OK, so Salamin wants a L + b R to be transformed to c L + d R where a' c + b' d = 0 where prime denotes complex conjugation. Because quantum mechanics is linear and unitary, (c,d) has to be got by multiplying (a,b) by a unitary 2x2 matrix U. So the question becomes: Does there exist a 2x2 unitary matrix U which converts any complex 2-vector to one complex-orthogonal to it? If a, b both real, then a solution is, as I'd said, c=-b, d=a and the matrix is 2x2 rotation by 90. If a,b both imaginary, then same solution also works. The trouble is if a,b general complex then the real-imag cross terms + ai br - ar bi + ai br - ar bi = 2 (ai br - ar bi) kill you (i.e. in general this is nonzero) hence the proposed solution fails because the imaginary part of the dot product, is in general nonzero. We can PROVE impossibility as follows. Wlog U is diagonal since we may take the eigendecomposition, because unitary matrices are normal hence this yields a diagonal because any normal matrix is unitarily similar to a diagonal matrix. (Note, the "complex dot product" is invariant under unitary transformations.) But obviously, no diagonal U can work. QED. On 6/4/15, math-fun-request@mailman.xmission.com <math-fun-request@mailman.xmission.com> wrote:
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Today's Topics:
1. Re: A polarization puzzle (Allan Wechsler) 2. Re: A polarization puzzle (Fred Lunnon) 3. Re: A polarization puzzle (James Propp) 4. Re: A polarization puzzle (Mike Stay)
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Message: 1 Date: Thu, 4 Jun 2015 09:38:02 -0400 From: Allan Wechsler <acwacw@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] A polarization puzzle Message-ID: <CADy-sGHBVqwuETFjYJB84biNFuL5VzmE_tyVH2JJd-a3uuMzmQ@mail.gmail.com> Content-Type: text/plain; charset=UTF-8
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincar? sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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Message: 2 Date: Thu, 4 Jun 2015 15:48:02 +0100 From: Fred Lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] A polarization puzzle Message-ID: <CAN57YqsE4xX5t3SyFuDRBniemVfzmmS7cEy5Gj5HoMkqvFW2Uw@mail.gmail.com> Content-Type: text/plain; charset=UTF-8
Join the club! I found this link enlightening, but it's surely only a start --- particularly regarding quantum-theoretical aspects (which I haven't even attempted to address yet).
https://en.wikipedia.org/wiki/Polarization_%28waves%29
WFL
On 6/4/15, Allan Wechsler <acwacw@gmail.com> wrote:
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincar? sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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Message: 3 Date: Thu, 4 Jun 2015 10:53:06 -0400 From: James Propp <jamespropp@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] A polarization puzzle Message-ID: <CA+G9J-cWv9XbZYFwjKKqKnY09BPZh83Bqr9BaQjJYP5Ke8c7OA@mail.gmail.com> Content-Type: text/plain; charset=UTF-8
While we're at it, Feynman figured out something about polarization of light and complex numbers that he seemed to view as the best illustration of why complex numbers are "real". An attempt at a lay explanation appears in Jim Ottaviani and Leland Myrick's 2011 graphic biography "Feynman". I couldn't make sense of it.
Jim Propp
On Thu, Jun 4, 2015 at 10:48 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Join the club! I found this link enlightening, but it's surely only a start --- particularly regarding quantum-theoretical aspects (which I haven't even attempted to address yet).
https://en.wikipedia.org/wiki/Polarization_%28waves%29
WFL
On 6/4/15, Allan Wechsler <acwacw@gmail.com> wrote:
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincar? sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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Message: 4 Date: Thu, 4 Jun 2015 08:47:33 -0700 From: Mike Stay <metaweta@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] A polarization puzzle Message-ID: <CAKQgqTZ9-Nt4StxNobOH+M36yK1jqWJhj4LeqOcB6PXyKv3-Lg@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
Spin is a property of individual quantum particles. It arises from considering angular momentum in a quantized setting. There are spin-0 particles like pions, spin-1/2 particles like electrons, protons, and neutrons, spin-1 particles like photons, spin-3/2 particles like Delta baryons, and spin-2 particles like gravitons. We can also treat collections of particles like nuclei as having spin.
A spin-0 particle has only one polarization. A spin-1/2 particle has two polarizations, +1/2 = "up" and -1/2 = "down". The Stern-Gerlach experiment shoots a beam of electrons through a magnetic field at a detector; you get two bright dots. A spin-1 particle has three polarizations, -1, 0, 1. In general, a spin s particle has 2s+1 polarizations: -s, -s+1, ..., s-1, s.
Pauli predicted the neutrino in part because a neutron can decay via the beta process to a proton and an electron---but that meant the total spin for the neutron should be 1, not 1/2 as had already been observed. Pauli added the neutrino so that the spin of a neutron plus a neutrino would equal the spin of a proton plus an electron.
In quantum mechanics, we have a Hilbert space with one dimension for each possible outcome. A photon is a spin-1 particle, but outside of a waveguide, we never see the 0 polarization, so the state of an individual photon lives in a 2-dimensional Hilbert space. We can write the polarization state of a free photon as a complex-weighted sum of the left and right circular polarizations: a|L> + b|R>. The condition |a|^2 + |b|^2 = 1 reduces the four degrees of freedom (two per complex coefficient) to three; this is what's known as the Riemann sphere.
The linear polarizations are |H> = (|L> + i|R>)/sqrt(2) and |V> = (|L> - i|R>)/sqrt(2). The inner product of the two is denoted <H|V>, where <H| = <L| - i*<R| denotes the conjugate transpose of |H>. The result is (1*1 + (-i)*(-i))/2 = 0, so they're orthogonal, whereas <H|H> = <V|V> = (1*1 + (-i)*i)/2 = 1.
On Thu, Jun 4, 2015 at 6:38 AM, Allan Wechsler <acwacw@gmail.com> wrote:
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincar? sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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