# Yes or No

1. Sep 21, 2004

### Alkatran

I've been reading through many of the posts in here, trying to get a sure-thing yes or no about uncertainty.

So here it is:
Are particles actually in different positions/speeds/what-have-you or is the uncertainty only because of the techniques used to measure these particles?

If it is the former, is there some variable we can't measure which affects it?
Wait, that question is stupid. How could we know of the existance of an immeasurable variable...

2. Sep 21, 2004

### Integral

Staff Emeritus
To the best of my knowledge, it is simply not correct to view subatomic particles as having a specific location or velocity. There is only the probability of them having these properties. That is we know (can compute) the probability that some particle will be at some point with some velocity. Upon observation the specific values are assumed. This is a basic property of subatomic particles not a facet of our ability to measure.

I believe that experiments have been done which rule out "hidden variables". Perhaps others can give more details on how this can be done.

3. Sep 21, 2004

### Gonzolo

It is not due to techniques. For as long as quantum mechanics hold in experiments (its going on about 80 years now), the Heisenberg principle forbids a technique allowing precise measurement of position and momentum beyond its statement : $$\Delta x \Delta p >= h/2$$.

This inequation is a consequence of any wave theory, not just quantum theory (which is itself a wave theory). It takes no more than 1 or 2 pages to demonstrate the inequation starting with the equation of a guitar string.

4. Sep 21, 2004

### humanino

If you like, it is derived [thread=39172]here[/thread] that the product of indeterminacies is given by the commutator :
$$(\Delta a)(\Delta b) \geq \frac{1}{2}\langle \psi |[\hat{A},\hat{B}]|\psi\rangle$$

It is just Schwartz inequality, nothing more, so if you know it, you can derive it in two lines !

5. Sep 21, 2004

### Alkatran

Thanks for your replies, but... I see not a yes nor a no yet. Looks like a yes so far.

6. Sep 21, 2004

### humanino

I could not just say : yes

7. Sep 22, 2004

### broegger

Could you please explain this? Do you mean that anything that we model as a wave phenomena (like macroscopic mechanical waves) has equivalent uncertainty principles or what? If so, how can you derive them?

8. Sep 22, 2004

### vanesch

Staff Emeritus
It's a basic property of Fourier transforms: the product of the standard deviations of a function (considered as a distribution) and its Fourier transform (also considered as a distribution) have a lower limit of the order of 1 (don't remember the exact value).

For instance, in signal theory there is this relation between the minimal duration of a signal and the minimum bandwidth a signal can occupy.

cheers,
Patrick.

9. Sep 22, 2004

### broegger

Interesting.. And this has nothing to do with quantum mechanics?

If I were to do some measurements involving let's say, water-waves propagating through some small slits in a pool would I encounter an uncertainty principle of some kind involving the momentum and position of the water-particles? I could imagine that because of interference patterns, but isn't that different - in principle - from Heisenbergs principle??

10. Sep 22, 2004

### humanino

I thought my "yes" would cause protestations. The flaw is, one is not supposed to talk about things outside measurements.

11. Sep 22, 2004

### Staff: Mentor

It isn't a yes or no question. If you cut it off with this:
...then its poorly worded, but the answer is yes. A better way to word it is "do particles have clearly defined (precise) positions and velocities?" (in which case the answer is no).
Yes, water waves are not the same as matter waves.

12. Sep 22, 2004

### Alkatran

The trick here is that if you said "no" everyone on the forum would have jumped on the thread. My proof was in the answer.

13. Sep 22, 2004

### humanino

:rofl:
I did not notice that !
I meant as an answer to the statement : A or only B ?
Yes <-> A
No <-> B
Sorry

14. Sep 22, 2004

### vanesch

Staff Emeritus
No, the interpretation of the conjugated fourier pair to be the distributions of position and momentum is typically quantum mechanics. But the mathematical property itself not, it is a property of fourier transforms ; only you don't interpret bandwidth and time duration as uncertainties on momentum or position in its classical applications.

cheers,
Patrick.

15. Sep 22, 2004

### Gonzolo

What vanesh said.

Heisenberg's principle is special because of what the inequation (= what wave theory) is applied to. Once Shrodinger and de Broglie came out with probability waves for matter (which is the real shocker), Heisenberg's principle was an inevitable consequence. Arriving to this consequence makes one question the validity of the postulates, but experimental facts shut up any debate.

Mathematically, water waves and probability waves are similar, but physically they are totally different. Heisenberg's principle is the mathematical Schwartz inequality specifically applied to Shrodinger's probability wave $$\psi$$, it's meaning, and the QM postulates.

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