[extropy-chat] FWD (SK) Heisenberg's Uncertainty Principle

[scerir]

From: "Alfio Puglisi"

> The net result is that, if the particle has any interaction 
> whatsoever with the rest of the world, its characteristics 
> are subject to uncertainty. And, if it doesn't interact, 
> there's no way to know its characteristic, so they are totally
> indeterminate :) you can't win.

The "weak measurement" technique, proposed by Aharonov (yes, that
Aharonov-Bohm effect) exploits quantum uncertainty. Aharonov's 
quantum detectors are so weakly linked to the experiment that any 
measurement moves the detector's "pointer" by less than the level 
of uncertainty. This also means "superpositions" are preserved.
There is a price to pay for these "delicate" readings: they are
inaccurate. But while this might appear to make the whole process 
pointless, Aharonov has calculated that when repeated many times, 
the average of these measurements approximates to the "true" value 
of the thing being measured.

A. Steinberg on weak measurements.
general paper
http://www.physics.utoronto.ca/~aephraim/Wheeler=0302003.pdf
lectures & presentations (many, long, and slow) here
http://www.physics.utoronto.ca/~aephraim/aephraim.html
experiments
http://xxx.lanl.gov/abs/quant-ph/0310091

Needless to say the quantum measurement/quantum weak measurement 
issue has to do with the irreversibility/reversibility issue, 
the collapse/superposition issue, and even with the consciousness/
unconsciousness issue of quantum automata, that is to say with
the D.Z.Albert/A.Peres debate on the weird behaviour of quantum
automata.

--------------
A quantum joke
--------------

Let us start from the most general question, which says:
is it possible to assign *definite* values of observables
to individual events?

Our usual *assumption* is that the result of the measurement
of a certain operator A depends only on the state ("psi")
of the quantum system we are measuring, and nothing else.

Let us consider (for simplicity, but the conceptual argument
does not need entanglements, necessarily) two spin 1/2 particles,
particle a and particle b, entangled in a singlet state.

We can measure   s(a,x)       = +/- 1      
we can measure   s(a,x)s(b,x) = -1
we can measure   s(a,y)s(b,y) = -1
we can measure   s(a,z)s(b,z) = -1

For a singlet state it is true that
[ s(a,x)s(b,y) , s(a,y)s(b,x) ] = 0

Thus it is possible to measure s(a,x)s(b,y) and s(a,y)s(b,x)
with no reciprocal disturbance.

We can write then
s(a,x)s(b,y)s(a,y)s(b,x) =
= s(a,x)s(a,y)s(b,y)s(b,x) =
= s(a,z)s(b,z) = -1

Thus s(a,x)s(b,y)s(a,y)s(b,x) = -1

Now let us *assume* that in s(a,x)s(b,y)
s(a,x) does not depend on s(b,y) and viceversa,
and let us assume that in s(a,y)s(b,x)
s(a,y) does not depend on s(b,x) and viceversa.

("Does not depend" just means that we can measure
s(a,x) and s(b,y) *separately*, and also
we can measure s(a,y) and s(b,x) *separately*).

With the above *assumption* we have that
s(a,x)s(b,y)s(a,y)s(b,x) = -1
which is in contradiction with
s(a,x)s(b,x) = -1
s(a,y)s(b,y) = -1

The above *assumption* must be wrong.

It is wrong (in general, but not when the "psi" of the quantum
system is an eigenstate of the operator A) our usual assumption
that the result of the measurement of a certain operator A
depends only on the state ("psi") of the quantum system
we are measuring, and nothing else.