Does the Uncertainty Principle restrict a particle's path?

[mike1000]

In the classical definition, velocity is the time derivative of the particles position curve (trajectory). The Uncertainty principle restricts the particles velocity from ever being zero. Doesn't this imply that a particles path is restricted to paths that do not have a zero time derivative at any point?. (ie implying that the path cannot reverse directions among other things). If this is true, does it place restrictions on the Path Integral Formulation, where all possible paths are assumed to be possible, but not likely? Or am I a victim of classical thinking once again?

 

[BvU]

I think you are mixing up concepts from the classical world with concepts from the quantum world, so : yes!. There certainly is no such restriction from the uncertainty principle (it doesn't even deal with $x$ and $p$, only with the product of $\Delta x$ and $\Delta p$ ).

 

[bhobba]

The Uncertainty principle restricts the particles velocity from ever being zero

Why do you think that?

Before posting that do a search on the uncertainty principle og this forum eg:
https://www.physicsforums.com/threa...out-the-uncertainty-principle-correct.825435/

In particular look at the correct statement of the principle:
The correct statement of the uncertainty principle is the following. Suppose you have a large number of similarly prepared systems ie all are in the same quantum state. Divide them into two equal lots. In the first lot measure position to a high degree of accuracy. QM places no limit on that accuracy - its a misunderstanding of the uncertainty principle thinking it does. The result you get will have a statistical spread. In the second lot measure momentum to a high degree of accuracy - again QM places no limit on that. It will also have a statistical spread. The variances of those spreads will be as per the Heisenberg Uncertainty principle.

The measurement can be zero - but please thing about why you believe it cant. It can't remain zero - but when you measure it it can be zero. The reason it can't remain zero is that would men position and momentum is known exactly at the same time.

Thanks
Bill

 

[BvU]

Perhaps studying a concrete example may be enlightening. Google 'expectation value harmonic oscillator' and pick one to your liking. Ground state has $<x>=0$ and $<p> = 0$ but still satisfies the HUP.

 

[mike1000]

Why do you think that?

Before posting that do a search on the uncertainty principle og this forum eg:
https://www.physicsforums.com/threa...out-the-uncertainty-principle-correct.825435/

In particular look at the correct statement of the principle:
The correct statement of the uncertainty principle is the following. Suppose you have a large number of similarly prepared systems ie all are in the same quantum state. Divide them into two equal lots. In the first lot measure position to a high degree of accuracy. QM places no limit on that accuracy - its a misunderstanding of the uncertainty principle thinking it does. The result you get will have a statistical spread. In the second lot measure momentum to a high degree of accuracy - again QM places no limit on that. It will also have a statistical spread. The variances of those spreads will be as per the Heisenberg Uncertainty principle.

The measurement can be zero - but please thing about why you believe it cant. It can't remain zero - but when you measure it it can be zero. The reason it can't remain zero is that would men position and momentum is known exactly at the same time.

Thanks
Bill

I do not know where I got the idea, but somehow I got the idea that in the quantum domain, momentum cannot be zero. I may have gotten that idea from the deBroglie equation, pλ = h. Or it may have come from my recent readings that indicated that the "quantum field" is always in motion.

If you are saying that momentum of an elementary particle can be zero then I am wrong.

 

[ZapperZ]

I do not know where I got the idea, but somehow I got the idea that in the quantum domain, momentum cannot be zero. I may have gotten that idea from the deBroglie equation, pλ = h. Or it may have come from my recent readings that indicated that the "quantum field" is always in motion.

If you are saying that momentum of an elementary particle can be zero then I am wrong.

I think you have the wrong understanding of the HUP. It has nothing to do with the absolute value of anything. It has everything to do with how much the value of an observable can be predicted, or how much repeated measurement of the value will spread out. I can have a non-zero value of momentum, and get a very tight, close-to-zero spread around some central value.

I suggest you look again at the physics behind the HUP.

Zz.

 

[mike1000]

I think you have the wrong understanding of the HUP. It has nothing to do with the absolute value of anything. It has everything to do with how much the value of an observable can be predicted, or how much repeated measurement of the value will spread out. I can have a non-zero value of momentum, and get a very tight, close-to-zero spread around some central value.

I suggest you look again at the physics behind the HUP.

Zz.

I just said that I did not get the idea from the Uncertainty Principle. I said I got it from DeBroglie's equation, pλ = h.

 

[ZapperZ]

I just said that I did not get the idea from the Uncertainty Principle. I said I got it from DeBroglie's equation, pλ = h.

That is even more puzzling since p=0 is a trivial solution of nothing happening.

No, I was referring to not just that post, but your topic in this thread.

Zz.

 

[mike1000]

That is even more puzzling since p=0 is a trivial solution of nothing happening.

No, I was referring to not just that post, but your topic in this thread.

Zz.

According to deBroglies equation p cannot equal zero.

(Yes I know you were referring to my original post. I incorrectly used the Uncertainty Principle in that post. However, having said that, I would not be surprised at all to find out that the Uncertainty Principle can be derived or implied, in some way, from DeBroglies equation.

 

[ZapperZ]

According to deBroglies equation p cannot equal zero.

(Yes I know you were referring to my original post. I incorrectly used the Uncertainty Principle in that post. However, having said that, I would not be surprised at all to find out that the Uncertainty Principle can be derived or implied, in some way, from DeBroglies equation.

This is going off-topic, but why can't it be zero? You are thinking of p being the only variable. This is not true. p and λ are linked. p is zero when λ→∞.

Zz.

 

[mike1000]

This is going off-topic, but why can't it be zero? You are thinking of p being the only variable. This is not true. p and λ are linked. p is zero when λ→∞.

Zz.

Yes, and, λ→∞ is a very special case and what does it mean?

 

[ZapperZ]

Yes, and, λ→∞ is a very special case and what does it mean?

It means nothing is moving, and nothing is happening, just like what I said earlier. What does this have anything to do with your claim that p can't be zero?

Zz.

 

[mike1000]

It means nothing is moving, and nothing is happening, just like what I said earlier. What does this have anything to do with your claim that p can't be zero?

Zz.

My claim is that there can be no trajectories in which, at some point in that trajectory, the first derivative with respect to time can be zero, implying that there can be no trajectories in which the particle reverses direction, among other things.

The only way that could occur, by your own analysis, is for the "wavelength" of the particle to approach infinity.

 

[ZapperZ]

My claim is that there can be no trajectories in which, at some point in that trajectory, the first derivative with respect to time can be zero, implying that there can be no trajectories in which the particle reverses direction, among other things.

Based on what? I've just shown you that the deBroglie equation which you have been citing CAN, in fact, yield a p=0 solution.

Secondly, "first derivative" of WHAT? What quantity are you taking the first derivative of?

Thirdly, you continue to misconstrue what the HUP is saying. It does NOT prohibit particles with zero velocities or momentum. Look for example, at the ARPES data that measure the momentum at the center of the band, the so-called Γ-point. What do you think is the momentum of the photoelectrons measured there?

http://www.bessy.de/rglab/pictures/highlights/quasifreestanding02.jpg

The HUP does NOT prohibit the zero absolute value of an observable. You are confusing this with the SPREAD in the value of the observable. It is not p, but Δp in the HUP equation.

Zz.

 

[Ken G]

Perhaps a simple way to answer it is to say that it is impossible to have knowledge that a particle has a state of zero momentum, say at some particular possible turnaround point. But that doesn't mean the particle cannot reverse direction, it just means you cannot know the exact place or time when it reversed direction. It's not even obvious that any such place or time needs to actually exist, all that is necessary is that the particle started out going one way and ended up going the opposite way.

Hence, I think the biggest jump in going from classical to quantum thinking is that we may need to relax the idea that motion is comprised of a series of definite events, but rather it is a behavior that emerges out of a combination of indefinite states (as per the path integral formulation mentioned). Motion may be inherently fuzzy, and even though we can have high-precision constraints on its initial and final states, the tiny uncertainties that will persist in those measurements will mean the exact nature of the motion may not only be unknowable, it may not actually exist at all.

I think Zeno's paradoxes of motion are informative here. Zeno felt that describing motion as a series of definite events led to paradoxes, so could not be a correct description of it. His brilliant insights there are often discounted because he reached what appears to be the wrong conclusion from his reasoning: he concluded that motion is merely an illusion, so cannot actually happen. It did not occur to him we are allowed to simply regard it as something inherently fuzzy, instead. (Though I do mention that Bohmians regard motion as precise and definite, but the fuzziness is enforced by a kind of quasi-magical pilot wave, which is a potentially allowable interpretation of the HUP but has its own sticky issues.)

 

[mike1000]

Based on what? I've just shown you that the deBroglie equation which you have been citing CAN, in fact, yield a p=0 solution.

Secondly, "first derivative" of WHAT? What quantity are you taking the first derivative of?

Thirdly, you continue to misconstrue what the HUP is saying. It does NOT prohibit particles with zero velocities or momentum. Look for example, at the ARPES data that measure the momentum at the center of the band, the so-called Γ-point. What do you think is the momentum of the photoelectrons measured there?

http://www.bessy.de/rglab/pictures/highlights/quasifreestanding02.jpg

The HUP does NOT prohibit the zero absolute value of an observable. You are confusing this with the SPREAD in the value of the observable. It is not p, but Δp in the HUP equation.

Zz.

I am not talking about the HUP. I mis-spoke in the first post. It is too late for me to go back and edit it now. I am talking about DeBroglie's equation. However, I would not be a bit surprised to find out that the HUP can be derived in some way from DeBroglies's equation. DeBroglies equation specifies that the particles momentum can only be zero in a very special case and that special case is when λ=∞. And there is no experimental proof that λ can ever be ∞.

Taking the first derivative of the particles position vs time curve (the particles trajectory), at some point, any point.

 

[ZapperZ]

I am not talking about the HUP. I mis-spoke in the first post. It is too late for me to go back and edit it now. I am talking about DeBroglie's equation. However, I would not be a bit surprised to find out that the HUP can be derived in some way from DeBroglies's equation.

Taking the first derivative of the particles position vs time curve (the particles trajectory), at some point, any point.

It doesn't matter. I've just shown you experimental results of a measurement that produces zero momentum.

Zz.

 

[mike1000]

It doesn't matter. I've just shown you experimental results of a measurement that produces zero momentum.

Zz.

I am not talking about measurements. I am talking about trajectories. Again, I am not talking about the HUP. I am talking about DeBroglie's equation. Doesn't the momentum operator incorporate DeBroglies equation?

You are avoiding the important part of this post, in my view.

 

[ZapperZ]

I am not talking about measurements. I am talking about trajectories. Again, I am not talking about the HUP. I am talking about DeBroglie's equation. Doesn't the momentum operator incorporate DeBroglies equation?

You are avoiding the important part of this post, in my view.

This is odd. You are arguing that a particle can't have p=0. Yet, I've shown experimental evidence that it can. Isn't this a "no momentum trajectory"?

Secondly, the crutch you've been using all along is deBroglie equation. I've shown you how it can have a p=0 solution. Do you still dispute this?

Thirdly, if p=0 solution is valid, then your whole premise here is wrong, i.e. I've destroyed your crutch. So what physics are you now using as a support?

Or have I avoided the "important part" of your post once more?

Zz.

 

[mike1000]

This is odd. You are arguing that a particle can't have p=0. Yet, I've shown experimental evidence that it can. Isn't this a "no momentum trajectory"?

Secondly, the crutch you've been using all along is deBroglie equation. I've shown you how it can have a p=0 solution. Do you still dispute this?

Thirdly, if p=0 solution is valid, then your whole premise here is wrong, i.e. I've destroyed your crutch. So what physics are you now using as a support?

Or have I avoided the "important part" of your post once more?

Zz.

Yes, you have avoided the important part once again.

You have not destroyed a thing. You are trying to debate me. I am not debating you. I am pointing out something, that I think needs to be addressed. You have not addressed it.

DeBroglies equation is clear, and you agree, the condition for p = 0 is λ=∞. Classically, the velocity of a particle is the first derivative of its position curve(trajectory) through space.

 

[ZapperZ]

Yes, you have avoided the important part once again.

You have not destroyed a thing. You are trying to debate me. I am not debating you. I am pointing out something, that I think needs to be addressed. You have not addressed it.

DeBroglies equation is clear, and you agree, the condition for p = 0 is λ=∞. Classically, the velocity of a particle is the first derivative of its position curve(trajectory) through space.

I don't get it. You are trying to apply classical trajectory to quantum mechanical situation?

I think you need to address THIS first on why you think this is a valid picture.

And I still do not understand why you are still clinging to the deBroglie equation. It is as if the rest of quantum mechanics doesn't even exist.

Zz.

 

[mike1000]

I don't get it. You are trying to apply classical trajectory to quantum mechanical situation?

I think you need to address THIS first on why you think this is a valid picture.

And I still do not understand why you are still clinging to the deBroglie equation. It is as if the rest of quantum mechanics doesn't even exist.

Zz.

I have stated several times what the claim is.

If momentum cannot be zero then that implies that the velocity of the particle cannot be zero. That implies there can be no point in the particles path in which the slope is zero. This implies that the particle cannot reverse direction, among other things.

Now, you have shown, that p can equal zero for the very special case when λ=∞. That may weaken the claim a tiny bit, but I do not think that it proves the claim to be false. There is no experimental proof that λ can equal ∞ for starters.

 

[ZapperZ]

I have stated several times what the claim is.

If momentum cannot be zero then that implies that the velocity of the particle cannot be zero. That implies there can be no point in the particles path in which the slope is zero. This implies that the particle cannot reverse direction, among other things.

Now, you have shown, that p can equal zero for the very special case when λ=∞. That may weaken the claim a tiny bit, but I do not think that it proves the claim to be false. There is no experimental proof that λ can equal ∞ for starters.

Oh, so now we go back to "experimental proof", are we? What happened to the experimental proof that *I* provided earlier. Or do you simply not know how "k" and "p" are related?

So what is deBroglie wavelength of a stationary particle, and what is this wavelength for a soccer ball sitting on the ground?

Zz.

 

[mike1000]

Oh, so now we go back to "experimental proof", are we? What happened to the experimental proof that *I* provided earlier. Or do you simply not know how "k" and "p" are related?

So what is deBroglie wavelength of a stationary particle, and what is this wavelength for a soccer ball sitting on the ground?

Zz.

There you finally said it. The particle can stop in its trajectory which would be a requirement to change direction.

This says to me that the "wavelength" of the particle is continually changing as it traverses its path. Is that a correct conclusion?

 

[ZapperZ]

There you finally said it. The particle can stop in its trajectory which would be a requirement to change direction.

This says to me that the "wavelength" of the particle is continually changing as it traverses its path. Is that a correct conclusion?

No. The photoelectrons measured in the ARPES experiment had p=0. These are electrons at the center of the band, the Γ-point. It did not have to change its "trajectory". It didn't have to 'stop'.

Zz.

 

[mike1000]

No. The photoelectrons measured in the ARPES experiment had p=0. These are electrons at the center of the band, the Γ-point. It did not have to change its "trajectory". It didn't have to 'stop'.

Zz.

How do you know that?

 

[ZapperZ]

How do you know that?

From the physics of band structure of solids.

Zz.

 

[mike1000]

From the physics of band structure of solids.

Zz.

You need to be able to explain it. I am pretty sure that if you started to explain it, we would find that your are not quite right about that. I am positive that a particle can stop and reverse directions at this point,

 

[ZapperZ]

You need to be able to explain it. I am pretty sure that if you started to explain it, we would find that your are not quite right about that. I am positive that a particle can stop and reverse directions at this point,

Most people who have studied QM would not have required an explanation, because you may not believe this, but this is NOT something new. We teach students the free-electron model of conduction electrons in solids, and this is often covered in the 1st 2 chapters in a standard solid state textbook!

But since you appear to not be aware of it, look at a lesson on the band structure of solids:

http://www.imprs-cs.mpg.de/pdf/Wolf_IMPRS_2013.pdf

Check out the dispersion relation shown in slide 17 for typical periodic potential. What do you think is going on at the center of the band in the First Brillouin zone? What is "k" or "p" there? And from the physics, are you able to make out the "trajectory" of it?

This is why I find your claim here to be very puzzling, considering that this is really a well-known physics.

I'm sorry I got into this thread. I should have left the mistakes alone, because it is very frustrating to go one step forward, and have to take 2 or 3 steps back. So I'm done.

Zz.

 

[PeterDonis]

The OP's questions have been answered repeatedly. The answers are:

In the classical definition, velocity is the time derivative of the particles position curve (trajectory). The Uncertainty principle restricts the particles velocity from ever being zero. Doesn't this imply that a particles path is restricted to paths that do not have a zero time derivative at any point?.

No.

Or am I a victim of classical thinking once again?

Yes.

Thread closed.