question about the uncertainty principle

[gib65]

I've been reading up on Heisenberg's Uncertainty Principle, and I've got a question about it. How does one measure the position of a particle such that it effects the momentum, and visa-versa? Does one effect the other regardless of the method used to measure them?

[swansont]

I've been reading up on Heisenberg's Uncertainty Principle, and I've got a question about it. How does one measure the position of a particle such that it effects the momentum, and visa-versa? Does one effect the other regardless[/i'] of the method used to measure them?

 

Yes, it is independent of the method. The HUP drops out of the mathematics - position and momentum wave functions are fourier transforms of each other (as are energy and time).

 

The physical mechanism is often given as an example. To know information, you have to bounce a photon off of your particle, and that changes the position and momentum of it. To localize the position more precisely requires a shorter wavelength, and that has a higher energy and momentum, and will cause a greater recoil, thus degrading the information you have about momentum.

[Johnny5]

I've been reading up on Heisenberg's Uncertainty Principle, and I've got a question about it. How does one measure the position of a particle such that it effects the momentum, and visa-versa? Does one effect the other regardless[/i'] of the method used to measure them?

 

Why don't you break up your question into two parts.

 

Part I: How can you measure the position of a particle?

Part II: How can you measure the momentum of a particle?

 

Swanson gave an answer to part I clearly. You fire a photon at the particle, it rebounds, and comes back to you. If you know the speed of the photon, and its travel time, you can compute the distance away that the particle WAS. That is information about the past, not the present of course.

[gib65]

So then, I presume longer wavelengths, though poor at measuring position, are better for measuring momentum. But please explain how this works. How do longer wavelengths tell us more about a particles momentum and how does this effect the position?

[Johnny5]

So then, I presume longer wavelengths, though poor at measuring position, are better for measuring momentum. But please explain how this[/i'] works. How do longer wavelengths tell us more about a particles momentum and how does this effect the position?

 

What is it that makes you say that longer wavelengths are poorer at measuring position, than shorter wavelengths. I presume your answer will make use of the relations:

 

[math] E = \hbar \omega [/math]

[math] p = \hbar k [/math]

[math] c = f \lambda [/math]

[math] k \equiv \frac {2\pi}{\lambda} [/math]

[math] \omega \equiv 2\pi f [/math]

[swansont]

So then, I presume longer wavelengths, though poor at measuring position, are better for measuring momentum. But please explain how this[/i'] works. How do longer wavelengths tell us more about a particles momentum and how does this effect the position?

 

If the photon has less momentum, it won't disturb the target particle much, so you can do multiple measurements and deduce its velocity. But your position information will be degraded.

[Daecon]

Why don't you just have 2 different people measure it at the same time, then they can measure one thing each?

[swansont]

Why don't you just have 2 different people measure it at the same time, then they can measure one thing each?

 

That doesn't solve the problem. Each measurement causes a perturbation.

[ydoaPs]

how do you know what interferes with each other? i know it is more than just position and velocity. i think another is the magnitude and direction of spin. also, what is the equation for the HUP

[swansont]

(delta X) (delta P) > hbar (also E and t work)

[cronxeh]

There is a workaround for HUP, but the science hasnt caught up yet.

 

I think, and Ive been thinking this for some time now, that if you can measure the tiny fluctuations around the particle's field, you can know its relative momentum, and as for position you can deduce those statistically later on, thus giving you a relative equation

 

Saying uncertainty is independent of method is ignorance. HUP is _exactly_ based on measurement flaws. It pertains to both position and energy measurements, and so far this unfortunate flaw in measurement has been delaying a lot of scientific progress

[Johnny5]

I have a question. In the following form of the HUP

 

[math] \Delta x \Delta p \ \underline{>} \ \frac{\hbar}{2} [/math]

 

can I replace p by Mv, or must I replace p by hbar divided by wavelength, or does it not matter? And I have further questions based upon the answer.

 

 

PS: I've seen the derivation of HUP from waveform analysis, but I don't remember it.

[swansont]

Saying uncertainty is independent of method is ignorance. HUP is _exactly_ based on measurement flaws. It pertains to both position and energy measurements, and so far this unfortunate flaw in measurement has been delaying a lot of scientific progress

 

This is spectacularly wrong. The HUP falls out of the math of QM.

[swansont]

I have a question. In the following form of the HUP

 

[math] \Delta x \Delta p \ \underline{>} \ \frac{\hbar}{2} [/math]

 

can I replace p by Mv' date=' or must I replace p by hbar divided by wavelength, or does it not matter? And I have further questions based upon the answer.

[/quote']

 

Whatever is appropriate. If you put in the wavelength, you basically get an identity.

[Johnny5]

Whatever is appropriate. If you put in the wavelength, you basically get an identity.

 

Thank you, can we try it?

[swansont]

Go right ahead.

[Johnny5]

Go right ahead.

 

Well in the case where we replace p by Planck's constant divided by wavelength, we get:

 

[math] \Delta x \Delta \frac{\hbar}{\lambda} \ \underline{>} \ \frac{\hbar}{2} [/math]

 

Now the uncertainty is defined using probability theory we have for example:

 

[math] \Delta P = <P^2> - <P>^2 [/math]

 

Realizing that they are integrals, we can just pull out planck's constant and then divide both sides of the inequality by it to obtain:

 

[math] \Delta x \Delta \frac{1}{\lambda} \ \underline{>} \ \frac{1}{2} [/math]

 

You say that if I put in wavelength, I basically get an identity. We have an inequality here not an identity, so you didn't say what you meant. But I am not sure how to interpret the inequality above, nor do I remember how to derive it.

[swansont]

rearrange it a little

 

[math] \Delta x {>} \ \frac{\lambda}{2} [/math]

 

 

IOW the wavelength is a good measure of the position uncertainty - you can't know a particle's position any better than ~the deBroglie wavelength. Which shouldn't be all that much of a revelation.

[Johnny5]

rearrange it a little

 

[math] \Delta x {>} \ \frac{\lambda}{2} [/math]

 

 

IOW the wavelength is a good measure of the position uncertainty - you can't know a particle's position any better than ~the deBroglie wavelength. Which shouldn't be all that much of a revelation.

 

How did you get lambda outside of the Uncertainty opertator.

[ydoaPs]

what is delta X? i assume p was momentum.

[Johnny5]

what is delta X? i assume p was momentum.

 

Delta X is uncertainty in position, and p is momentum. The thing is there are two relationships for momentum, the classical one p=mv, and the quantum mechanical one p = h/ 2 p l .

[swansont]

How did you get lambda outside of the Uncertainty opertator.

 

I cheated. I wasn't being rigorous, but having two spatial uncertainties isn't particularly helpful. The spatial - momentum uncertainty is tied up in the wave nature, so it's kind of incestuous to use the deBroglie relationship. So it's generally mv that you use, though there may be exceptions I can't think of off the top of my head.

 

And I don't know that I'd call ^{[math]\Delta[/math]} an operator.

 

If you'd rather, you can say

 

[math] \frac {\Delta x}{\Delta \lambda} > \frac {1}{2} [/math]

 

which tells you exactly the same thing

[Johnny5]

I cheated. I wasn't being rigorous

 

 

 

Yeah I figured as much. Sometimes rigor is good, sometimes not. Depends.

 

Let me ask you this, do you personally know how to derive the uncertainty principle from wave analysis? It was in a book of mine, but I confess I didn't follow the argument.

[ydoaPs]

so, how would you calculate the uncertainty of the curvature in a certain area? i think you would start by finding the uncertainty of energy, but where do you go from there?

[Johnny5]

so, how would you calculate the uncertainty of the curvature in a certain area? i think you would start by finding the uncertainty of energy, but where do you go from there?

 

Can you explain your question more please?