# Is Heisenberg uncertainty principle a problem of our measuring techniques?

1. Oct 10, 2011

### valekovski

Hello

I know this topic must have been done to death already, but i can't seem to find a satisfying answer.

As the title suggests, my question is, what experiment proves that the uncertainty principle is not just a result of our flawed measuring techniques? From what i understand, when we probe a particle with a photon of a given wavelength, we change it's momentum (trajectory). The momentum then becomes a wave (probability) function, so we don't know exactly how much the momentum was changed, hence the uncertainty. If we probe the particle with a more charged photon (lower wavelength), the particle's position can be determined with greater accuracy, but it's momentum was changed even more (still not known for exactly how much). Now this experiment only proves the lack of our measuring abilities.

I've also found this youtube link http://www.youtube.com/watch?v=KT7xJ0tjB4A" with the laser through a single slit experiment. Isn't this simple diffraction? So is Heisenberg's principle just a way to describe how we see the physical model or is it a property of nature?

Sorry for the long explanation, i just wanted to give some context to my questions. Thank you for any help. Cheers, Val

Last edited by a moderator: Apr 26, 2017
2. Oct 10, 2011

### atyy

It is not a question of our experimental incompetence.

It states that exact values of canonically conjugate position and momentum do not simultaneously exist, so they cannot be simultaneously measured.

This guy says it more properly than me:

Last edited by a moderator: Sep 25, 2014
3. Oct 10, 2011

### dextercioby

Surely they exist, surely they can be simultaneously measured, it's just that by performing simultaneous measurements on an infinity of identically prepared systems, there's not just a unique value measured for x and one for p, respectively, but at least 2 of them (one can't say how many, though), so that the inequality containing $\hbar$/2 in the rhs be satisfied.

4. Oct 10, 2011

### tom.stoer

The Heisenberg uncertainty principle (HUP) is not a problem due to insufficient precision of measuring decives, but it is a fundamental principle of nature which "exists" in nature w/o any measurement, interaction etc. It follows from the abstract mathematical properties of quantum mechanics (Hilbert space geometry).

But I think you cannot prove this statement using a measurent!

What the "experimental explanations" do (and this is misleading) is to "explain" the HUP in terms of limited accuracy of measurements. Now one can argue that this "explanation" is wrong b/c of the above mentioned fundamental properties of qm, but one cannot prove this mathematical statement using measurements.

Think about a red ball. Think about a measuring device which is able to detect red or green objects "in principle". How can you convince me that 'a ball you have detected is red b/c it is red' and not b/c 'the device only says that it's red even if it's green'? You have to convince me that the device works correctly, i.e. that there is no single detection where it says 'red' in case of a green ball. But the HUP claims that whatever you do with the device and independent from the balls you are using in the experiment the outcome will always be 'red' b/c all balls are red. That means you cannot make one single experiment that proves that the device works with arbitrary high precision and is able to say 'green'. It will never say 'green'.

The conclusion of most scientists is to believe in Heisenberg, forget about the device and believe in all balls being 'red'. But logically there is no problem in believing in some green balls and some incorrect measurements.

5. Oct 10, 2011

### atyy

How can you simultaneously accurately measure position and momentum?

6. Oct 10, 2011

### dextercioby

As far as I'm aware, there's an example in Ballentine's book on quantum mechanics inspired from his famous 1970 RMP article.

7. Oct 10, 2011

### atyy

I believe Ballentine was wrong. Although he does assign simultaneous position and momentum, the position and momentum he gets are not canonically conjugate.

8. Oct 10, 2011

### kith

This is not a question of measurement, but a question of interpreting measurement results. In the statistical interpretation, quantum mechanical states describe ensembles of systems and not single systems themselves. In such an interpretation, the HUP is a statement about these ensembles of systems and there is nothing to prevent you from knowing the exact position and momentum of a single particle.

This is independent from Ballentine's claim, which was discussed in the other thread.

9. Oct 10, 2011

### atyy

Hmmm, what is the statistical interpretation? I thought it meant that only ensembles have meaning, so the momentum of a single particle doesn't make sense, let alone its momentum and position?

10. Oct 10, 2011

### kith

Well, there is the minimal statistical interpretation, which is roughly what you describe. However, additional assumptions can be made. An ensemble of particles with position distribution ψ(x) and momentum distribution ψ(p) can't be experimentally distinguished from an ensemble of particles with individual wavefunctions ψ(x).

11. Oct 10, 2011

### atyy

So to each particle, one assigns two wavefunctions (ψx(x),ψp(p))?

12. Oct 10, 2011

### kith

No, the wavefunctions are obtained in the usual way and are related by the fourier transform.

The minimal interpretation says: when we measure an ensemble, we get the predicted outcome. We don't make statements about the reality of single systems.

A possible extended version says: when we measure an ensemble, we get the predicted outcome. This is because the systems in our ensemble have well-defined properties λ at all times, distributed according to <λ|ψ>.

13. Oct 10, 2011

### atyy

Hmm, does that lead to a unique assignment of position and momentum for each member of the ensemble?

14. Oct 10, 2011

### kith

Yes.

15. Oct 10, 2011

### atyy

Does it mean "identically" prepared systems don't have identical properties?

16. Oct 10, 2011

### kith

In terms of such an interpretation, you don't prepare individual systems. You prepare ensembles.

17. Oct 10, 2011

### atyy

18. Oct 10, 2011

### eaglelake

Consider an interference experiment where a particle has passed through a slit, or a system of slits, and is detected on a distant screen. We measure the scattering angle $$\theta$$.
Then we know simultaneousely the values of both position and momentum: $$y = L\tan \theta$$, where $$L$$ is the distance from slits to screen and $$p_y = p\sin \theta$$.
The accuracy of the measurements is as good as allowed by the instruments used to measure the scattering angle $$\theta$$.
Best wishes

19. Oct 10, 2011

### kith

Sneaky? Although few phycisists stick to it, the ensemble interpretation is pretty standard. The version you are interested in was first advocated by Einstein and more recently by Ballentine (see his 1970 paper).

An extreme version of the ensemble interpretation was proposed this year by Smolin:
http://arxiv.org/abs/1104.2822

20. Oct 11, 2011

### atyy

The assigned position and momentum are not canonically conjugate.