Tracks in particle detectors and quantum paths

[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.

 

[TrickyDicky]

A wave packet is identified with a particle in QM.

Hmmm, so what was Born's discrepancy with Schrödinger about wave packets?

 

[Nugatory]

Hmmm, so what was Born's discrepancy with Schrödinger about wave packets?

The modern understanding of "particles", based on QFT, still hadn't been hashed out. They aren't around to ask, but it's likely that they would have found common ground there.

(and it's important to remember that the straightest path to a clean formulation of a theory is almost never the historical route by which the theory was first reached. It's a lot easier to chart a course when you already know where your destination is).

 

[atyy]

Hmmm, so what was Born's discrepancy with Schrödinger about wave packets?

Here's the history according to Wikipedia http://en.wikipedia.org/wiki/Schrödinger_equation#Historical_background_and_development

According to Born, the square of the wave function gives the probability of a particle's position. That's the Born rule that's made it into canonical quantum mechanics.

So a wave packet does represent a particle, except that it does not have a definite position and momentum at all times.

 

[TrickyDicky]

So a wave packet does represent a particle, except that it does not have a definite position and momentum at all times.

Ok, doesn't that mean it can't have a classical trajectory?

 

[WannabeNewton]

A wave packet is identified with a particle in QM.

Where in QM is a wave packet identified with a particle? It is well known that such an interpretation is highly limited and as such rather useless beyond visualization. Not only is such an interpretation restricted to single-particle systems, but also it only holds for those systems wherein the wave-packet does not spread on time scales comparable to the time evolution under Schrödinger's equation so it will work for the harmonic oscillator but not for the free particle.

 

[atyy]

Where in QM is a wave packet identified with a particle? It is well known that such an interpretation is highly limited and as such rather useless beyond visualization. Not only is such an interpretation restricted to single-particle systems, but also it only holds for those systems wherein the wave-packet does not spread under the Schrödinger equation so it will work for the harmonic oscillator but not for the free particle.

ψ(x₁) is identified with 1 particle.

ψ(x₁,x₂) is identified with 2 particles.

ψ(x₁,x₂,x₃) is identified with 3 particles.

 

[WannabeNewton]

ψ(x₁) is identified with a particle.

ψ(x₁,x₂) is identified with two particles.

ψ(x₁,x₂,x₃) is identified with 3 particles.

A wave-packet is a Gaussian wave-form propagating through configuration space. The wave-function is a much more general concept and the wave-function of a multi-particle system certainly cannot be identified with a configuration space wave-form since the wave-function of such a system lives in a higher dimensional space.

This is exactly why TrickyDicky referred to the history behind Born's interpretation of the wave-function in light of Schrödinger's incorrect interpretation of the wave-function as a wave-packet representing a particle in configuration space.

 

[atyy]

A wave-packet is a Gaussian wave-form propagating through configuration space. The wave-function is a much more general concept and the wave-function of a multi-particle system certainly cannot be identified with a configuration space wave-form since the wave-function of such a system lives in a higher dimensional space.

OK, if that's what you mean by wave packet, I agree (I thought "wave packet" was just another way to say " wave function"). You can see my comments on Gaussian wave functions several posts up. There, in the free particle case, we can even associate classical trajectories with the Gaussian wave function.

 

[atyy]

This is exactly why TrickyDicky referred to the history behind Born's interpretation of the wave-function in light of Schrödinger's incorrect interpretation of the wave-function as a wave-packet representing a particle in configuration space.

But his post #99 replied to my post #98, where his use of the term "wave packet" would make more sense if it referred to what I called the "wave function". It is clear in my post #98 that "wave function" and "wave packet" are meant to be the same thing.

 

[WannabeNewton]

There, in the free particle case, we can even associate classical trajectories with the Gaussian wave function.

Thank you. Do you have any further reading on that?

 

[atyy]

Thank you. Do you have any further reading on that?

As far as I know, it only works for single particle free Gaussian wave functions. The idea is that the Wigner function is the quantum analogue of the classical joint probability of position and momentum. But it is not the same because in general, the Wigner function has negative bits, whereas a classical probability distribution is positive. Also, the quantum time evolution is derived from Schroedinger's equation, whereas we need the Liouville equation for the classical case. In the special case of a Gaussian wave function, the Wigner function is positive. The free particle evolution preserves Gaussianity, and surprisingly (to me) also results in the quantum evolution being the same as the classical Liouville equation. So in this special case, we can have trajectories even in Copenhagen, without a Bohmian interpretation. There's an explanation somewhere in Ganguly's essay http://dspace.mit.edu/bitstream/handle/1721.1/49800/50586846.pdf .

However, the derivation of momentum measurements from position measurements of single particles at far field using a single slit set up hold more generally (ie. even if the initial wave function is not Gaussian). Essentially, this is because the far field wave function is the Fourier transform of the initial wave function, analogous to the Fraunhofer limit for classical waves. http://www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/ungrouped%20papers/wigner%20function/PFK97.pdf

 

[TrickyDicky]

But his post #99 replied to my post #98, where his use of the term "wave packet" would make more sense if it referred to what I called the "wave function". It is clear in my post #98 that "wave function" and "wave packet" are meant to be the same thing.

It wasn't so clear to me, so I referred to a wave packet explicitly.

Still I'm not able to conclude from your explanations or your references that single particle free Gaussian wave functions have classical trajectories.
"Free particles" are known not to exist in the quantum world in any case, they are just practical idealizations.

 

[atyy]

It wasn't so clear to me, so I referred to a wave packet explicitly.

Still I'm not able to conclude from your explanations or your references that single particle free Gaussian wave functions have classical trajectories.
"Free particles" are known not to exist in the quantum world in any case, they are just practical idealizations.

OK, so everywhere that I say "wave packet", I mean "wave function" (I didn't know there was a difference till now).

Free particles don't exist, so this is just an approximation. However, as long as we are just doing quantum mechanics with a fixed number of particles, these two cases in which classical trajectories seem to have some meaning are treated differently. In the Mott cloud chamber case we have decoherence throughout or multiple measurements, whereas in the case of momentum measurement from the flight of a free particle, we only have decoherence at the end of the path, or a single measurement of position at the far field location. That these are approximations ultimately mean that neither position nor momentum are perfectly accurately measured (in fact, in quantum field theory, there isn't a relativistic position operator), but they are good enough.

 

[atyy]

BTW, there is a very interesting discussion about "free particles" in this thread https://www.physicsforums.com/showthread.php?t=749257 .

 

[mfb]

Ok, doesn't that mean it can't have a classical trajectory?

Right, but the deviations from a classical trajectory can be negligible. The result is a trajectory that looks classical.

 

[TrickyDicky]

Free particles don't exist, so this is just an approximation. However, as long as we are just doing quantum mechanics with a fixed number of particles, these two cases in which classical trajectories seem to have some meaning are treated differently. In the Mott cloud chamber case we have decoherence throughout or multiple measurements, whereas in the case of momentum measurement from the flight of a free particle, we only have decoherence at the end of the path, or a single measurement of position at the far field location. That these are approximations ultimately mean that neither position nor momentum are perfectly accurately measured (in fact, in quantum field theory, there isn't a relativistic position operator), but they are good enough.

Right, but the deviations from a classical trajectory can be negligible. The result is a trajectory that looks classical.

We had all agreed that an approximation to a classical trajectory is possible and it is good enough in practice, still that approximation is not a quantum microparticle's classical trajectory(and if it were it wouldn't be the trajectory of a quantum microparticle) in the rigorous sense I referred to in #97 last sentence.

I think it is important to remember here that even in classical mechanics a classical trajectory is based on the idealization of extended bodies as point-like particles, in that sense it is not possible to exactly measure classical trajectories in practice either, they are just possible in the theory. In QM in the case of elementary particles they are actually considered point particles, so no true classical trajectory is possible not only in practical measurable terms but also in principle theoretically due to the uncertainty relations, that's why even in theory only approximations to a classical trajectory are allowed in QM.
The Wigner quasiprobabilistic distribution is no different, in fact it is an approximation to a classical probabilistic distribution and to use wikipedia words only "vestiges of local trajectories are normally barely discernible in the evolution of the Wigner distribution function".

 

[mfb]

What is a "quantum microparticle" - or what is not one?
You can neglect QM in the same way you can neglect the influence of gravity on particles in the bubble chamber - it is there, but you just don't (have to) care.

Nonrelativistic QM itself is just an approximation of QFT, and that might be an approximation of some more fundamental theory. So what? Does that change our view on the bubble chamber in any way?

 

[WannabeNewton]

We had all agreed that an approximation to a classical trajectory is possible and it is good enough in practice, still that approximation is not a quantum microparticle's classical trajectory(and if it were it wouldn't be the trajectory of a quantum microparticle) in the rigorous sense I referred to in #97 last sentence.

Here's a question for you: on what length scales and time scales is the classical trajectory of the particle in the bubble chamber being realized?

 

[WannabeNewton]

There's an explanation somewhere in Ganguly's essay http://dspace.mit.edu/bitstream/handle/1721.1/49800/50586846.pdf .

Brilliant, thank you. Chapter 4 was quite lucid.

 

[TrickyDicky]

Here's a question for you: on what length scales and time scales is the classical trajectory of the particle in the bubble chamber being realized?

It was already discussed previously that length scales of the tracks (or the beams in a CRT) are big enough so that no problem with the HUP ever arises, the particles have room to be sufficiently blurred to avoid it.

So it is impossible both theoretically(taking Planck scale as a theoretic limit) or in practice(much bigger limit with present technology) to realize or probe a true classical trajectory.