# How to devise an experiment that violates uncertainty principle

1. Jan 28, 2015

### Palpatine

Lets suppose device A measures the position of a particle very accurately. Device B now cannot measure its momentum to high accuracy due to the uncertainty principle.

But lets suppose that neither A nor B can ever communicate their findings to the rest of the world. Now the uncertainty principle can be violated because no observer can ever put the two pieces of information together and know more about the state of the particle than is usually allowed.

So how does one devise an experiment such that a scientist knows that device A measured a particle's position very accurately and that device B measured its momentum very accurately but without being able to know the actual values recorded by both.

2. Jan 28, 2015

### phinds

You misunderstand the HUP. Observation has nothing to do with it, although that canard is understandable since Heisenberg himself started it with what turned out to be a mistaken way of stating his own principle.

Also, FYI, there is significant contention, even on this forum, as to whether or not it is in fact possible to simultaneously measure conjugate pairs such as position/momentum of quantum objects. There is clear agreement, however, that the HUP says that regardless of what you can measure, it is not possible to set up exactly the same starting conditions for a quantum object and then get the same subsequent results for measurements. In the classical, deterministic, world, exactly the same starting conditions lead to exactly the same results. This is not true in the quantum world, which is probabilistic.

EDIT: Oh, and by the way ... if you ever create an experiment that truly violates the HUP, I can pretty well guarantee you a Nobel Prize since you will have discovered completely new science.

3. Jan 28, 2015

### bhobba

That's not what the uncertainty principle says. It places no limitation on the accuracy you can measure anything. Its a statistical statement about measurements on large ensembles of systems.

The exact statement is suppose you have a large number of similarly prepared systems. Divide it into two lots. In the first lot you measure position. You can get an answer you want for each measurement as accurately you like. Then in the second lot you measure momentum, and again you get an answer as accurately as you like. But when you compare the standard deviations of the results (ie their statistical spread) they are as per the Heisenberg uncertainty relations.

Your confusion however is understandable since some (possibly even most) popularisations and beginner textbooks are rather careless in explaining it. The explanation I gave above is exactly what you find in my go-to textbook Ballentine.

Conciousness, communicating results etc has nothing to do with it - you could imagine the experiment done by robots, analysed by a computer, and the result stored in memory then looked at later. It makes no difference. This conciousness stuff you read in popularisations is a very backwater idea these days - and it never really caught on. In the standard Copenhagen interpretation QM is a theory about observations that appear here in a common-sense classical world of everyday experience that exists regardless of if you or anyone observes it or not. The issue with QM is how a theory that assumes such a world explains it - but that is another story.

And Phinds is correct - according to QM you can't violate the uncertainty principle. It follows from the very foundations of the theory. If it's wrong QM is wrong. That of course may be true, that all knowledge is provisional is the hallmark of science and can be overthrown at any time by experiment, but such would be an earth shattering revelation revolutionising our understanding of the world - they happen very very rarely.

Thanks
Bill

Last edited: Jan 28, 2015
4. Jan 28, 2015

### jfizzix

The modern formulation of the uncertainty principle is that you can never prepare a particle to be in such a state that you can reliably predict the outcome of both a position measurement, and a momentum measurement.

Alternatively, this can also be stated as: no matter what sort of measurement you do, the remaining position uncertainty plus the remaining momentum uncertainty will always be above some positive constant value.

This can also be formulated in terms of information: the more a single measurement reduces your uncertainty in the position of a particle, the less that measurement can reduce your uncertainty in the momentum of a particle, and vise versa.

Hypothetically there are measurements that give you a little bit of information about the position of a particle, while allowing you to then gather a lot of subsequent information about the momentum of a particle; you just can't can't gather lots of information about both in a single measurement.

5. Jan 29, 2015

### vanhees71

For a kind of "violation" of the Heisenberg-Robertson uncertainty principle, just have a look at:

http://phys.org/news/2015-01-popper-againbut.html

That there is no true violation of the HRUP would mean that quantum theory itself is proven wrong. That would for sure be worth a Nobel prize, but unfortunately for the researchers that's not the case, as is written in the phys.org article and also in their paper.

6. Jan 29, 2015

### Demystifier

This is not possible in quantum mechanics, essentially because both devices measure the same object. Nevertheless, there is a speculation that something similar might be possible in quantum gravity when an outside and inside observer measure the black hole:
http://en.wikipedia.org/wiki/Black_hole_complementarity

Last edited: Jan 30, 2015
7. Jan 29, 2015

### jk22

You could take the epr argument : 2 particles fly away at the same unknown opposite momentum. If you measure position of a you can predict with certainty position of b since it is entangled. The same for momentum measured on b. Hence you know exactly momentum of b and its position by deduction. I suppose this is what was meant.

8. Jan 29, 2015

### bhobba

At the same time? That's some trick and would violate basic principles of QM.

I suggest you detail the exact set-up for some of the experimental guys that frequent here to peruse - if it stands up a Nobel prize is likely yours.

The uncertainty principle places no limitation of the accuracy of either, but it does say you cant measure both exactly and simultaneously - that's because no state is simultaneously an eigenstate of position and momentum - entangled or not.

Doing an observation breaks the entanglement and if what you suggested occurred it would mean it was in an eigenstate of both position and momentum which since that's impossible means you cant do it.

Thanks
Bill

Last edited: Jan 29, 2015
9. Jan 29, 2015

### vanhees71

No it's not a trick, and it doesn't violate anything within QT. The short version of the resolution of this seemingly paradox statement is that the uncertainty principle is about a single particle. But here we deal with two particles, and there single-particle observables of different particles always commute. So you can determine Particle 1's momentum and indpendently from Particle 2's position etc. In addition there's a momentum-position entanglement for two particles originating from the decay of another particle, which has been considered by EPR in their original paper.

It was just in the news for photons (here, the issue with the photon's position is more subtle, but just take photon-detection-event position, and then it's correct):

http://phys.org/news/2015-01-popper-againbut.html

10. Jan 29, 2015

### bhobba

Interesting.

I am however unsure exactly what its measuring - from what I can see its measuring correlations rather than individual properties.

Thanks
Bill

Last edited: Jan 29, 2015
11. Jan 29, 2015

### jk22

In fact it is simply that after the measurements one particle is in a position eigenstate and the other in momentums so we cannot deduce the position of the second even if we know the first's. We could think that this breaks a conservation law though since total momentum should be 0 ?

12. Jan 29, 2015

### bhobba

But isn't entanglement broken so it's rather moot anyway? By this I mean once one actual measurement occurs entanglement no longer exists so measuring the second particle no longer correlates

Thanks
Bill

13. Jan 29, 2015

### jk22

Yes or in other words the system is not closed and the interaction with the measurement apparatuses could bring energy or momentum to it...

14. Jan 29, 2015

### jk22

In fact this is true for quantum but if we were classical ?

15. Jan 29, 2015

### bhobba

Then, since entanglement is a quantum feature, you wouldn't have such strange correlations in the first place.

Thanks
Bill

16. Feb 3, 2015

### Mark Harder

I like to think of the effect of observation on uncertainties as one mechanism through which uncertainty is realized. But the principle itself is about the state of a quantum system, not necessarily its observation. The quantum system IS uncertain, with or without you. Suppose you confine a particle in a very narrow well with high walls. The particle's position is precisely given, therefore its momentum is very poorly delimited. Time and energy are also mutually uncertain. So, if some activated state has an extremely short lifetime, then the energy of that state is very poorly determined, independently of any measurement. This is the guess of an amateur but bear with me. How can a particle pair pop out of a vacuum? The mass-energy should be conserved and nothing should yield nothing, right? But the point I have in mind is the that particle pair exists for an extremely short time, so the energy of that bit of space is so poorly determined, that non-zero energy and zero energy states are both possible, in the extremely short term, that is. Of course, the particle pair will become a vacuum again, reversing their creation. Over the slightly longer time scale of this creation-annihilation event, zero energy has in fact remained zero energy!

17. Feb 3, 2015

### lutz kayser

Mark Hader is correct. But the notion that one parameter is precisely known and the other not is wrong. Neither can be precisely measured or is knowable.We can only determine the product of two conjugates approximately.

18. Feb 3, 2015

### Mark Harder

That's right. If the uncertainty in one variable is zero then the product of uncertainties would be zero! Which is too small, regardless of the value of the other factor.

19. Feb 4, 2015

### vanhees71

The Robertson-Heisenberg uncertainty relation for any pair of observables and for any pure or mixed state reads
$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$
where $\Delta A$ and $\Delta B$ are the standard deviations of the observables and $\hat{A}$ and $\hat{B}$ the representing self-adjoint operators.

It's a property of the states of the quantum system, i.e., a property due to state preparation. It has nothing to do with the disturbance by measurement. This implies the operational description of how to empircally check the uncertainty relation: As for any probabilistic notion you have to check it statistically, i.e., prepare many system independently in a given fixed state and then measure $A$ at an accuracy much higher than suggested by $\Delta A$ to significantly measure this standard deviation. With another equally prepared ensemble you repeart this procedure with $B$. You never measure $A$ and $B$ in some way simultaneously on one and the same system.

It's an open heavily discussed question, how to define the older (and wrong!) idea by Heisenberg about disturbance of a quantum system by measurement, who interpreted wrongly that the uncertainty relation is due to disturbance by measuring one observable on the statistical properties of the other. This wrong interpretation was corrected by Bohr immediately after Heisenberg's first paper, where he used the famous "Heisenberg microscope" on measuring a particles position and "disturbing" its momentum. Bohr clarified on this example that Heisenberg's interpretation is wrong and that the uncertainty relation refers to properties of quantum systems according to quantum theory and not to measurement-disturbance relations. If in doubt about interpretational issues it's better to trust Bohr than Heisenberg anyway ;-).