Violation of Heisenberg’s Measurement-Disturbance Relationship

http://prl.aps.org/abstract/PRL/v109/i10/e100404

by soupboy • 14 years ago

54 11 14 years ago

prl.aps.org

fjorder• 14 years ago

This is a really well written paper. I especially like their review of Heisenberg's original uncertainty relation, which had already been shown to be incorrect in theory under certain conditions, and the modern uncertainty principle that still holds. (They then experimentally violate the former, not the latter.)

Why should anyone care? Well, some security proofs for quantum cryptographical systems out there could conceivably assume the original relation holds, and would therefore require correction. Not to go into too much detail, this might have an impact on the bit-rate that some quantum crypto systems can securely run at.

Daniel_Newby2• 14 years ago

I don't have free access to the paper. Could you tell me if the measurement-disturbance violation they show is a version of the quantum nondemolition measurement approach?

cgs1019• 14 years ago

Here's the preprint: http://arxiv.org/pdf/1208.0034.pdf

Daniel_Newby2• 14 years ago

Here's a more accessible article from the Beeb: http://www.bbc.co.uk/news/science-environment-19489385

Xcelerate2• 14 years ago

(I posted this in another thread, but nobody read that person's submission so I'm reproducing it here. The BBC article made this much more sensationalistic than it was, implying the Uncertainty Principle had been overthrown):

Nobody has cast doubt on anything in modern quantum mechanics. The abstract makes this much clearer than the BBC article does:

"While there is a rigorously proven relationship about uncertainties intrinsic to any quantum system, often referred to as “Heisenberg’s uncertainty principle,” Heisenberg originally formulated his ideas in terms of a relationship between the precision of a measurement and the disturbance it must create. Although this latter relationship is not rigorously proven, it is commonly believed (and taught) as an aspect of the broader uncertainty principle. Here, we experimentally observe a violation of Heisenberg’s “measurement-disturbance relationship”, using weak measurements to characterize a quantum system before and after it interacts with a measurement apparatus. Our experiment implements a 2010 proposal of Lund and Wiseman to confirm a revised measurement-disturbance relationship derived by Ozawa in 2003. Its results have broad implications for the foundations of quantum mechanics and for practical issues in quantum measurement."

In other words, Heisenberg originally thought that the inability to measure two incompatible observables (like momentum and position) was because of something that is now called the "observer effect": http://en.wikipedia.org/wiki/Observer_effect_(physics) The observer effect applies even to classical systems without quantum mechanics! You just can't measure something without affecting it in some way [1].

The modern version of Heisenberg's uncertainty principle says nothing about measurement disturbing the system. It instead says that there is an inherent uncertainty to the system. I've been seeking clarification on what exactly this means for a long time and have never really gotten a satisfactory answer. I've tried Physics Stack Exchange, my QM professor, even Hacker News and most people either confuse HUP with the observer effect or they reply in illy-defined terms that don't help me any.

The best that I can figure out on my own -- the true HUP -- is thus: You can prepare a system in a certain state. You then take a position measurement and a momentum measurement at the same time. You get two real numbers. Now, you repeat the experiment. Create a system just like you did before and take measurements of r and p. After multiple repetitions of the same experiment over and over, you'll get two long lists of position and momentum measurements. Take the standard deviation of the position measurements, multiply that value by the standard deviation of the momentum measurements, and HUP guarantees that the resulting real number will be ≥ ħ/2 (reduced Planck's constant over 2).

[1] Well, there's a few loopholes if anyone is curious.

shardling• 14 years ago

The general HUP has to do with the eigenstates of noncommuting operators. I feel like you've probably seen this before, but one more hand waving try! :)

In QM the state of a system will be a solution to Schroedingers equation for that system. The system is characterized by a Hamiltonian H.

If an operator commutes with H, then we can find a basis for the solution space such that every basis state is an eigenstate of that operator. (i.e. it has a definite eigenvalue for that operator; applying the operator to an eigenstate gives back the state times a constant.)

If two operators commute, we can even find a basis such that that each basis state is an eigenstates of both operators.

But if they don't commute, this is impossible.[1] That means that if a basis state has a definite eigenvalue for one operator, it cannot for the other.

What does it mean for a particle to have a well defined momentum? That means it is an eigenstate of the momentum operator. Likewise for position, and the position and momentum operators don't commute. This shows that if a particle is in a state of well-defined momentum, it must be a mixture of states in the basis for position.

With more work you can translate this into a well-defined lower bound on the product of the uncertainties of the two operators.

>HUP guarantees that the resulting real number will be ≥ ħ/2 (reduced Planck's constant over 2).

Well, not quite. You're making an observation of the standard deviation, and it might not accurately measure the actual standard deviation.

So we know that if enough measurements are taken, the measured value will converge to a number greater than /hbar / 2. That's different than guaranteeing that any particular such measurement will obey this.

Xcelerate2• 14 years ago

You know, I think this is the best explanation I've seen yet. I've done plenty of calculations for my classes on which operators commute and which don't. And if [A, B] ≠ 0, then the two are incompatible observables. And I already knew the mathematics behind the eigenstates/eigenfunctions.

But I think the key insight you provided to me is what most people mean when they say "well-defined" -- that the definition for "certainty" implies multiple measurements.

In other words, I was thinking: okay, you can measure a particle's position and momentum. You'll get real numbers for both to as much precision as your equipment allows. And I assumed that this was the definition of position and momentum -- the instantaneous values you get for the two measurements are defined to be the particle's position and momentum at time t. So I didn't understand why everyone was saying you can't find both simultaneously -- from my definition, of course you can find them both!

But from the definition you've given me, I am guessing most other people define position and momentum as consistent quantities, meaning they have to have the same values upon each new measurement of the system in order to be called "position" or "momentum". Is this what you're saying? If so, that makes a lot of sense.

But I would think textbooks would clarify that more, because there are a lot of people who, like me, get confused by this type of wording. You can measure a car's instantaneous speed and that is THE speed at time t. Not "let's measure an identically prepared car multiple times and call the average of its dx/dt measurements 'speed".

With regard to your last point about the standard deviations product always being greater than h_bar / 2, I realized my fallacy after I posted it, but didn't think anyone would catch it. Good eye for subtlety!

wisty• 14 years ago

As a loose analogy, could you say:

People think that your calculator can only show 12 significant figures because it's only got a small screen. But, the actually numbers it works with are also limited, and you can get even weirder things happening due to its internal representation being inexact.

That's only an analogy for the sort of confusion people have, not an analogy for how quantum physics actually works.

wamatt2• 14 years ago

> The observer effect applies even to classical systems. You just can't measure something without affecting it in some way [1].

What I have never understood about the observer effect, is how it applies to the observation of blackbody radiation.

For example say we observe the photons emitted from the sun discreetly on earth, how does that alter the state of the sun itself?

zmmz• 14 years ago

You have not disturbed the sun. You have disturbed the photons.

You disturb the thing that you are measuring, not the source source "higher up the foodchain".

Violation of Heisenberg’s Measurement-Disturbance Relationship

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