

Violation of Heisenberg’s Measurement-Disturbance Relationship - soupboy
http://prl.aps.org/abstract/PRL/v109/i10/e100404

======
fjorder
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_Newby
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
Here's the preprint: <http://arxiv.org/pdf/1208.0034.pdf>

------
Xcelerate
(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
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.

1\. If a state |n> is an eigenstate of two operators A and B, such that A|n> =
_a_ |n> and B|n> = _b_ |n> ( _a_ and _b_ are eigenvalues), then necessarily
AB|n> = BA|n> = ab|n>. This implies that [A, B]|n> = 0 which generally
contradicts the idea that A and B don't commute; certainly when the commutator
is just a constant.

~~~
Xcelerate
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!

