fractured pottery}

[Old Man]

André Michaud