Tracks in particle detectors and quantum paths

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The discussion centers on the relationship between the tracks left by electrons in cloud chambers and their undefined quantum trajectories. Participants highlight the Mott problem, which addresses why these tracks appear linear despite the quantum nature of particles. There is debate over the applicability of classical physics to electron trajectories in particle detectors, with some arguing that classical models are sufficient in many contexts. The conversation also touches on the measurement problem and wave function collapse, emphasizing the complexity of reconciling classical observations with quantum mechanics. Ultimately, the discussion underscores the ongoing challenges in understanding the behavior of particles at quantum scales.
  • #31
StrangeCoin said:
I am referring to instrumentation uncertainty being misinterpreted as an actual property of what is being measured. I believe that's exactly what the OP question is about. You don't say electron position is undefined because you measured that to be true, but because you couldn't measure any better. Isn't that right?

No, not really. You seem to be suggesting that quantum uncertainty only reflects our ignorance of the true position of particles. That was certainly what Einstein thought. But subsequent experiments and analysis show that that claim is doubtful. That's a hidden-variables theory, which isn't COMPLETELY ruled out, but there are reasons to be skeptical. It seems that there is no such hidden-variables theory that doesn't involve strange instantaneous interactions between distant particles.

So now on one hand we have the thing we can actually measure, the bubble chamber trajectories, which are apparently continuous and very well defined. And on the other hand we have QM theory which wants us to believe the opposite, that electron is actually doing something else, but it sneakily only does so when we are not looking. Apart from being funny, is there any actual reason to believe this theory based on unsuccessful measurements, rather than to believe what is obvious from those measurements that were successful?

Yes, because even though quantum mechanics seems weird, it CORRECTLY describes every experiment ever performed, while the common-sensical belief that holds that particles really do have precise positions, we just don't know what those are unless we measure them, has not been successful.

If it were really the case that bubble-chamber results contradicted quantum mechanics, that would be big news, and we would throw out quantum mechanics. But they don't contradict quantum mechanics (even though it is a little work to see why not).
 
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  • #32
Nugatory said:
What does the uncertainty principle have to do with the observation that electrons do not "crash into the nucleus"?
I'm just referring to the usual argument that HUP is incompatible with well defined orbits to begin with so it wouldn't allow even the starting point for considering that an "orbiting" electron could crash into the nucleus since it doesn't follow a defined path, just has a probability cloud for its position in the atom, which might even coincide with the nucleus position with some not vanishing probability.(See PF FAQ on this https://www.physicsforums.com/showthread.php?t=511179 )

And how does applying "this same logic" about including the environment in the Hamiltonian of a bound electron lead to the conclusion that the electron should follow a classical trajectory?
If one includes the electron's environment(photon cloud, nucleus, polarized vacuum, other quantum particles...) in the analysis or uses these elements surrounding the electron in the atom as classical measurement devices and performs the same operations Mott does for the setting of the quantum particle in the cloud chamber, one might naively expect to also compute a classical trajectory for the electron in a stable ground state atom. It would be interesting to list the main differences in the scenarios that prevent this.
 
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  • #33
Nugatory said:
What does the uncertainty principle have to do with the observation that electrons do not "crash into the nucleus"?

I consider that one of the primary issues that led to the development of quantum mechanics: A classical model of a positive electron orbiting a negatively charged nucleus is unstable; the electron would radiate and fall into the nucleus. The uncertainty principle implies a minimum value for the electron's energy. So there is a connection between the two. I assume that's what TrickyDicky meant.
 
  • #34
TrickyDicky said:
Yes, that's the premise, Mott et al. are trying to explain, how do we go from individual imprecise measurements that should lead to not definite trajectories to a classical trajectory. The arguments used by these authors: the classical detectors and the environment. If they are valid it seems like they could be also applied in principle to situations where we don't observe clasical trajectories, like electrons in an atom...

I don't think there is any necessary conflict between the two. The environment of an electron within an atom is very different from the environment of a high-energy particle in a bubble or cloud chamber. The way I understand it, the appearance of definite trajectories in a bubble/cloud chamber is due to collisions of the particle in question with much more massive atoms.
 
  • #35
TrickyDicky said:
I'm just referring to the usual argument that HUP is incompatible with well defined orbits to begin with so it wouldn't allow even the starting point for considering that an "orbiting" electron could crash into the nucleus since it doesn't follow a defined path, just has a probability cloud for its position in the atom, which might even coincide with the nucleus position with some not vanishing probability.(See PF FAQ on this https://www.physicsforums.com/showthread.php?t=511179 )
ZapperZ edited that FAQ, so maybe you should listen to him not me... But it seems pretty clear to me that you're misunderstanding it. The uncertainty principle is (literally) just a footnote and the essential part of the explanation is the QM prediction of stable states as the solution to Schodinger's equation.

If one includes the electron's environment(photon cloud, nucleus, polarized vacuum, other quantum particles...) in the analysis or uses these elements surrounding the electron in the atom as classical measurement devices and performs the same operations Mott does for the setting of the quantum particle in the cloud chamber, one might naively expect to also compute a classical trajectory for the electron in a stable ground state atom. It would be interesting to list the main differences in the scenarios that prevent this.

You would have to be very naive indeed to expect that. The main difference is that in the case of a bound electron, the contribution of these environmental factors to the Hamiltonian is small compared with the central force. In the cloud chamber case the interaction of the particle with the environment isn't negligible - it's the only interaction present for the unbound particle.
 
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  • #36
stevendaryl said:
If it were really the case that bubble-chamber results contradicted quantum mechanics, that would be big news, and we would throw out quantum mechanics.

This way of putting it I think it dramatizes it unnecessarily. Certainly it is not the purpose of this thread to throw out anything, on the contrary QM as a theory is robust enough to tackle all these only apparent contradictions. It makes little sense that particle detectors and scattering in general contradicts in any way QM, particle physicists would be quite amused.

It is in the nature of the theory not to cling to a unique graphical representation like it was the case in classical mechanics, and to allow different sometimes apparently contradictory approximate and relative representations that together with the mathematical formalism conform QM.

The environment of an electron within an atom is very different from the environment of a high-energy particle in a bubble or cloud chamber. The way I understand it, the appearance of definite trajectories in a bubble/cloud chamber is due to collisions of the particle in question with much more massive atoms.
Yes, it is different in many aspects, I'm not sure that the mass of the environment microparticles is the main one though.
 
  • #37
Nugatory said:
ZapperZ edited that FAQ, so maybe you should listen to him not me... But it seems pretty clear to me that you're misunderstanding it. The uncertainty principle is (literally) just a footnote and the essential part of the explanation is the QM prediction of stable states as the solution to Schrodinger's equation.

It might be a matter of semantics, but I don't consider the Schrodinger equation to be an alternative to reasoning using the uncertainty principle. It's a mathematically precise way of doing that reasoning. The Schrodinger equation in some sense already incorporates the uncertainty principle, since the primary mathematical object, the wave function, is interpreted probabilistically.
 
  • #38
TrickyDicky said:
This way of putting it I think it dramatizes it unnecessarily. Certainly it is not the purpose of this thread to throw out anything, on the contrary QM as a theory is robust enough to tackle all these only apparent contradictions. It makes little sense that particle detectors and scattering in general contradicts in any way QM, particle physicists would be quite amused.

I'm agreeing with you--bubble chamber physics certainly doesn't contradict QM.

Yes, it is different in many aspects, I'm not sure that the mass of the environment microparticles is the main one though.

I'm not sure. I was thinking that the particle collides with much more massive (and therefore, more localized) atoms, and that was responsible for the appearance of tracks, but the masses of the atoms might not be important.
 
  • #39
Instead of electrons or alpha particles, you can shoot a grain of sand through the detector, physics will be the same: quantum effects are negligible. Even between bubbles, the particle interacts so much with the environment that quantum effects vanish.
 
  • #40
mfb said:
Instead of electrons or alpha particles, you can shoot a grain of sand through the detector, physics will be the same: quantum effects are negligible. Even between bubbles, the particle interacts so much with the environment that quantum effects vanish.

As I said, the original puzzle about bubble chambers was why a spherically symmetrical situation (the emission of alpha particles by a nucleus is approximately modeled as spherically symmetric) should lead to linear tracks, which are certainly not spherically symmetric.

If the alpha particle were initially given an approximately definite trajectory (a wave packet centered on a classical trajectory), it wouldn't be too surprising to find linear tracks. Of course, the uncertainty principle limits how classical its trajectory can be, but in the case of sand, its path can be pretty close to classical.
 
  • #41
I just don't see what is puzzling about it. You measure the approximate position and momentum of the particle with every single atom that is close to it. Of course repeated measurements give a consistent result.
 
  • #42
Nugatory said:
You would have to be very naive indeed to expect that. The main difference is that in the case of a bound electron, the contribution of these environmental factors to the Hamiltonian is small compared with the central force. In the cloud chamber case the interaction of the particle with the environment isn't negligible - it's the only interaction present for the unbound particle.
I really doubt this is a problem of free(unbound) vs bound particles. First of all a quantum particle is not exactly free in the sense a classical particle is, no matter how small the environment factors they can't be neglected in the same way in the quantum system case as in the classical. And in the case of the cloud/bubble chamber one could set up a strong central potential that would make the environment contribution small in comparison and still have a classical path result.
 
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  • #43
mfb said:
I just don't see what is puzzling about it. You measure the approximate position and momentum of the particle with every single atom that is close to it. Of course repeated measurements give a consistent result.
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's why it is not so straightforward to give the answer you give,(wich is by the way the first thing that comes to mind or at least it was for me). One of the reasons it is not so straight forward I supposed is related to the fact that the measurement problem is still considered an usolved one in QM.
 
  • #44
TrickyDicky said:
I really doubt this is a problem of free(unbound) vs bound particles. First of all a quantum particle is not exactly free in the sense a classical particle is, no matter how small the environment factors they can't be neglected in the same way in the quantum system case as in the classical. And in the case of the cloud/bubble chamber one could set up a strong central potential that would make the environment contribution small in comparison and still have a classical path result.

I'm not clear about whether people are using the word "environment" with respect to a bubble/cloud chamber to mean the stuff filling the chamber (water vapor, or whatever it is), or the rest of the universe that is outside of the chamber. If the interaction with the stuff inside the chamber were negligible, then you wouldn't see tracks, I don't think. That's the whole point of filling the chamber with stuff.
 
  • #45
mfb said:
I just don't see what is puzzling about it. You measure the approximate position and momentum of the particle with every single atom that is close to it. Of course repeated measurements give a consistent result.

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?

It's hard to know exactly how puzzled to be about something until after you thoroughly understand it (at which point, you're no longer puzzled, I guess).
 
  • #46
TrickyDicky said:
The track describes the classical trajectory of a quantum system, this is what Mott, Heisenberg, Born and Darwin as explained in the rerence provided by stevendaryl were trying to explain within the quantum formalism. A classical trajectory has well defined momentum and position, that is what's classical about it and what seems to clash with quantum theory in the context of the yet unsolved measurement problem.

Heruristically, the atoms of the cloud chamber are the "pointer" of the instrument. These give a well-defined reading. But this reading is not necessarily correct or "accurate". For example, let's take a spin pointing up. A measurement will always yield a definite answer - either "up" or "down". But since the spin is in a definite up state, these well-defined readings are inaccurate if the measurement yields either answer with some non-zero probability. In contrast, an accurate measurement will always tell you the particle is in an up state.

In the same way, the definite trajectory you assign is only the definite reading of the instrument. It does not tell you that the measurement was accurate. This distinction is important, because definite but inaccurate measurements do not leave the quantum system in the definite state indicated by the measurement outcome.

TrickyDicky said:
Yes, that's the premise, Mott et al. are trying to explain, how do we go from individual imprecise measurements that should lead to not definite trajectories to a classical trajectory. The arguments used by these authors: the classical detectors and the environment. If they are valid it seems like they could be also applied in principle to situations where we don't observe clasical trajectories, like electrons in an atom, but here it is not used basically on the grounds that the HUP forbids well defined orbits(see for instance PF physics FAQ footnote on why electrons don't crash into the nucleus).

In quantum mechanics, the answer you get depends on what you measure. Heuristically, if you were to measure the position of the electrons in an atom, you would localize the electron. (I say heuristically, because I don't know if this can be done non-relativistically, and relativistically, position is not a good measurement operator).

TrickyDicky said:
Too heuristic, as I said in as much as it is considered a classical trajectory the track can be considered to have position as precisely measured as momentum.

The mathematical details behind the idea of continuous imprecise measurement are given in http://arxiv.org/abs/math-ph/0512069. This article also mentions the cloud chamber.

A general introduction to continuous imprecise measurement is given in http://arxiv.org/abs/quant-ph/0611067.

Note on my sloppy terminology: Here I have used "precise" and "accurate" to mean the same thing, ie. "correct". Others would prefer to reserve "precise" to mean "well-defined", and "accurate" to mean "correct".
 
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  • #47
mfb said:
Instead of electrons or alpha particles, you can shoot a grain of sand through the detector, physics will be the same: quantum effects are negligible.
We know quantum effects are negligible for macroscopic objects(sand grains, pick-up trucks,... but not for say buckyballs). But they shouldn't be for electrons or alpha particles, or else we wouldn't need quantum theory to explain them. Besides the base of the solutions given to the Mott problem relies on the quantum effects of the environment rather than they're being negligible.
Even between bubbles, the particle interacts so much with the environment that quantum effects vanish.
Again, I think all quantum systems interact significatively with the environment.
 
  • #48
TrickyDicky, what you wrote is utterly, completely and totally false. I write this in such a strong manner because you will be tempted to hold on to as much of that as you can, and so long as you do, you will never, ever understand the right answer. You have to let go.

Quantum mechanics applies to everything: sand grains, baseballs and pickup trucks. In the limit of large quantum number n, the behavior approaches the classical limit. But what matters is n, not whether this is a baseball or an electron. An electron in a bubble chamber is at very large n - millions or more - so its behavior is very close to classical.

What matters is n.
 
  • #49
Vanadium 50 said:
TrickyDicky, what you wrote is utterly, completely and totally false. I write this in such a strong manner because you will be tempted to hold on to as much of that as you can, and so long as you do, you will never, ever understand the right answer. You have to let go.

Quantum mechanics applies to everything: sand grains, baseballs and pickup trucks. In the limit of large quantum number n, the behavior approaches the classical limit. But what matters is n, not whether this is a baseball or an electron. An electron in a bubble chamber is at very large n - millions or more - so its behavior is very close to classical.

What matters is n.

I don't quite understand that. The quantum number n isn't well-defined for an unbound particle. Or maybe you want to say that n is infinite in that case? What is n for a baseball?
 
  • #50
Vanadium 50 said:
TrickyDicky, what you wrote is utterly, completely and totally false. I write this in such a strong manner because you will be tempted to hold on to as much of that as you can, and so long as you do, you will never, ever understand the right answer. You have to let go.

Quantum mechanics applies to everything: sand grains, baseballs and pickup trucks. In the limit of large quantum number n, the behavior approaches the classical limit. But what matters is n, not whether this is a baseball or an electron. An electron in a bubble chamber is at very large n - millions or more - so its behavior is very close to classical.

What matters is n.
Hmmm, I'm ready to let go of anything but I have not made any claim in the sense that QM doesn't apply to everything, in the post you seem to refer to as being utterly wrong I was quoting mfb to disagree with him that quantum effects are negligible, maybe that confused you.

On the other hand my layman understanding coincides with stevendaryl's about free particles not having the defined quantum numbers of particles in an atom, is this not correct?
 
  • #51
stevendaryl said:
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
 
  • #52
Vanadium 50 said:
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?
 
  • #53
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?
 
  • #54
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  • #55
Vanadium 50 said:
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.
 
  • #56
stevendaryl said:
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.
 
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  • #57
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.
 
  • #58
Haelfix said:
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.
 
  • #59
Haelfix said:
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?
 
  • #60
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.
 

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