Uncertainty principle and momentum

1. Feb 10, 2014

shreder

if you know the position why cant you know the momentum

2. Feb 10, 2014

Staff Emeritus

3. Feb 10, 2014

Drakkith

Staff Emeritus
Imagine a perfect sine wave of a single frequency and amplitude. The frequency represents the momentum so this perfect sine wave represents knowing the momentum of the particle with perfect accuracy. Such a wave extends to infinity because it is a single frequency. If the amplitude represents the possibility of finding the particle in a certain position in space, then there is an equal possibility of finding the particle everywhere in space because the wave extends to infinity. So we have complete uncertainty in the position. In order to localize the particle it is necessary to add more waves of different frequencies so that the interference of the waves causes the amplitude to increase in a certain area. But, adding these different frequency waves means that we no longer have a perfect sine wave, and no longer know the momentum with perfect accuracy.

Does that make sense?

4. Feb 10, 2014

ZapperZ

Staff Emeritus
This is where you need to describe the SCENARIO in which you made the measurements. It makes a difference in QM, because QM is all about CONTEXT!

If I have just ONE particle, let's say I want to measure its position. What do I do? In QM this makes a boatload of difference on HOW you measure something. I can, for example,make it pass a very tiny slit, so that when it pass through it, I know that it had that position. Call this position x1, and the uncertainty in that position is Delta(x), corresponding to the width of the slit.

Now, there's nothing to prevent me from measuring, as accurately as I want, the momentum px AFTER it passes through the slit. How do I measure this? I can do that by

(i) putting a detector/screen after the slit;

(ii) measure the location that the particle hit the detector/screen;

(iii) measure how much it has moved transversely (i.e. in the x-direction) when it was moving from the slit to the screen)

(iv) calculate the x-component of the momentum since I know the speed that this particle moves and its mass;

(v) include the detector uncertainty that will give me the uncertainty in the momentum that I just measured. I will provisionally call this DDelta(px) - the double "D" is not a typo.

Now, I've just demonstrated that, in this case, I had just measured both position, and the corresponding momentum, to arbitrary accuracy limited by my instrument. In fact, I will argue that those two uncertainties (the Delta's) have no correlation with each other at all! I can increase the detector's accuracy, say by reducing the cross-talk between each of the detector channel, and thus, make the "spot" that I see on the detector or screen smaller, thus reducing the momentum uncertainty. This has ZERO effect on the width of the slit that I had the particle passed through.

So I've just showed you a way to measure position and momentum of a particle, and the HUP doesn't come into play... or does it?

What if I shoot a SECOND identical particle with the identical initial condition, at the identical slit? THIS is where it gets interesting!

After passing through the slit, do you think the particle will land on the detector at the same location? Classical mechanics will say that it should. After all, all the conditions are the same. But this is not true in quantum particles and in a quantum system.

Where the particle will land depends on how small the width of the slit is, i.e. how small Delta(x) is! The smaller Delta(x) is, the LARGER the spread in where the particle will hit the detector/screen. Remember, this also means that there will be a larger spread in the value of the momentum of the particle after it passes through the slit!

To know the nature of this spread, you have to send many more particles through the slit. It is ONLY when you have a statistically significant number will you see that there is a "central" value of the most probable momentum, and there is a spread (similar to the Standard Deviation in statistics) in the value of the momentum. Now this spread in momentum IS the "Delta(px) in the HUP!

So what does that tell you? You can't get the HUP from just ONE single measurement of x and px. If you look at the definition of the HUP, you'll see <x> and <p> and <x2> and <p2>. These are averages, which makes very little sense when you make just ONE measurement of each! So the uncertainties here is the SPREAD in the values that one would have upon REPEATED measurement of the identical system.

Secondly, it also tells you how well you can predict the outcome of the next measurement. The smaller the slit width, the less likely you will be able where the next particle will hit the screen. In other words, your uncertainty of the momentum will be larger with smaller slit!

This, in essence, is the HUP, and I had just described to you the single-slit diffraction.
https://www.physicsforums.com/blog.php?b=4364 [Broken]

Zz.

Last edited by a moderator: May 6, 2017
5. Feb 17, 2014

atyy

In this method to measure the position and momentum of a single particle, it seems that you assume classical equations of motion of the particle from the slit to the screen. However, with Bohmian mechanics, the particle does not follow classical equations of motion. Why are the classical equations of motion privileged over the Bohmian ones?

Last edited by a moderator: May 6, 2017
6. Feb 17, 2014

ZapperZ

Staff Emeritus
Because it works!

Zz.

7. Feb 17, 2014

atyy

But does your picture say that single particles have trajectories?

8. Feb 17, 2014

ZapperZ

Staff Emeritus
It has after it landed at the detector.

The same way that I can say that the particle that passed through the slit had that position at that slit BECAUSE it passed through it, the particle that made the "spot" at the detector had a definite trajectory from the slit to the detector AFTER it has been detected. It is why, from my avatar, the ARPES data can measure its momentum and energy.

Zz.

9. Feb 17, 2014

atyy

It had a trajectory from the slit to the detector after it was detected, but not before it was detected?

10. Feb 17, 2014

ZapperZ

Staff Emeritus
It had a range of possible momenta before it was measured. That is why there is a momentum spread upon repeated measurement.

Zz.

11. Feb 18, 2014

atyy

@ZapperZ, in this method of assigning a trajectory to a single particle, it seems that one is doing two successive measurements of transverse position. If the screen is very close to the first particle, because the particle is in a position eigenstate just after the slit, one would expect the transverse position on the screen to be very close to the first position, is that right?

12. Feb 18, 2014

ZapperZ

Staff Emeritus
Of course. At some point, if you are too close, you will not get any transverse resolution, and thus, you will not get good information about the momentum.

I forget if there is a near-field solution to this, i.e. when the screen distance is comparable to, say, the slit width or the wavelength. I think there is. When you are that close, the detector is doing nothing more than confirming the position of the particle, rather than a detector where you can extract the momentum.

Zz.

13. Feb 18, 2014

curious bishal

You know, position means where it is right now. Let's take an example of electron. There are many factors which accounts for the uncertainity:-
1) If you calculate the position of the electron, it escapes out from the present position to the next one while you are calculating and it always escapes.
2) For calculating the velocity, you have to take account a certain distance and time to measure the velocity and as you measure the velocity , electron slips more.......
3) There is a fine delay coming the light waves from the electron to you so that there is always a remarkable error measuring the position and velocity
and there are others, too

But in the question, where you fails is that, measuring the position needs two co-ordinates i.e it can be measured at once. But measuring dx needs two position ( dx is the difference between two positions ), so as you measure dx , you will miss its position. And dt is a duration of time and requires two different point of time to measure it and as you measure dt, your position slips,
This is where you did mistake.
Hence, two canonical conjugate variables ( like position and velocity, time and energy, etc) can't be measured simultaneously with absolute accuracy.......

Last edited: Feb 18, 2014
14. Feb 18, 2014

ZapperZ

Staff Emeritus
What you said here made very little sense. A "calculation" is a mathematical operation. How could an electron "escapes" while one is calculating? Furthermore, in a calculation, you can make your detection as accurately as you want! You have no instrument resolution to deal with such as in an actual experiment!

The rest of what you wrote is equally puzzling.

Zz.

15. Feb 18, 2014

I think you have to add: with arbitrary precision, simultaneously.

Short answer: Because in classical mechanics 0 = 0 but in QM 0 ≠ 0.

Long answer: Naturally, we tend to think of "QM stuff" as spheres or marbles, traveling in classical trajectories. This is wrong. No one has seen a single electron traveling in space and besides, there are different interpretations whether the electron exists or not, before measurement. There are also different approaches in calculating expected outcomes, however, everybody get the same results in the end.

One way is to treat a QM particle as a "traveling wave packet of probability", i.e. interpreted as a "probability wave", giving the probability for a particle being measured with a given position and momentum.

From this it's easy to see that the uncertainty principle actually states a fundamental property of quantum systems, which is inherent in all wave-like systems, including a classical sound wave – it's impossible to measure both the frequency (pitch) and position of a classical sound wave with arbitrary precision, simultaneously. As explained:

Piece of cake, end of story!

Almost... what if... we could 'cool down' the electron, not to be so 'wavy'? Could that work, maybe? Sorta $x=0,p=0$ huh? :uhh:

Well no, we're stuck again because; yes we can cool the electron to its lowest energy state, i.e. the ground state, but as already pointed out in QM 0 ≠ 0, which means that in quantum zero-point energy, there is always some tiny potential energy left for fluctuations. The wavefunction of the ground state is a half-period sine wave which goes to zero at the two edges, whereas this is not the case in a classical oscillator:

And the energy of the quantum ground state is $E_0 = \hbar \omega / 2$

Where $\omega$ is the angular frequency at which the system oscillates and $\hbar$ is Planck's constant $h$ dived by $2\pi$, which is an extremely tiny number ≈ 1.05457×10−34 J s.

Thus, Planck's constant is one reason this is not a 'problem' for classical objects like tennis balls...

Hope it helped!

16. Feb 19, 2014

atyy

If the classical equations fail when the screen is near the slit, why is it justified to use the classical equations to assign the entire trajectory from the slit to the distant screen, since the trajectory must start near the slit? Shouldn't at least the initial part of the trajectory near the slit be described by non-classical equations?

Last edited: Feb 19, 2014
17. Feb 20, 2014

freshman123

Anyhow, it is in excellent agreement with experimental results, I think, which is the most significant reason. I believe that there will be some certain new theories taking place of quantum mechanics.To be frank, I am more inclined to the idea of determination even though quantum mechanics is used as the main tool in my research.

18. Feb 20, 2014

ZapperZ

Staff Emeritus
Fail? Who said it failed?

Remember, in classical E&M, there is a such a thing as a near and far-field solutions! The near solutions often looks different than the far solution. Furthermore, at some point, if you make the slit too big, you will no longer see the diffraction pattern, and thus, no more spread in the transverse momentum! This is the same thing as making the screen too close to the slit!

Do you now claim that the HUP "failed"?

You seem to have ignore a very important point that I made in one of my responses - IT WORKS! We have been using it for ALL ARPES measurement, whereby the we can gather information about the electron's transverse momentum when it makes an image on our CCD screen. There is no better verification of the validity of anything than that!

Zz.

19. Feb 20, 2014

atyy

You yourself said it fails in near field.

20. Feb 20, 2014

ZapperZ

Staff Emeritus
No, I said that the information about the momentum may not have the same resolution as the far field!

Did you read what I actually wrote?

Zz.