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[OT[: Heisenberg Uncertainty Principle

G

Guy Macon

Jan 1, 1970
0
There is an interesting thread that I started in sci.physics with
the title "Heisenberg Uncertainty Principle." Think of it as a
non-expert getting a second opinion... :)
 
J

Jonathan Kirwan

Jan 1, 1970
0
There is an interesting thread that I started in sci.physics with
the title "Heisenberg Uncertainty Principle." Think of it as a
non-expert getting a second opinion... :)

I read it. So, have you learned anything from it?

Jon
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Guy said:
There is an interesting thread that I started in sci.physics with
the title "Heisenberg Uncertainty Principle." Think of it as a
non-expert getting a second opinion... :)

You have told us where your article is, but not when you posted it. ;-)
 
J

Jonathan Kirwan

Jan 1, 1970
0

It would be interesting to know exactly what. I'll be more generous about my
reply than you:

I noted Gregory's response was congruent to the one I chose to write. Old Man
got into some other details (but not incompatible ones, I believe.) If you are
interested in the math, I'd recommend "Basic Quantum Mechanics" from the Oxford
Physics Series. And for a general comment on what HUP means here, see pages 22
and 23. Another place to go is Feynman undergrad Lectures on Physics, vol III,
chapter 2, page 2-2ff. (And most of the rest of that volume, as well.)

There are two different 'uncertainty' concepts being discussed, by the way. One
is the uncertainty principle itself, which can be seen as a necessary outcome of
applying the correspondence principle to Newtonian and quantum mechanics. The
other is the uncertainty relation. It is here that is found the uncertainty
inequality I mentioned, which is imposed because for any real numbered
wavelength, the squared length vector is always real and non-negative. The
quadratic solution then forces the important further restriction on the
uncertainties of two observables (X and Y) for the same state. Here, generally,
if the two observables do not commute (XY-YX is non-zero), then the uncertainty
of X and the uncertainty of Y cannot both be made arbitrarily small
__simultaneously__.

Jon
 
K

Kevin Aylward

Jan 1, 1970
0
Jonathan said:
It would be interesting to know exactly what. I'll be more generous
about my reply than you:

I noted Gregory's response was congruent to the one I chose to write.
Old Man got into some other details (but not incompatible ones, I
believe.) If you are interested in the math, I'd recommend "Basic
Quantum Mechanics" from the Oxford Physics Series. And for a general
comment on what HUP means here, see pages 22 and 23. Another place
to go is Feynman undergrad Lectures on Physics, vol III, chapter 2,
page 2-2ff. (And most of the rest of that volume, as well.)

There are two different 'uncertainty' concepts being discussed, by
the way. One is the uncertainty principle itself, which can be seen
as a necessary outcome of applying the correspondence principle to
Newtonian and quantum mechanics. The other is the uncertainty
relation. It is here that is found the uncertainty inequality I
mentioned, which is imposed because for any real numbered wavelength,
the squared length vector is always real and non-negative. The
quadratic solution then forces the important further restriction on
the uncertainties of two observables (X and Y) for the same state.
Here, generally, if the two observables do not commute (XY-YX is
non-zero), then the uncertainty of X and the uncertainty of Y cannot
both be made arbitrarily small __simultaneously__.

I will have to disagree here. My Bible is the Ballentine text book. In
this book it explains in detail, that HUP isn't even about simultaneous
measurements at all.

"Quantum Mechanics, A Modern Development" ISBN981-02-4105-4 - Leslie E.
Ballentine,

p.225
"...One must have a repeatable preparation procedure corresponding to
the state p which is to be studied. Then on each one of a large number
of similarly prepared systems, one performs a single measurement (either
Q od O). The statistical distributions of the results are shown as
histograms, and the root mean square half-widths or the two
distributions deltaQ and deltaP, are indicated in fig. 8.2. The theory
predicts that the product of these two half-widths can never be less
then hbar/2, no matter what state is considered."

The issue with much of what is written about HUP, is that it is usually
based on un-rigorous definitions, and hand waving.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
G

Guy Macon

Jan 1, 1970
0
Jonathan Kirwan said:
It would be interesting to know exactly what. I'll be more
generous about my reply than you:

I was terse for four reasons:

[1] I couldn't tell whether what you wrote was an honest question
or a subtle put-down.

[2] I believe that any recounting of what has been posted in the
sci.physics thread by a non-expert such as myself will be
inferior to reading the thread itself.

[3] I am asking questions, not providing answers.

[4] Those who are providing answers appear to contracdict each other.

..if the two observables do not commute (XY-YX is non-zero), then
the uncertainty of X and the uncertainty of Y cannot both be made
arbitrarily small __simultaneously__.

In the case of the position and velocity of a single subatomic
particle, do the two observables commute? Is XY-YX zero in
that case?
 
K

Kevin Aylward

Jan 1, 1970
0
N

Norm Dresner

Jan 1, 1970
0
Guy Macon said:
In the case of the position and velocity of a single subatomic
particle, do the two observables commute? Is XY-YX zero in
that case?

We live in a 3-D space. That means that there are three independent,
mutually perpendicular coordinate axes that you can construct at a point.
Actually there are an infinite number of these triads, all related by
various rotations. Choose any one of them. The HUP implies that for each
of these directions individually, the components of position and velocity of
a single particle do not commute. That is, for, say, an electron, the
position along the, say, X-axis and the velocity component along the X-axis
are non-commuting observables. Same with Y-axis and Z-Axis. But the
inter-axis variables are commuting. That is, you can simultaneously measure
the positions on the three coordinate axes with unlimited precision -- or
the three velocities; or the X-position and the Y-velocity, or ... It's the
corresponding velocities and positions that don't commute and hence are
limited by the HUP.

Norm
 
K

Kevin Aylward

Jan 1, 1970
0
Norm said:
We live in a 3-D space. That means that there are three independent,
mutually perpendicular coordinate axes that you can construct at a
point. Actually there are an infinite number of these triads, all
related by various rotations. Choose any one of them. The HUP
implies that for each of these directions individually, the
components of position and velocity of a single particle do not
commute.

The observable operaters, X an P don't commute, this is not the same.
According to the ensemble interpretation, X and P only have relevance
statistically, not to individual particles, so HUP says nothing about
individual particles.
That is, for, say, an electron, the position along the,
say, X-axis and the velocity component along the X-axis are
non-commuting observables. Same with Y-axis and Z-Axis. But the
inter-axis variables are commuting. That is, you can simultaneously
measure the positions on the three coordinate axes with unlimited
precision -- or the three velocities; or the X-position and the
Y-velocity, or ... It's the corresponding velocities and positions
that don't commute and hence are limited by the HUP.

I think you have missed the point of the thread. Its about the *correct*
interpretation of HUP. The summary is here,
http://www.anasoft.co.uk/quantummechanics/index.html.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
N

Norm Dresner

Jan 1, 1970
0
Kevin Aylward said:
The observable operaters, X an P don't commute, this is not the same.
According to the ensemble interpretation, X and P only have relevance
statistically, not to individual particles, so HUP says nothing about
individual particles.

The observables X & P for a single particle are non-commuting.
I think you have missed the point of the thread. Its about the *correct*
interpretation of HUP. The summary is here,
http://www.anasoft.co.uk/quantummechanics/index.html.

I know the correct interpretation of HUP. I studied it in graduate school.

Norm
 
J

Jonathan Kirwan

Jan 1, 1970
0
I will have to disagree here. My Bible is the Ballentine text book. In
this book it explains in detail, that HUP isn't even about simultaneous
measurements at all.

"Quantum Mechanics, A Modern Development" ISBN981-02-4105-4 - Leslie E.
Ballentine,

Well, Ballentine does have a perspective. Excellent book, though, I suspect.

I think this whole thing may be leading us right into the EPR/Bohr debates,
Bohm's early thinking on what then led to Bell's inequality, John Clauser's
non-definitive experiments to test Bell's at Berkeley, later Aspect's much more
conclusive work, and perhaps leading to the re-unearthing of the many-worlds
viewpoint.

It's my understanding that whether you choose to give up on locality, or the
reality principle, or choose 'many world's as your forte', it's all the same
thing -- 1:1 correspondences, regardless of the insight you choose. The
mathematics may be easier for one problem or other, one choice versus another,
but there is nothing yet to distinguish their predictions. Wheeler may lead you
one way, Everett another, but so far as I'm aware none wins over the other.

More specifically, position and momentum are fundamentally incompatible
observables, in the sense that they do not share a complete orthogonal set of
eigenvectors and therefore cannot be simultaneously measured. If you measure
one, the system goes into some eigenstate. Then measure another, that into yet
another eigenstate. But since they __do not__ commute (observables X and Y have
such a complete set of orthogonal eigenvectors if and only if they commute),
they are fundamentally incompatible and cannot be measured simultaneously. So
what is the meaning, then, of making measurements arbitrarily small? It simply
cannot be done.

That was the question I'd been responding to, when I talked about the viewpoint
of these kinds of dynamic attributes, which are just two faces or facets of some
higher dimensioned volume.

The inequality I mentioned resides at the heart of a debate over interpretation
about the condition of a particle *before* measurement. But there is no way to
measure both position and momentum simultaneously, as they do not commute. One
could say it's a meaningless question and then just rely on the statistical
interpretation. One could suggest that there is a deeper, reality position that
implies more (a path I like to take, at times.) So I think this leads to
cross-purposes types of answers -- and perhaps that is my fault alone. I can't
say, for sure.

Been interesting,
Jon
 
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