Questions for quantum physicists only!
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ilper@abv.bg - 11 Nov 2008 17:56 GMT 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. 2. Maybe this should be understood as a private case of the Mean Value Theorem of Ehrenfest? 3. Of course it shouldn’t be accepted that part of the impulse is transferred to the material of the slit holder because the particle doesn’t interact with it (or am I wrong?) 4. On the other side MCL is a direct consequence of the homogeneity of space. So QM and in particular HUP do insist that space is not homogeneous? (in presence of a slit?) So is the space homogeneous in QM or not? I suppose this is connected to the creation- annihilation of the virtual particles?? But then HUP must automatically imply their existence as necessary???? Although I have read many QM textbooks I didn’t learn such a thing so far.
Thank you in advance for your considerations and discussion! Ilian
Glen Herrmannsfeldt - 12 Nov 2008 23:12 GMT > 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. Feynman's quote:
"I think I can safely say that nobody understands quantum mechanics."
http://en.wikiquote.org/wiki/Quantum_mechanics#Sourced
> One of these came to me soon as I meticulously stared upon the > Heisenberg Uncertainty Principle. I would be pleased to hear your [quoted text clipped - 6 lines] > doesn’t apply. Nevertheless I haven’t read about this in any QM > textbook. 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
Igor Khavkine - 13 Nov 2008 05:00 GMT > il...@abv.bg wrote: > > I'm learning QM for quite a time. Nevertheless there are still many [quoted text clipped - 3 lines] > > "I think I can safely say that nobody understands quantum mechanics." That quote has probably done more harm than good to education in quantum mechanics. Nearly a century of successful, even spectacularly so, application of QM speaks otherwise.
Now, I know that Feynman had a particular idea in mind having to do with intuition and realistic interpretations, and he may even have had a point. Unfortunately, that's not how his words are generally understood.
As a service to the physics community at large, I'll start a meme that I hope will propagate far and wide:
Screw Feynman! Anyone who works at it can understand quantum mechanics.
Igor
Glen Herrmannsfeldt - 13 Nov 2008 21:38 GMT (snip, I wrote)
>>Feynman's quote:
>>"I think I can safely say that nobody understands quantum mechanics."
> That quote has probably done more harm than good to education in > quantum mechanics. Nearly a century of successful, even spectacularly > so, application of QM speaks otherwise.
> Now, I know that Feynman had a particular idea in mind having to do > with intuition and realistic interpretations, and he may even have had > a point. Unfortunately, that's not how his words are generally > understood. I would say that we mostly understand it, but not completely, maybe asymptotically is about right.
For one, what ever happened to quantum gravity?
http://en.wikipedia.org/wiki/Interpretation_of_quantum_mechanics
There was an article a few years ago, something like the greatest hoax ever pulled on students is teaching the Copenhagen interpretation of quantum mechanics. I don't remember where I saw it, and some web searching didn't find it.
-- glen
J. J. Lodder - 12 Nov 2008 23:15 GMT > Hi to all quantum physicists! > I'm learning QM for quite a time. Nevertheless there are still many [quoted text clipped - 9 lines] > 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).
> 2. Maybe this should be understood as a private case of the Mean Value > Theorem of Ehrenfest? No.
> 3. Of course it shouldn't be accepted that part of the impulse is > transferred to the material of the slit holder because the particle > doesn't interact with it (or am I wrong?) Yes, you are wrong here. 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. According to QM this destroys the interference pattern.
> 4. On the other side MCL is a direct consequence of the homogeneity of > space. So QM and in particular HUP do insist that space is not > homogeneous? (in presence of a slit?) No, see above.
> So is the space homogeneous in QM or not? Yes.
> I suppose this is connected to the creation- annihilation > of the virtual particles?? But then HUP must automatically imply their > existence as necessary???? Although I have read many QM textbooks I > didn't learn such a thing so far. Virtual particle have no physical reality. They are an artifact of the way we do perturbation theory,
Oh No - 13 Nov 2008 04:26 GMT Thus spake J. J. Lodder <nospam@de-ster.demon.nl>
>> Hi to all quantum physicists! >> I'm learning QM for quite a time. Nevertheless there are still many [quoted text clipped - 12 lines] >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.
Regards
 Signature Charles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces)
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J. J. Lodder - 19 Nov 2008 17:32 GMT > Thus spake J. J. Lodder <nospam@de-ster.demon.nl> > > [quoted text clipped - 22 lines] > 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
johndevers@iprimus.com.au - 14 Nov 2008 22:53 GMT > Virtual particle have no physical reality. I seem to be able to feel the virtual photons being emitted and absobed when I hold 2 magnets and face them at each other, if I am not feeling the physical reality of virtual particles then what is it?
J. J. Lodder - 17 Nov 2008 11:06 GMT > > Virtual particle have no physical reality. > > I seem to be able to feel the virtual photons being emitted and > absobed when I hold 2 magnets and face them at each other, if I am not > feeling the physical reality of virtual particles then what is it? You can feel nothing of the kind. All you can feel is a force. The rest is (mistaken) interpretation. You should try not to confuse observation and interpretation,
Jan
Oh No - 13 Nov 2008 22:35 GMT Thus spake Igor Khavkine <igor.kh@gmail.com>
>On Nov 12, 6:12 pm, Glen Herrmannsfeldt <g...@ugcs.caltech.edu> wrote: >> il...@abv.bg wrote: [quoted text clipped - 19 lines] > Screw Feynman! Anyone who works at it can understand quantum > mechanics. Anyone can understand the mathematical structure of quantum mechanics, and anyone can put their head in the sand by claiming that interpretation is not the business of physics, but so long as interpretation is an unresolved issue, I do not find it reasonable to dismiss what Feynman said.
Regards
 Signature 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
Glen Herrmannsfeldt - 14 Nov 2008 22:53 GMT > Thus spake Igor Khavkine <igor.kh@gmail.com> (snip)
>> Screw Feynman! Anyone who works at it can understand quantum >> mechanics.
> Anyone can understand the mathematical structure of quantum mechanics, > and anyone can put their head in the sand by claiming that > interpretation is not the business of physics, but so long as > interpretation is an unresolved issue, I do not find it reasonable to > dismiss what Feynman said. Yeah, what he said.
Also, being able to solve the equations is one thing, explaining the meaning is another.
-- glen
ilper@abv.bg - 14 Nov 2008 22:53 GMT 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. [quoted text clipped - 4 lines] >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 [quoted text clipped - 3 lines] >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.
================================================ 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).
Regards: Ilian
Igor Khavkine - 16 Nov 2008 20:06 GMT On Nov 14, 5:53 pm, il...@abv.bg wrote:
> There two opposite views on this thread. > > 1. momentum comes from/to the slit:
> 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. You are right in general, but not in the idealized experiment being considered here. The interaction with the slit is considered elastic, hence energy and momentum magnitude conserving.
> 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. Your reasoning is flawed here, and your conclusion is false. Of course the interaction changes when the slit is narrowed; there is now more barrier material for the electron to bounce off of than before. And delta-p increases because both the Schroedinger equation and the Heisenberg uncertainty principle say so.
> 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). Again, your argument and conclusion here are flawed. For an isolated system, even in special relativity, 4-momentum is conserved as a whole, including its individual components. The key word here is isolated.
Hope this helps.
Igor
ilper@abv.bg - 17 Nov 2008 14:15 GMT On 16 Ðоем, 22:06, Igor Khavkine <igor...@gmail.com> wrote:
> On Nov 14, 5:53 pm, il...@abv.bg wrote: > [quoted text clipped - 10 lines] > considered here. The interaction with the slit is considered elastic, > hence energy and momentum magnitude conserving. Well I admit the change in momentum would be inversly proportional to the mass of the material of the slit holder and so it is normal to be neglected. (Nevertheless some inside dissatisfaction about making this QM an approximation to reality on its very beginning creeps in).
So more important is your secondary notice.
Igor Khavkine <igor...@gmail.com> wrote:
> On Nov 14, 5:53 pm, il...@abv.bg wrote:
> > 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.
> Your reasoning is flawed here, and your conclusion is false. Of course > the interaction changes when the slit is narrowed; there is now more > barrier material for the electron to bounce off of than before. And > delta-p increases because both because both the Schroedinger equation and the Heisenberg uncertainty principlethe espesially say so.
I dont agree. Consider a flux of just one particle. It sees just the edge of the slit and it changes its direction. But what if the other edge is near or far? I think this changes nothing (espesially in an idealised picture of absolutely rigid walls). I dont thik there would be more barrier material for the electron to bounce off. Note that this is just a classical consideration of the slit. I'm not arguing against what the Schroedinger equation and the Heisenberg uncertainty principle say. Alternatively one can think that the wave function now sees more barrier material. But as it is (WF) unphysical (just a probability) there can not be maintained that it makes an interraction leading to broadening of delta p.
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
noshellswill - 18 Nov 2008 09:06 GMT On Mon, 17 Nov 2008 09:15:31 -0500, ilper wrote:
[Moderator's note: Quoted text snipped. -P.H.]
> Of course the percentage of the electrons deflected towards electrons > not deflected would be greater for a narrow slit but this has nothing to [quoted text clipped - 3 lines] > > Regards: Ilian Rigid or not, if the (finite length) slit-walls are metallic, wouldn't the passing electron generate a current leading to (transient) dipole radiation from the slit?
nss ******
Rich L. - 19 Nov 2008 10:47 GMT > 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 [quoted text clipped - 3 lines] > > 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.
Glen Herrmannsfeldt - 20 Nov 2008 18:13 GMT (snip of electron and slit question)
> 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). 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.
> If > you do the path integral calculations described by QED, you find that [quoted text clipped - 11 lines] > 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. True for any slit localized well enough to see the electron interference pattern.
> It is possible, of course, that the electron might scatter from an > individual electron or atom in the slit. In that case the exchanged [quoted text clipped - 3 lines] > electrons do. This is a different process with a different > distribution of results. 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
Igor Khavkine - 18 Nov 2008 23:53 GMT On Nov 17, 9:15 am, il...@abv.bg wrote:
> On 16 Ðоем, 22:06, Igor Khavkine <igor...@gmail.com> wrote: > > On Nov 14, 5:53 pm, il...@abv.bg wrote:
> > You are right in general, but not in the idealized experiment being > > considered here. The interaction with the slit is considered elastic, [quoted text clipped - 4 lines] > neglected. (Nevertheless some inside dissatisfaction about making this > QM an approximation to reality on its very beginning creeps in). This idealization is no worse than any other. Do you think neglecting air drag unreasonably impacts your understanding of projectile motion?
> > > 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.
> > Your reasoning is flawed here, and your conclusion is false. Of course > > the interaction changes when the slit is narrowed; there is now more > > barrier material for the electron to bounce off of than before. And > > delta-p increases because both because both the Schroedinger equation > > and the Heisenberg uncertainty principle espesially say so.
> I dont agree. Consider a flux of just one particle. It sees just the > edge of the slit and it changes its direction. But what if the other [quoted text clipped - 14 lines] > slit. I cant see classicaly why there should be greater deviation for > some electrons when the walls are closer. Your reasoning is still flawed. There are no individual trajectories that pass near or far to the slit edges. QM only makes predictions about statistical distribution of repeated measurements. A trajectory is not a measurement outcome, hence not predicted by QM.
Whether you think the wave function is physical or not, it's amplitude squared is what predicts the measurement distribution. 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).
Igor
Oh No - 19 Nov 2008 17:32 GMT Thus spake Igor Khavkine <igor.kh@gmail.com>
>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). 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
 Signature Charles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces)
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ilper@abv.bg - 20 Nov 2008 18:13 GMT > > Igor Khavkine <igor...@gmail.com> wrote: > .. [quoted text clipped - 35 lines] > > Rich L. 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!
Igor Khavkine - 21 Nov 2008 12:26 GMT > 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
ilper@abv.bg - 22 Nov 2008 08:36 GMT > > I think what you describe is just a very real wave (the electron as > > wave) which does not correspond to Copenhagen probabilistic [quoted text clipped - 8 lines] > > Igor 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 - 23 Nov 2008 20:19 GMT On Nov 22, 3:36 am, il...@abv.bg 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. True enough.
> It can not interract with matter > (other pointlike particles). It determines the whole picture according > to the arrangement given but not for individual particles. 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).
> 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. 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.
> 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. 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
Glen Herrmannsfeldt - 24 Nov 2008 05:11 GMT (snip of statements agree with)
> 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. This makes some assumptions about the slit(s).
I would say that they are true for most slits, but that isn't required. It might be that one can make slits that are sensitive to the momentum change and do result in a macroscopic change. In that case, one would know which slit the particle went through, and no interference pattern would be observed.
-- glen
scerir - 24 Nov 2008 20:35 GMT [ Mod. note: The following quote is from Ilian. Please make sure to attribute quotes correctly. -ik ]
Igor writes:
> 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. Individual, trajectory, ensembles, ... The point seems deeper, according to mr. Copenhagen in person:
'However, since the discovery of the quantum of action, we know that the classical ideal cannot be attained in the description of atomic phenomena. In particular, any attempt at an ordering in space-time leads to a break in the causal chain, since such an attempt is bound up with an essential exchange of momentum and energy between the individuals and the measuring rods and clocks used for observation; and just this exchange cannot be taken into account if the measuring instruments are to fulfil their purpose. Conversely, any conclusion, based in an unambiguous manner upon the strict conservation of energy and momentum, with regard to the dynamical behaviour of the individual units obviously necessitates a complete renunciation of following their course in space and time'. - N.Bohr, 'Atomic Theory and the Description of Nature', pp. 97-8, Cambridge University Press, 1934
ilper@abv.bg - 20 Nov 2008 21:14 GMT On 19 Ноем, 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. [quoted text clipped - 14 lines] > > Jan You have omitted the impulse of the source.
Ilian
ilper@abv.bg - 21 Nov 2008 12:26 GMT > On 19 <garbled>, 19:32, nos...@de-ster.demon.nl (J. J. Lodder) wrote: > [quoted text clipped - 14 lines] > > The plate must have acquired a transverse momentum, > > from the transverse momentum component of the photon. 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 - 21 Nov 2008 18:00 GMT > On 19 ????, 19:32, nos...@de-ster.demon.nl (J. J. Lodder) wrote: > > > Thus spake J. J. Lodder <nos...@de-ster.demon.nl> [quoted text clipped - 20 lines] > > You have omitted the impulse of the source. 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 - 21 Nov 2008 02:59 GMT Thus spake ilper@abv.bg
>There two opposite views on this thread. > [quoted text clipped - 47 lines] >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.
>================================================ >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.
Regards
 Signature Charles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces)
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ilper@abv.bg - 21 Nov 2008 12:26 GMT > >2. momentum comes from the measurement apparatus > [quoted text clipped - 13 lines] > >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.
> >================================================= > >I'm still thinking the momentum conservation law doesnt hold for a [quoted text clipped - 4 lines] > particles. They are usually regarded as off mass shell, but conserving > energy. I'm thinking of Ehrenfest theorem.
Regards: Ilian
Oh No - 22 Nov 2008 08:14 GMT Thus spake ilper@abv.bg
>> >2. momentum comes from the measurement apparatus >> [quoted text clipped - 21 lines] >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. 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.
>> >================================================= >> >I'm still thinking the momentum conservation law doesnt hold for a [quoted text clipped - 6 lines] > >I'm thinking of Ehrenfest theorem. Ehrenfest's theorem applies to the expectation of the measured value.
Regards
 Signature Charles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces)
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ilper@abv.bg - 21 Nov 2008 03:28 GMT On 19 Ðоем, 12:47, "Rich L." <ralivings...@sbcglobal.net> wrote:
> > Igor Khavkine <igor...@gmail.com> wrote: > .. [quoted text clipped - 35 lines] > > Rich L. 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
Oh No - 24 Nov 2008 03:12 GMT Thus spake ilper@abv.bg
>On November 21, 14:26, Igor Khavkine <igor...@gmail.com> wrote: >> [quoted text clipped - 23 lines] >slit and than with just one atom in a photomultipier or in a Compton >effect scattering. Copenhagen contains the notion of duality, in which the wave function is a property of matter. It does not allow that particles have individual paths. This is altered in the orthodox interpretation, in which the wave function is just a means of calculation of probabilities, and in which the notion of a path is also meaningless in the quantum domain. You have a problem if you try to add in to any interpretation of qm the notion that paths exist and are meaningful, since then one would have a classical probability theory and no interference patterns.
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ilper@abv.bg - 24 Nov 2008 03:13 GMT On 22 Ðоем, 10:14, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Ehrenfest's theorem applies to the expectation of the measured value. Yes the measured quantity is the impulse of the particles. <p>=0 Why does Ehrenfest's theorem not apply to the impulse conseravation law?? Are there exceptions to Ehrenfest theorem??
I dont feel it right also that the wave function interracts with the slit. I've put this question on the thread but there is no reply so far from those who have put this mechanism forward.(The question beeing: How can the WF interract with the whole material of the slit and then with just one atom when later registered by say scattering on the screen?)
Regards:Ilian
Oh No - 24 Nov 2008 20:35 GMT Thus spake ilper@abv.bg
>On 22 Ð?оÐμм, 10:14, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote: > [quoted text clipped - 4 lines] >Why does Ehrenfest's theorem not apply to the impulse conseravation >law?? Are there exceptions to Ehrenfest theorem?? When the particle passes through the slit, it acquires uncertainty in p. Momentum is conserved in the quantum domain, even as it applies to uncertainty. As discussed on s.p.f. the uncertainty in p is counterbalanced by an increase in uncertainty of the momentum of the slit. However, the mass of the slit is large, and the corresponding uncertainty in velocity of the slit is immeasurably small, and is absorbed into the classical uncertainty in any measurement. If you could mount the slit in such a way that its momentum could be measured, (for example on a perfectly frictionless rail) then a measurement of the momentum of the slit (even if it were to take place some time later) would constitute a measurement of which slit the particle came through, and there would be no observed interference pattern).
>I dont feel it right also that the wave function interracts with the >slit. I've put this question on the thread but there is no reply so >far from those who have put this mechanism forward.(The question >beeing: How can the WF interract with the whole material of the slit >and then with just one atom when later registered by say scattering on >the screen?) As you will have gathered on this and the corresponding thread in s.p.f. there is no agreement among physicists wrt the interpretation of quantum mechanics. I am convinced that there is no interaction with the slit. Imv, the uncertainty in momentum reflects uncertainty in the structure of space, which itself is due, not to interaction, but to lack of interactions, .
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ilper@abv.bg - 24 Nov 2008 03:14 GMT On November 21, 14:26, Igor Khavkine <igor...@gmail.com> wrote:
> > I believe you have a mistaken understanding of the Copenhagen interpretation. The wave function represents what the 'picture' would be after many experiments (even when there is one particle at a time). That's why it is imposible and unreasonable to speak about trajectories in QM (even when there are trajectories). There just aren't individual particles as an object of QM.
Regards:Ilian
Igor Khavkine - 24 Nov 2008 13:05 GMT On Nov 23, 10:14 pm, il...@abv.bg wrote:
> On November 21, 14:26, Igor Khavkine <igor...@gmail.com> wrote: > [quoted text clipped - 6 lines] > when there are trajectories). There just aren't individual particles > as an object of QM. OK, this is all nice and correct. The obvious followup question: if you do have a decent understanding of the role of the wave function, as evidenced above, why do you insist on bringing particle trajectories into your questions elsewhere in this thread?
Igor
Murray Arnow - 24 Nov 2008 17:13 GMT >il...@abv.bg wrote: >>Igor Khavkine wrote: [quoted text clipped - 12 lines] >as evidenced above, why do you insist on bringing particle >trajectories into your questions elsewhere in this thread? It's comments like this that get me wondering if I misunderstand what the wave function is. I went back to Landau & Lifshitz, Messiah, Schiff and even Wikipedia, and just as I thought, it is the the amplitude-squared of the WF that is related to what gives the statistical results of measurements, and the WF is postulated to completely describe the dynamical properties of a system at any given time, in rough analogy to the classical Hamiltonian or Lagrangian. I don't understand the WF to be the picture described above.
It is getting the information out of the WF that is the tough part. All the states available to a system can be found, but exactly which state the system is in cannot be known precisely at a given time and must be determined probabilistically. There are two references I think give thoughtful descriptions of the WF: Messiah's Quantum Mechanics, Ch. II, Sec. 10 and Landau & Lifshitz's Quantum Mechanics - Non-Relativistic Theory, Ch. 1 - particularly section 2.
Aside: the Dirac formalism is more sophisticated (system states defined by vectors in Hilbert space), but it does lead us back to the same WF with the above same properties.
I don't see why it is impossible to speak of QM trajectories. Trajectories are described through scattering theory; the paths of particles are calculated probabilistically--a non-scattered particle travels in a straight line. A bubble chamber photograph verifies the existence of these particle trajectories.
Igor Khavkine - 25 Nov 2008 17:01 GMT > >il...@abv.bg wrote:
> >> The wave function represents what the 'picture' would be after many > >> experiments (even when there is one particle at a time). That's why it [quoted text clipped - 15 lines] > time, in rough analogy to the classical Hamiltonian or Lagrangian. I > don't understand the WF to be the picture described above. Perhaps there is slight disagreement about terminology here. Yes, it is the *absolute value squared* of the wave function that gives probabilities for measurement outcomes. I believe that is what Ilian meant by his colloquial reference to the wave function. I hope that the simple omission of the fact that it has to be squared does not result in a major misunderstanding.
Also, Murray, I don't really see what is the analogy that you draw between the wafe function and the classical Hamiltonian or Lagrangian. Could you elaborate?
> It is getting the information out of the WF that is the tough part. All > the states available to a system can be found, but exactly which state [quoted text clipped - 9 lines] > by vectors in Hilbert space), but it does lead us back to the same WF > with the above same properties. Just a note. The Dirac formalism of states as elements of abstract Hilbert spaces and observables as linear operators on them is the standard modern picture of quantum mechanics. Although it may in many cases be reduced to the wave function description, it is more general. Moreover, reasoning with abstract states and operators is sometimes easier, as it rids the thinker of preconcieved notions about about functions on space and space-time, which may creep in when reasoning with wave functions.
> I don't see why it is impossible to speak of QM trajectories. > Trajectories are described through scattering theory; the paths of > particles are calculated probabilistically--a non-scattered particle > travels in a straight line. A bubble chamber photograph verifies the > existence of these particle trajectories. The trajectories photographed in bubble chambers are much like this line of dots:
.................................................
Clearly, a straight line can be drawn passing through each of these dots. Equally, many curved paths can also be drawn that pass through each point as well. All we can say for sure from a bubble chamber photograph is that a passing particle has interacted sufficiently with the detection medium to create bubbles (a macroscopic change, hence a measurement). So instead of seeing particle trajectories, we see bubble sequences. It is perfectly possible to use quantum mechanics to try to predict these bubble sequences, though ultimately only probabilistically.
Lastly, your comment about scattering theory is not completely clear to me. Scattering in quantum mechanics does not speak of trajectories, it only describes the evolution of incoming nearly-free states to outgoing nearly-free states.
Hope this helps.
Igor
Arnold Neumaier - 27 Nov 2008 00:50 GMT Igor Khavkine wrote [in ''Questions for quantum physicists only!'']:
>> I don't see why it is impossible to speak of QM trajectories. What one cannot speak about consistently is of exact QM trajectories.
What one can speak about consistently is of _approximate_ QM trajectories, trajectories defined only within an inherent uncertainty given by the uncertainty principles.
These approximate trajectories are observed in bubble chambers, etc., and calculated from photographs of collision events such as http://www.bnl.gov/bnlweb/history/Omega-minus.asp These trajectories are to a high accuracy helices, whose defining parameters are used to calculate masses, charges, and initial momenta of the particles. These in turn are used to determine the kind of particles involved and (using many such trajectories) associated reaction cross sections.
>> Trajectories are described through scattering theory; the paths of >> particles are calculated probabilistically--a non-scattered particle [quoted text clipped - 10 lines] > each point as well. All we can say for sure from a bubble chamber > photograph I think we can be fairly sure that each photographed particle follows a helical path to high accuracy.
At least this is universally assumed in the analysis of individual scattering events. R. Frühwirth, M. Regler, R. K. Bock, H. Grote, D. Notz, Data Analysis Techniques for High-energy Physics Cambridge Univ. Press, Cambridge 2002. http://www.cambridge.org/catalogue/catalogue.asp?isbn=0521635489
That the path is not infinitely precise is clear from the way the data were collected. But nothing in QM forbids the existence of the approximate path.
For an interpretation of quantum mechanics in which approximate paths are explicitly accounted for, see Chapter 7 of my book Arnold Neumaier and Dennis Westra, Classical and Quantum Mechanics via Lie algebras, Cambridge University Press, to appear (2009?). http://www.mat.univie.ac.at/~neum/papers/physpapers.html#QML arXiv:0810.1019
> is that a passing particle has interacted sufficiently with > the detection medium to create bubbles (a macroscopic change, hence a [quoted text clipped - 7 lines] > it only describes the evolution of incoming nearly-free states to > outgoing nearly-free states. The transition from scattering states to probabilities for definite particle paths is analyzed in detail in the classical paper N.F. Mott, The Wave Mechanics of Alpha-Ray Tracks Proc. Roy. Soc. London A, 126, Issue 800, pp. 79-84 http://www.jstor.org/pss/95407 He treats only the case without external fields, in which case he gets essentially straight paths. The direction is random, but the shape of the path is determined to high accuracy. Adding an external magnetic field would produce by a similar analysis helical paths.
Hendrik van Hees - 27 Nov 2008 15:31 GMT > These approximate trajectories are observed in bubble chambers, > etc., and calculated from photographs of collision events such as [quoted text clipped - 4 lines] > kind of particles involved and (using many such trajectories) > associated reaction cross sections. This is a bit misleading. As one can calculate pretty easily, a free particle's wave function broadens with time (just take a Gaussian wave function as initial state and propagate it with the Schroedinger time evolution which results again in a Gaussian with the width increasing with time). This is not observed in the cloud-chamber pictures of the good old days (as you see at the link above). The same holds also true for particles in external fields like here where a magnetic field is present to measure the momentum of the particles with help of the curvature of the cloud-chamber traces.
The reason is that the "trajectory" depicted by the trace is not that of particles in an external field alone, but is caused by the interaction of the particles with the vapor in the cloud chamber, ionizing the molecules. The basic theory explanation, why we see rather clear "trajectories" here is indeed given by Mott in the reference quoted below.
> The transition from scattering states to probabilities for definite > particle paths is analyzed in detail in the classical paper [quoted text clipped - 6 lines] > shape of the path is determined to high accuracy. Adding an external > magnetic field would produce by a similar analysis helical paths.
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Arnold Neumaier - 28 Nov 2008 01:27 GMT Hendrik van Hees schrieb:
>> These approximate trajectories are observed in bubble chambers, >> etc., and calculated from photographs of collision events such as [quoted text clipped - 19 lines] > interaction of the particles with the vapor in the cloud chamber, > ionizing the molecules. A more detailed model for an observed particle trajectory would be a stochastic differential equation, which leads to a probabilistic broadening with time. But as far as I know, actual particle tracks are always analyzed by matching with deterministic helices using a least squares approach. The speeds are far too high for the stochastic broadening to have a noticeable effect.
In any case, a particle must have an approximate trajectory close to the ionization tracks, since a ionization can happen only when a particle is close by.
Talking of particle tracks rather than waves or fields is appropriate once the assumptions for geometric optics (which turns fields into particles) are satisfied.
[[Mod. note -- Excess quoted text deleted. -- jt]]
Murray Arnow - 28 Nov 2008 08:36 GMT >Igor Khavkine wrote >[in ''Questions for quantum physicists only!'']: [quoted text clipped - 15 lines] >particles involved and (using many such trajectories) associated >reaction cross sections. I know you are referring to the necessity of a magnetic field to produce helical paths, but you have neglected to say we that we can only see the paths of charged particles directly. Uncharged particle paths are determined indirectly; those paths are linear.
Of course we have uncertainty in the calculation of a particle path, but I was saying that uncertainties do not make particle trajectories meaningless. The thing is, we can no longer use classical methods to determine trajectories, but including uncertainties does not mean we cannot calculate trajectories to a high accuracy. I don't think you are in disagreement with me (the snipped portion confirms this, as well); I am trying to further clarify what I was saying.
Arnold Neumaier - 29 Nov 2008 11:22 GMT Murray Arnow schrieb:
>> Igor Khavkine wrote >> [in ''Questions for quantum physicists only!'']: [quoted text clipped - 19 lines] > paths of charged particles directly. Uncharged particle paths are > determined indirectly; those paths are linear. Yes, uncharged particles don't ionize anything, hence cannot bee seen. And they are not deflected by the magnetic field, hence follow straight paths.
The analysis of a complex picture with multiple helices and intermediate neutral particles is nontrivial, since one needs to find out what belongs to what. Especially when (as in modern detectors) the recordings are available only at discrete positions.
> Of course we have uncertainty in the calculation of a particle path, but > I was saying that uncertainties do not make particle trajectories [quoted text clipped - 3 lines] > in disagreement with me (the snipped portion confirms this, as well); I > am trying to further clarify what I was saying. The uncertainty principle and Ehrenfest's theorem would not make any sense if there were no mean particle paths.
The mean xbar = <x> cannot be a mean over many particles, since these all have different paths, and must therefore be understood as the approximate time-dependent location of the particle, accurate to a small multiple of the standard deviation sigma(x)=sqrt(<(x-xbar)^2>). Thus the particle is localized in the tube swept out by the moving ball {xi in R^3 | |xi-xbar(t)|<=3*sigma(x(t))}, and any more stringent localization is meaningless.
In some cases (e.g., a fast moving free particle in an external field), the paths are very well determined, i.e., sigma << |xbar|. In others (e.g., an electron bound in an atom at rest in the origin), the paths are very poorly determined, i.e., |xbar| << sigma. It depends very much on the state the particle is in.
Arnold Neumaier
p.kinsler@ic.ac.uk - 25 Nov 2008 18:05 GMT > I don't see why it is impossible to speak of QM trajectories. You can certainly generate quantum trajectories using quantum Monte Carlo techniques; however they only have physical meaning when an average is taken over an ensemble of all possible trajectories.
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Oh No - 28 Nov 2008 08:36 GMT Thus spake Murray Arnow <arnow@iname.com>
>I don't see why it is impossible to speak of QM trajectories. >Trajectories are described through scattering theory; the paths of >particles are calculated probabilistically--a non-scattered particle >travels in a straight line. A bubble chamber photograph verifies the >existence of these particle trajectories. The path in a bubble chamber is not a quantum path. It is a sequence of points defined in classical measurement. Quantum mechanics only describes probabilities from an initial measurement to a final measurement. It does not describe a sequence. Between the initial and final measurements, we do not have a definition of a particle trajectory.
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ilper@abv.bg - 28 Nov 2008 14:15 GMT Arnold Neumaier wrote
> The transition from scattering states to probabilities for definite > particle paths is analyzed in detail in the classical paper > N.F. Mott, > The Wave Mechanics of Alpha-Ray Tracks > Proc. Roy. Soc. London A, 126, Issue 800, pp. 79-84 Could you describe briefly the main point of Mott analysis. I dont have access to that paper.
Ilian
[[Mod. note -- The first page of Mott's paper is freely available at http://www.jstor.org/pss/95407 Alas, you'll need to be a JSTOR subscriber (or more likely, affiliated with an institution that's a subscriber) to see the rest of that paper. -- jt]]
Arnold Neumaier - 29 Nov 2008 21:51 GMT ilper@abv.bg schrieb:
> Arnold Neumaier wrote > [quoted text clipped - 6 lines] > Could you describe briefly the main point of Mott analysis. > I dont have access to that paper. The paper is also reprinted in the reprint volume J.A. Wheeler and W. H. Zurek, Quantum theory and measurement. Princeton Univ. Press, Princeton 1983. which you might have access to. It is very valuable reading for the foundations of quantum physics up to around 1980.
The main point of Mott's analysis is that the angle alpha_1 of the first ionization as seen from the source is random. But the conditional probability of the angle alpha_{k+1} of the (k+1)st ionization given the kth ionization can be computed from first principles (under slightly simplifying assumptions) and is very strongly peaked around alpha_k. This leads to an essentially straight line in a random direction from the source.
Arnold Neumaier
ilper@abv.bg - 01 Dec 2008 16:21 GMT On 29 Nov, 23:51, Arnold Neumaier <Arnold.Neuma...@univie.ac.at> wrote:
> The main point of Mott's analysis is that the angle alpha_1 of the > first ionization as seen from the source is random. But the [quoted text clipped - 5 lines] > > Arnold Neumaier Thanks a lot. I have found the article.
The question I raised in my first post on the thread 'For quantum physicist only' is in my opinion far from being answered. It was argued here that the impulse of a deflected particle is compensated for by the impulse of the material of the slit (by almost all of the authors except Charles Francis).
As I see there are 2 possibilities:
1. The particle on its 'trajectory' interacts with just one atom from the edge of the slit (unknowingly why with the edge exactly). The particle changes its impulse and the atom takes an opposite impulse.
However this opportunity is impossible because than the minimal change of the impulse wouldn’t depend on the width of the slit.
In this sense if there are approximate trajectories (and if they are many of them thru the slit delta x < lambda) the particle could not 'see' the other edge of the slit. As I am not well acquainted with de Mott analysis I don’t know if there are approximate trajectories in the case of empty space.
In this connection another question arises: if we can observe the trajectories in a slit (in bubble chamber) would there be broadening of delta p observed after the slit?
2.The particle being a wave (or if one prefers showing its wave properties) interacts in the range of its wavelength (unknowingly why but never mind) with the material of the slit.
A) this interaction must happen very selectively - just with atoms from the edge of the surface (dp depends only on the utmost outside width of a real slit). This selectivity needs also to be accounted for.
B) it must be accepted that WF is not associated with a group of undependable particles and has no probability meaning but some real physical nature.
C) even if WF has real physical nature (or even not) I’m very disturbed by the fact that WF will interact (or better impart interaction of the order lambda) on the slit edges but wont show a trace of such on the screen. Igor Khavkine had answered to this that the change in the screen is macroscopic whereas this (these) on the slit is not. I would be indebted if he can explain what is meant by this (maybe macroscopic=observable). And than where is the boundary between macroscopic and microscopic?
D) One should explicitly note that from the point of view of classical wave theory there is no at any means exchange of impulse between the wave and the walls of the slit (good old Huygens principle). So Schrödinger function also doesn’t take/give any impulse to/from the slit (if it is to be attributed any physical reality to it). The material doesn’t interact with the wave it just blocks the waves and don’t let them interfere further and restore the falling plane wave
Igor Khavkine - 03 Dec 2008 07:29 GMT On Dec 1, 11:21 am, il...@abv.bg wrote:
> As I see there are 2 possibilities:
> 2.The particle being a wave (or if one prefers showing its wave > properties) interacts in the range of its wavelength (unknowingly why > but never mind) with the material of the slit. The particle is not a wave. The particle is. The possible outcomes of its impact on a screen after passing through a slit are described by solutions of the Schroedinger equation (an equation that does possess wave properties). If you want to get to the nitty-gritty details, please try to use the correct terminology.
This caveat aside, yes your possibility 2. is the one closest to reality.
> A) this interaction must happen very selectively - just with atoms > from the edge of the surface (dp depends only on the utmost outside > width of a real slit). This selectivity needs also to be accounted > for. The interaction happens with all atoms, everywhere. The result of *all* of these interactions is what is observed. This result can be predicted by solving the Schroedinger equation. I repeat, not by calculating a trajectory, but by solving the Schroedinger equation.
> B) it must be accepted that WF is not associated with a group of > undependable particles and has no probability meaning but some real > physical nature. Since you keep coming back to this point, I presume it is important to you. From previous conversations I think you have a decent understanding of how the wave function is used to predict measurement outcomes. So, what can you gain on top of that by convincing yourself that the wave function either does or does not have "some real physical nature"? Assuming that this "physical nature" can even be unambiguously defined.
> C) even if WF has real physical nature (or even not) I’m very > disturbed by the fact that WF will interact (or better impart > interaction of the order lambda) on the slit edges but wont show a > trace of such on the screen. Unfortunately, I don't really understand your assertion here, nor why it would be disturbing for anyone. It might even be based on erroneous assumptions. As I pointed out above, the electron interacts with all atoms, everywhere, including all the atoms of the slit as well as those of the screen.
> Igor Khavkine had answered to this that the change in the screen is > macroscopic whereas this (these) on the slit is not. > I would be indebted if he can explain what is meant by this (maybe > macroscopic=observable). > And than where is the boundary between macroscopic and microscopic? Yes, I used the term "macroscopic change" to indicated where one would declare a measurement and apply the probabilistic interpretation of the wave function. However, the line between macroscopic and microscopic changes is not sharp; it is a continuum.
> D) One should explicitly note that from the point of view of classical > wave theory there is no at any means exchange of impulse between the [quoted text clipped - 3 lines] > material doesn’t interact with the wave it just blocks the waves and > don’t let them interfere further and restore the falling plane wave Unfortunately, you're quite incorrect here. Whether classical or not, a wave bumping into a wall will exchange momentum with it. If the wall is infinitely massive, it absorbs the momentum without motion. I believe this was pointed out in the very first responses to your original post.
Hope this helps.
Igor
Arnold Neumaier - 05 Dec 2008 23:24 GMT ilper@abv.bg schrieb:
> On 29 Nov, 23:51, Arnold Neumaier <Arnold.Neuma...@univie.ac.at> > wrote: [quoted text clipped - 16 lines] > for by the impulse of the material of the slit (by almost all of the > authors except Charles Francis).
> In this sense if there are approximate trajectories (and if they are > many of them thru the slit delta x < lambda) the particle could not > 'see' the other edge of the slit. As I am not well acquainted with de > Mott analysis I don't know if there are approximate trajectories in > the case of empty space. In empty space behind the slit one cannot talk of a particle path since the wave function is a spherical wave and hence completely smeared in space.
> I'm very > disturbed by the fact that WF will interact (or better impart > interaction of the order lambda) on the slit edges but wont show a > trace of such on the screen. The screen absorbs the particle with high probability; it is just very very difficult to observe this effect.
Passing the screen must be viewed in practice as a dissipative process when viewed as a process of the particle only. In dissipative processes, the conservation laws do not hold. They only hold for the combined system (particle + environment, here the screen), but this is a complex many-particle process.
Arnold Neumaier
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