1. Aug 2, 2007

### Klaus_Hoffmann

Let be a particle, we measure the position at time 't' $$x(t)$$

We measure the position after an infinitesimal chnage t-->t+dt so $$x(t+dt)$$ is obtained.

then as an approximation we can make (definition of derivative and velocity)

$$x(t+dt)-x(t)=dtp(t)$$ so we can obtain the momentum at time 't' $$p(t)$$ avoiding uncertainty

2. Aug 2, 2007

### HallsofIvy

Where is the paradox? You now know the average velocity between t and t+ dt. You do NOT know the velocity before t because in measuring the postion you have changed the velocity when you measured the postion. You do NOT know the velocity after t+ dt because you changed the velocity when you measured the position. The average velocity you have contains no useful information.

3. Aug 2, 2007

### ZapperZ

Staff Emeritus
The HUP says nothing about the measurement of a single observable. you can measure the position and momentum of one particle as accurately as you are capable of.

The HUP comes in when you make the NEXT AND SUBSEQUENT x(t) and x(t+dt) measurements. The p(t) that you measure will depend very much on how well you measured x(t). The more certain you are of x(t), the larger a range of values you will measure for x(t+dt), which will result in a larger range of p(t) values. This will occur even if you have an identical initial conditions and you simply repeat the identical experiment. This is the HUP.

Zz.

Last edited: Aug 2, 2007
4. Aug 2, 2007

### StatusX

Incidentally, in relativistic quantum mechanics, if you carry out the procedure you described, you're guaranteed to measure the particle travelling at an average velocity of c. This is becausen the momentum is uniformly distributed after the first measurement of position, and so "most" of the possible states lie at very high momenta, whose cooresponding velocities all approach the speed of light.

5. Aug 2, 2007

### nrqed

Yes, but the key point is that the HUP syas something about the spread of values in the observables. It does not say anything about a single measurement.

The whole point is that if you start with a second particle with exactly the same quantum state (same wavefunction) and you repeat exactly the procedure you just described, you will get a completely different result, in general. That's the key point. In classical physics, of course, one would get reproducible results (the same conditions will yield the same results, within th euncertainty of the measurements performed, which may in principle be reduced without any limit). In QM, repeating the same experiment to the same quantum system will yield different results every time, in general. The spread of the values in position and momenta will satisfy the HUP.

6. Aug 2, 2007

### Klaus_Hoffmann

then you mean that there's also some uncertainty involving

[x(t),x(t+dt)]=ih where [] is the commutator

7. Aug 2, 2007

### nrqed

Well, I don't think it is possible to express it this way because I don't see what would be the meaning of putting them in the order x(t) x(t+dt). What would that mean?

What I *am* saying is that the value of x(t+dt) is not uniquely defined even for a given x(t) and a specific initial wavefunction.

But to get to the uusal form of the HUP with x and p_x andto relate this to a comutator [x,p], the value of "p" must be obtained by different means, by using momentum conservation in a collision, say. That's at least my understanding, maybe others will have some clarifying comments to make.

regards

8. Aug 2, 2007

### ueit

There is no dificulty in finding both position and momentum of a particle with any accuracy you like. The problem is that you cannot use the result to make predictions, because the measurement alters the system. As Heisenberg said, HUP does not apply to the past.

9. Aug 2, 2007

### meopemuk

I think that this conclusion is basically correct, although the actual form of the commutator is different. The operator x(t) is connected to x(t=0) = x by the time evolution equation

$$x(t) = e^{\frac{i}{\hbar}Ht} x e^{-\frac{i}{\hbar}Ht}$$

where H is the Hamiltonian whose commutator with x is $[x,H] = i \hbar v$, where v = p/m is the operator of velocity. (I assumed that there are no interactions.) Using this commutator one can obtain, as expected

$$x(t) = x + vt$$

so that

$$[x(0), x(t)] = [x, v]t = \frac{i t \hbar}{m}$$

Eugene.

10. Aug 3, 2007

### lightarrow

Can you please explain this better? HUP is often expressed saying: "it's not possible to measure x and p simultaneously with arbitrary accuracy".

11. Aug 3, 2007

### MarkeD

You can measure only one varaible at a time to any degree of accuracy. IF you get an accurate measurement of momentum, you will have a large uncertainty of position.

A nice metaphor is that the HUP is like taking a photograph of a moving car; you can have a fast shutter speed and accurately measure the position f the car but know little on how fast it was going, if you use a slower shutter speed you can measure the speed via the blur on the photo but have little knowledge on its exact position.

12. Aug 3, 2007

### ZapperZ

Staff Emeritus
Er.. this is not quite right. The problem in your example is that you are using inappropriate method to measure different things. The HUP never say anything about using the SAME technique to measure something.

First of all, there is a sequence of measurement, i.e. you measure position first, and then momentum, or you measure momentum first and then position. Both do not give you the same result because of non-commutation. But more importantly, the HUP, if you look carefully at the expression, requires the knowledge of the AVERAGE values of each of the observable, ie. an ensemble of measurements, not just one. This means that for a single measurement of the position, followed by the single measurement of the momentum, there's nothing that restricts the accuracy of each of those measurements. The accuracy of each of those measurements depends on the accuracy of the instruments, and they are independent of each other. I could use a slit of size $\Delta (x)$ to measure the position and then look at where the particle hit the detector/screen behind the slit to measure the transverse momentum. The size of the slit and the size of the spot on the detector are independent of each other and are not governed by the HUP. Both of these sizes determine the uncertainty of that single measurement of position and single measurement of the momentum. These are not the HUP.

Zz.

Last edited: Aug 4, 2007
13. Aug 4, 2007

### country boy

This "paradox" can be posed just as well without the infinitesimal part. Make the first position measurement and then later make the second position measurement. You can make each of these measurements as accurately as you want to and, with the times, you also have the velocity (and momentum) to arbitrary accuracy. The catch is that each position measurement introduces its own momentum uncertainty (the HUP position uncertainty is not the distance between the measurements). The momentum uncertainty introduced by the second measurement is independent of the first measurement.

It is easy to see this with waves and diffraction. If a wave encounters a slit of dimension similar to its wavelength, it will spread out (diffract). If another slit is placed downstream, the small flat portion of the wavefront that encounters the second slit will spread out again. The diffraction at each slit does not depend on what happened previously. [Note that if the slits are large compared with the wavelength, the wave will be collimated, not spread out. That corresponds to the classical limit.]

14. Aug 4, 2007

### country boy

This can be also be explained in terms of diffraction at a slit (please see my previous post). The width of the slit is the uncertainty in position lateral to the direction of travel. The angular spread (diffraction width) of the emerging wave is given by the ratio of the wavelength to the slit width. This angle corresponds to the lateral uncertainty of the momentum. So the uncertainties of both the position and momentum are determined "simultaneously" at the slit.

15. Aug 4, 2007

### lightarrow

Yes, but ( probably my english wasn't very good), I intended to ask ueit to explain his phrase:

"There is no dificulty in finding both position and momentum of a particle with any accuracy you like. "

since it would seem in contradiction with the fact we can't simultaneously measure position and momentum with arbitrary accuracy.

16. Aug 4, 2007

### ZapperZ

Staff Emeritus
This is NOT a fact. It is a misunderstanding of the HUP. We can "simultaneously" measure position and momentum with arbitrary accuracy (up to our present-day technology), if by "simultaneous", you mean a single measurement of position, and then followed by a single measurement of the momentum of that particle. The HUP never stated anything about an instantaneous and simultaneous measurement of both position and momentum. If it does, then the order of the commutation relation won't make a difference.

Zz.

17. Aug 4, 2007

### jostpuur

The way you use term "simultaneous" seems confusing. If you first measure the position, and after it the momentum, then the measurements were not simultaneous.

You can find the position with arbitrary accuracy, and after it you can find the momentum with arbitrary accuracy, but you cannot know them both with arbitrary accuracy simultaneously.

18. Aug 4, 2007

### ueit

Sure. Just place a detector at a very large distance from a particle source and switch on the source for a small time. The accuracy in position is only limited by the detector resolution while the accuracy in momentum can be made arbitrary small by increasing the distance between the source and detector and/or decreasing the time when the source is on. However, the determined values cannot be used to make predictions about how the particle will evolve after detection, but you know them how they were in the past.

19. Aug 4, 2007

### ZapperZ

Staff Emeritus
That is why I clarified what I meant. Note that an instantaneous, simultaneous measurement is meaningless and irrelevant as far as the HUP is concerned. So this issue doesn't even matter. That is why I stated clearly what I meant as "simultaneous". In addition, I'm not the one who brought up the "simultaneous" issue.

Zz.

20. Aug 4, 2007

### country boy

Sorry, I'm not sure what you mean either. The textbooks I've looked at refer to either "simultaneously" or "at the same instant." The wording may be a little imprecise, but the uncertainty principle does make a statement about the degree to which two quantities can be measured accurately at the same time.