# I Does the Uncertainty Principle restrict a particle's path?

1. Feb 27, 2017

### 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?

2. Feb 27, 2017

### 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 $\Delta p$ ).

3. Feb 27, 2017

### Staff: Mentor

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 cant 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

4. Feb 28, 2017

### 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.

5. Feb 28, 2017

### mike1000

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.

6. Feb 28, 2017

### ZapperZ

Staff Emeritus
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.

7. Feb 28, 2017

### mike1000

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.

8. Feb 28, 2017

### ZapperZ

Staff Emeritus
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.

9. Feb 28, 2017

### mike1000

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.

10. Feb 28, 2017

### ZapperZ

Staff Emeritus
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.

11. Feb 28, 2017

### mike1000

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

12. Feb 28, 2017

### ZapperZ

Staff Emeritus
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.

13. Feb 28, 2017

### mike1000

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.

14. Feb 28, 2017

### ZapperZ

Staff Emeritus
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.

15. Feb 28, 2017

### 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.)

Last edited: Feb 28, 2017
16. Feb 28, 2017

### mike1000

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.

17. Feb 28, 2017

### ZapperZ

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

Zz.

18. Feb 28, 2017

### mike1000

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.

19. Feb 28, 2017

### ZapperZ

Staff Emeritus
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.

20. Feb 28, 2017

### mike1000

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.