Uncertainty principle and momentum

[shreder]

if you know the position why can't you know the momentum

isnt p=mv and v=dx/dt its the only doubt i have so far please help

 

[Vanadium 50]

What reading have you already done on this?

 

[Drakkith]

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?

 

[ZapperZ]

if you know the position why can't you know the momentum

isnt p=mv and v=dx/dt its the only doubt i have so far please help

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 p_x 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(p_x) - 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(p_x) in the HUP!

So what does that tell you? You can't get the HUP from just ONE single measurement of x and p_x. 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

Zz.

 

[atyy]

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 p_x 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(p_x) - 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(p_x) in the HUP!

So what does that tell you? You can't get the HUP from just ONE single measurement of x and p_x. 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

Zz.

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?

 

[ZapperZ]

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?

Because it works!

Zz.

 

[atyy]

Because it works!

Zz.

But does your picture say that single particles have trajectories?

 

[ZapperZ]

But does your picture say that single particles have trajectories?

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.

 

[atyy]

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.

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

 

[ZapperZ]

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

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

Zz.

 

[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?

 

[ZapperZ]

@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?

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.

 

[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...

 

[ZapperZ]

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

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.

 

[DevilsAvocado]

if you know the position why can't you know the momentum

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:

https://www.youtube.com/watch?v=7vc-Uvp3vwg
http://www.youtube.com/embed/7vc-Uvp3vwg

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?

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!

 

[atyy]

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.

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?

 

[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.

 

[ZapperZ]

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?

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.

 

[atyy]

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.

You yourself said it fails in near field.

 

[ZapperZ]

You yourself said it fails in near field.

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

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.

Did you read what I actually wrote?

Zz.

 

[atyy]

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

If the slit and screen are close, and I register the position of the particle on the screen with high resolution, did the particle have a trajectory from the slit to the screen that is described by the classical equations of motion?

 

[ZapperZ]

If the slit and screen are close, and I register the position of the particle on the screen with high resolution, did the particle have a trajectory from the slit to the screen that is described by the classical equations of motion?

Yup.

Zz.

 

[atyy]

Yup.

Zz.

So if the screen and slit are near, and I have accurate positions of the particle at the slit and at the screen, then I can assign an accurate momentum to the particle along its entire trajectory using the classical equations, since the classical equations hold, right? Let's call this the single particle momentum.

However, we also know that when the screen is near the slit, the ensemble of single particle momenta obtained this way does not accurately reflect the momentum distribution we expect for an accurate momentum measurement of the wave function at the slit.

So would you say here we have accurate single particle momenta (since the classical equations hold, and we have two accurate position measurements), but the ensemble of these accurate single particle momenta are not accurate momentum distributions?

 

[DevilsAvocado]

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.

Yes, it would be quite astonishing if QM is the final answer to everything (except maybe gravity ). If science continues to develop as it has – there will be two new questions to solve for every answer we find. Personally I think it's great, who wants to live in a "100% explained universe"... booooooring. :zzz:

But "personal preferences" are probably not fruitful in the search for new knowledge. I think most of us, including myself, would prefer a "common sense" world that works the way we are used to, however there is no way to dispute already proven facts. Even if Newton wasn't 100% correct about gravity – the apple will always fall down to Sir Isaac's head, whether we like it or not.

Same thing with determinism/definiteness, if we finally finds out that definite fundamental values are a fact, then locality has to go (unless we give up free will or split the universe in a gazillion forks for every event).

Anyway, the old classical world of local realism is as dead as the Norwegian Blue Parrot; no new discoveries can ever change this fact, thanks to John[/PLAIN] Stewart Bell.

No no, local realism is not dead... it's just resting...

 

[ZapperZ]

So if the screen and slit are near, and I have accurate positions of the particle at the slit and at the screen, then I can assign an accurate momentum to the particle along its entire trajectory using the classical equations, since the classical equations hold, right? Let's call this the single particle momentum.

However, we also know that when the screen is near the slit, the ensemble of single particle momenta obtained this way does not accurately reflect the momentum distribution we expect for an accurate momentum measurement of the wave function at the slit.

So would you say here we have accurate single particle momenta (since the classical equations hold, and we have two accurate position measurements), but the ensemble of these accurate single particle momenta are not accurate momentum distributions?

Think about this: what is the difference between a slit of width 1 micron, and a screen placed 1 micron away, versus a slit of width 1 meter, and a screen placed 1 meter away?

Can you get accurate momentum spread and momentum information out of that? Or did you just make Δx so big that there are no more momentum spread to be had?

Zz.

 

[atyy]

Think about this: what is the difference between a slit of width 1 micron, and a screen placed 1 micron away, versus a slit of width 1 meter, and a screen placed 1 meter away?

Can you get accurate momentum spread and momentum information out of that? Or did you just make Δx so big that there are no more momentum spread to be had?

Zz.

Of course the momentum distribution when the screen is near the slit is not accurate. But the position measurements on the screen can be accurate whether or not the screen is near or far. Since there are two accurate positions - one at the slit and one at the screen - then if the classical equations hold near when the screen is near, I should be able to use them to get an accurate measurement of the single particle momentum.

 

[DevilsAvocado]

I should be able to use them to get an accurate measurement of the single particle momentum.

But never simultaneously, right?

 

[atyy]

But never simultaneously, right?

You are quoting me out of context.

 

[ZapperZ]

Of course the momentum distribution when the screen is near the slit is not accurate. But the position measurements on the screen can be accurate whether or not the screen is near or far. Since there are two accurate positions - one at the slit and one at the screen - then if the classical equations hold near when the screen is near, I should be able to use them to get an accurate measurement of the single particle momentum.

No you can't, because if it is too close, the particle can come from ANY part of the slit! You now have an ambiguous origin on how to calculate your classical trajectory!

You have no way of assigning a unique momentum to that spot on the screen anymore!

Zz.

 

[DevilsAvocado]

You are quoting me out of context.

Sorry, what I meant is; when the particle hit the screen momentum will be = 0, so what you get is the calculated momentum between the slit and the screen, not the 'actual' momentum, right?

(But this is probably wrong, looking at Zz's picture...)

 

[atyy]

No you can't, because if it is too close, the particle can come from ANY part of the slit! You now have an ambiguous origin on how to calculate your classical trajectory!

You have no way of assigning a unique momentum to that spot on the screen anymore!

I see. And as the width of the slit gets smaller, the state at the slit is closer and closer to a position eigenstate, then even for any finite distance, the momentum will become more and more accurate. So is your basic argument that if we do know the particle is in an eigenstate of one observable, we can make simultaneous accurate measurements of non-commuting observables on it, because we can measure one observable without disturbing the eigenstate, leaving the same state available for a measurement of the non-commuting observable?

 

[DevilsAvocado]

What am I missing in this screen/momentum scenario...??

Take two identical cars, traveling 100 km on a test track, ending in a rock wall. Car #1 makes the distance on exactly 60 min, and gets completely demolished when crashing into the rock wall; average speed 100 km/h.

Car #2 also makes the distance on exactly 60 min, and gets a tiny dent when bumping into the wall; average speed 100 km/h.

What happened?

Car #1 went 100 km/h from start to finish. Car #2 went 110 km/h for 54 min, hit the brakes and then went 10 km/h for 6 min.

So how on Earth can a smash into a screen tell us anything about simultaneous position and momentum?? (unless you check the 'recoil' of the screen)

I don't get it...

 

[Jilang]

What am I missing in this screen/momentum scenario...??
...

I don't get it...

Are you perhaps mixing up your components of momentum and position? The transverse momentum and the transverse position are the non-commuting variables. Px and Y will commute so no issue.

 

[ZapperZ]

What am I missing in this screen/momentum scenario...??

Take two identical cars, traveling 100 km on a test track, ending in a rock wall. Car #1 makes the distance on exactly 60 min, and gets completely demolished when crashing into the rock wall; average speed 100 km/h.

Car #2 also makes the distance on exactly 60 min, and gets a tiny dent when bumping into the wall; average speed 100 km/h.

What happened?

Car #1 went 100 km/h from start to finish. Car #2 went 110 km/h for 54 min, hit the brakes and then went 10 km/h for 6 min.

So how on Earth can a smash into a screen tell us anything about simultaneous position and momentum?? (unless you check the 'recoil' of the screen)

I don't get it...

OK, I didn't realize this is that difficult to understand, considering that this is almost first-year physics.

v is known, since you can prepare the particles to be of a certain speed/momentum, or in the case of light, this is c.

The time taken for the particle to go from the slit to the screen is

t = d/v_y = d/(v cosΘ)

For most cases, the slit is very small, and the screen is very far, so cosΘ ≈ 1 is a good approximation.

In that same time, the particle has a transverse x-component movement, which means that it has a transverse velocity/momentum, i.e.

v_x = x/t = xv/d.

Since v and d are know, this means that knowing x, i.e. where it landed on the screen transversely, give you the velocity, and thus, the momentum.

I've done this for a non-relativistic case, since this gives a clearer illustration how we detect momentum in practically ALL cases. Note that you cannot detect "collision", i.e. you don't measure it. You measure where the colliding particle landed at detectors, and via reconstruction, you arrive at the momentum. This is how we measure energy and momentum of particles in high energy physics. We DETECT where the particles are when they hit the detector. All we see are locations of these particles at the detector. We do NOT see a "collision"!

Zz.

 

[DevilsAvocado]

OK, I didn't realize this is that difficult to understand, considering that this is almost first-year physics.

OMG, c, why didn't I think of that... (maybe because my Volvo has a hard time keeping the pace in winter time < 100 km/h)… I get it (please don't laugh).

It's quite amazing that classical equations hold so nice. Who needs path integrals?