# Heisenberg Uncertainty Principle

1. May 15, 2005

### tom_the_cowboy

The Heisenberg Uncertainty Principle states that both the exact (as near as we can get to thusfar) position AND momentum of a particle cannot be obtained because in learning its exact position we alter its momentum and vice versa. Does this mean we only have rough measurements of particles produced in particle accelerators? If so, why can't two particles be produced from a collision and the momentum of one measured and the position of the other measured thereby giving exact measurements of both position and momentum for each (albeit after the measurements were taken both variables would have changed, but at least exact measurements at the point of collision could be obtained). Can this be done? Has it? Is there some basic physics principle I've forgotten about?

2. May 15, 2005

Staff Emeritus
The uncertainty principle states that for "non-commuting observables" like position and momentum, The product of the uncertainty in one times the uncertainty of the other has a constant lower bound.

Note that you can get position as accurate as you like (and your experimental techniques can support), or you can do momentum as accurately as you want. You just can't do them BOTH with arbitrary precision at the same time. Classical readouts from accelerators, like bubble chambers, tried for position accuracy and simply didn't look at momentum. You got photographs of trajectories.

On the other hand "calorimeters", essentially stacked sheets of armor plate that measure momentum by seeing how many sheets the particle can penetrate, look only at momentum. So you see they DO obey the Uncertainty Principle - they have to! But they do manage to collect the information they need by looking at different events with different particles.

3. May 15, 2005

### cartuz

Dear Cowboy,
You are right! Your idea is the same which you can read in A. Einstein, P. Podolsky, R Rosen "Can the Quantum-Mechanical Description be Complete?" in Phys. Rew., 1935, p.777, Number I do not remember.
This paper is very often citation. But the EPR idea is change and now employ to the spin-states. In this paper you cannot see particles with SPIN. In the same time Bell's experiments was found on the this paper but Bell employ EPR idea to spin-statistics.
Best, Cartuz.

4. May 15, 2005

### marlon

TO THE OP : Though this is very correct, i never understood why people explained the HUP like this. I find it very misleading. As a matter of fact you can get both position and momentum completely accurate, or at least as accurate as your measuring device alows you to acquire these data. What i mean is that for a given fixed position, you can get a definite momentum value. It is just that if you were to measure the momentum value again at that same position, you will acquire another value. This way of looking at the HUP also gives a clear view as to how you need to look at the calorimeter thing.

regards
marlon

5. May 15, 2005

### dextercioby

And how do u define that using physical & mathematical terms...?

Daniel.

6. May 15, 2005

### marlon

:rofl:

just assume that the uncertainty of position is zero and the uncertainty of p is infinite. :rofl: That's the whole point

marlon

7. May 15, 2005

### dextercioby

U can't assume the uncertainty in position is zero.It's a nonphysical state.

Daniel.

8. May 15, 2005

### cartuz

I can to clear HUP by my way.

$$xp=\hbar/2$$

$$xm\overset{\cdot}{x}=x\overset{\cdot}{x}\frac{F}{\overset{\cdot\cdot}{x}}=\hbar/2$$

It is the equation which I can name the quantum oscillation equation

$$\overset{\cdot\cdot}{x}-(\frac{2F\overset{\cdot}{x}}{\hbar})x=0$$
$$\omega^{2}=\frac{2F\overset{\cdot}{x}}{\hbar}$$

Than HUP is equal to oscillation equation! Because x and p not independent.

Last edited: May 15, 2005
9. May 15, 2005

### dextercioby

It doesn't make any sense what u wrote there,i sincerely hope u realize that,too.

Daniel.

10. May 15, 2005

### marlon

I think you are missing the point. If you don't want to take a zero uncertainty, then just take a small one...Even you can take it as big as you want, it really does not matter to what i am trying to say. There is nothing that says you cannot place a detector at one fixed position and check out if a particle passed through it.

marlon

11. May 15, 2005

### dextercioby

Maybe...

Oh,but I do,but it's IMPOSSIBLE. :surprised

Daniel.

12. May 15, 2005

### caribou

Actually it's not even about measurement, as no particle exists in the theory with the properties of both a precise position and a precise momentum, so if quantum mechanics is the correct way to describe nature, then both can't be assigned to any particle in the way we might like.

Last edited: May 15, 2005
13. May 15, 2005

### wangyi

I think EPR is the key to your question, as posts has discussed before, but i still have one point to say. at colliders, this is not a problem in experiment (although it is still a problem in theory), because the momentum is very large, and a uncertainty delta p can be egnored for practical insterest. That is why
at colliders it *seemed* to be able to measure x and p together.