Three Prism Problem
ca314159
As a variation on the three polarizer / Stern Gerlach filter problem,
does the same principle apply in the case of prisms ?
http://www.bestweb.net/~ca314159/QM3.GIF
The Red/Green state is assumed to be in between the red and green
frequencies.
Do the shields prevent the quantum mechanical effect ?
Does replacing the white light source with a red light source prevent
the effect ?
--
rober...@citicorp.com
In article <3503E4...@bestweb.net>,
The implications indicated are that if the green light is present
in the PM1 case of three prisms (presumably with the shields removed)
and much less so in the case of a typical red light source (such as
a HeNe) then the degree of 'purity' of the frequency of the light
source can be determined by this means.
Also indicated is that the unpolarized light having passed through
a linear polarizer is not in dispersion free state (of course) but
that such dispersion free state (as Bell refers to them) are
possible as also implied by the Kwiat, Weinfurter, Zeilinger quantum
zeno experiments.
This might then seem to clinch the EPR debate in favour of
Einstein's position and provide a valuable model ?
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading
ca31...@bestweb.net
Answering the many questions put to the modelling of
a three prism analog of the three polarizer problem,
I have clarified many of the points in the comparison:
The general idea is that random polarization states
of single photons are dispersive through a single linear
filter just as random frequency states of white light are
dispersive through a prism.
http://www.bestweb.net/~ca314159/QMHUP.HTM
--
http://www.bestweb.net/~ca314159/
rober...@citicorp.com
In article <6e3f3l$era$1...@nnrp1.dejanews.com>,
rober...@citicorp.com wrote:
>
> In article <3503E4...@bestweb.net>,
> ca314159 <ca31...@bestweb.net> wrote:
> >
> > As a variation on the three polarizer / Stern Gerlach filter problem,
> > does the same principle apply in the case of prisms ?
> >
> > http://www.bestweb.net/~ca314159/QM3.GIF
> >
> > The Red/Green state is assumed to be in between the red and green
> > frequencies.
> >
> > Do the shields prevent the quantum mechanical effect ?
> >
> > Does replacing the white light source with a red light source prevent
> > the effect ?
> > --
> >
> > http://www.bestweb.net/~ca314159/
> >
>
> The implications indicated are that if the green light is present
"Green" light isn't derived from the "red" light.
The implication is that the variance of these macrostates
represents as dispersion which in the case of polarization
operator:
/delta A = A - <A> = 1 - cos(theta)
and:
</delta A^2> = <A^2> -<A>^2 = 1 - cos^2(theta)
defines the dispersion of the polarization states.
rober...@citicorp.com
Does sandwiching a yellow filter between a red and green amount to the
same thing ?
Jim Carr
ca31...@bestweb.net writes:
>
>The general idea is that random polarization states
>of single photons are dispersive through a single linear
>filter just as random frequency states of white light are
>dispersive through a prism.
>
> http://www.bestweb.net/~ca314159/QMHUP.HTM
Even a cursory reading of this spotted several problems, not the
least of which was a statement about a polarizer not changing the
polarization of a photon. You can send 100% linearly polarized
light onto a polarizer and get out light that is also 100% linearly
polarized (just not as intense) but with a different orientation.
There is no dispersion assuming ideal polarizers.
The comparable idea that a similarly ideal red filter would change
the color of photons does not make sense.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
ca314159
Jim Carr wrote:
>
> ca31...@bestweb.net writes:
> >
> >The general idea is that random polarization states
> >of single photons are dispersive through a single linear
> >filter just as random frequency states of white light are
> >dispersive through a prism.
> >
> > http://www.bestweb.net/~ca314159/QMHUP.HTM
>
> Even a cursory reading of this spotted several problems, not the
> least of which was a statement about a polarizer not changing the
> polarization of a photon. You can send 100% linearly polarized
> light onto a polarizer and get out light that is also 100% linearly
> polarized (just not as intense) but with a different orientation.
it through a typical polarizer. What comes out, you call 100%
in regards to intensity. Then you send this through another
identical filter oriented at an angle relative to the first,
and note the intensity is less. Less than 100%. Some photons
are missing. The beam energy is less. And then you say this
beam is composed of photons which are 100% polarized in the
direction of the second filter (less than 100% of the photons
have come through the second filter)
But then you note that some of these will get through
a third filter at a different angle from the second.
It sounds like there is dispersion in each filter so far.
That is, if the polarizers are *filters* and not rotators
like 1/4 wave plates.
Suppose the third filter is the same orientation of the
second. Then 100% of the photons (that came out of the
second filter) will pass through. I'm guessing this is
what you mean when you say they must all be perfectly
oriented with the third filter ? Now this, doesn't sound
like there's dispersion from the second filter otherwise
you'd expect some of the photons to get filtered out.
The beam intensity is less though. You seem to be saying
in an 'ideal' filter that the beam intensity would be the
same. Would ideal filters exhibit the three polarizer
effect that Feynman calls an exhibit of the mysterious
quantum behaviour ?
> There is no dispersion assuming ideal polarizers.
> The comparable idea that a similarly ideal red filter would change
> the color of photons does not make sense.
No. You're right. It only -filters-. It doesn't change the
frequencies of individual photons but it does change the
frequency state of the beam. The macrostate.
I'm thinking that polarizers also don't change the polarization
state of individual photons but only filters them. It does
however also change the state of the *beam* like the red
filter.
--
ca314159
One of the problems I'm having with the QM description of polarization is
that on the one hand, polarizers are treated as filters (they
either absorb or transmit) and on the other hand they are treated as
actually changing individual photon polarization states. These two
approaches do not seem compatible.
--
Michael Weiss
The comparable idea that a similarly ideal red filter would change
the color of photons does not make sense.
A point already made by Newton in his _Opticks_ (though with different
terminology): colored glass, colored liquids, prisms, reflection off
of colored objects: in no case does light acquire color "by new
modification of the rays", which was the prevailing theory before Newton.
Goethe never accepted this, and devoted a decade of his life to developing
his own theory, published as _Farbenlehre_ (Theory of Colors).
You can send 100% linearly polarized
light onto a polarizer and get out light that is also 100% linearly
polarized (just not as intense) but with a different orientation.
There is no dispersion assuming ideal polarizers.
Experimental results similar to this are *also* reported in the
_Opticks_, in the section on Iceland spar. As Newton puts it, the
rays have "sides". I don't recall the theoretical spin Newton put on
this, though (pardon the pun). Are the sides of the rays twisted
about, or does the polarizer act purely as a filter? (I'm asking what
Newton thought, not what we know now.) I'll have to take a second
look.
me...@cars3.uchicago.edu
In article <3507FD...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>Jim Carr wrote:
>>
>> ca31...@bestweb.net writes:
>> >
>> the color of photons does not make sense.
>
> frequencies of individual photons but it does change the
> frequency state of the beam. The macrostate.
>
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
ca314159
me...@cars3.uchicago.edu wrote:
>
> In article <3507FD...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
> >Jim Carr wrote:
> >>
> >> ca31...@bestweb.net writes:
> >> >
> >> There is no dispersion assuming ideal polarizers.
> >> The comparable idea that a similarly ideal red filter would change
> >> the color of photons does not make sense.
> >
> > No. You're right. It only -filters-. It doesn't change the
> > frequencies of individual photons but it does change the
> > frequency state of the beam. The macrostate.
> >
> No, there is no such thing as a "frequency state of the beam".
Hmm. If you pass a beam of white light through a red filter,
how do you describe the state of the beam afterwards ?
Thermodynamics refers to macrostates vs microstates.
A macrostate in this case being what I would call the beam state
or a particular spectral decomposition of the beam: the
frequency state of the beam.
How should I have described this ?
Partovi's article in Zurek's book describes QM in terms of the Jaynesian
max-entropy formalism so this is probably affecting my terminology here.
--
ca314159
A monochromatic beam can be viewed as having a specific
polarization since a monochromatic beam can be represented as the
superposition of many waves with various orientations
which, as a superposition is also a single wave with a
unique polarization direction.
This implies that a prism will disperse the polarization states
of the various spectral components of a beam.
[1] Theory of Partial Coherence by Mark J. Beran and George B. Parrent
Jr.
--
Uncle Al
As polarizing beam splitters based upon two cemented prisms are standard
fare, I'd say that has a pretty good probability pf being at least
partially correct.
--
Uncle Al Schwartz
Uncl...@ix.netcom.com ("zero" before @)
http://pw2.netcom.com/~uncleal0/uncleal.htm
http://www.ultra.net.au/~wisby/uncleal.htm
http://www.guyy.demon.co.uk/uncleal/uncleal.htm
(Toxic URLs! Unsafe for children, Democrats, and most mammals)
"Quis custodiet ipsos custodes?" The Net!
ca314159
Uncle Al wrote:
> As polarizing beam splitters based upon two cemented prisms are standard
> fare, I'd say that has a pretty good probability pf being at least
> partially correct.
Yes. I guess that was a big non-statement I made, but I was toying
with the idea of polarization tagging the photons, and the idea
that resultants do not have unique decompositions.
--
me...@cars3.uchicago.edu
In article <350944...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>me...@cars3.uchicago.edu wrote:
>>
>> In article <3507FD...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>> >Jim Carr wrote:
>> >>
>> >> ca31...@bestweb.net writes:
>> >> >
>> >> There is no dispersion assuming ideal polarizers.
>> >> The comparable idea that a similarly ideal red filter would change
>> >> the color of photons does not make sense.
>> >
>> > No. You're right. It only -filters-. It doesn't change the
>> > frequencies of individual photons but it does change the
>> > frequency state of the beam. The macrostate.
>> >
>> No, there is no such thing as a "frequency state of the beam".
>
> Hmm. If you pass a beam of white light through a red filter,
> how do you describe the state of the beam afterwards ?
>
> Thermodynamics refers to macrostates vs microstates.
> A macrostate in this case being what I would call the beam state
> or a particular spectral decomposition of the beam: the
> frequency state of the beam.
>
> How should I have described this ?
>
Well, the thing is that your white beam is not a superposition but a
mixture. Which is really what macrostates are. The beam has a
frequency distribution but it is a property of the ensamble, not of
any individual photon (as opposed to polarization which is an
individual photon property).
So, when you've a photon with linear polarization at 45 degrees
(relative to the X axis, in some coordinate system) it can also be
viewed as a superposition between the sates of 0 and 90 degrees
polarization. And when you pass it through a polarization filter at o
degrees the photon may be either stopped or passed as 0 degrees
polarization. So here you're right, the terminology is misleading and
the polarization filter is not exactly a filter. It does more than
just pass-stop, it also changes the state of what's passed.
On the other hand your red filter is a true filter. It doesn't change
the state of any photon that's passed, it just stops those that are
not within range. And it is applied at the level of the ensamble, not
an individual photon.
> Partovi's article in Zurek's book describes QM in terms of the Jaynesian
> max-entropy formalism so this is probably affecting my terminology here.
>--
Speaking in terms of entropy (I'm going out on the limb here) I would
say that the polarization filter doesn't change the normalized
(normalized to the number of photons) entropy of the beam. The red
filter, on the other hand, changes the normalized entropy, thus the
change introduced is not reversible.
ca31...@bestweb.net
In article <EpsA9...@midway.uchicago.edu>,
The frequencies don't add (normally).
> The beam has a frequency distribution but it is a property of the
> ensamble, not of any individual photon
> (as opposed to polarization which is an individual photon property).
> So, when you've a photon with linear polarization at 45 degrees
> (relative to the X axis, in some coordinate system) it can also be
> viewed as a superposition between the sates of 0 and 90 degrees
> polarization. And when you pass it through a polarization filter at o
> degrees the photon may be either stopped or passed as 0 degrees
> polarization. So here you're right, the terminology is misleading and
> the polarization filter is not exactly a filter. It does more than
> just pass-stop, it also changes the state of what's passed.
I was talking to someone about vectors and arbitrary decomposition and
a sailboat analogy came up. In this case the wind direction can be
determined by the superposition of forces on a sail. Yet the wind
direction itself is fundemental. You don't change the winds direction
just by "viewing" it in terms of a different sailboat oriented
differently.
>
> On the other hand your red filter is a true filter. It doesn't change
> the state of any photon that's passed, it just stops those that are
> not within range. And it is applied at the level of the ensamble, not
> an individual photon.
>
> > Partovi's article in Zurek's book describes QM in terms of the
Jaynesian
> > max-entropy formalism so this is probably affecting my terminology
here.
> >--
> Speaking in terms of entropy (I'm going out on the limb here) I would
> say that the polarization filter doesn't change the normalized
> (normalized to the number of photons) entropy of the beam. The red
> filter, on the other hand, changes the normalized entropy, thus the
> change introduced is not reversible.
I'm reading an article now called Entropy and Indistinguishability
by David Hestenes Am. Journal Phys July 1970 which seems relevant
to this.
me...@cars3.uchicago.edu
In article <6eds2h$sc0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>In article <EpsA9...@midway.uchicago.edu>,
> I was talking to someone about vectors and arbitrary decomposition and
> a sailboat analogy came up. In this case the wind direction can be
> determined by the superposition of forces on a sail. Yet the wind
> direction itself is fundemental. You don't change the winds direction
> just by "viewing" it in terms of a different sailboat oriented
> differently.
>
That's exactly the business of distinguishing a vector from its
representation. A good analogy.
ca31...@bestweb.net
In article <6eds2h$sc0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>In article <EpsA9...@midway.uchicago.edu>,
> I was talking to someone about vectors and arbitrary decomposition and
> a sailboat analogy came up. In this case the wind direction can be
> determined by the superposition of forces on a sail. Yet the wind
> direction itself is fundemental. You don't change the winds direction
> just by "viewing" it in terms of a different sailboat oriented
> differently.
>
That's exactly the business of distinguishing a vector from its
representation. A good analogy.
Now consider that these force vectors on the sailboat are not definite
quantities and you can compare them to polarizers. Any one sailboat only
determines only a probabilistic picture of the wind vector. A chain of
sailboats each resolving the probabilistic force vectors of the one before
it will have a diminished representation of the wind's vector.
Consider then the idea of reconstruction using such sailboats of a
classical deterministic vector. The probabilistic components of these
sailboat forces are superposed in a chain to reconstruct a deterministic
vector by converging on a definite state. Not a probabilistic one.
This is what Kwiat, Weinfurter, and Zeilinger seem to be doing by
passing a beam many times through a polarizer rotater pair.
The synthesis results in such a classical determinstic vector of
the polarization state.
How is the trace representation invariant. What's a good analogy here ?
The July 1970 Am.J.Phys. has an article on the Lorentz invariance of
the Heisenberg Uncertainty. Hestenes' article discusses properties
of filters and mixers in QM.
--
http://www.bestweb.net/~ca314159/
ca314159
Michael Weiss wrote:
>
> The comparable idea that a similarly ideal red filter would change
> the color of photons does not make sense.
>
> terminology): colored glass, colored liquids, prisms, reflection off
> of colored objects: in no case does light acquire color "by new
> modification of the rays", which was the prevailing theory before Newton.
>
> Goethe never accepted this, and devoted a decade of his life to developing
> his own theory, published as _Farbenlehre_ (Theory of Colors).
>
> You can send 100% linearly polarized
> light onto a polarizer and get out light that is also 100% linearly
> polarized (just not as intense) but with a different orientation.
The renormalizing above, evades the description of the polarizer
as a filter. In a conceptual "filter" the intensity before
and afterwards is related as a "percentage".
Then you say " X percent will always get through" and this is
deterministic for all cases.
But QM treats a polarizer in terms of the probability
that a photon will pass through it. This is a
non-deterministic filtering ? The opposite would be
non-deterministic mixing.
>
> Experimental results similar to this are *also* reported in the
> _Opticks_, in the section on Iceland spar. As Newton puts it, the
> rays have "sides". I don't recall the theoretical spin Newton put on
> this, though (pardon the pun). Are the sides of the rays twisted
> about, or does the polarizer act purely as a filter? (I'm asking what
> Newton thought, not what we know now.) I'll have to take a second
> look.
--
me...@cars3.uchicago.edu
In article <6egk0h$4i0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>In article <6eds2h$sc0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>>In article <EpsA9...@midway.uchicago.edu>,
>
>> I was talking to someone about vectors and arbitrary decomposition and
>> a sailboat analogy came up. In this case the wind direction can be
>> determined by the superposition of forces on a sail. Yet the wind
>> direction itself is fundemental. You don't change the winds direction
>> just by "viewing" it in terms of a different sailboat oriented
>> differently.
>>
>That's exactly the business of distinguishing a vector from its
>representation. A good analogy.
>
> Now consider that these force vectors on the sailboat are not definite
> quantities and you can compare them to polarizers. Any one sailboat only
> determines only a probabilistic picture of the wind vector.
Nah, the analogy breaks here. There is no probabilistic picture of
the wind vector. There is the wind vector and there are its
components, relative to an arbitrary basis, that's it.
>
> Consider then the idea of reconstruction using such sailboats of a
> classical deterministic vector. The probabilistic components of these
> sailboat forces are superposed in a chain to reconstruct a deterministic
> vector by converging on a definite state. Not a probabilistic one.
> This is what Kwiat, Weinfurter, and Zeilinger seem to be doing by
> passing a beam many times through a polarizer rotater pair.
All they're doing is keep projecting a vector on a subspace. There is
no mystery here and no need for profound interpretations.
me...@cars3.uchicago.edu
In article <350BA...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>Michael Weiss wrote:
>>
> The renormalizing above, evades the description of the polarizer
> as a filter.
So don't call it a "filter", call it "green tomato":-) Stop getting
hung on words and check the mathematical description of what polarizer
does.
> In a conceptual "filter" the intensity before
> and afterwards is related as a "percentage".
This is a wrong description, even for a plain filter. A filter is
defined in terms of bandwidth that's passed and bandwidth that's
blocked, not any percentage.
ca314159
me...@cars3.uchicago.edu wrote:
>
> In article <6egk0h$4i0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
> >In article <6eds2h$sc0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
> >>In article <EpsA9...@midway.uchicago.edu>,
> >
> >> I was talking to someone about vectors and arbitrary decomposition and
> >> a sailboat analogy came up. In this case the wind direction can be
> >> determined by the superposition of forces on a sail. Yet the wind
> >> direction itself is fundemental. You don't change the winds direction
> >> just by "viewing" it in terms of a different sailboat oriented
> >> differently.
> >>
> >That's exactly the business of distinguishing a vector from its
> >representation. A good analogy.
> >
> > Now consider that these force vectors on the sailboat are not definite
> > quantities and you can compare them to polarizers. Any one sailboat only
> > sailboats each resolving the probabilistic force vectors of the one before
>
> Nah, the analogy breaks here. There is no probabilistic picture of
> the wind vector. There is the wind vector and there are its
> components, relative to an arbitrary basis, that's it.
The polarization state of a single prepared photon at zero degrees is
passed through a polarizer at 45 degrees and the transmission of that
single photon through the 45 degree filter is described by QM
probabilistically.
But the prepared polarization state of that same photon at 0 degrees is
an assumption and that requires the transmission probability through
the 45 degree filter to be probabilistic.
> > Consider then the idea of reconstruction using such sailboats of a
> > classical deterministic vector. The probabilistic components of these
> > sailboat forces are superposed in a chain to reconstruct a deterministic
> > vector by converging on a definite state. Not a probabilistic one.
> > This is what Kwiat, Weinfurter, and Zeilinger seem to be doing by
> > passing a beam many times through a polarizer rotater pair.
> > The synthesis results in such a classical determinstic vector of
> > the polarization state.
>
> All they're doing is keep projecting a vector on a subspace. There is
> no mystery here and no need for profound interpretations.
Why do you suppose they are doing this ? And why do they want to
do this many times ?
--
ca314159
me...@cars3.uchicago.edu wrote:
>
> In article <350BA...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
> > In a conceptual "filter" the intensity before
> > and afterwards is related as a "percentage".
>
> This is a wrong description, even for a plain filter. A filter is
> defined in terms of bandwidth that's passed and bandwidth that's
> blocked, not any percentage.
What's the bandwidth of coffee grounds ? :)
For single photons, the transmission through a polarizer is
given quantum mechanically as a probabilistic value which
translates into a percentage of the total number of
photons incident on the polarizer to the total number
transmitted by the polarizer, which is a percentage
based on the intensities before and after.
But I agree, were also talking about a bandwidth of probability
here.
--
Michael Weiss
Mati, Mati, Mati! And ca31...@bestweb.net, ca31...@bestweb.net,
ca31...@bestweb.net!!
Here's what I see on my screen:
In article <350BA...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>Michael Weiss wrote:
>>
> The renormalizing above, evades the description of the polarizer
> as a filter.
Please, have a little care with attributions!
me...@cars3.uchicago.edu
In article <350C91...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>me...@cars3.uchicago.edu wrote:
>>
>> In article <6egk0h$4i0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>> >In article <6eds2h$sc0$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>> >>In article <EpsA9...@midway.uchicago.edu>,
>> >
>> > quantities and you can compare them to polarizers. Any one sailboat only
>> > determines only a probabilistic picture of the wind vector. A chain of
>> > sailboats each resolving the probabilistic force vectors of the one before
>> > it will have a diminished representation of the wind's vector.
>>
>> Nah, the analogy breaks here. There is no probabilistic picture of
>> the wind vector. There is the wind vector and there are its
>> components, relative to an arbitrary basis, that's it.
>
> The polarization state of a single prepared photon at zero degrees is
> passed through a polarizer at 45 degrees and the transmission of that
> single photon through the 45 degree filter is described by QM
> probabilistically.
>
filter does simply "pass" or "stop". The polarizer projects the state
vector on the direction of the polarizer. the photon is either
passed, in a 45 degrees polarized state, or not passed at all. There
is no "passed as is" option.
>>
>> All they're doing is keep projecting a vector on a subspace. There is
>> no mystery here and no need for profound interpretations.
>
> Why do you suppose they are doing this ? And why do they want to
> do this many times ?
>
didn't).
me...@cars3.uchicago.edu
In article <350C92...@bestweb.net>, ca314159 <ca31...@bestweb.net> writes:
>me...@cars3.uchicago.edu wrote:
>>
>> > and afterwards is related as a "percentage".
>>
>> This is a wrong description, even for a plain filter. A filter is
>> defined in terms of bandwidth that's passed and bandwidth that's
>> blocked, not any percentage.
>
> What's the bandwidth of coffee grounds ? :)
>
> For single photons, the transmission through a polarizer is
> given quantum mechanically as a probabilistic value which
> translates into a percentage of the total number of
> photons incident on the polarizer to the total number
> transmitted by the polarizer, which is a percentage
> based on the intensities before and after.
>
what is confusing you.
me...@cars3.uchicago.edu
In article <ys067le...@nifty.camb.opengroup.org>, Michael Weiss <colu...@nifty.camb.opengroup.org> writes:
>Mati, Mati, Mati! And ca31...@bestweb.net, ca31...@bestweb.net,
>ca31...@bestweb.net!!
>
>Here's what I see on my screen:
>
>>>
>> The renormalizing above, evades the description of the polarizer
>> as a filter.
>
>Please, have a little care with attributions!
Ooops. Sorry. When attributions pile up to that many layers, it is
easy to miss a bit. My apologies.
ca31...@bestweb.net
In article <EpxFJ...@midway.uchicago.edu>,
me...@cars3.uchicago.edu wrote:
>
> In article <350C91...@bestweb.net>, ca314159 <ca31...@bestweb.net>
writes:
> > The polarization state of a single prepared photon at zero degrees is
> > passed through a polarizer at 45 degrees and the transmission of that
> > single photon through the 45 degree filter is described by QM
> > probabilistically.
> >
> Yes and no. That's why the term "filter" here is misleading. A plain
> filter does simply "pass" or "stop". The polarizer projects the state
> vector on the direction of the polarizer. the photon is either
> passed, in a 45 degrees polarized state, or not passed at all. There
> is no "passed as is" option.
> >>
> >> All they're doing is keep projecting a vector on a subspace. There is
> >> no mystery here and no need for profound interpretations.
> >
> > Why do you suppose they are doing this ? And why do they want to
> > do this many times ?
> >
> For this I would've to read in detail their motivation (which I
> didn't).
This is what I have from our discussion so far:
There are two descriptions of polarization. The first involves plane
waves, superposition and identical photons incident on the polarizer.
The identical partical assumption leads to the ability of the photons to
interfere, or equivalently to be treated in terms of superpositions. The
transmission is deterministic,(0% or 100%) because if one photon gets
through they all must and vice versa; being identical) The intensity
of the light that passed through is determined by interference or
superposition depending on how you look at the problem.
The second is the quantum mechanical description which treats the
incident photons as non-identical and so they individually have
transmission probabilities. The photons are non-identical and
cannot be treated as superpositions. They are treated probabilistically.
There is no superposition of photon states in this case since
we are dealing with non-identical particles (they can't interfere)
A good reference is "Can individual particals have individual properties?"
by Frederik J. Belinfante, Am.J.Phys.46,4
Part of the abstract is posted at: http://www.bestweb.net/~ca314159/QMHUP.HTM
ca31...@bestweb.net
In article <6elm5t$tlk$1...@nnrp1.dejanews.com>, ca31...@bestweb.net wrote:
> There are two descriptions of polarization. The first involves plane
> waves, superposition and identical photons incident on the polarizer.
> The identical partical assumption leads to the ability of the photons to
> interfere, or equivalently to be treated in terms of superpositions. The
> transmission is deterministic,(0% or 100%) because if one photon gets
> through they all must and vice versa; being identical) The intensity
> of the light that passed through is determined by interference or
> superposition depending on how you look at the problem.
>
> The second is the quantum mechanical description which treats the
> incident photons as non-identical and so they individually have
> transmission probabilities. The photons are non-identical and
> cannot be treated as superpositions. They are treated probabilistically.
> There is no superposition of photon states in this case since
> we are dealing with non-identical particles (they can't interfere)
>
> A good reference is "Can individual particals have individual properties?"
> by Frederik J. Belinfante, Am.J.Phys.46,4
>
> Part of the abstract is posted
> at: http://www.bestweb.net/~ca314159/QMHUP.HTM
Non-orthogonal states can overlap. If A and B are non-orthogonal states
then "A and B" is the coincidence of the states A and the state B.
Let's say we score a class's exam and 48% of the students fail the
first question and 37% fail the second question. Then what is the
percentage of students failing "A and B" ? It is not A + B.
The number of questions on the test was not indicated here, so
the percentage is not of a mixture of A's and B's, but percentages
of an ensemble which may posess one or more these "states" on
an individual level in the ensemble.
Orthogonal states do not overlap. The percentages add and
represent the percent of a specific orthogonal state in
the whole ensemble (the mixture of these orthogonal states).
If A="Failed the test" and B="Passed the test" are the orthogonal
states relative to the ensemble of students that may take the test.
The total number of students that took test is A + B.
In the non-orthogonal case, are A and B pure states ? Surely they
look like discrete entities. Either you pass or fail the first
question. That's not a fuzzy probability. But the fact that
probabilities and percentages get confused is why Edwin T. Jaynes
was very annoyed with the frequentist interpretation of
probability theory.
ca31...@bestweb.net
ca314159
and guess what the value will be, in a maximum overlap, the total
number of failures seems to be 1-(A^2+B^2) or more generally
1- sigma_i (A_i)^2 for i questions on the test. In the maximum
overlap case the "interference term" drops out ?
--
ca31...@bestweb.net
In article <6emq3t$81m$1...@agate.berkeley.edu>,
ca31...@bestweb.net wrote:
>
>
> Non-orthogonal states can overlap. If A and B are non-orthogonal states
> then "A and B" is the coincidence of the states A and the state B.
> Let's say we score a class's exam and 48% of the students fail the
> first question and 37% fail the second question. Then what is the
> percentage of students failing "A and B" ? It is not A + B.
> The number of questions on the test was not indicated here, so
> the percentage is not of a mixture of A's and B's, but percentages
> of an ensemble which may posess one or more these "states" on
> an individual level in the ensemble.
>
> Orthogonal states do not overlap. The percentages add and
> represent the percent of a specific orthogonal state in
> the whole ensemble (the mixture of these orthogonal states).
> If A="Failed the test" and B="Passed the test" are the orthogonal
> states relative to the ensemble of students that may take the test.
> The total number of students that took test is A + B.
>
> In the non-orthogonal case, are A and B "pure" states ? Surely they
> look like discrete entities. Either you pass or fail the first
> question. That's not a fuzzy probability. But the fact that
> probabilities and percentages get confused is why Edwin T. Jaynes
> was very annoyed with the frequentist interpretation of
> probability theory.
>
[moderator: I believe, an important elaboration on the previous submission]
Since the expectation of the total failures given by "A and B"
can be expressed analogously in terms of orthogonal vectors
( A*B*cos(theta) ) as the maximum overlap case:
1- p(A)^2 + p(B)^2
where p(A) is the percentage of fails for question A.
this implies the orthogonality and independence of the "eigenvectors"
(the questions) hence the term cos(theta) which we cannot asscribe
in the case of exam questions drops out.
This does not eliminate the possibility of a less than maximal
overlap but we do not have an analogous expression for the
"intereference term" (cos(theta)) for the case of exam questions
so instead we assume that we can maximize the orthogonality of
the eigenstates (the independence of the exam questions) by
increasing the number of questions in the exam (available states)
which gives:
<total fails> := 1 - sigma_i (p(A_i)^2)
for i - > infinity.
Filling in the analogy, this directly relates the Jayesian
max entropy formalism with the quantum theory.
ca31...@bestweb.net
This article attempts to display the meaning of the phase information
in quantum mechanics, when applied to a common probability problem.
TWO DESCRIPTIONS OF POLARIZATION
A) There are two descriptions of polarization. The first involves plane
waves, superposition and identical photons incident on the polarizer.
The identical partical assumption leads to the ability of the photons to
interfere, or equivalently to be treated in terms of superpositions. The
transmission is deterministic,(0% or 100%) because if one photon gets
through they all must and vice versa; being identical) The intensity
of the light that passed through is determined by an interference term or
superposition depending on how you look at the problem.
B) The second is the quantum mechanical description which treats the
incident photons as non-identical and so they individually have
transmission probabilities. The photons are non-identical and
cannot be treated as superpositions. They are treated probabilistically.
There is no superposition of photon states in this case since
we are dealing with non-identical particles (they can't interfere)
see reference [1]
TWO KINDS OF PROBABILITIES
A) Non-orthogonal states can overlap. If A and B are non-orthogonal states
then "A and B" is the coincidence of the states A and the state B.
Let's say we score a class's exam and 48% of the students fail the
first question and 37% fail the second question. Then what is the
percentage of students failing "A and B" ? It is not p(A) + p(B).
where p(A) is the percent of students failing question A, p(B)
percent failing B.
The number of questions on the test was not indicated here, so
the percentage is not of a mixture of A's and B's, but percentages
of an ensemble which may posess one or more these "states" on
an individual level in the ensemble.
B) Orthogonal states do not overlap. The percentages add and
represent the percent of a specific orthogonal state in
the whole ensemble (the mixture of these orthogonal states).
If A="Failed the test" and B="Passed the test" are the orthogonal
states relative to the ensemble of students that may take the test.
The total number of students that took test is p(A) + p(B).
PHASE INFORMATION IN QM
In a two slit experiment the probabilities of identical particles
do not add in the same way as non-dentical particles.
A) In the case of identical particles, phase information is contained in the
in the interference term [2]:
I_x= 2 * I_a * I_b * cos (theta)
of the more general equantion:
I_ab = (I_a)^2 + (I_b)^2 + I_x
in terms of the intensities.
B) When orthogonality is applied, theta=90 and the interference term is
zero.In this case, the relation above is equivalent to that of simply
adding the probabilities as in a mixture.
THE MEANING OF PHASE INFORMATION IN THE CASE OF EXAM QUESTIONS
A) In the non-orthogonal case, are A and B pure states ? They
look like discrete entities. Either you pass or fail the first
question. That's not a fuzzy probability. But the fact that
probabilities and percentages get confused is why E. T. Jaynes
was very annoyed with the frequentist interpretation of
probability theory.
B) If the percent of the total failures given by "A and B"
can be expressed analogously in terms of *orthogonal* vectors,
if the exam question A is orthogonal to the exam question B
(A is independant of B) then any "phase angle" we might conceive
of in this analogy must be theta=90. In this case the "vectors"
A, B are expressed as a resultant:
A*B*cos(theta) = 0
but this is the intereference term in the quantum mechanical expression
so that we expect in this case the squares of the probabilities add and
the probability of "A and B" is
P(A and B) = 1 - p(A)^2 + p(B)^2
where p(A) is the percentage of fails for question A, and p(B) for B.
this implies the orthogonality and independence of the "eigenvectors"
(the questions) hence the hypothetical interference term in the
case of exam questions, being zero implies no correlation between
questions A and B.
In the quantum mechanics, this is also the case when we distinguish
particle paths. The paths themselves can be treated as independent
orthogonal eigenvectors when we know which path a particle took.
The probabilities also add here and the interefence term is zero.
In the exam problem, P(A and B) is the statistical maxmimal overlap
of probabilities A and B.
A) This does not eliminate the possibility of a less than maximal
overlap but we do not have an analogous expression for the
"interference term" for the case of exam questions so instead we
assume that we can maximize the orthogonality of the eigenstates
(the independence of the exam questions) by increasing the number
of questions in the exam (available states) which gives:
P(A) = 1 - /sigma_i p(A_i)^2 =
percent of students failing all questions
for i -> infinity. Where P(A) is the maximum percent of all the
A_i questions being failed, given the percentages of failures of
each individual question.
This is same for P(A and B) for many questions on the assumption that
each question is totally independent from another.
A) If we now assume that A and B are correlated, then our hypothetical
0<=theta<90, and we would expect a hypothetical interference term to
be representable as a statistical correlation measure between these
"states" or exam questions. In this case, in which the questions can
no longer be regarded as distinct they begin to take on the
analogous physical meaning as particles paths which cannot be
distinguished.
The actual percentage of "fails all questions" (or subsets of
questions) will typically vary over ensembles. Part of this is
due to the differences between students, but there is also this
hypothetical interference term or statistical correlation between
questions that is the root of the interference term in the
quantum mechanical expression for Young's experiment.
The quantum mechanical "effect" is seen here in a non-quantum mechanical
analogy.
The introduction of phase information to probability theory has already
been suggested [3] to illustrate the comparison of quantum mechanical
effects with a complex-valued probability theory.
[1] "Can individual particals have individual properties?" by
Frederik J. Belinfante, Am.J.Phys.46,4
Part of the abstract is posted at:
http://www.bestweb.net/~ca314159/QMHUP.HTM
[2] Feynman Lectures, vol 3.
[3] Youssef, Saul Quantum Mechanics as an Exotic Probability Theory,
quant-ph/9509004, http://xxx.lanl.gov
[4] E.T. Jaynes,
ftp://bayes.wustl.edu/pub/Jaynes/book.probability.theory/
Michael Weiss
ca314159: are you under the impression that you are describing what
standard QM says? Or are you presenting your own personal theory,
different from QM? Or do you believe your description is equivalent
to QM, though expressed in a non-standard form? Or are you just
trolling?
ca314159
ca31...@bestweb.net wrote:
> A) In the case of identical particles, phase information is contained in the
> in the interference term [2]:
>
> I_x= 2 * I_a * I_b * cos (theta)
>
> of the more general equantion:
>
> I_ab = (I_a)^2 + (I_b)^2 + I_x
>
> in terms of the intensities.
Correction: These should be amplitudes, not intensities.
and the term I_x= 2 * I_a * I_b * cos (theta)
should read: I_x= 2 * sqrt(I_a * I_b) * cos (theta)
ca314159
I have tried to integrate the ideas and results from Jaynes,
Belinfante, Kwiat, Weinfurter, Zeilinger, Youssef, and others
into one picture.
When the idea, of what QM refers to as phase information or
an interference term, can be introduced into a common place
probability problem like a test exam, it only points to a
much broader application for quantum mechanical principles.
Jaynes, much earlier, indicated that QM was much more general in
this way in his paper "Generalized Scattering" from
Maximum-Entropy and Bayesian Methods in Inverse Problems
edited by Smith and Grandy.
I may flounder in my presentation but I always appreciate
constructive criticism. Many people in this newsgroup have
contributed towards sharpening the picture I am presenting.
--
ca31...@bestweb.net
Actually, the results are even more elegant if the
quantum mechanical equation for two slit interference
is given in terms of the probability equation:
P(A or B) = P(A) + P(B) - 2*A*B*cos(theta)
where the interference term is associated with
a correlation (overlap) between A and B p(A and B)
http://www.bestweb.net/~ca314159/OVERLAP.HTM
Michael Weiss
ca314159:
I have tried to integrate the ideas and results from Jaynes,
Belinfante, Kwiat, Weinfurter, Zeilinger, Youssef, and others
into one picture.
I've looked at Paul Kwiat's website, and everything there fits in
perfectly with standard QM (nor does he claim otherwise).
I was not able to download Youssef's article in a readable form.
I haven't looked at Belinfante's article yet, just the snippet you
have on your website.
I haven't tried to download Jaynes' book yet.
I may flounder in my presentation but I always appreciate
constructive criticism. Many people in this newsgroup have
contributed towards sharpening the picture I am presenting.
I hope the following criticism is constructive. Your post contains
several statements that appear to contradict QM. This suggests that
either, (a) you believe QM is wrong, or (b) you believe the
contradiction is only apparent.
Which is it, (a) or (b)? Answering this question would sharpen the
picture you are presenting.
Jim Carr
ca314159 <ca31...@bestweb.net> writes:
>
> Can you clairify this for me.
I will try. Note the emphasis on experimental observation, except
for the assumption of 100% efficiency for convenience.
> You prepare a beam by sending
> it through a typical polarizer. What comes out, you call 100%
> in regards to intensity.
I do more than that. I can test this by sending the new beam
through another polarizer that has the same orientation. The
100% transmission tells me this beam has that polarization.
By the way, lets start with I0 originally, and call the amount that
gets through our initial polarizer (with orientation A) I1. Normally
I1 = I0/2. What I am saying here is that another polarizer set with
orientation A, used to analyze this beam, will transmit I1. As a
check, I can set the analyzer to 90 degrees from A and see zero
transmission in the orthogonal direction.
> Then you send this through another
> identical filter oriented at an angle relative to the first,
> and note the intensity is less. Less than 100%.
Right, but not zero, unless this one is at 90 degrees to the first.
Lets call this orientation B, and the new intensity I2 < I1.
> Some photons
> are missing. The beam energy is less. And then you say this
> beam is composed of photons which are 100% polarized in the
> direction of the second filter (less than 100% of the photons
> have come through the second filter)
Yes. Since I send this beam through a polarizer with that new
orientation B, now used as a analyzer, and 100% of I2 will be
transmitted. According to our criteria, all of the photons in
the beam have the polarization of orientation B. We check this
by setting the analyzer to 90 degrees from B, and see nothing.
> But then you note that some of these will get through
> a third filter at a different angle from the second.
Correct. I set a third filter at orientation C, and some fraction
is transmitted, say I3 < I2. All of this assumes that C is not at
90 degrees to B, of course. Now an analyzer set at C transmits I3.
> It sounds like there is dispersion in each filter so far.
> That is, if the polarizers are *filters* and not rotators
> like 1/4 wave plates.
Except that it is not like there are some photons with the
original polarization that slipped through. Recall our set up:
==I0=> A ==I1=> A ==I1=> B ==I2=> B ==I2=> C ==I3=> C ==I3=>
filter analyze filter analyze filter analyze
> Suppose the third filter is the same orientation of the
> second. Then 100% of the photons (that came out of the
> second filter) will pass through. I'm guessing this is
> what you mean when you say they must all be perfectly
> oriented with the third filter ?
Yes. See above.
> Now this, doesn't sound
> like there's dispersion from the second filter otherwise
> you'd expect some of the photons to get filtered out.
Exactly.
> The beam intensity is less though. You seem to be saying
> in an 'ideal' filter that the beam intensity would be the
> same. Would ideal filters exhibit the three polarizer
> effect that Feynman calls an exhibit of the mysterious
> quantum behaviour ?
Of course. The three polarizer effect is simply when C is at
90 degrees to A in the above example, and B is anything inbetween.
Note well that this is real stuff, not textbook stuff.
You can do this experiment yourself, although real polaroid sheets
are not 100% efficient. But since the eye is used to thicker stuff
being dimmer than thin stuff, our eye makes a good detector for the
experiment. Then the "magic trick" is to put C on A and get black.
Show that if you rotate C it goes from clear to black. Then put B
on top of C; still black. Slip B between A and C; ah hah! Some
light gets through. Best of all, unlike the example above, we can
see that light that bypasses B (which is only inserted part way)
is still absorbed by C. C has not been bumped or disturbed.
Try it. You can order big sheets of polaroid from Edmunds, suitable
for making art works by inserting cool stuff between the crossed
polaroids and/or messing around with Karo syrup -- which *does*
rotate the polarization. (Comparing the ABC experiment to Karo
syrup shows this, since the transmission is very high.)
> I'm thinking that polarizers also don't change the polarization
> state of individual photons but only filters them. It does
> however also change the state of the *beam* like the red
> filter.
Except that you cannot put a green filter after the red filter
and get green. There is a very basic difference here.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Michael Weiss
Very clear explanation, Jim, nice!
ca314159 (Robert Fung) wrote:
> I'm thinking that polarizers also don't change the polarization
> state of individual photons but only filters them. It does
> however also change the state of the *beam* like the red
> filter.
a historically interesting point of view, since I *think* it is what
Newton believed (with respect to Iceland spar aka birefringent calcite).
At least Newton's discussion of this in the _Opticks_ (Queries 24 and 25,
IIRC) parallels his discussion of splitting white light with prisms.
But as Jim notes:
Except that you cannot put a green filter after the red filter
and get green. There is a very basic difference here.
Newton seems to have missed recognizing this basic difference.
Perhaps this is because he didn't get around to doing the crucial
experiments. Indeed, he says that the Queries address topics where
he had not the time to do all the experiments he wanted to.
ca31...@bestweb.net
In article <ys0soo5...@nifty.camb.opengroup.org>,
Michael Weiss <colu...@nifty.camb.opengroup.org> wrote:
>
> Very clear explanation, Jim, nice!
>
> ca314159 (Robert Fung) wrote:
>
> > state of individual photons but only filters them. It does
> > however also change the state of the *beam* like the red
> > filter.
>
> Newton believed (with respect to Iceland spar aka birefringent calcite).
> At least Newton's discussion of this in the _Opticks_ (Queries 24 and 25,
> IIRC) parallels his discussion of splitting white light with prisms.
>
> But as Jim notes:
>
> and get green. There is a very basic difference here.
>
> Perhaps this is because he didn't get around to doing the crucial
> experiments. Indeed, he says that the Queries address topics where
> he had not the time to do all the experiments he wanted to.
>
We're assuming perfect filters here. If polarizers were perfect
filters you'd not get light from a 45 deg after a 0 deg. Given
imperfect color filters you'd expect some green light through.
The polarization filters are probably best looked at in the context
of impedance matching of partially coherent signals.
me...@cars3.uchicago.edu
In article <6feoph$g4a$1...@nnrp1.dejanews.com>, ca31...@bestweb.net writes:
>In article <ys0soo5...@nifty.camb.opengroup.org>,
>
> We're assuming perfect filters here. If polarizers were perfect
> filters you'd not get light from a 45 deg after a 0 deg. Given
> imperfect color filters you'd expect some green light through.
>
Jim Carr
> waves, superposition and identical photons incident on the polarizer.
I would say the first one involves just waves, and gives the
usual transmission rules -- and works just fine as long as you
don't use a detector that counts individual photons.
> The identical partical assumption leads to the ability of the photons to
> interfere, or equivalently to be treated in terms of superpositions. The
> transmission is deterministic,(0% or 100%) because if one photon gets
> through they all must and vice versa; being identical)
No. Same probability does not mean same outcome.
> The second is the quantum mechanical description which treats the
> incident photons as non-identical and so they individually have
> transmission probabilities.
That is like a description of a Bohmian hidden variable theory,
not standard QM, and even there it has a probability that hidden
variable theories lack. Standard QM says they individually have
identical transmission probabilities, and that is why only some
of them are transmitted.
Jim Carr
ca31...@bestweb.net writes:
>
> This article attempts to display the meaning of the phase information
> in quantum mechanics, when applied to a common probability problem.
However, it seems to be missing the essential point of one of the
references you cite
> [3] Youssef, Saul Quantum Mechanics as an Exotic Probability Theory,
> quant-ph/9509004, http://xxx.lanl.gov
which is that it is an _exotic_ probability theory where the
probability amplitudes are complex rather than real numbers.
And for the person who had trouble downloading the PostScript file
from LANL, you might look at Modern Physics Letters A 6, 225 (1991);
loc cit. 9, 2571 (1994); Phys. Lett. A 204, 181 (1995).
> B) The second is the quantum mechanical description which treats the
> incident photons as non-identical ...
Wrong
> ... and so they individually have
> transmission probabilities.
Correct, but not for the reason you give.
> The photons are non-identical and
> cannot be treated as superpositions. They are treated probabilistically.
> There is no superposition of photon states in this case since
> we are dealing with non-identical particles (they can't interfere)
But they can interfere, because the probability theory allows it.
It is not the probability theory of dice.
ca31...@bestweb.net
In article <6fpfb2$l0l$1...@news.fsu.edu>,
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
>
> ca31...@bestweb.net writes:
> >
> > This article attempts to display the meaning of the phase information
> > in quantum mechanics, when applied to a common probability problem.
>
> references you cite
>
> > quant-ph/9509004, http://xxx.lanl.gov
>
> probability amplitudes are complex rather than real numbers.
The phase information is, from these complex numbers.
> > B) The second is the quantum mechanical description which treats the
> > incident photons as non-identical ...
>
> Wrong
>
> > ... and so they individually have
> > transmission probabilities.
>
> Correct, but not for the reason you give.
Both correct.
If you treat the individual members of some set as identical
particles you must always talk about them in a macroscopic sense.
Once you start separting out individuals they are no longer identical.
If we can assign individual probabilities to photons then
they are distinct. You should not be able to do this with
identical particles.
> > The photons are non-identical and
> > cannot be treated as superpositions. They are treated probabilistically.
> > There is no superposition of photon states in this case since
> > we are dealing with non-identical particles (they can't interfere)
>
> But they can interfere, because the probability theory allows it.
> It is not the probability theory of dice.
Quantum mechanics uses two probability theories;
they are technically equivalent, but not the same.
And not realizing this, and the distinction is why you get all
that interpretive hype about many-worlds and schrodinger cats...
Saul Youssef makes a similar point in his articles, that the
incomplete picture of probability theory is causing the
interpretive problems. He didn't however point out that there
is a probability theory based on discrete non-identical objects
and one based on wave-like identical objects and a connection
between these two approaches.
http://www.bestweb.net/~ca314159/
Jim Carr
Michael Weiss <colu...@nifty.camb.opengroup.org> wrote:
|
| ca314159 (Robert Fung) wrote:
| > I'm thinking that polarizers also don't change the polarization
| > state of individual photons but only filters them. It does
| > however also change the state of the *beam* like the red
| > filter.
|
| Newton believed (with respect to Iceland spar aka birefringent calcite).
| At least Newton's discussion of this in the _Opticks_ (Queries 24 and 25,
| IIRC) parallels his discussion of splitting white light with prisms.
|
| But as Jim notes:
|
| and get green. There is a very basic difference here.
|
| Perhaps this is because he didn't get around to doing the crucial
| experiments. Indeed, he says that the Queries address topics where
| he had not the time to do all the experiments he wanted to.
ca31...@bestweb.net writes:
>
> We're assuming perfect filters here.
I was assuming both perfect filters and perfect polarizers.
> If polarizers were perfect
> filters you'd not get light from a 45 deg after a 0 deg.
Incorrect for two reasons.
1) It is an observable fact that the effect is most dramatic
when you use high quality polarizers. That makes it sharp
and the colors transmitted when Karo syrup is introduced
particularly pure.
2) Your hypothesis would say that only a very tiny fraction of
unpolarized light (the miniscule fraction that had exactly
45 degree polarization within the tolerance of the filter)
would get through. I'm not sure if you would also keep all
circularly polarized light from going through a linear polaroid.
Michael Weiss
I suspect that ca314159 will not be convinced, Jim.
A more interesting direction to head in is the difference between
prisms and birefringent crystals, in terms of the relevant vector
spaces.
A birefringent crystal resolves a polarization vector into two
orthogonal components, both linearly polarized. In other words, a
vector is resolved into its components in a particular basis. By
changing the orientation of the crystal, you change the choice of
basis.
A prism resolves a vector into an *infinite* number of components,
corresponding to the continuum of pure frequencies; more precisely, it
expresses the state-vector as an integral. One can think of this
integral as a kind of generalized sum over a generalized basis for the
relevant vector space.
Unlike the birefringent crystal case, changing the orientation of the
prism doesn't change the choice of generalized basis--- it's still the
continuum of pure frequencies.
One can certainly imagine using other generalized bases (resolutions
of the identity, in functional-analysis-lingo). This is a
mathematical fact, but I don't know how you'd perform this
experimentally.
So the "pure frequency coordinate system" seems kind of special, more
fundamental somehow than other generalized bases that one can cook
up. Ultimately I guess this comes from the central importance of
energy in physics.
Thus Newton could build a well-nigh irrefutable experimental case for
the statement, "White light is a mixture of colors".
If it works once, try it again, right? So Newton initially approached
polarization the same way. But here one has an infinity of bases, all
equally natural and equally accessible to experiment. Newton seems to
have dropped the investigation just as it started to run into this
point.
Finally, let me draw attention to this diagram on ca314159's website:
http://www.bestweb.net/~ca314159/QM3.GIF, referred to in:
http://www.bestweb.net/~ca314159/QMHUP.HTM
a "what if" picture--- what if prisms behaved like polarizers?
ca314159 writes:
Figure 3 (below) - The Three Prism Problem, Functionally,
in terms of abstract states, why is a prism different than a polarizer ?
---a good question. But note that his diagram does NOT correspond to
the way prisms really behave!
ca31...@bestweb.net
In article <ys07m51...@nifty.camb.opengroup.org>,
Michael Weiss <colu...@nifty.camb.opengroup.org> wrote:
> ---a good question. But note that his diagram does NOT correspond to
> the way prisms really behave!
>
Stay above the first <HR> line: http://www.bestweb.net/~ca314159/
the early stuff was the thrashing, I keep it there to remind me
of where I've been. Sometimes we have to imagine things under
extremes to find out more about how they work or don't work.
Think of it as abstract art.
Saul Youssef
http://www.scri.fsu.edu/~youssef/packages/talks/ud0/ud0.html
Bart Kosko (fuzzy subsets)
http://sipi.usc.edu/~kosko/
John Preskill
http://www.theory.caltech.edu/people/preskill/
Hartle
http://jovian.physics.uoguelph.ca/~droz/uni/gr/FileH.html
G. t'Hooft
http://www.fys.ruu.nl/wwwboz/studgids/vakken34.htm
Steven B Giddings
http://www.nsf.gov/awards/awards_1991/awd_1991_57/a9157463.txt
Michael Weiss
ca314159 writes:
Stay above the first <HR> line: http://www.bestweb.net/~ca314159/
the early stuff was the thrashing, I keep it there to remind me
of where I've been. Sometimes we have to imagine things under
extremes to find out more about how they work or don't work.
Think of it as abstract art.
It is nonetheless an interesting and instructive diagram, and shows
one important way in which prisms are *not* like polarizers.
The chart on the "overlap" page also has some important lessons, which
(IMHO) you have not fully absorbed.
You ask some good questions. Now just listen more closely to some of
the good answers....
ca31...@bestweb.net
In article <ys04t04...@nifty.camb.opengroup.org>,
Michael Weiss <colu...@nifty.camb.opengroup.org> wrote:
(in the sci.physics thread "Random polarization state...")
> It is nonetheless an interesting and instructive diagram, and shows
> one important way in which prisms are *not* like polarizers.
Polarizers can act like prisms:
If one looks at a white light source reflected from a
a CD-ROM, the frequency dispersion is evident as one
would obtain from a prism, and we describe this in terms
of diffraction.
Yet now look at a white light source -through- the CD-ROM
and one first notes its transmission of the light is poor.
Then, looking at this transmitted light with a typical
polarizer, one can see that it is faintly polarized.
Looking at the -reflected- light from a polarizer does not
generate a frequency dispersion. The polarizer was not
fabricated to do this and a diffraction grating is not
in general fabricated to polarize light, yet the both
seem to have something in common; specifically, when
their transmittivity and reflectivity are equal.
I'm not sure how one would describe a prism in these terms
except perhaps as a mixture of scattering and transmission
within the prism ?
Jim Carr
Sigh.
> If one looks at a white light source reflected from a
> a CD-ROM, the frequency dispersion is evident as one
> would obtain from a prism, and we describe this in terms
> of diffraction.
That is acting as a diffraction grating. Not the same thing.
ca31...@bestweb.net
In article <6grl8j$otl$1...@news.fsu.edu>,
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
>
> ca31...@bestweb.net writes:
> >
> > Polarizers can act like prisms:
>
> Sigh.
> > If one looks at a white light source reflected from a
> > a CD-ROM, the frequency dispersion is evident as one
> > would obtain from a prism, and we describe this in terms
> > of diffraction.
>
> That is acting as a diffraction grating. Not the same thing.
Why ? Because the velocities are all the same for each frequency ?
That's not how Feynman describes a diffraction grating.
Because there's no intervening media of specific refractive index ?
I'm sure there's a generalized refractive index applicable to both cases
and we can use virtual media in the diffraction case if that makes it
more amicable a comparison.
But if you mean the words are not the same, or how we now describe them
in text books is not the same, then I agree.
Jim Carr
| > in quantum mechanics, when applied to a common probability problem.
|
| references you cite
|
| > quant-ph/9509004, http://xxx.lanl.gov
|
| probability amplitudes are complex rather than real numbers.
ca31...@bestweb.net writes:
>
> The phase information is, from these complex numbers.
Of course. The point is that your fuzzy logic analogy runs
aground because that stuff all assumes real probability amplitudes.
Most of your intuition based on classical probabilities runs
into trouble when it faces complex amplitudes.
| > B) The second is the quantum mechanical description which treats the
| > incident photons as non-identical ...
|
| Wrong
|
| > ... and so they individually have
| > transmission probabilities.
|
| Correct, but not for the reason you give.
> Both correct.
The photons are treated as identical particles with identical
initial wave functions. There are no hidden variables.
> If you treat the individual members of some set as identical
> particles you must always talk about them in a macroscopic sense.
Not in quantum mechanics, and certainly not in what Youssef does,
which is completely equivalent to QM.
> Once you start separting out individuals they are no longer identical.
They have identical wavefunctions and are otherwise identical in all
ways. You are not separating out a _particular_ individual. You
are separating out a single photon.
> If we can assign individual probabilities to photons then
> they are distinct. You should not be able to do this with
> identical particles.
You certainly can. The probabilities are all the same.
| > The photons are non-identical and
| > cannot be treated as superpositions. They are treated probabilistically.
| > There is no superposition of photon states in this case since
| > we are dealing with non-identical particles (they can't interfere)
|
| But they can interfere, because the probability theory allows it.
| It is not the probability theory of dice.
> Quantum mechanics uses two probability theories;
> they are technically equivalent, but not the same.
You have missed the whole point of Youssef's paper.
> Saul Youssef makes a similar point in his articles, that the
> incomplete picture of probability theory is causing the
> interpretive problems.
I would say his point concerns understanding probability theory,
specifically that you do not have to break the problem into two
parts only one of which involves probabilities, if you break away
from the assumption that classical probability theory is based on.
> He didn't however point out that there
> is a probability theory based on discrete non-identical objects
> and one based on wave-like identical objects and a connection
> between these two approaches.
That is because that would be wrong. You have missed his point.
There are only objects and probabilities, no waves.
ca31...@bestweb.net
In article <6gu493$fd3$1...@news.fsu.edu>,
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:>
> Of course. The point is that your fuzzy logic analogy runs
> aground because that stuff all assumes real probability amplitudes.
> Most of your intuition based on classical probabilities runs
> into trouble when it faces complex amplitudes.
Like ?
Complex numbers are already used by electricians.
They know when to throw out the imaginary parts after
they're done turning the crank. I've heard of uses for
complex numbers where the imaginary values actually have
a physical interpretation in actual measurements.
That's because all the nice rotational properties were
mapped over onto the physical problem.
Complex numbers are a tool. We can use all or part of them.
There's an old statistics joke:
"If you torture the data long enough,
it will say anything you like."
But specifically I think Bonferroni's Inequality applies here
since we are talking the physical application of probability theory.
CRC Math Glossary:
http://www.astro.virginia.edu/~eww6n/math/math0.html
> The photons are treated as identical particles with identical
> initial wave functions.
You state this globally, but it is true only if we do not
look at what paths they took.
> There are no hidden variables.
Of course not. If we saw them they wouldn't be hidden :)
> > Once you start separting out individuals they are no longer identical.
>
> They have identical wavefunctions and are otherwise identical in all
> ways. You are not separating out a _particular_ individual. You
> are separating out a single photon.
You describe a two-slit this way when the paths of the photon
are not observed. And you then say "a single photon went through
both slits" if you like. Again, this is not the same situation as when
the paths of that photon are observed.
> > If we can assign individual probabilities to photons then
> > they are distinct. You should not be able to do this with
> > identical particles.
>
> You certainly can. The probabilities are all the same.
You're not assigning individual or distinguishable probabilities
then. They are all the same.
> > Saul Youssef makes a similar point in his articles, that the
> > incomplete picture of probability theory is causing the
> > interpretive problems.
>
> I would say his point concerns understanding probability theory,
> specifically that you do not have to break the problem into two
> parts only one of which involves probabilities, if you break away
> from the assumption that classical probability theory is based on.
>
> > He didn't however point out that there
> > is a probability theory based on discrete non-identical objects
> > and one based on wave-like identical objects and a connection
> > between these two approaches.
>
> That is because that would be wrong. You have missed his point.
> There are only objects and probabilities, no waves.
Do you mean objects and "complex" probabilities ?
No waves ? What happened to the celebrated
Euler's relation ? Spiral synthesis ...
This is a real good book:
http://www.midi-classics.com/p5526.htm
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:>
|
| Of course. The point is that your fuzzy logic analogy runs
| aground because that stuff all assumes real probability amplitudes.
| Most of your intuition based on classical probabilities runs
| into trouble when it faces complex amplitudes.
ca31...@bestweb.net writes:
>
> Like ?
Like every example you have been writing about.
> Complex numbers are already used by electricians.
> They know when to throw out the imaginary parts after
> they're done turning the crank.
And so do physicists -- only your analogy is inexact since that is
not an example of probability amplitudes and probabilities.
> I've heard of uses for
> complex numbers where the imaginary values actually have
> a physical interpretation in actual measurements.
Certainly. Still not relevant.
> That's because all the nice rotational properties were
> mapped over onto the physical problem.
And sometimes you get interference effects. Still not relevant.
> There's an old statistics joke:
> "If you torture the data long enough,
> it will say anything you like."
However, you will not get dice, or the associated standard probability
theory, to exhibit interference effects.
> But specifically I think Bonferroni's Inequality applies here
> since we are talking the physical application of probability theory.
>
> CRC Math Glossary:
> http://www.astro.virginia.edu/~eww6n/math/math0.html
Looks to me like it concerns real probabilities, not an exotic
probability theory with complex amplitudes.
} The photons are treated as identical particles with identical
} initial wave functions.
> You state this globally, but it is true only if we do not
> look at what paths they took.
If you do that, then you have altered the probabilities by making
that measurement. Have you read Saul's papers?
} > If we can assign individual probabilities to photons then
} > they are distinct. You should not be able to do this with
} > identical particles.
}
} You certainly can. The probabilities are all the same.
> You're not assigning individual or distinguishable probabilities
> then. They are all the same.
Well, duh. That is the point. The particles are not distinquishable,
and the probabilities are the same. It does not matter if you study
them one by one or as an ensemble.
> Do you mean objects and "complex" probabilities ?
Yes.
> No waves ? What happened to the celebrated
> Euler's relation ? Spiral synthesis ...
Whatever. Have you read Saul's papers? You don't even need waves
in standard quantum mechanics unless you choose to use the
Schroedinger picture.
Jim Carr
I wrote:
|
|> Do you mean objects and "complex" probabilities ?
|
| Yes.
Correction. No. Probability amplitudes.
ca31...@bestweb.net
In article <6h134n$fa0$1...@news.fsu.edu>,
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
>
> I wrote:
> |
> |> Do you mean objects and "complex" probabilities ?
> |
> | Yes.
>
> Correction. No. Probability amplitudes.
>
So you get something like 2*sqrt(imag^2 + real^2),
where from the Fourier transform the phase angle:
theta = arctan(imag/real) instead of 2*sqrt(P(A)*P(B))*cos(theta),
which realtes to the dot/inner product of the amplitudes 2*|A||B|cos(theta)
with a phase difference theta, the angle that we use to describe
analoguously as the set overlap (orthogonality) of A and B, or in
Dirac's notation <A|B>=0 if theta=90. But that is not in general
the case as the cross-correlation |A><B| is the value we often look for:
http://aurora.phys.utk.edu/~forrest/papers/fourier/index.html#correlation
And:
"If the input to the FFT is real, the Nyquist frequency index n/2 in the
output will always have a real value (meaning the imaginary part will be
zero, or something really close to zero due to floating-point roundoff
errors).
This is because, using the formula above, i' = n - n/2 = n/2, so the
Nyquist frequency is its own negative frequency counterpart. Therefore, it
must equal its own complex conjugate, which in turn forces it to be a real
number."
from: http://www.intersrv.com/~dcross/fft.html
where we get all that stuff about Zero-Point energy (h * f/2) ?
But Bonferroni's Inequality just states P(A union B) = 1
so 1 >= P(A) + P(B) - P(A and B). But these are the probabilities that
are the squared modulus of the amplitudes which can be positive or
negative. In classical probability theory, I'm thinking the
negative amplitudes may be best introduced in terms of the Fourier
transform applied to the maximum entropy formalism over ensembles
(identically prepared systems).
--
http://www.bestweb.net/~ca314159/
Jim Carr
Michael Weiss <colu...@nifty.camb.opengroup.org> writes:
>
>I suspect that ca314159 will not be convinced, Jim.
Suspect?
>Unlike the birefringent crystal case, changing the orientation of the
>prism doesn't change the choice of generalized basis--- it's still the
>continuum of pure frequencies.
The key difference; thanks for pointing it out succinctly.
>So the "pure frequency coordinate system" seems kind of special, more
>fundamental somehow than other generalized bases that one can cook
>up. Ultimately I guess this comes from the central importance of
>energy in physics.
For photons, frequency is energy so it is more than a guess. Of
course, the energy is not invariant so you could use doppler shifts
to alter the spectrum. You could think of that as a basis change,
but it remains a poor analogy because you can't rotate energy
through 2 pi.
ca31...@bestweb.net
In article <6h8gnf$8gc$1...@news.fsu.edu>,
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
>
> Michael Weiss <colu...@nifty.camb.opengroup.org> writes:
> >
> >I suspect that ca314159 will not be convinced, Jim.
>
> Suspect?
>
> >Unlike the birefringent crystal case, changing the orientation of the
> >prism doesn't change the choice of generalized basis--- it's still the
> >continuum of pure frequencies.
In the continuum of frequencies, each specific frequency is a
pure state, but polarizers are not treated this way. The 'pure
states' there, are not really pure states but aggregates of
the pure polarization states of individual photons which you
cannot observe with a typical polarizer. You need the quantum zeno
effect to get at these really pure polarization states. That's
why the quantum non-demolition works when photons are prepared
from the QZE. Because these photons are in a pure polarization state
in the same sense as we talk about pure frequency states.
> The key difference; thanks for pointing it out succinctly.
>
> >So the "pure frequency coordinate system" seems kind of special, more
> >fundamental somehow than other generalized bases that one can cook
> >up. Ultimately I guess this comes from the central importance of
> >energy in physics.
>
> For photons, frequency is energy so it is more than a guess. Of
> course, the energy is not invariant so you could use doppler shifts
> to alter the spectrum. You could think of that as a basis change,
> but it remains a poor analogy because you can't rotate energy
> through 2 pi.
Are sound frequencies superpositions ? You can separate them
out with a fourier transform just like a prism. Same with
radio waves.
If you are to show that light is different from radio waves,
then why is light considered a mixture of frequencies
while radio waves are considered a superposition ?
If you call radio waves a mixture of in-phase coherent photons
that brings you back to the dualism where you can treat them
as mixtures and as superposition depending on how you look at
the problem in terms of statistical identicality and indistinguishability
of both the individual particles and their macro and micro states.
You bring up the doppler shift as a kind of basis change but that's
really a translation. If you group the frequencies into buckets
that would a basis change, but each bucket would have a range of
frequencies. The resulting subspace, begins to look like the
subspaces you get from polarizers. It is an orthogonal basis if there is
no probability of a photon with a specific frequency to wind up
in more than one bucket (macrostate).
What's interesting though is that the doppler shift has a corrsponding
aberrration which is angular. So that relativity theory itself seems
to have a built in dispersion (frequency change+angular change).
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
|
| Michael Weiss <colu...@nifty.camb.opengroup.org> writes:
| >I suspect that ca314159 will not be convinced, Jim.
|
| Suspect?
|
| >Unlike the birefringent crystal case, changing the orientation of the
| >prism doesn't change the choice of generalized basis--- it's still the
| >continuum of pure frequencies.
ca31...@bestweb.net writes:
>
> In the continuum of frequencies, each specific frequency is a
> pure state, but polarizers are not treated this way.
Yes. I know this, Michael knows this, and you _say_ this.
Then you go off and construct some prism diagram that contradicts
what you say. This tells me that you do not understand what you
say up above, despite our attempts to explain this point, and
others. When you say "Z", and then we patiently explain "A",
only to have you reply "Of course, A" but follow it with "not A,
Z", I see little point in the effort.
> If you are to show that light is different from radio waves, ...
Why would anyone want to do that? Sheesh.
Janne Pesonen
ca31...@bestweb.net wrote:
> Complex numbers are already used by electricians.
> They know when to throw out the imaginary parts after
> they're done turning the crank. I've heard of uses for
> complex numbers where the imaginary values actually have
> a physical interpretation in actual measurements.
> That's because all the nice rotational properties were
> mapped over onto the physical problem.
> Complex numbers are a tool. We can use all or part of them.
Cheers,
Following articles of the complex numbers in QM written by David Hestenes might
interest you:
"Vectors, spinors and complex numbers in classical and quantum physics" (Am. J.
Phys. 39 (1971) pgs. 1013-1027)
"Observables, operators and complex numbers in Dirac theory" (J. Math. Phys. 16
(1975) pgs. 556-572)
"Spin and uncertainty in the interpretation of quantum mechanics" (Am. J. Phys.
47 (1979) pgs. 399-415)
Yours,
Janne Pesonen
http://fkmarilyn.pc.helsinki.fi/Janne/index.html
Jim Hunter
Janne Pesonen wrote:
>
> ca31...@bestweb.net wrote:
>
> > Complex numbers are already used by electricians.
> > They know when to throw out the imaginary parts after
> > they're done turning the crank. I've heard of uses for
> > complex numbers where the imaginary values actually have
> > a physical interpretation in actual measurements.
> > That's because all the nice rotational properties were
> > mapped over onto the physical problem.
> > Complex numbers are a tool. We can use all or part of them.
All numbers are tools.
Take 1 wire and place it next to another wire. That makes
2 wires. Put electro-juice in both of them and you have
a complex voltage.
---
Jim