Is Heisenberg uncertainty principle a problem of our measuring techniques?

[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

 

[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:

 

[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.

 

[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.

 

[atyy]

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.

How can you simultaneously accurately measure position and momentum?

 

[dextercioby]

How can you simultaneously accurately measure position and momentum?

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

 

[atyy]

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

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

 

[kith]

How can you simultaneously accurately measure position and momentum?

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.

 

[atyy]

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.

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?

 

[kith]

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?

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).

 

[atyy]

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).

So to each particle, one assigns two wavefunctions (ψ_x(x),ψ_p(p))?

 

[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>.

 

[atyy]

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>.

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

 

[kith]

Yes.

 

[atyy]

Yes.

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

 

[kith]

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

 

[atyy]

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

Sneaky! Ok, point me to some good references so I can read about this stuff.

 

[eaglelake]

How can you simultaneously accurately measure position and momentum?

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

 

[kith]

Sneaky! Ok, point me to some good references so I can read about this stuff.

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

 

[atyy]

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

The assigned position and momentum are not canonically conjugate.

 

[atyy]

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

Thanks!

 

[Runner 1]

Perhaps take a look at my post I wrote a few days ago:

https://www.physicsforums.com/showpost.php?p=3544398&postcount=46

 

[Fredrik]

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

There's an example in the 1970 article. Was it also included in the book? I tried to find it there a few months ago, but was unsuccessful. (I just quickly flipped through the pages of the sections where I thought it might appear, and didn't see anything that looked like it).

Regarding the example in the 1970 article, there was some debate about it in that thread from a few months ago, but we were unable to determine if that experiment should be considered a momentum measurement or not. atyy argued that it shouldn't. I don't understand why, but I still haven't read the articles that inspired his thoughts on the matter.

 

[Fredrik]

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

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

Since the preparation procedure is applied to an individual system, I would answer atyy's question with "yes". (This is assuming that we're talking about the statistical interpretation described by Ballentine's 1970 article, according to which all particles have well-defined positions...even when their wavefunctions are spread out over some region of space). For example, the preparation may consist of sending a particle through a small hole. This preparation is represented by a wavefunction that's more or less constant over the hole and rapidly falls to zero at its edges. So in Ballentine's interpretation, one particle that's subjected to that preparation procedure may end up in the "left" part of the hole while the next one may end up in the "right" part of the hole. They wouldn't have the same properties, in spite of their identical preparations. Since the wavefunction represents the preparation (actually it represents an equivalence class of preparations), they would still have the same wavefunction.

This would imply that wavefunctions don't represent the properties of the individual systems, but that may very well be true even if particles don't have well-defined positions.

 

[dextercioby]

The assigned position and momentum are not canonically conjugate.

Can you prove that ?

 

[Demystifier]

Regarding the example in the 1970 article, there was some debate about it in that thread from a few months ago, but we were unable to determine if that experiment should be considered a momentum measurement or not. atyy argued that it shouldn't. I don't understand why, but I still haven't read the articles that inspired his thoughts on the matter.

I haven't seen the debate on this forum, but I can present a simple reason why the Ballentine's experiment is NOT a measurement of momentum p_y. The point is that in this experiment p_y is NOT MEASURED but CALCULATED. Measurement and calculation are not the same. A calculation always contains an additional theoretical assumption which a true measurement does not need to use.

More specifically, p_y is calculated as
p_y = p sin theta
and this equation is correct only if one ASSUMES that the particle has been moving along a definite straight trajectory and with a constant momentum before it hitted the detector (at position y = L tg theta). However, since such a trajectory has not been measured, there is no theory-independent justification for such an assumption.

If, on the other hand, one accepts the Ballentine's experiment as a valid measurement of momentum, then it is equivalent to an acceptance of the idea that particles may have trajectories even when they are not measured. That's fine, as long as one is aware that we do not have a direct experimental confirmation that this idea is correct. Indeed, the trajectory tacitly assumed in the Ballentine's experiment exactly coincides with the Bohmian trajectory. It is well known that Bohmian trajectories are consistent with QM, and yet that a direct experimental verification of their reality does not exist.

So, loosely speaking, one could say that the Ballentine's measurement of p_y assumes that the Bohmian interpretation is right, even if he is not aware of it.

 

[Fra]

I haven't seen the debate on this forum

I think it was this thread: https://www.physicsforums.com/showthread.php?t=516224&page=18

/Fredrik

 

[Demystifier]

I think it was this thread: https://www.physicsforums.com/showthread.php?t=516224&page=18

/Fredrik

Thanks, now I have linked my post above in that thread as well.

 

[atyy]

I think it was this thread: https://www.physicsforums.com/showthread.php?t=516224&page=18

/Fredrik

Thanks for the link!

Can you prove that ?

Not personally. My understanding is based on http://tf.nist.gov/general/pdf/1283.pdf and http://www.mpq.mpg.de/qdynamics/publications/library/Nature395p33_Duerr.pdf . The position at "infinity" reflects the momentum immediately after the particle has passed through the slit.

There are elements of Demystifier's argument in post #26 which I think should also come into play, since my naive expectation is that if the position measurement at a late time measures the momentum at an earlier time, then the collapse should be to the early time momentum eigenstate, rather than a late time position eigenstate.

 

[Demystifier]

The assigned position and momentum are not canonically conjugate.

Can you prove that ?

Not personally.

Isn't the proof trivial?
Assume that they are canonically conjugate. Then, by definition,
[x,p]=i hbar
There is a well known theorem saying that this commutation relation implies
Δx Δp ≥ hbar/2
But this is in contradiction with the result that Δx Δp may be arbitrarily small, which implies that the assumption above is not true. Hence, they are not canonically conjugate. Q.E.D.