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Rich L. · Guest

[quote]Igor Khavkine <igor...@gmail.com> wrote:
...
Of course the percentage of the electrons deflected towards electrons
not deflected would be greater for a narrow slit but this has nothing
to do with the imergence of electrons with greater deviation after the
slit. I cant see classicaly why there should be greater deviation for
some electrons when the walls are closer.

Regards: Ilian
[/quote]
You are trying to think of the electron as a physical object that is
at some definable (if not known) position at any time between emission
from the source and detection by the detector. In a very real sense
the electron does not scatter from just one edge of the slit, but from
the entire slit (or slits in the case of multi slit interference). If
you do the path integral calculations described by QED, you find that
to get from the source to the detector via any of the available paths
results in a net action (plus or minus 2nPi) that is within the
uncertainty limits. Not only can you not determine the exact path of
the "electron", but in a very real sense the "electron" traverses all
paths available to it. In particular for your question, the
scattering of the electron DOES depend on the position of BOTH sides
of the slit. You cannot say the electron was scattered by one side or
the other. It is scattered by both at the same time. The resulting
momentum transfer is not from a localized electron to a specific atom
(or electron) in the slit, it is to the entire slit as a unit. This
is similar to the Mossbauer effect where the recoil momentum of the
emitted gamma ray is taken up not by the nucleus emitting the gamma
ray, but by the entire crystal in which the nucleus is embedded.

It is possible, of course, that the electron might scatter from an
individual electron or atom in the slit. In that case the exchanged
momentum would end up in the scattering electron (or atom) and may
result in an ejected electron. The scattered electron would also not
end up in the same distribution past the slit as the diffracted
electrons do. This is a different process with a different
distribution of results.

Rich L.

J. J. Lodder · Guest

[quote]Thus spake J. J. Lodder <nospam@de-ster.demon.nl
ilper@abv.bg> wrote:

Hi to all quantum physicists!
I>m learning QM for quite a time. Nevertheless there are still many
unclear items especially when I look back at them after some years.
One of these came to me soon as I meticulously stared upon the
Heisenberg Uncertainty Principle. I would be pleased to hear your
opinion about my suggestions. They are logically connected and
formulated as questions so I put numbers to the 'separate' items.

1. Does the Momentum (impulse) Conservation Law hold true for a
particle going thru a single slit? I think that as the particle can
change its impulse (in fact velocity direction) obviously the Law
doesn>t apply. Nevertheless I haven>t read about this in any QM
textbook.

Momentum conservation always holds.
Your missing momentum is taken up by the material around the slit(s).

At the slits, there is only a change in the wave function. This leads to
no missing momentum, or contribution from the slits. If there is a later
measurement of momentum, the wave function collapses, to a value
typically different from the value prior to passing through the slits.
But this change in momentum takes place at time of measurement. One may
conclude that the change in momentum comes from the measurement
apparatus, not from the material around the slit.
[/quote]
Both, in fact.

Consider the following thought experiment:

Let one photon pass through the slits,
and register on the photographic plate (off centre).

The plate must have acquired a transverse momentum,
from the transverse momentum component of the photon.
The screen with the double slits must have acquired
an equal and opposite transverse momentum,
since momentum is conserved.
(and the photon is no longer present)

Jan

Oh No · Guest

Thus spake Igor Khavkine <igor.kh@gmail.com>
[quote]And the wave function amplitude is determined by the Schroedinger
equation. All other reasoning is subsumed by this fact (as long as the
situation does not warrant switching to the Dirac equation or other
alternative).
[/quote]
Dirac himself considered the Dirac equation to be a Schroedinger
equation. Does not Stone>s theorem tell us that all possibilities 's are
Schroedinger equations?

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

Guest

On 19 Nov 2008, 12:47, "Rich L." <ralivings...@sbcglobal.net> wrote:
[quote]Igor Khavkine <igor...@gmail.com> wrote:
..
Of course the percentage of the electrons deflected towards electrons
not deflected would be greater for a narrow slit but this has nothing
to do with the imergence of electrons with greater deviation after the
slit. I cant see classicaly why there should be greater deviation for
some electrons when the walls are closer.

Regards: Ilian

You are trying to think of the electron as a physical object that is
at some definable (if not known) position at any time between emission
from the source and detection by the detector. In a very real sense
the electron does not scatter from just one edge of the slit, but from
the entire slit (or slits in the case of multi slit interference). If
you do the path integral calculations described by QED, you find that
to get from the source to the detector via any of the available paths
results in a net action (plus or minus 2nPi) that is within the
uncertainty limits. Not only can you not determine the exact path of
the "electron", but in a very real sense the "electron" traverses all
paths available to it. In particular for your question, the
scattering of the electron DOES depend on the position of BOTH sides
of the slit. You cannot say the electron was scattered by one side or
the other. It is scattered by both at the same time. The resulting
momentum transfer is not from a localized electron to a specific atom
(or electron) in the slit, it is to the entire slit as a unit. This
is similar to the Mossbauer effect where the recoil momentum of the
emitted gamma ray is taken up not by the nucleus emitting the gamma
ray, but by the entire crystal in which the nucleus is embedded.

It is possible, of course, that the electron might scatter from an
individual electron or atom in the slit. In that case the exchanged
momentum would end up in the scattering electron (or atom) and may
result in an ejected electron. The scattered electron would also not
end up in the same distribution past the slit as the diffracted
electrons do. This is a different process with a different
distribution of results.

Rich L.
[/quote]
I think what you describe is just a very real wave (the electron as
wave) which does not correspond to Copenhagen probabilistic
interpretation.

Why then the electron wave interacts with the whole slit (I dont know
how many atoms) but on the screen with just one atom.

Ilian

============ Moderator>s remark ==========================

It is well known that Schroedinger>s initial interpretation of
the particle-wave function does lead to severe contradictions to
experiment which can only be resolved by Born>s interpretation
of the wave function as a probability amplitude!

Glen Herrmannsfeldt · Guest

Rich L. wrote:
(snip of electron and slit question)

[quote]You are trying to think of the electron as a physical object that is
at some definable (if not known) position at any time between emission
from the source and detection by the detector. In a very real sense
the electron does not scatter from just one edge of the slit, but from
the entire slit (or slits in the case of multi slit interference).
[/quote]
Well, you can just take the problem back one level. You can>t
determine the position and momentum of the slit (or one side
of the slit) exactly. If you have a slit such that you can measure
the momentum transfer, you won>t know the position of the slit
very accurately. In such case, you could know which side of
the slit the electron scattered from, but the uncertainty
in scattering position would destroy any interference
pattern from the electron wave.

[quote]If
you do the path integral calculations described by QED, you find that
to get from the source to the detector via any of the available paths
results in a net action (plus or minus 2nPi) that is within the
uncertainty limits. Not only can you not determine the exact path of
the "electron", but in a very real sense the "electron" traverses all
paths available to it. In particular for your question, the
scattering of the electron DOES depend on the position of BOTH sides
of the slit. You cannot say the electron was scattered by one side or
the other. It is scattered by both at the same time. The resulting
momentum transfer is not from a localized electron to a specific atom
(or electron) in the slit, it is to the entire slit as a unit. This
is similar to the Mossbauer effect where the recoil momentum of the
emitted gamma ray is taken up not by the nucleus emitting the gamma
ray, but by the entire crystal in which the nucleus is embedded.
[/quote]
True for any slit localized well enough to see the electron
interference pattern.

[quote]It is possible, of course, that the electron might scatter from an
individual electron or atom in the slit. In that case the exchanged
momentum would end up in the scattering electron (or atom) and may
result in an ejected electron. The scattered electron would also not
end up in the same distribution past the slit as the diffracted
electrons do. This is a different process with a different
distribution of results.
[/quote]
In that case, you could detect the ejected electron, determine
which part of the slit it was from, and again lose the
interference pattern.

-- glen

Guest

On 19 îÏÅÍ, 19:32, nos...@de-ster.demon.nl (J. J. Lodder) wrote:
[quote]Thus spake J. J. Lodder <nos...@de-ster.demon.nl
[/quote]
One may
[quote]conclude that the change in momentum comes from the measurement
apparatus, not from the material around the slit.

Both, in fact.

Consider the following thought experiment:

Let one photon pass through the slits,
and register on the photographic plate (off centre).

The plate must have acquired a transverse momentum,
from the transverse momentum component of the photon.
The screen with the double slits must have acquired
an equal and opposite transverse momentum,
since momentum is conserved.
(and the photon is no longer present)

Jan
[/quote]
You have omitted the impulse of the source.

Ilian

Oh No · Guest

Thus spake ilper@abv.bg
[quote]There two opposite views on this thread.

1. momentum comes from/to the slit:

Glen Herrmannsfeldt wrote:

Yes, but momentum can transfer to/from the slit. If the slit
can move you have changed the problem from the uncertainty of
the particle to the uncertainty of the slit. If it can>t, then
you can>t measure the momentum transfer.

-- glen

J. J. Lodder wrote:
Momentum conservation always holds.
Your missing momentum is taken up by the material around the slit(s)

All matter consists of charged particles,
which do interact with photons.
If the photons didn>t interact with the slitted screen
they would all go straigh ahead as if in vacuum,
and produce no interference pattern.
Conversely, you could make the screen very light,
and measure the momentum transferred to it
to predict where the photons will go.

I have some objectioin to this.

If there is an interraction between the material of the slit and the
particles moving tru the slit then there must be loss not only in the
direction but also in the magnitude of the impulse. Secondly if we
narrow the slit this wouldn>t increase delta p because this would not
change the interraction in order to take effect and broaden delta p.

2. momentum comes from the measurement apparatus

Charles Francis wrote

At the slits, there is only a change in the wave function. This leads to
no missing momentum, or contribution from the slits. If there is a later
measurement of momentum, the wave function collapses, to a value
typically different from the value prior to passing through the slits.
But this change in momentum takes place at time of measurement.
One may conclude that the change in momentum comes from the
measurement apparatus, not from the material around the slit.

I have the following objections:

It turns out that the wave function is something which carry the
momentum to the measurement apparatus. But WF is not real (in
Copenhagen interpretatiion) so it can not do this.
[/quote]

It is the particle which carries momentum. The only change in momentum
here is due to the uncertainty in a measurement of momentum, which can
be understood as being caused by the interaction of the measurement
apparatus with the particle.
[quote]
================================================
I>m still thinking the momentum conservation law doesnt hold for a
single particle. Why not if this is permitted for the energy (at the
end they form a four momentum in SR).
[/quote]
energy is conserved in measurement. I think you are thinking of virtual
particles. They are usually regarded as off mass shell, but conserving
energy.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

Guest

On 19 ÐÐ¾ÐµÐ¼, 12:47, "Rich L." <ralivings...@sbcglobal.net> wrote:
[quote]Igor Khavkine <igor...@gmail.com> wrote:
..
Of course the percentage of the electrons deflected towards electrons
not deflected would be greater for a narrow slit but this has nothing
to do with the imergence of electrons with greater deviation after the
slit. I cant see classicaly why there should be greater deviation for
some electrons when the walls are closer.

Regards: Ilian

You are trying to think of the electron as a physical object that is
at some definable (if not known) position at any time between emission
from the source and detection by the detector. In a very real sense
the electron does not scatter from just one edge of the slit, but from
the entire slit (or slits in the case of multi slit interference). If
you do the path integral calculations described by QED, you find that
to get from the source to the detector via any of the available paths
results in a net action (plus or minus 2nPi) that is within the
uncertainty limits. Not only can you not determine the exact path of
the "electron", but in a very real sense the "electron" traverses all
paths available to it. In particular for your question, the
scattering of the electron DOES depend on the position of BOTH sides
of the slit. You cannot say the electron was scattered by one side or
the other. It is scattered by both at the same time. The resulting
momentum transfer is not from a localized electron to a specific atom
(or electron) in the slit, it is to the entire slit as a unit. This
is similar to the Mossbauer effect where the recoil momentum of the
emitted gamma ray is taken up not by the nucleus emitting the gamma
ray, but by the entire crystal in which the nucleus is embedded.

It is possible, of course, that the electron might scatter from an
individual electron or atom in the slit. In that case the exchanged
momentum would end up in the scattering electron (or atom) and may
result in an ejected electron. The scattered electron would also not
end up in the same distribution past the slit as the diffracted
electrons do. This is a different process with a different
distribution of results.

Rich L.
[/quote]
I think what you describe is just a very real wave (the electron as
wave) which does not correspond to Copenhagen probabilistic
interpretation.
Why than the electron-wave interracts with the whole slit (I dont know
how many atoms) but on the screen with just one atom.

Ilian

Igor Khavkine · Guest

On Nov 20, 1:13 pm, il...@abv.bg wrote:

[quote]I think what you describe is just a very real wave (the electron as
wave) which does not correspond to Copenhagen probabilistic
interpretation.

Why then the electron wave interacts with the whole slit (I dont know
how many atoms) but on the screen with just one atom.
[/quote]
I believe you have a mistaken understanding of the Copenhagen
interpretation. Would you mind stating what you believe it to be? Then
it can be corrected if necessary.

Igor

Guest

On 21 <garbled>, 04:59, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:

[quote]2. momentum comes from the measurement apparatus

Charles Francis wrote

At the slits, there is only a change in the wave function. This leads to
no missing momentum, or contribution from the slits. If there is a later
measurement of momentum, the wave function collapses, to a value
typically different from the value prior to passing through the slits.
But this change in momentum takes place at time of measurement.
One may conclude that the change in momentum comes from the
measurement apparatus, not from the material around the slit.

I have the following objections:

It turns out that the wave function is something which carry the
momentum to the measurement apparatus. But WF is not real (in
Copenhagen interpretatiion) so it can not do this.

It is the particle which carries momentum. The only change in momentum
here is due to the uncertainty in a measurement of momentum, which can
be understood as being caused by the interaction of the measurement
apparatus with the particle.
[/quote]
But this change depends not on the apparatus but ONLY on the width of
the slit. So the change of p must have happened on the slit.

[quote]

=================================================
I>m still thinking the momentum conservation law doesnt hold for a
single particle. Why not if this is permitted for the energy (at the
end they form a four momentum in SR).

energy is conserved in measurement. I think you are thinking of virtual
particles. They are usually regarded as off mass shell, but conserving
energy.
[/quote]
I>m thinking of Ehrenfest theorem.

Regards: Ilian

Guest

On 20 <garbled>, 23:14, il...@abv.bg wrote:
[quote]On 19 <garbled>, 19:32, nos...@de-ster.demon.nl (J. J. Lodder) wrote:

Thus spake J. J. Lodder <nos...@de-ster.demon.nl

One may

conclude that the change in momentum comes from the measurement
apparatus, not from the material around the slit.

Both, in fact.

Consider the following thought experiment:

Let one photon pass through the slits,
and register on the photographic plate (off centre).

The plate must have acquired a transverse momentum,
from the transverse momentum component of the photon.
[/quote]
The question is how? Is there interraction on the slit or not.
I>m holding that the photon looses or gains impulse as in HU relations
(though in the mean value for many photons change of impulse is zero)
because the classical laws apply only as mean values (Eherfest
theorem).

Suppose there is a slit made of two separate plates. If the plates
acquire a transverse momentum as after Newton they must move inwards.
Strange - isnt it!! By passing a material true a hole we cause the
hole to shrink!!

Rich L. said: the electron interracts with the slit as a whole. But
I>m asking then why interracts the electron with the slit as a whole
(with trillions..of......trillions of molecules) but when it reaches
the screen it suddenly decides to interract with just one atom??

Ilian

J. J. Lodder · Guest

<ilper@abv.bg> wrote:

[quote]On 19 ????, 19:32, nos...@de-ster.demon.nl (J. J. Lodder) wrote:
Thus spake J. J. Lodder <nos...@de-ster.demon.nl

One may
conclude that the change in momentum comes from the measurement
apparatus, not from the material around the slit.

Both, in fact.

Consider the following thought experiment:

Let one photon pass through the slits,
and register on the photographic plate (off centre).

The plate must have acquired a transverse momentum,
from the transverse momentum component of the photon.
The screen with the double slits must have acquired
an equal and opposite transverse momentum,
since momentum is conserved.
(and the photon is no longer present)

Jan

You have omitted the impulse of the source.
[/quote]
No, we are talking transverse components only.
The incoming beam is perpendicular to the screen,
so no transverse momentum component in the source

Jan

Oh No · Guest

Thus spake ilper@abv.bg
[quote]On 21 <garbled>, 04:59, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:

2. momentum comes from the measurement apparatus

Charles Francis wrote

At the slits, there is only a change in the wave function. This leads to
no missing momentum, or contribution from the slits. If there is a later
measurement of momentum, the wave function collapses, to a value
typically different from the value prior to passing through the slits.
But this change in momentum takes place at time of measurement.
One may conclude that the change in momentum comes from the
measurement apparatus, not from the material around the slit.

I have the following objections:

It turns out that the wave function is something which carry the
momentum to the measurement apparatus. But WF is not real (in
Copenhagen interpretatiion) so it can not do this.

It is the particle which carries momentum. The only change in momentum
here is due to the uncertainty in a measurement of momentum, which can
be understood as being caused by the interaction of the measurement
apparatus with the particle.

But this change depends not on the apparatus but ONLY on the width of
the slit. So the change of p must have happened on the slit.
[/quote]
You are right. The uncertainty in momentum of the particle is
counterbalanced with an uncertainty (normally unmeasurably small) of the
momentum of the slit material. When either is measured, the uncertainty
in the other is removed. This reduces the problem to an EPR type
entanglement problem. The measurement is required to be able to define a
state of definite momentum, and to thereby to apply the law of
conservation of momentum in a meaningful way. As with ordinary EPR, it
is impossible to think of this classically in any clear way.
[quote]
=================================================
I>m still thinking the momentum conservation law doesnt hold for a
single particle. Why not if this is permitted for the energy (at the
end they form a four momentum in SR).

energy is conserved in measurement. I think you are thinking of virtual
particles. They are usually regarded as off mass shell, but conserving
energy.

I>m thinking of Ehrenfest theorem.

Ehrenfest>s theorem applies to the expectation of the measured value.[/quote]

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

Guest

On November 21, 14:26, Igor Khavkine <igor...@gmail.com> wrote:
[quote]On Nov 20, 1:13 pm, il...@abv.bg wrote:

I think what you describe is just a very real wave (the electron as
wave) which does not correspond to Copenhagen probabilistic
interpretation.

Why then the electron wave interacts with the whole slit (I dont know
how many atoms) but on the screen with just one atom.

I believe you have a mistaken understanding of the Copenhagen
interpretation. Would you mind stating what you believe it to be? Then
it can be corrected if necessary.

Igor
[/quote]
The wave function gives the amplitude of the probability of say the
position of pointlike particles. It can not interract with matter
(other pointlike particles). It determines the whole picture according
to the arrangement given but not for individual particles.
Out of Copenhagen I think that paticles must have individual paths -
one path for one particle. I can not believe that a particle
desintegrates as it moves and than integrates again when it makes a
real interraction.
If you think that wave function is conected with interraction would
explain why the electron interracts with the material of the whole
slit and than with just one atom in a photomultipier or in a Compton
effect scattering.
Regards: Ilian

Igor Khavkine · Guest

On Nov 22, 3:36 am, il...@abv.bg wrote:
[quote]On November 21, 14:26, Igor Khavkine <igor...@gmail.com> wrote:

I believe you have a mistaken understanding of the Copenhagen
interpretation. Would you mind stating what you believe it to be? Then
it can be corrected if necessary.

The wave function gives the amplitude of the probability of say the
position of pointlike particles.
[/quote]
True enough.

[quote]It can not interract with matter
(other pointlike particles). It determines the whole picture according
to the arrangement given but not for individual particles.
[/quote]
Umm, don>t really see what you mean by this statement. However, unless
what you mean to say is equivalent to the following, it>s likely to be
wrong: the interaction between particles is determined by the
Hamiltonian and the Hamiltonian determines the time evolution of the
system>s state (which is described by wave functions).

[quote]Out of Copenhagen I think that paticles must have individual paths -
one path for one particle. I can not believe that a particle
desintegrates as it moves and than integrates again when it makes a
real interraction.
[/quote]
Just as I suspected. That is *not* what the Copenhagen interpretation
says. The particle neither dis- nor re-integrates. Also, there is no
definite individual trajectory. The only thing that can be concluded
from a given wave function is the distribution of measurement outcomes
obtained in repeated experiments. The position of an electron as
registered by a screen after it passes through a slitted screen is an
example of a measurement. No trajectories involved.

BTW, whether you are comfortable with this kind of picture or not is
much less important that it works with high accuracy when compared to
experiments.

[quote]If you think that wave function is conected with interraction would
explain why the electron interracts with the material of the whole
slit and than with just one atom in a photomultipier or in a Compton
effect scattering.
[/quote]
The difference is that the electron>s interaction with the detector
screen results in a macroscopic change in the screen>s state, while
the interaction with the slit doesn>t. Hence the interaction with the
screen constitutes a measurement (implying an a la Copenhagen
collapse), while the interaction with the slit doesn>t.

Hope this helps.

Igor


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