Tracks in particle detectors and quantum paths

[Vanadium 50]

What is n for a baseball?

This is an exercise in, I believe, French and Taylor. But it is very, very large. 10^30? 10^40? It doesn't matter. The key is that it's big. And that's what matters.

Cheers,

Tom

 

[stevendaryl]

This is an exercise in, I believe, French and Taylor. But it is very, very large. 10^30? 10^40? It doesn't matter. The key is that it's big. And that's what matters.

Cheers,

Tom

What do you mean by n for a baseball? What is the definition of n? I know what it is for a bound particle, but what does it mean for a particle that isn't bound?

 

[Vanadium 50]

A baseball is bound. It's bound in the Earth's gravitational field, if you want to do a calculation. But you are quibbling - do you really think that the thing that matters is whether n is a million or a bejillion?

 

[atyy]

Also, as one can see in the beta particles do not have straight line tracks in the cloud chamber.

http://www.ap.smu.ca/demos/index.php?option=com_content&view=article&id=114&Itemid=85
http://www.nuffieldfoundation.org/practical-physics/alpha-particle-tracks-showing-their-short-range

In the examples above, alpha particles typically leave thick lines, which correspond intuitively to imprecise position measurements.

 

[stevendaryl]

A baseball is bound. It's bound in the Earth's gravitational field, if you want to do a calculation. But you are quibbling - do you really think that the thing that matters is whether n is a million or a bejillion?

I don't know what you mean by n when talking about a baseball.

 

[TrickyDicky]

I don't know what you mean by n when talking about a baseball.

I think he is just referring to how we can in principle calculate n for any macroscopic object, I think there is an example in Shankar about an oscilating mass where you just plug the frequency and the energy in the formula for n=E/hw and obtain n about 10^27. I'm not sure how to exactly do this for a baseball(what do you take as its oscillating frequency), but I guess it ¡s doable, although quite useless in practice. It just shows that macroscopic situations are not out of the scope of a quantum analysis. Wich is fine but it is totally irrelevant and unrelated to what is being discussed in this thread, and besides nobody has said anything even close to questioning it(I'm not totally sure about mfb's post but I doubt that he argued it).

We are instead dealing with a microscopic quantum system that shows classical trajectories in some instances but not in others, it is kind of the opposite case.

 

[Haelfix]

It gives a result that is perfectly consistent and *expected* by quantum mechanics. The fact that it leaves a straight line is a prediction of that particular system. Unfortunately it is not obvious to see this, unless you go through the calculation (and then you see that the combined bubble chamber, alpha particle system forms a pointer state etc). The spherical symmetry of the original problem is only apparent when you repeat the same measurement hundreds of thousands of times, and then you see that indeed there is no angular dependance.

More generally, it is a beginner mistake to believe that a continuous spectrum, or a set of discrete lumps implies either classical or quantum behavior. The mistake is attributed to the fact that simple free systems often show that type of behaviour, but in general for more complicated many body systems (like molecules) you will often see both behaviours simultaneously (recall the generalized spectral theorem).

Even more amusing, classical physics can sometimes produce discrete phenomena... For instance a classical vibrating string.

Anyway, all this to say, read up on those papers at the beginning of the thread. The Mott phenomenon is an important example of how a complicated interacting quantum system can yield simple emergent behaviour.

 

[StrangeCoin]

More generally, it is a beginner mistake to believe that a continuous spectrum, or a set of discrete lumps implies either classical or quantum behavior. The mistake is attributed to the fact that simple free systems often show that type of behaviour, but in general for more complicated many body systems (like molecules) you will often see both behaviours simultaneously (recall the generalized spectral theorem).

I'm not sure what is it exactly you are suggesting, but we can bend those trajectories with magnetic fields and they go right where they are supposed to go, according to classical prediction.

So how is that possible if the electron was not exactly on the continuous path between every two successive bubbles? If it was going anywhere else its acceleration wouldn't be uniform, it's velocity would vary which in turn would cause magnetic force to vary and it would not pass through all the expected bubble "check-points" as it does.

 

[TrickyDicky]

It gives a result that is perfectly consistent and *expected* by quantum mechanics. The fact that it leaves a straight line is a prediction of that particular system. Unfortunately it is not obvious to see this, unless you go through the calculation (and then you see that the combined bubble chamber, alpha particle system forms a pointer state etc). The spherical symmetry of the original problem is only apparent when you repeat the same measurement hundreds of thousands of times, and then you see that indeed there is no angular dependance.

I'm afraid this is too vague and handwavy, can you elaborate?
If your refernce to pointer states refers to "environment-induced-superselection", the wikipedia page says: "the question of whether the 'einselection' account can really explain the phenomenon of wave function collapse remains unsettled"
No angular dependance(angular invariance) is equivalent to spherical symmetry so in what sense it is only apparent?

 

[atyy]

Whatever way one does it, a collapse is needed.

There are two alternatives, both are mentioned in the review that stevendaryl posted.

1) Treat the particles of the cloud chamber and particle as quantum, followed by a sngle measurement and collapse. This is the decoherence, einselection and pointer states solution that Mott, stevendaryl, haelfix have mentioned.

2) Treat it as many successive measurements, and many successive collapses. This is Belavkin's approach.

Both approaches give the same result, corresponding to our ability to place the Heisenberg cut in any reasonable place.

 

[mfb]

Generating a multitude of different trajectories(one for each interaction) would be the expected outcome for a spherical wave function, and it would be an equally consistent result, but it is not what is observed.

That would violate quantum mechanics, energy conservation, and probably some more laws, at the same time. Each measurement localizes the particle, you can't have the particle jumping around like crazy between measurements.
In many worlds, you get many trajectories at the same time, but in different branches of the world, so you always see just one trajectory.

We know quantum effects are negligible for macroscopic objects(sand grains, pick-up trucks,... but not for say buckyballs).

There is no fixed size limit. Just a decoherence limit that depends on the experiment.
By the way, quantum effects are not negligible in macroscopic objects - without quantum mechanics, chemistry would not work.

Again, I think all quantum systems interact significatively with the environment.

Not significantly in the sense of decoherence within the relevant timescale, otherwise we would not see quantum effects at all.

Well, sort of. If the atoms themselves have a definite location, then interacting with the atoms would localize the particle. But why should the atoms themselves have definite locations?

Decoherence/Measurements on a timescale of picoseconds.

Also, as one can see in the beta particles do not have straight line tracks in the cloud chamber.

Sure, because they interact with the atoms, and some interactions change the momentum significantly.

Whatever way one does it, a collapse is needed.

Let's say decoherence is needed. There are interpretations of QM that do not need collapses.

I'm not sure what is it exactly you are suggesting, but we can bend those trajectories with magnetic fields and they go right where they are supposed to go, according to classical prediction.

So how is that possible if the electron was not exactly on the continuous path between every two successive bubbles? If it was going anywhere else its acceleration wouldn't be uniform, it's velocity would vary which in turn would cause magnetic force to vary and it would not pass through all the expected bubble "check-points" as it does.

The uncertainty is negligible compared to the size of the bubbles.

 

[atyy]

Let's say decoherence is needed. There are interpretations of QM that do not need collapses.

Yes. My point is that we are just doing quantum mechanics here, so we are not trying to get rid of collapse. Asking for a solution to the measurement problem is beyond the scope of this question.

 

[vanhees71]

It's very important to "get rid of the collapse" by just not introducing it. It only causes a lot of trouble, including the whole EPR debate etc. The good thing is that it's not needed at all. Instead we can simply take Born's postulate serious and take the Minimal Statistical Interpretation. That's how, in fact, quantum theory is used in practice, when real-world experiments are made in the labs and described with help of quantum theory.

 

[kith]

No angular dependance(angular invariance) is equivalent to spherical symmetry so in what sense it is only apparent?

You cannot observe the spherical symmetry in a single run of the emission experiment. In a simplified situation where the alpha particle is emitted into a perfect vacuum and detected by a 4π array of position detectors, only one of the position detectors will click. In order to conclude that the emission process is spherically symmetric you have to repeat the experiment many times and note that all position detectors click equally often. Analogously, the spherical symmetry of the problem at hand manifests in the fact that all linear tracks are equally likely.

 

[stevendaryl]

[about why baseballs seem to have definite positions] Decoherence/Measurements on a timescale of picoseconds.

That's probably true, but it seems like overkill to explain a baseball.

If you prepare a baseball at a position x with position uncertainty of \Delta x, then the expected length of time required for it to be at a location significantly different from that (just taking into account QM) is on the order of \dfrac{m \Delta x^2}{\hbar}. For baseball-sized values of m and \Delta x, that can be a huge length of time. So once prepared in a more-or-less definite position, a baseball will keep to an approximately classical path for a long time without the need to invoke decoherence.

 

[Vanadium 50]

but it is totally irrelevant and unrelated to what is being discussed in this thread.

Do you know the answer or don't you? If you don't, perhaps it is unwise to be telling people that what they are posting is irrelevant.

We are instead dealing with a microscopic quantum system that shows classical trajectories in some instances but not in others, it is kind of the opposite case.

This is why I made a big deal of pointing out where you were going wrong. You are still holding on to that wrong viewpoint, which is why you will never understand it.

 

[TrickyDicky]

That would violate quantum mechanics, energy conservation, and probably some more laws, at the same time.

How so?, be specific.
When I mention multiple trajectories I obviously don't mean simultaneously, but sequentially as it interacts with the chamber environment.

Each measurement localizes the particle, you can't have the particle jumping around like crazy between measurements.

I'm not sure what you are calling measurements in this context, if every interaction with the environment was a measurement in the sense of collapse then we would not see quantum effects at all, it would all be classical. Or as you say:

Not significantly in the sense of decoherence within the relevant timescale, otherwise we would not see quantum effects at all.

If all interaction with the environment is a measurment with decoherence we would indeed not see them.

 

[TrickyDicky]

This is why I made a big deal of pointing out where you were going wrong. You are still holding on to that wrong viewpoint, which is why you will never understand it.

You sure made a big deal,(it would be nice if you toned down your emotional reactions if only to comply with the civility values posted by Greg in PF's main page), what you haven't even hinted at is what you consider my wrong viewpoint so I can correct it if necessary. Have you even considered that you might be misunderstanding my viewpoint?

 

[stevendaryl]

If all interaction with the environment is a measurment with decoherence we would indeed not see them [quantum effects].

There is a meta-question about your questions: Are you asking for detailed calculations that show why an alpha particle in a bubble chamber has an apparently definite trajectory, while an electron in a hydrogen atom apparently doesn't? Without doing the calculations, I feel that the two cases are different enough that you shouldn't expect them to be similar. The detailed calculations have been done showing how the tracks in a bubble chamber are consistent with quantum mechanics. In the case of a hydrogen atom, the effects of the environment would certainly be tiny compared with the effects of the coulomb attraction. So I certainly wouldn't expect the environment to be important.

In this paper:
http://arxiv.org/ftp/quant-ph/papers/0306/0306072.pdf
there is a formula giving an order-of-magnitude estimate of the decoherence time. The authors claim:

For microscopic systems and, occasionally, even for very macroscopic ones, the deco-
herence times are relatively long. For an electron...[the decoherence times] can be much
larger than the other relevant time scales on atomic and larger energy and distance scales.

 

[TrickyDicky]

There is a meta-question about your questions: Are you asking for detailed calculations that show why an alpha particle in a bubble chamber has an apparently definite trajectory, while an electron in a hydrogen atom apparently doesn't? Without doing the calculations, I feel that the two cases are different enough that you shouldn't expect them to be similar. The detailed calculations have been done showing how the tracks in a bubble chamber are consistent with quantum mechanics. In the case of a hydrogen atom, the effects of the environment would certainly be tiny compared with the effects of the coulomb attraction. So I certainly wouldn't expect the environment to be important.

In this paper:
http://arxiv.org/ftp/quant-ph/papers/0306/0306072.pdf
there is a formula giving an order-of-magnitude estimate of the decoherence time. The authors claim:

Yes, that is part of what I'm after with my questions. Maybe not so much a detailed comparative calculation between the two cases(wich I think it's much to ask in a forum, besides my math wouldn't be up to par to seriously analyze them), but more of conceptual explanation of the key differences wrt definite trajectories between those cases.
I know the effects of the environment are tiny compared to the coulomb atraction, but I'm having difficulties seeing how that is different from the case of a particle in a chamber with say a very powerful magnetic field deflecting the trajectory of the charged particle but not making it any less definite.
I'll take a look at the reference, thanks Daryl.

 

[stevendaryl]

It's very important to "get rid of the collapse" by just not introducing it. It only causes a lot of trouble, including the whole EPR debate etc. The good thing is that it's not needed at all. Instead we can simply take Born's postulate serious and take the Minimal Statistical Interpretation. That's how, in fact, quantum theory is used in practice, when real-world experiments are made in the labs and described with help of quantum theory.

In practice, there is not much difference between all the various interpretations of quantum mechanics. The Von Neuman recipe that measurement collapses the wave function works perfectly well. The various debates are really trying to understand what the quantum recipe means.

You say that in practice, you don't need anything like collapse, but I don't see that that's completely true. What you do in performing an experiment is to prepare a system in a particular state, let it evolve, the perform a measurement. But how do you prepare a system in a particular state, in the first place? Well, one approach is to use measurement: If you want to prepare electrons in the spin-up state, you start with a source of electrons, and measure the spins (via Stern-Gerlach, or whatever). Then you only use those that have spin-up. But why does measuring spin-up mean that the electron is in the spin-up state after the measurement? Isn't that a collapse-type assumption?

 

[stevendaryl]

Yes, that is part of what I'm after with my questions. Maybe not so much a detailed comparative calculation between the two cases(wich I think it's much to ask in a forum, besides my math wouldn't be up to par to seriously analyze them), but more of conceptual explanation of the key differences wrt definite trajectories between those cases.
I know the effects of the environment are tiny compared to the coulomb atraction, but I'm having difficulties seeing how that is different from the case of a particle in a chamber with say a very powerful magnetic deflecting the trajectory of the charged particle but not making it any less definite.
I'll take a look at the reference, thanks Daryl.

I'm curious: how did you know that my name was "Daryl", when my user name is stevendaryl? (Daryl is actually my middle name, and Steven is my first name, but I go by Daryl)

 

[TrickyDicky]

I'm curious: how did you know that my name was "Daryl", when my user name is stevendaryl? (Daryl is actually my middle name, and Steven is my first name, but I go by Daryl)

Didn't even notice it, I just used it for short, elsewhere I think I used Steven,

 

[atyy]

For the question of why decoherence doesn't decohere everything, the conceptual answer is that there are different strengths of interactions. From the measurement point of view, this corresponds to measuring position with different degrees of precision. For specific examples of how even in the presence of decoherence leading to non-unitary dynamics, there can be subsystems which are decoherence free and continue to evolve unitarily, one review is http://arxiv.org/abs/quant-ph/0301032 .

There's an analogous problem in the quantum Zeno effect, in which continuous accurate measurement freezes a system. However, once again, the measurement does not have to be accurate, and the system does not have to freeze completely. There can even be unitary dynamics in a subsystem. http://arxiv.org/abs/0711.4280

One of the authors on the linked papers above, Pascazio, has addressed decoherence and particle tracks in a cloud chamber. http://www.ba.infn.it/~pascazio/publications/Particle_tracks_and_the_mechanis.pdf

 

[TrickyDicky]

Thanks, atyy, I'll take a look at those references.

 

[TrickyDicky]

Ok, I can see how not having classical orbits for electrons in atoms doesn't have any bearing on having classical trajectories for electrons in cloud/bubble chambers or the Mott problem. There is a key difference in the size vs momentum imparted by environment between the tiny atom and the chamber. The "resolution" needed is much bigger for the atom and the act of measurement completely discards even the idea of a trajectory.
Thinking it over I would say that it is deceiving to think in terms of trajectories even in the case of the tracks in particle detectors. One just observes the result of interactions and constructs something like a classical trajectory.

So if one thinks about the wave function of the whole environment+original source of interactions be it alpha decay or any other process interacting with the environment, the spherical symmetry is still there and there is no paradox at all. It is only when one clings to a discontinuous view of the wave function that one gets in trouble with spherical waves vs. "linear" tracks.

Comments, criticisms?

 

[TrickyDicky]

Wave descriptions (not necessarily spherical) of individual microscopic particles may be where quantum mechanics started, but it long ago grew beyond that early formulation of the problem. Quantum mechanically two entangled particles in the singlet state are not two microscopic particles; they're a single quantum system with a single wavefunction and two sets of observables on that system. Including the environment increases the complexity of the system (enough that completely different computational methods may be needed) but even before we include it, we've lost any sense of individual microscopic particles.

(my bold)

I guess I should have payed more attention to this.

 

[Jilang]

Thinking it over I would say that it is deceiving to think in terms of trajectories even in the case of the tracks in particle detectors. One just observes the result of interactions and constructs something like a classical trajectory.

So if one thinks about the wave function of the whole environment+original source of interactions be it alpha decay or any other process interacting with the environment, the spherical symmetry is still there and there is no paradox at all. It is only when one clings to a discontinuous view of the wave function that one gets in trouble with spherical waves vs. "linear" tracks.

Comments, criticisms?

Bravo, I feel we are getting there! I am totally happy with the first paragraph, not so sure about the second paragraph though. What is this spherical symmetry idea? All backwards moving waves are self-cancelling (although actually it was not Huygens who proved this, it came later).

 

[atyy]

Ok, I can see how not having classical orbits for electrons in atoms doesn't have any bearing on having classical trajectories for electrons in cloud/bubble chambers or the Mott problem. There is a key difference in the size vs momentum imparted by environment between the tiny atom and the chamber. The "resolution" needed is much bigger for the atom and the act of measurement completely discards even the idea of a trajectory.
Thinking it over I would say that it is deceiving to think in terms of trajectories even in the case of the tracks in particle detectors. One just observes the result of interactions and constructs something like a classical trajectory.

So if one thinks about the wave function of the whole environment+original source of interactions be it alpha decay or any other process interacting with the environment, the spherical symmetry is still there and there is no paradox at all. It is only when one clings to a discontinuous view of the wave function that one gets in trouble with spherical waves vs. "linear" tracks.

Comments, criticisms?

I think the idea of whether the trajectory is continuous or not is very technical, and not very conceptual. Conceptually, one can always assume a discretization that is much finer than the spatial resolution of one's measurements, and "construct" the continuous track from these low resolution readouts. It's in the same spirit as lattice simulations of quantum field theory, which cannot have true Lorentz invariance, but can be Lorentz invariant for all practical purposes, as long as the discretization is much finer than what current experiments can see. Mathematically, it does matter, and one can worry about whether the continuum limit really exists, but I would say that at the physics "conceptual" level, there is no big distinction between continuous and discrete (there are exceptions to the rule of thumb, eg. chiral interactions on a lattice, but let's worry about that elsewhere).

In the 2 methods of analysis the conceptual puzzles are different.

1) In decoherence followed by a single measurement and collapse, the puzzle is indeed why decoherence localizes objects. The general answer is that interactions are local in space, with nearby objects interacting more strongly, and distant objects interacting more weakly. This is just a fact of nature that we incorporate into our models of decoherence. Incorporating this fact leads to localization by decoherence.

2) In the case of multiple successive measurements, it's no puzzle why localization occurs, since each position measurement will collapse the wave function into something like a definite position. The puzzle is why in many cases the track is straight enough to get a pretty accurate momentum measurement, which seems to violate the restriction on simultaneous accurate measurement of position and momentum. The answer is that in many cases, the position measurement is inaccurate. An inaccurate position measurement gives a definite position readout, but it does not localize the wave function very tightly. Once localization occurs, then given a spherical wave function, the spherical symmetry is broken on any single measurement trial, but preserved over multiple measurement trials.

 

[TrickyDicky]

Bravo, I feel we are getting there! I am totally happy with the first paragraph, not so sure about the second paragraph though. What is this spherical symmetry idea? All backwards moving waves are self-cancelling (although actually it was not Huygens who proved this, it came later).

Oh, the spherical symmetry(isotropy) idea is specific to those systems with such symmetry like it is the case for high energy physics experiments in particle colliders and bubble/cloud chambers. Not meant as something general.
It didn't occurr to me to think of the linear tracks as related to the Huygens principle but it is a nice way to get an understanding of the Mott problem, relating it to the path integral approach, for instance this quote from wikipedia seems pertinent if one trades isotropic space/medium and real waves for isotropic system wavefunction that includes the environment:

"Huygens' principle can be seen as a consequence of the isotropy of space—all directions in space are equal. Any disturbance created in a sufficiently small region of isotropic space (or in an isotropic medium) propagates from that region in all radial directions. The waves created by this disturbance, in turn, create disturbances in other regions, and so on. The superposition of all the waves results in the observed pattern of wave propagation.

Isotropy of space is fundamental to quantum electrodynamics (QED) where the wave function of any object propagates along all available unobstructed paths. When integrated along all possible paths, with a phase factor proportional to the path length, the interference of the wave-functions correctly predicts observable phenomena. Every point on wave front acts as the source of secondary wavelets that spread out in the forward direction with the same speed as the wave. The new wave front is found by constructing the surface tangent to the secondary wavelets."

 

[TrickyDicky]

I think the idea of whether the trajectory is continuous or not is very technical, and not very conceptual. Conceptually, one can always assume a discretization that is much finer than the spatial resolution of one's measurements, and "construct" the continuous track from these low resolution readouts. It's in the same spirit as lattice simulations of quantum field theory, which cannot have true Lorentz invariance, but can be Lorentz invariant for all practical purposes, as long as the discretization is much finer than what current experiments can see. Mathematically, it does matter, and one can worry about whether the continuum limit really exists, but I would say that at the physics "conceptual" level, there is no big distinction between continuous and discrete (there are exceptions to the rule of thumb, eg. chiral interactions on a lattice, but let's worry about that elsewhere).

In the 2 methods of analysis the conceptual puzzles are different.

1) In decoherence followed by a single measurement and collapse, the puzzle is indeed why decoherence localizes objects. The general answer is that interactions are local in space, with nearby objects interacting more strongly, and distant objects interacting more weakly. This is just a fact of nature that we incorporate into our models of decoherence. Incorporating this fact leads to localization by decoherence.

2) In the case of multiple successive measurements, it's no puzzle why localization occurs, since the position measurement will collapse the wave function into something like a definite position. The puzzle is why in many cases the track is straight enough to get a pretty accurate momentum measurement, which seems to violate the restriction on simultaneous accurate measurement of position and momentum. The answer is that in many cases, the position measurement is inaccurate. An inaccurate position measurement gives a definite position readout, but it does not localize the wave function very tightly.

According to the Copenhagen interpretation both methods are equivalent, or so It claims the first refence given by stevendaryl when analyzing the early works of Heisenberg and Born.

I myself see too many holes in 2) to agree they are equivalent, in any case I prefer 1), and I don't quite agree with you that there is no big distinction in this respect between a discrete and a continuous model, in fact a continuous field model easily solves the puzzle you mention for decoherence and localization for 1)

 

[Jilang]

I see some analogies here to the quantum Zeeman effect, with continuous measurement surpressing the random outcomes...

 

[TrickyDicky]

I see some analogies here to the quantum Zeeman effect, with continuous measurement surpressing the random outcomes...

You lost me here, what is the analogy?

 

[Lapidus]

How are the track leftt say by an electron in a cloud chamber and its wave function undefined trajectory related exactly?

I have not read all posts in the thread, so I'm not sure if someone gave a satisfying answer.

The uncertainty of postion-momnetum of particles is of order of the Planck constant, but the momentum of high-energy particles is of course much higher than that. So you have more "leeway" for those particles.

The same as when you squeeze a particle in tiny box, it gains higher and higher momentum in the box which in turn allows it to have higher momentum spreads according to HUP.

 

[TrickyDicky]

I have not read all posts in the thread, so I'm not sure if someone gave a satisfying answer.

The uncertainty of postion-momnetum of particles is of order of the Planck constant, but the momentum of high-energy particles is of course much higher than that. So you have more "leeway" for those particles.

The same as when you squeeze a particle in tiny box, it gains higher and higher momentum in the box which in turn allows it to have higher momentum spreads according to HUP.

Thanks, I basically got around to it in #76.

 

[atyy]

According to the Copenhagen interpretation both methods are equivalent, or so It claims the first refence given by stevendaryl when analyzing the early works of Heisenberg and Born.

I myself see too many holes in 2) to agree they are equivalent, in any case I prefer 1), and I don't quite agree with you that there is no big distinction in this respect between a discrete and a continuous model, in fact a continuous field model easily solves the puzzle you mention for decoherence and localization for 1)

What holes do you see in 2?

(I agree with the Born and Heisenberg views in the reference given by stevendaryl that both methods should be equiavlent in Copenhagen. There are some problems, since placing the Heisenberg cut requires common sense, but I don't believe the problems occur in this case. For an example of problems with wrong placement of the Heisenberg cut, see http://arxiv.org/abs/quant-ph/9712044)

I see some analogies here to the quantum Zeeman effect, with continuous measurement surpressing the random outcomes...

You lost me here, what is the analogy?

Jilang probably meant quantum Zeno effect.

 

[TrickyDicky]

What holes do you see in 2?

I generally dislike the Copenhagen interpretation, especially the artificial separation between quantum systems and classical apparatus to observe them. In this particular case of the Mott problem I find more elegant the decoherence view.

Jilang probably meant quantum Zeno effect.

Ah,ok then. I think I even mentioned the effect at the start of the thread.

 

[atyy]

I generally dislike the Copenhagen interpretation, especially the artificial separation between quantum systems and classical apparatus to observe them. In this particular case of the Mott problem I find more elegant the decoherence view.

Yes, one can take MWI and decoherence. I'm not sure MWI totally works, but let's assume it does. In that case, 2) can be rephrased as mutiple decohering events, one for each of the successive measurements. So 2) should also have an analogue in MWI.

 

[mfb]

Well, sort of. If the atoms themselves have a definite location, then interacting with the atoms would localize the particle. But why should the atoms themselves have definite locations?

Decoherence/Measurements on a timescale of picoseconds.

That's probably true, but it seems like overkill to explain a baseball.

No, this was about the atoms in the bubble chamber.

You say that in practice, you don't need anything like collapse, but I don't see that that's completely true.

All collapse-free interpretations give you a way to prepare a pure, well-known state. Even the procedure to do this in the lab is the same, just the interpretation how it happens differs.

How so?, be specific.
When I mention multiple trajectories I obviously don't mean simultaneously, but sequentially as it interacts with the chamber environment.

Sorry, how can I be more specific than "particles teleporting randomly around in a bubble chamber of arbitrary size (even with superluminal speeds if it is large enough) are unphysical"? This is contrary to all physics we know of.

I'm not sure what you are calling measurements in this context, if every interaction with the environment was a measurement in the sense of collapse then we would not see quantum effects at all, it would all be classical. Or as you say:

I never said "every interaction". Many interactions of a fast charged particle with an atom in a liquid medium count as measurement.

So if one thinks about the wave function of the whole environment+original source of interactions be it alpha decay or any other process interacting with the environment, the spherical symmetry is still there and there is no paradox at all. It is only when one clings to a discontinuous view of the wave function that one gets in trouble with spherical waves vs. "linear" tracks.

Right.
If you use the many-worlds interpretation, for example, the wave function of the whole system keeps the spherical symmetry of the initial particle. But the system quickly decomposes to many different branches that will never "see" (influence) each other again afterwards. Each branch just sees one (nearly) classical trajectory.

 

[TrickyDicky]

Yes, one can take MWI and decoherence. I'm not sure MWI totally works, but let's assume it does. In that case, 2) can be rephrased as mutiple decohering events, one for each of the successive measurements. So 2) should also have an analogue in MWI.

I don't have a single favourite interpretation but I dislike MWI even more than Copenhagen.

 

[atyy]

I don't have a single favourite interpretation but I dislike MWI even more than Copenhagen.

What interpretation are you using that has no collapse if it is not MWI?

 

[TrickyDicky]

Not sure, a strange mix of ensemble and consistent histories?

 

[StrangeCoin]

Not sure, a strange mix of ensemble and consistent histories?

As I see it there are only three options: electrons always move in continuous trajectories, sometimes, or never. Since the second one lacks logical consistency, I suppose you are investigating the possibility of the first one. But no matter how bubble chamber trajectories are compelling, you are still left with double-slit experiments and such. If you are to ever confirm those classical trajectories you have to move away from bubble chambers and grapple with those experiments that indicate otherwise, and I'm afraid there are just too many of them. Still, I'd like to see that, I never liked QM explanations myself, way too esoteric and uncomfortably paranormal.

 

[vanhees71]

That you see tracks from single particles (!) in a detector like a cloud chamber has nothing to do with the interpretation you use for quantum theory but is a well-understood phenomenon (the minimal representation is sufficient ;-)). What you see is, of course, not the particle, but a macroscopic track of the particle, due to the interactions with the gas molecules in the cloud chamber. It's a very coarse-grained picture of the particle not the particle itself!

That you see tracks as if the particle was a classical particle has been explained already very early by Mott in a famous publication

Mott, N. The Wave Mechanics of alpha-Ray Tracks. Proceedings of the Royal Society of London. Series A 126, 800 (1929), 79-84.

 

[TrickyDicky]

As I see it there are only three options: electrons always move in continuous trajectories, sometimes, or never. Since the second one lacks logical consistency, I suppose you are investigating the possibility of the first one. But no matter how bubble chamber trajectories are compelling, you are still left with double-slit experiments and such. If you are to ever confirm those classical trajectories you have to move away from bubble chambers and grapple with those experiments that indicate otherwise, and I'm afraid there are just too many of them.

Not exactly, I think you misunderstood the key point made that it is misleading to think about trajectories in all cases, then you don't have any problems either with bubble chamber tracks, electrons in atoms or double slit behaviour. It helps getting acquainted with Feynman's sum over all possible paths aproach.

Still, I'd like to see that, I never liked QM explanations myself, way too esoteric and uncomfortably paranormal.

The math formulism of QM is not esoteric or paranormal per se, certain interpretation have some of that. And in any case you should know that most of the theoretical physicists working with QM towards a quantum gravity theory beyond the Standard model naturally consider it (together with GR) as a very good approximation to the next theory and therefore incomplete as we know it.

 

[atyy]

OK, so now that we agree on the basic approaches, I have a technical question (Tricky Dicky, let me know if this is hijacking). In Mott's paper, as described by Figari and Teta's http://arxiv.org/abs/1209.2665v1 which stevendaryl linked to above, only the time-independent Schroedinger equation is considered. Why is this permitted?

I see that Figari and Teta are co-authors on an analysis that uses the full Schroedinger equation.

http://arxiv.org/abs/0907.5503
A time-dependent perturbative analysis for a quantum particle in a cloud chamber
G. Dell'Antonio, R. Figari, A. Teta
Annales Henri Poincaré
August 2010, Volume 11, Issue 3, pp 539-564

 

[TrickyDicky]

In Mott's paper, as described by Figari and Teta's http://arxiv.org/abs/1209.2665v1 which stevendaryl linked to above, only the time-independent Schroedinger equation is considered. Why is this permitted?
I see that Figari and Teta are co-authors on an analysis that uses the full Schroedinger equation.

I find this an interesting question, maybe some of the experts might give it a try. My take is that the original paper by Mott is centered on obtaining a straitgh track in the context of a spherical wave function, and for that he just has to show that in a system with an alpha-particle and two atoms the 2 atoms can only be excited if they lie in a line, so for this kind of "geometrical" solution he doesn't need to introduce any time-dependence for that function, a stationary solution is enough to show there is no problem regarding spherical vs linear.

Actually my OP was a bit beyond the specific Mott problem, it was more related to the problem of considering classical trajectories like chamber tracks(but it could equally applied to electrons trajectories in a TV CRT or rays in any vacuum tube). In all these cases the path is considered of infinitesimal width, it is not the macroscopic width of the chamber tracks or of the beam in a CRT, as it is sometimes stated to justify that the microparticle trajectory doesn't compromise the HUP.
As commented above, in these examples one either has to renounce to referring to what is observed as a trajectory or as a microparticle, whatever is psychologically less difficult, calling it both is not QM.

 

[atyy]

Actually my OP was a bit beyond the specific Mott problem, it was more related to the problem of considering classical trajectories like chamber tracks(but it could equally applied to electrons trajectories in a TV CRT or rays in any vacuum tube). In all these cases the path is considered of infinitesimal width, it is not the macroscopic width of the chamber tracks or of the beam in a CRT, as it is sometimes stated to justify that the microparticle trajectory doesn't compromise the HUP.
As commented above, in these examples one either has to renounce to referring to what is observed as a trajectory or as a microparticle, whatever is psychologically less difficult, calling it both is not QM.

In these other cases, the flight of the particle is "free". The most common use of a particle-like derivation is to show that given an initial wave function, when the final position of the particle is measured on a screen a large distance away, that final position can be used to accurately measure the initial momentum of the particle. In fact, there is a strict quantum mechanical derivation that does not involve any assumption of a classical trajectory. The basic idea is that the initial wave from a slit is Fourier transformed (momentum is the Fourier transform of position), analogous to the Fraunhofer or far field limit in classical waves.
http://www.rodenburg.org/theory/y1200.html
http://people.ucalgary.ca/~lvov/471/labs/fraunhofer.pdf
http://www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/ungrouped%20papers/wigner%20function/PFK97.pdf

Nonetheless, a classical derivation with trajectories works. This is strictly correct, even from the quantum mechanical point of view, if the initial wave function is Gaussian. This is because the Wigner function, which is the quantum analogue of the classical joint distribution for momentum and position, is positive for Gaussian wave functions and can be interpreted as a classical probability distribution. Furthermore, the Schroedinger equation for a free Gaussian wave function leads to the classical Liouville equation for the Wigner function. So in this special case of a free Gaussian wave packet, even without a Bohmian interpretation, Copenhagen does permit classical trajectories.

I believe it is a matter of luck that the quantum formula remains the same, whether or not the initial wave packet is Gaussian. So I believe that for non-Gaussian wave packets, a strictly correct derivation does not involve trajectories. I think this luck is analogous to that in Rutherford scattering, where classical and quantum derivations give the same formula for inverse squared potentials.

 

[TrickyDicky]

The point remains that you cannot identify a wave packet with a particle in QM.

 

[atyy]

A wave packet is identified with a particle in QM.