Polarization questions
|Agent
individual photon may be polarized with a left spin or a right spin.
En masse, a beam of light may be unpolarized, circularly polarized to
the left or right in various degrees, linearly polarized along some
axis perpendicular to its direction of travel, or some combination if
a beam of light hits a polarizer at an oblique angle.
My first question is about how individual photons' polarizations
relate to their polarization en masse. I am told (by [1] and more
briefly in [2]) that a linearly polarizing film produces photons that
are split evenly between left and right spin. I expect that the
circular polarizations have different ratios between the two. What
does it mean if the photons are almost all right-spun, or almost all
left-spun? Is it nothing more than a really tight circular
polarization of the beam?
My second question is about unpolarized light, as from the sun. Let's
limit this to a specific frequency of light. I was originally going to
suppose that each photon generated by the sun had a 50/50 chance of
being left- or right-polarized, so then the population of photons in a
beam of light would be similarly 50/50 split, and thus "unpolarized"
light is actually linearly polarized. But photons are quantum
mechanical and their spin polarity won't be settled until it needs to
be. If you pass a beam through a linearly polarizing filter, the
photons will take on a 50/50 split, and if you pass it through a
circularly polarizing filter, it will take on some other ratio. It
would be more correct to say that natural light is not "unpolarized,"
but rather of "undetermined polarity." Is this right?
My third question relates to the experiment we've all done (or at
least heard of) where you pass light through a linearly polarizing
film and then through second film. If the second film is oriented one
way, you see the light; if you rotate it by 90°, it goes dark. The
classical explanation says that the first film filters out the
vertical components of the electromagnetic wave and the second film
filters out the horizontal components, leaving nothing. But I cannot
relate that to photons; they have spin polarities, not horizontal or
vertical components. How does that work?
Maybe it has nothing to do with their spin or polarity in the quantum
sense. Maybe it is a question of their position. If the polarizing
film absorbs photons in parallel bands, then it is like looking
through a picket fence: you can only see between the slats, and if you
put another fence behind that with perpendicular slats, you cannot see
anything.
But that cannot be the explanation, because if it were, why would
light reflected from water be linearly polarized?
[1] http://www.mathpages.com/rr/s9-04/9-04.htm
[2] http://www.physicsforums.com/showthread.php?t=180282
Autymn D. C.
a diaelèctrèt) is a group of straiht antennæ absorbent elèctric waves
straihtways (rather than thwartways), so it works against your picket
fence analoghy. A wave's (fotòn's) orientation follows its mote's
(elèctròn's); a charge a'bobbing up and down will induce charges in
other materials to do the same.
-Aut
Tom Roberts
> I have a few questions about polarized photons. As I understand it, an
> individual photon may be polarized with a left spin or a right spin.
Not quite. An individual photon can be in a quantum superposition of both left-
and right-handed spin states. This is the key idea you missed.
> En masse, a beam of light may be unpolarized, circularly polarized to
> the left or right in various degrees, linearly polarized along some
> axis perpendicular to its direction of travel, or some combination if
> a beam of light hits a polarizer at an oblique angle.
Each individual photon can be so polarized, depending on the relative amplitude
and phase of the two basis spin states (left- and right-handed) that make up its
actual spin state.
> My first question is about how individual photons' polarizations
> relate to their polarization en masse. I am told (by [1] and more
> briefly in [2]) that a linearly polarizing film produces photons that
> are split evenly between left and right spin. I expect that the
> circular polarizations have different ratios between the two. What
> does it mean if the photons are almost all right-spun, or almost all
> left-spun? Is it nothing more than a really tight circular
> polarization of the beam?
In a 100% right-handed circularly polarized beam, all photons have their spin
state 100% in right-handed spin. In a 100% linearly-polarized beam along X, all
photons are in a superposition of right- and left-handed spin states with equal
amplitudes and phases such that the result is linear polarization along the X
axis. In an unpolarized beam, the individual photons are in a multitude of spin
states randomly distributed among all possibilities.
In general one must use a density matrix to describe such a beam.
> My second question is about unpolarized light, as from the sun. Let's
> limit this to a specific frequency of light. I was originally going to
> suppose that each photon generated by the sun had a 50/50 chance of
> being left- or right-polarized, so then the population of photons in a
> beam of light would be similarly 50/50 split, and thus "unpolarized"
> light is actually linearly polarized. But photons are quantum
> mechanical and their spin polarity won't be settled until it needs to
> be. If you pass a beam through a linearly polarizing filter, the
> photons will take on a 50/50 split, and if you pass it through a
> circularly polarizing filter, it will take on some other ratio. It
> would be more correct to say that natural light is not "unpolarized,"
> but rather of "undetermined polarity." Is this right?
No. It's more complicated than that. In an unpolarized beam, the individual
photons' spin states are uncorrelated with each other and in general there are
photons present with all possible spin states. Putting such a beam through a
perfectly polarizing filter projects all photons' spin states onto the axis of
the filter; for an ideal filter this results in 50% of the photons being
absorbed by the filter, and those photons not absorbed have spin states
corresponding to the filter's projection.
> My third question relates to the experiment we've all done (or at
> least heard of) where you pass light through a linearly polarizing
> film and then through second film. If the second film is oriented one
> way, you see the light; if you rotate it by 90°, it goes dark. The
> classical explanation says that the first film filters out the
> vertical components of the electromagnetic wave and the second film
> filters out the horizontal components, leaving nothing. But I cannot
> relate that to photons; they have spin polarities, not horizontal or
> vertical components. How does that work?
It works in an inherently quantum-mechanical way: the first filter projects the
photons' spin states onto its axis, and the second does the same. The net
transmission depends on the angle between the filters' axes and the initial
polarization of the beam.
The usual proof of this: take two linearly polarizing filters
and orient them 90 degrees apart so no light gets through. Now
insert a third filter between them at 45 degrees, and the
transmission increases greatly. This is easily understood in
terms of the projection of spin states; it cannot be explained
classically.
> Maybe it has nothing to do with their spin or polarity in the quantum
> sense. Maybe it is a question of their position. If the polarizing
> film absorbs photons in parallel bands, then it is like looking
> through a picket fence: you can only see between the slats, and if you
> put another fence behind that with perpendicular slats, you cannot see
> anything.
No. It is INHERENTLY quantum mechanical. Position has nothing to do with it, and
this is not like a picket fence at all.
Also: for a picket fence with 50% slats and 50% gaps, crossing
two of them merely reduces the transmission to 25%, not zero --
there remain square holes between the pickets. But by using
two PARALLEL fences and aligning slat to gap you can reduce
transmission to zero. You can also get Moiré patterns. But this
is purely classical, and is unrelated to the quantum mechanics
of polarizing filters.
Tom Roberts
|Agent
> |Agent wrote:
> > I have a few questions about polarized photons. As I understand it, an
> > individual photon may be polarized with a left spin or a right spin.
>
> Not quite. An individual photon can be in a quantum superposition of both left-
> and right-handed spin states. This is the key idea you missed.
>
> > limit this to a specific frequency of light. I was originally going to
> > suppose that each photon generated by the sun had a 50/50 chance of
> > being left- or right-polarized, so then the population of photons in a
> > beam of light would be similarly 50/50 split, and thus "unpolarized"
> > light is actually linearly polarized. But photons are quantum
> > mechanical and their spin polarity won't be settled until it needs to
> > be. If you pass a beam through a linearly polarizing filter, the
> > photons will take on a 50/50 split, and if you pass it through a
> > circularly polarizing filter, it will take on some other ratio. It
> > would be more correct to say that natural light is not "unpolarized,"
> > but rather of "undetermined polarity." Is this right?
>
> No. It's more complicated than that. In an unpolarized beam, the individual
> photons' spin states are uncorrelated with each other and in general there are
> photons present with all possible spin states. Putting such a beam through a
> perfectly polarizing filter projects all photons' spin states onto the axis of
> the filter; for an ideal filter this results in 50% of the photons being
> absorbed by the filter, and those photons not absorbed have spin states
> corresponding to the filter's projection.
I see. So there is another layer of probability that lies within an
individual photon's spin state that I missed. Even after being passed
through a linearly polarizing filter, the photon's spin is still not
definitely determined; it has 50/50 odds of being left-spun or right-
spun if it were ever required to be one or the other. Whereas, an
unpolarized photon has unknown odds of ending up as one or the other.
But I am not sure how a spin state can correspond to an axis, and if
or how it is important that 50% of the photons are absorbed by an
ideal filter.
As an aside from the main question, am I correct in believing that a
population of unpolarized photons, if all were required to pick a
spin, would end up in a 50/50 ratio anyway? Though I understand that
THAT 50/50 ratio would be consequence of population probability, and
not borne of any quantum interactions of the individual photons of the
population, if there can even be said to BE individual photons before-
hand.
> > My third question relates to the experiment we've all done (or at
> > least heard of) where you pass light through a linearly polarizing
> > film and then through second film. If the second film is oriented one
> > way, you see the light; if you rotate it by 90°, it goes dark. The
> > classical explanation says that the first film filters out the
> > vertical components of the electromagnetic wave and the second film
> > filters out the horizontal components, leaving nothing. But I cannot
> > relate that to photons; they have spin polarities, not horizontal or
> > vertical components. How does that work?
>
> It works in an inherently quantum-mechanical way: the first filter projects the
> photons' spin states onto its axis, and the second does the same. The net
> transmission depends on the angle between the filters' axes and the initial
> polarization of the beam.
>
> The usual proof of this: take two linearly polarizing filters
> and orient them 90 degrees apart so no light gets through. Now
> insert a third filter between them at 45 degrees, and the
> transmission increases greatly. This is easily understood in
> terms of the projection of spin states; it cannot be explained
> classically.
Let me get this straight. A photon's spin state can only be left or
right in some superposition of the two. I see 1 qubit of information
there; I do not see room for arbitrary axis information. But I suppose
there is room for axis 1 and perpendicular axis 2, if each axis maps
to one of the two available spin states.
Now let's say we have a beam of light traveling along Z and two
filters oriented on an arbitrary X and perpendicular Y. This process
of projecting on an axis…are you saying that the way it works out, the
X-axis filter will take on a correspondence to one of the two spin
states, and the Y-axis filter will take on a correspondence to the
other? And each filter will depress the odds of its corresponding spin
state, so that with both filters present the odds of a photon with any
spin state at all approach 0 and thus almost all the photons get
absorbed?
If so, I have trouble with this. How does this two filters…I find I
cannot avoid terms of agency here…how does the Y-axis filter know that
the X-axis filter has taken one of the spin states so that the Y-axis
filter has to take the other one? They don't have agency, obviously,
nor does the two-filter-and-light-beam system as a whole, but how does
the system assign spin states to filters such that one filter gets one
and a perpendicular filter gets the other?
Tom Roberts
> On Aug 13, 9:22 am, Tom Roberts <tjrob...@sbcglobal.net> wrote:
>> An individual photon can be in a quantum superposition of both left-
>> and right-handed spin states. This is the key idea you missed.
>
> individual photon's spin state that I missed.
No. This is not "probability", this is a mixed state. That cannot be represented
by mere probability, one must use the superposition of basis states; these are
complex amplitudes, the square of which gives probability.
> Even after being passed
> through a linearly polarizing filter, the photon's spin is still not
> definitely determined; it has 50/50 odds of being left-spun or right-
> spun if it were ever required to be one or the other.
This depends on what you mean. After an ideal linear-polarizing filter, the
photon's spin state is completely determined to be the mixed state corresponding
to the filter. But yes, if you then measure the individual photon spins in the
beam, you get 50% left and 50% right, because that's the composition of that
mixed state projected onto that helicity basis.
> Whereas, an
> unpolarized photon has unknown odds of ending up as one or the other.
Well, no. When you measure individual photon spins in an unpolarized beam of
photons you also get 50% left and 50% right.
The difference between an unpolarized beam and a linearly-polarized beam is that
in the former the photons' spins are uncorrelated, while in the latter they are
100% correlated with each other: every one is the particular mixed state
corresponding to the axis of polarization.
This is a correlation between the mixed spin states, not between
left- and right- helicity states -- those are not correlated
among the photons.
> As an aside from the main question, am I correct in believing that a
> population of unpolarized photons, if all were required to pick a
> spin, would end up in a 50/50 ratio anyway?
Yes. (Assuming that by "pick a spin" you mean a measurement of their spin.)
> Though I understand that
> THAT 50/50 ratio would be consequence of population probability, and
> not borne of any quantum interactions of the individual photons of the
> population, if there can even be said to BE individual photons before-
> hand.
Yes, the genesis of the 50/50 left/right ratio in an unpolarized beam is indeed
different from that in a linearly-polarized beam.
>> The usual proof of this: take two linearly polarizing filters
>> and orient them 90 degrees apart so no light gets through. Now
>> insert a third filter between them at 45 degrees, and the
>> transmission increases greatly. This is easily understood in
>> terms of the projection of spin states; it cannot be explained
>> classically.
>
> Let me get this straight. A photon's spin state can only be left or
> right in some superposition of the two.
No. A photon's spin can point in any direction except perpendicular to its
trajectory, and can be a complex superposition of any set of basis states you
choose (two basis states span the spin space of a single photon). The usual
basis uses left- and right-hand helicity (spin along the direction of motion),
but that is not at all the only possibility; for definiteness I'm only using
that basis here.
> I see 1 qubit of information
> there; I do not see room for arbitrary axis information.
But there is axis information for linear polarization. It's more complicated
than you think.
> Now let's say we have a beam of light traveling along Z and two
> filters oriented on an arbitrary X and perpendicular Y. This process
> of projecting on an axis…are you saying that the way it works out, the
> X-axis filter will take on a correspondence to one of the two spin
> states, and the Y-axis filter will take on a correspondence to the
> other?
No. The first filter selects out a specific mixture of the two basis states,
left- and right-handed helicity. The second selects out a different mixture;
since their axes are 90 degrees apart, these two selections are orthogonal and
the result is zero transmission.
> And each filter will depress the odds of its corresponding spin
> state, so that with both filters present the odds of a photon with any
> spin state at all approach 0 and thus almost all the photons get
> absorbed?
For ideal filters all photons get absorbed, but your description is incorrect.
> If so, I have trouble with this. How does this two filters…I find I
> cannot avoid terms of agency here…how does the Y-axis filter know that
> the X-axis filter has taken one of the spin states so that the Y-axis
> filter has to take the other one? They don't have agency, obviously,
> nor does the two-filter-and-light-beam system as a whole, but how does
> the system assign spin states to filters such that one filter gets one
> and a perpendicular filter gets the other?
They need know nothing about each other. The first (ideal) filter passes only
one particular spin state, corresponding to linear polarization along its axis.
The second does the same, but with a different axis. When their axes are 90
degrees apart, the overlap between the states they pass is zero, so no photons
get through them both.
With three filters, the outer two 90 degrees apart and the middle
one at 45 degrees, the first passes the spin state corresponding
to its axis. The second one does so too, and because they are 45
degrees apart the overlap is nonzero. After this filter all photons
are polarized along its axis. The third filter does the same, but
now it is 45 degrees from the incoming photons, so again the
overlap is nonzero. Exiting photons are polarized along the axis
of the last filter; the orientations of the earlier filters have
been "forgotten", except for their effect on intensity.
This is subtle and counter-intuitive, because your intuition was developed in
the classical realm, not with quantum objects like this. The only way to truly
understand it is to study quantum mechanics. There is no shortcut.
Tom Roberts
|Agent
> |Agent wrote:
> > On Aug 13, 9:22 am, Tom Roberts <tjrob...@sbcglobal.net> wrote:
> > Let me get this straight. A photon's spin state can only be left or
> > right in some superposition of the two.
>
> No. A photon's spin can point in any direction except perpendicular to its
> trajectory, and can be a complex superposition of any set of basis states you
> choose (two basis states span the spin space of a single photon). The usual
> basis uses left- and right-hand helicity (spin along the direction of motion),
> but that is not at all the only possibility; for definiteness I'm only using
> that basis here.
>
> > I see 1 qubit of information
> > there; I do not see room for arbitrary axis information.
>
> But there is axis information for linear polarization. It's more complicated
> than you think.
Ah, okay. This makes much more sense. I hadn't heard much about spin
axis; I thought they simply meant axis of travel. I was not aware it
could differ. Thanks for your explanations!