QM and electron orbits

(Andrew Usher) wrote in message . com...
This message is a continuation of the discussion in the thread
'Neutrino mass'.

I admit to not being formally educated in QM. I am nevertheless trying
to criticise a belief normally taught in such education. 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.

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, 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.

This explains the solar system, for example, where the quantum numbers
are very, very large and thus quantum effects are unobservable. The
solar system obeys the same physical laws as the atom. There is one
important difference, though: the electrons interact with each other
to such an extent that their orbits would be chaotic even with the
pure Bohr-Sommerfeld orbits.

Andrew Usher


Quantum theory is a theory for extending the macrostructure of the
universe to its submicroscopic structure. Gravitational and other
three force fields have to be minimally diverse for the universe to be
visible. And so, parallelism between the force fields or their
unification is neither expected nor achieved.

S S Shastry

(Andrew Usher) wrote in message om...
"Old Man" wrote in message ...

[snip]

Shabby metaphysics, at best. Here's some Gems:


Is anything beyond the equations 'metaphysics' to you? Should we not
try to actually understand the physics?

The equations are all that is necessary to relate measurable physical
variables. They *are* the physics, though there are other ways of
expressing the concepts besides the equations. Equations are very
useful for making logical deductions, but aren't always the best
explanatory tool.

But anything beyond these core concepts, however we choose to express
them (a physical concept is the same whether expressed as an equation,
a paragraph of text or a diagram), is metaphysics. By definition.

I posted to sci.physics recently about the fallacy of considering the
equations as something that stands apart from the actual physics. It
seems a regrettably common misconception among certain sections of the
newsgroup.

Placed properly, classical physics is isn't incorrect:
The Hamiltonian of Quantum Mechanics contains all
of classical physics.

True. Classical physics is the limit for systems in which Planck's
constant can be considered negligibly small. The important thing with
any physical theory is to understand the conditions under which it
gives a reliable prediction of the results of an experiment.

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

I admit to not being formally educated in QM. I am nevertheless
trying
to criticise a belief normally taught in such education.
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.

The idea that the Bohr-Sommerfeld orbits are an incorrect and an
obsolete model of the atom is in fact quite correct. That
approach
led to a dead end and was entirely superceded by quantum mechanics
about seventy years ago.


This is only because the mathematics are easier in the wavfunction
picture.

No. You suffer from a fundamental misunderstanding of the
relationship between quantum mechanics and the Sommerfeld orbits.

Similarly, light is still modeled as a wave in most situations,
though
it is truly quantised.

In QED it is always modelled as a particle.

The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct description.

There is no such thing as a time-independent wave function. A
wave is
by defnition a function of space and time. The only reason why
the
orbitals are written as time-independent functions is because
each of
them has an additional implied factor
exp(iEt/hbar), where E is the energy. So the orbitals are
actually
standing waves.


Yes, of course. I meant that the probability density is
time-independent in that picture.

The probability density is also time-independent in the case of the
travelling wave corresponding to a free particle.

From the uncertainty principle (which states
that particles occupy h^3 in phase space), this can only be
strictly
true for the 1s orbital.

Unless you are trying to be clever by making a statement about the
fact that excited states are not stable, I can't follow what you
are
driving at.


Only in the 1s orbital can the position and momentum uncertainties,
respectively, be equal to the whole radius and momentum, meaning
that
the electron's state is maximally random.

I don't understand what you are driving at. The HUP, in the form you
are invoking it, involves the product of the position and momentum
uncertainties. Also, what does "the whole radius" mean? The 1s
wavefunction is finite everywhere, tending to zero as the radius tends
to infinity.

For all higher n, the relative uncertainty
becomes smaller,

The uncertainty relation relating momentum and position errors is
independent of the quantum number.

and the classical orbit becomes an increasingly
better approximation.

This explains the solar system, for example, where the quantum
numbers
are very, very large and thus quantum effects are unobservable.
The
solar system obeys the same physical laws as the atom. There is
one
important difference, though: the electrons interact with each
other
to such an extent that their orbits would be chaotic even with
the
pure Bohr-Sommerfeld orbits.

Yes, the so-calle Rydberg states, with very high quantum numbers,
do
begin to approximate to the classical situation. This is a
consequence of the corespondence principle.

This, basically, acknowledges my point.

Then you should have made your point by using correct arguments.

Franz

"Andrew Usher" wrote in message
om...
"Old Man" wrote in message
...
"Andrew Usher" wrote in message
m...

Is anything beyond the equations 'metaphysics' to you? Should we not
try to actually understand the physics?

One is cautioned against the attempt to venture beyond
what is sufficient to Nature. The atmosphere of meta-
physics causes many to seek necessary conditions, that is,
to seek "the truth". Physics is unable to find a complete set
of alternate explanations from which the "true" explanation
is selected via the application of impeccable logic. Physics
can only identify that which is sufficient to Nature.

Placed properly, classical physics is isn't incorrect:
The Hamiltonian of Quantum Mechanics contains all
of classical physics.


That this must be so is what I have been saying.

Clearly, Andrew would have liked to have said it, but
then, what excuse is offered for Andrew's verbosity ?

[Old Man]


Well, the double posting is explained in the thread 'My computer is
cursed'.

My purpose was to attempt to get people to understand the implications
of this unity, not simply to state it.

It may well be that this dispute in a philosophical matter, not a
scientific one. But philosophical approaches are a valid method of
seeking truth, especially where science fails. Empirical science has
its limits, but logic and reason have none.

Andrew Usher

Physics contains logic. That's what makes physical theories
logically self-consistent. Those theories can't be falsified
from within, nor can they be falsified by other self-contained
and self-consistent theories. Each theory is tested separately
against Nature via experiment.

The process leaves no room for philosophy. Physics doesn't
seek necessary causes ("the truth"), and it's unable to identify
a complete set of theories that would exhaust the various
possibilities.

Science generates the greatest plausible stories ever told:
evolution of the Universe; evolution of galaxies and stars,
evolution of the Earth; evolution of life. They might even
be true stories, but we'll never know for sure.

[Old Man]

"Franz Heymann" wrote in message ...
"Andrew Usher" wrote in message
om...
"Franz Heymann" wrote in message
...


The idea that the Bohr-Sommerfeld orbits are an incorrect and an
obsolete model of the atom is in fact quite correct. That
approach
led to a dead end and was entirely superceded by quantum mechanics
about seventy years ago.


This is only because the mathematics are easier in the wavfunction
picture.

No. You suffer from a fundamental misunderstanding of the
relationship between quantum mechanics and the Sommerfeld orbits.


The correct interpretation is to add the quantum uncertainty to the
Sommerfeld orbits. This has, as I recall, been done for hydrogen and
helium, but becomes too complex after that, even with computers.

However the wave function method, though it also becomes too complex
to analyse directly, is amenable to iterative approximation. This is
impossible in the orbit model as it is chaotic.

Similarly, light is still modeled as a wave in most situations,
though
it is truly quantised.

In QED it is always modelled as a particle.


QED is not used in real world calculations.

The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct description.

There is no such thing as a time-independent wave function. A
wave is
by defnition a function of space and time. The only reason why
the
orbitals are written as time-independent functions is because
each of
them has an additional implied factor
exp(iEt/hbar), where E is the energy. So the orbitals are
actually
standing waves.


Yes, of course. I meant that the probability density is
time-independent in that picture.

The probability density is also time-independent in the case of the
travelling wave corresponding to a free particle.


This is a bizarre assertion. Propagation velocity for a particle is
well defined. Perhaps you are thinking of an infinite wave train,
which is the typical example in electromagnetics.

From the uncertainty principle (which states
that particles occupy h^3 in phase space), this can only be
strictly
true for the 1s orbital.

Unless you are trying to be clever by making a statement about the
fact that excited states are not stable, I can't follow what you
are
driving at.


Only in the 1s orbital can the position and momentum uncertainties,
respectively, be equal to the whole radius and momentum, meaning
that
the electron's state is maximally random.

I don't understand what you are driving at. The HUP, in the form you
are invoking it, involves the product of the position and momentum
uncertainties. Also, what does "the whole radius" mean? The 1s
wavefunction is finite everywhere, tending to zero as the radius tends
to infinity.


Take the product of the mean radius and mean momentum (given by sqrt(2
m E)). This is minimal in the 1s orbit.

For all higher n, the relative uncertainty
becomes smaller,

The uncertainty relation relating momentum and position errors is
independent of the quantum number.

and the classical orbit becomes an increasingly
better approximation.

This explains the solar system, for example, where the quantum
numbers
are very, very large and thus quantum effects are unobservable.
The
solar system obeys the same physical laws as the atom. There is
one
important difference, though: the electrons interact with each
other
to such an extent that their orbits would be chaotic even with
the
pure Bohr-Sommerfeld orbits.

Yes, the so-calle Rydberg states, with very high quantum numbers,
do
begin to approximate to the classical situation. This is a
consequence of the corespondence principle.

This, basically, acknowledges my point.

Then you should have made your point by using correct arguments.

Franz

Andrew Usher

Bjoern Feuerbacher wrote in message ...
Andrew Usher wrote:
This message is a continuation of the discussion in the thread
'Neutrino mass'.

I admit to not being formally educated in QM. I am nevertheless trying
to criticise a belief normally taught in such education. 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.

No, that idea is not false, but entirely correct.


Argument by assertion.


As we know, the idea of fixed orbits is
not exactly correct, but that does not make it useless.

I know of no sensible use of the Bohr-Sommerfeld orbits.


Why, to make pretty models


The orbitals are conventionally given as time-independent
wavefunctions,

The orbitals *are* the wave functions.


Meaningless.


and that is held to be the correct description.

If you know of any observations which contradicts this, feel free to
share it.


This
leads to the false belief that an electron's position is smeared out
over the orbital,

No. This is sometimes used as a nice picture, a convenient visualization
- but I don't think that this is actually *taught* to *be* the behaviour
of the electron.


and that the probability function is independent of
earlier observations.

No. Every physicist gets taught that after an observation, the wave
function will be in an eigenstate of the corresponding observable. So
obviously the probability function is *dependent* on earlier observations.


This is why the time-independent wave functions are physically
meaningless - they can not be observed as such.


From the uncertainty principle (which states
that particles occupy h^3 in phase space),

It states that they occupy *at least* that much space in phase space.


this can only be strictly
true for the 1s orbital.

Huh? How did you arrive at this conclusion?


The uncertainty is momentum can not be greater than the mean momentum,
which goes as 1/n. This implies the uncertainty of position goes as n,
yet the size of the orbital goes as n^2, so the relative uncertainty
is less.


For all higher n, the relative uncertainty becomes smaller,

Why do you think so?


and the classical orbit becomes an increasingly
better approximation.

In some aspects, yes. But not in all. E.g. even for n and l going to
infinity,
there will be still an uncertainty in the variable phi - i.e. you could
only say that the electron will be *very likely* be found *somewhere* on
that circle - but saying that it moves around on that circle still does
not make sense.


This is only true for the meaningless wave function.



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

No, In order to understand the solar system, the correspondence
principle is *not* enough. You need to take decoherence into account to
explain why macroscopic objects have clearly defined positions.


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

Gravity is the same as electrostatic attraction? ;-)


There is one
important difference, though: the electrons interact with each other
to such an extent that their orbits would be chaotic even with the
pure Bohr-Sommerfeld orbits.

Right. So, let's concentrate on the hydrogen atom, where we only have
one electron, o.k.?


Bye,
Bjoern

Andrew Usher

Franz Heymann wrote:
"Andrew Usher" wrote in message
om...

"Franz Heymann" wrote in message
...

[snip]


Similarly, light is still modeled as a wave in most situations,
though bit is truly quantised.


In QED it is always modelled as a particle.

I don't agree totally with that. In QED, photons are modelled as
excitations of electromagnetic waves, which show properties of particles
- but calling this "modelled as a particle" is IMO misleading.


The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct description.

There is no such thing as a time-independent wave function. A
wave is
by defnition a function of space and time. The only reason why
the orbitals are written as time-independent functions is because
each of them has an additional implied factor
exp(iEt/hbar), where E is the energy. So the orbitals are
actually standing waves.


Yes, of course. I meant that the probability density is
time-independent in that picture.


The probability density is also time-independent in the case of the
travelling wave corresponding to a free particle.

Yes. But why is that relevant here???


[snip]


Bye,
Bjoern

Andrew Usher wrote:
"Franz Heymann" wrote in message ...

"Andrew Usher" wrote in message
.com...

"Franz Heymann" wrote in message

...


The idea that the Bohr-Sommerfeld orbits are an incorrect and an
obsolete model of the atom is in fact quite correct. That

approach

led to a dead end and was entirely superceded by quantum mechanics
about seventy years ago.


This is only because the mathematics are easier in the wavfunction
picture.

No. You suffer from a fundamental misunderstanding of the
relationship between quantum mechanics and the Sommerfeld orbits.



The correct interpretation is to add the quantum uncertainty to the
Sommerfeld orbits.

Err, how would you do that???


This has, as I recall, been done for hydrogen and
helium, but becomes too complex after that, even with computers.

Err, no. What has been done for the hydrogen atom is solving its stationary
wave functions with the Schroedinger equation. That has nothing to do
with "add the quantum uncertainty to the Sommerfeld orbits".

For helium, one can solve the problem only approximately. And your claim
that it "becomes too complex after that, even with computers" is also
wrong. AFAIK, for *all* elements the atomic wave functions are known,
at least approximately. Try looking he
http://www.emsl.pnl.gov/forms/basisform.html

There you can get different basis sets (i.e. different approximations)
for *all* elements. (well, not all basis sets support all elements -
choose a set from the list, then mark the box "Sjow supported elements",
and then click on "Submit" to see for which elements a basis set of this
type is available).


However the wave function method, though it also becomes too complex
to analyse directly, is amenable to iterative approximation. This is
impossible in the orbit model as it is chaotic.

The orbit model simply gives wrong results. E.g. it tells us that
the lowest energy state should have a non-vanishing angular momentum
- and that's simply wrong. The ground state has angular momentum zero!


Similarly, light is still modeled as a wave in most situations,
though it is truly quantised.

In QED it is always modelled as a particle.

QED is not used in real world calculations.

So calculating electron-positron scattering is not a "real world
calculation"?

What *is* a "real world calculation"?



[snip]


Yes, of course. I meant that the probability density is
time-independent in that picture.

The probability density is also time-independent in the case of the
travelling wave corresponding to a free particle.



This is a bizarre assertion.

No, it is entirely correct. For a free particle, the complete,
time-dependent wave function is a normalization constant
times e^(i(kx-omega t)) - a travelling wave. And the density
corresponding to this wave function is obviously time-independent.


Propagation velocity for a particle is well defined.

Err, travelling waves for a free particle have little to do with
"propagation velocity for a particle".


Perhaps you are thinking of an infinite wave train,

Yes. As he said: the case of the travelling wave corresponding to a free
particle. That *is* an "infinite wave train".


which is the typical example in electromagnetics.

And also in QM when talking about free particles with a definite momentum.


[snip]

Only in the 1s orbital can the position and momentum uncertainties,
respectively, be equal to the whole radius and momentum, meaning
that the electron's state is maximally random.

I don't understand what you are driving at. The HUP, in the form you
are invoking it, involves the product of the position and momentum
uncertainties. Also, what does "the whole radius" mean? The 1s
wavefunction is finite everywhere, tending to zero as the radius tends
to infinity.



Take the product of the mean radius and mean momentum (given by sqrt(2
m E)). This is minimal in the 1s orbit.

That's probably right. But what has that to do with your original claim
above about "the whole radius and momentum"? And why does this imply
that the electron's state is "maximally random"?


[snip]

Bye,
Bjoern

Andrew Usher wrote:
Bjoern Feuerbacher wrote in message ...

Andrew Usher wrote:

This message is a continuation of the discussion in the thread
'Neutrino mass'.

I admit to not being formally educated in QM. I am nevertheless trying
to criticise a belief normally taught in such education. 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.

No, that idea is not false, but entirely correct.



Argument by assertion.

The Bohr-Sommerfeld orbits give false predictions, i.e. for the
angular momentum of the ground state. Do you call this statement
also "argument by assertion"?


As we know, the idea of fixed orbits is
not exactly correct, but that does not make it useless.

I know of no sensible use of the Bohr-Sommerfeld orbits.



Why, to make pretty models

A model in physics should make as many as possible right predictions.
Since the Bohr-Sommerfeld model makes (far) less right predictions
than the solutions of the Schroedinger equation, I see no reason to use
it anymore.


The orbitals are conventionally given as time-independent
wavefunctions,

The orbitals *are* the wave functions.

Meaningless.

Err, no. Simply a statement of fact. What does "orbital" mean
in *your* opinion, if not the atomic or molecular wave functions?


and that is held to be the correct description.

If you know of any observations which contradicts this, feel free to
share it.

Hello?



This
leads to the false belief that an electron's position is smeared out
over the orbital,

No. This is sometimes used as a nice picture, a convenient visualization
- but I don't think that this is actually *taught* to *be* the behaviour
of the electron.

Did you get this?



and that the probability function is independent of
earlier observations.

No. Every physicist gets taught that after an observation, the wave
function will be in an eigenstate of the corresponding observable. So
obviously the probability function is *dependent* on earlier observations.



This is why the time-independent wave functions are physically
meaningless - they can not be observed as such.

They can. Simply take a (large) ensemble of atoms and measure the same
observable for all of them. The time-independent wave function tells
you then the distribution of the possible results of measurements.

The same as for the double-slit experiment with electrons: only if you
take a large number of electrons and shoot them at the slit, you can
observe the predictions of QM for the distribution of probability, and
hence for the distribution of the electrons.

For someone who doesn't know such elementary stuff, you make really
broud pronouncements here!


From the uncertainty principle (which states
that particles occupy h^3 in phase space),

It states that they occupy *at least* that much space in phase space.

Did you get that?



this can only be strictly true for the 1s orbital.

Huh? How did you arrive at this conclusion?



The uncertainty is momentum can not be greater than the mean momentum,
which goes as 1/n.

What do you mean by "mean momentum"? The expectation value of the
momentum vector is exactly zero for the hyrogen eigenstates. Do you
mean the square root of the expectation value of the momentum vector
squared, or what? If yes, then that *is* the uncertainty in momentum,
hence your statement that the uncertainty can not be greater
than the mean momentum becomes rather meaningless (since they are
identical, by *definition*).


This implies the uncertainty of position goes as n,
yet the size of the orbital goes as n^2, so the relative uncertainty
is less.

What do you mean by "size of the orbital"? The radius? I.e. the square
root of the expectation value of the position vector squared? If yes,
then that is again *equal* to the uncertainty in position (since the
expectation value of the position vector is also zero), and hence
your claim here that one goes with n, while the other goes with n^2,
can't be right.

Show your calculations, please.



[snip a bit]


and the classical orbit becomes an increasingly
better approximation.

In some aspects, yes. But not in all. E.g. even for n and l going to
infinity,
there will be still an uncertainty in the variable phi - i.e. you could
only say that the electron will be *very likely* be found *somewhere* on
that circle - but saying that it moves around on that circle still does
not make sense.



This is only true for the meaningless wave function.

Why on earth do you call the wave function "meaningless"? Using it, one
can make quite precise predictions for any type of measurement.

The wave function isn't simply a handy mathematical tool which is used
because it makes the calculations easier than in the Bohr-Sommerfeld
model (actually I would think that the calculations using the wave
function are harder!). It makes predictions *different* from the
Bohr-Sommerfeld model - and, big surprise, it turned out that the
predictions of the Bohr-Sommerfeld model were wrong, while those of
the Schroedinger equation are right.



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

No, In order to understand the solar system, the correspondence
principle is *not* enough. You need to take decoherence into account to
explain why macroscopic objects have clearly defined positions.

Did you get this?


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

Gravity is the same as electrostatic attraction? ;-)

Hello?



There is one
important difference, though: the electrons interact with each other
to such an extent that their orbits would be chaotic even with the
pure Bohr-Sommerfeld orbits.

Right. So, let's concentrate on the hydrogen atom, where we only have
one electron, o.k.?

Agreed?


Bye,
Bjoern

"Andrew Usher" wrote in message
m...
"Franz Heymann" wrote in message
...
"Andrew Usher" wrote in message
om...
"Franz Heymann" wrote in
message
...


The idea that the Bohr-Sommerfeld orbits are an incorrect and
an
obsolete model of the atom is in fact quite correct. That
approach
led to a dead end and was entirely superceded by quantum
mechanics
about seventy years ago.


This is only because the mathematics are easier in the
wavfunction
picture.

No. You suffer from a fundamental misunderstanding of the
relationship between quantum mechanics and the Sommerfeld orbits.


The correct interpretation is to add the quantum uncertainty to the
Sommerfeld orbits. This has, as I recall, been done for hydrogen and
helium, but becomes too complex after that, even with computers.

Please do try to understand that the concept of the Sommerfeld orbits
led to a dead end 80 years ago. They were totally abandoned when QM
came to the fore.

However the wave function method, though it also becomes too complex
to analyse directly, is amenable to iterative approximation. This is
impossible in the orbit model as it is chaotic


Similarly, light is still modeled as a wave in most situations,
though
it is truly quantised.

In QED it is always modelled as a particle.


QED is not used in real world calculations.

The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct
description.

There is no such thing as a time-independent wave function. A
wave is
by defnition a function of space and time. The only reason
why
the
orbitals are written as time-independent functions is because
each of
them has an additional implied factor
exp(iEt/hbar), where E is the energy. So the orbitals are
actually
standing waves.


Yes, of course. I meant that the probability density is
time-independent in that picture.

The probability density is also time-independent in the case of
the
travelling wave corresponding to a free particle.


This is a bizarre assertion.

No, it is quite correct.

Propagation velocity for a particle is
well defined. Perhaps you are thinking of an infinite wave train,
which is the typical example in electromagnetics.

The wavefunction for a free particle is
Psi = A exp i (k dot x - wt)

The probability density is
Psi*Psi = A

Therefore the probability density is not only time independent, it is
actually also independent of position.

From the uncertainty principle (which states
that particles occupy h^3 in phase space), this can only be
strictly
true for the 1s orbital.

Unless you are trying to be clever by making a statement about
the
fact that excited states are not stable, I can't follow what
you
are
driving at.


Only in the 1s orbital can the position and momentum
uncertainties,
respectively, be equal to the whole radius and momentum, meaning
that
the electron's state is maximally random.

I don't understand what you are driving at. The HUP, in the form
you
are invoking it, involves the product of the position and momentum
uncertainties. Also, what does "the whole radius" mean? The 1s
wavefunction is finite everywhere, tending to zero as the radius
tends
to infinity.


Take the product of the mean radius and mean momentum (given by
sqrt(2
m E)).

Rubbish. The mean momentum of the electron is zero.

This is minimal in the 1s orbit.

[snip]

Franz