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I learned that light can be linearly and circularly polarized. It can also be a mix of several simultaneously. Is there also a polarization state that is neither linear nor circular? (ie not polarized?) If so, what is it?
An unpolarized photon is not in a pure state of polarization but in a mixture of different states, e.g. 50% V and 50% H, which is quite different from V+H (For example, V+H will pass with 100% probability through a polarizer at 45 degrees, but an unpolarized photon has 50% probability to pass) Mathematically a pure state is described by a vector in a "Hilbert space", but a mixed state is described by a density matrix.
I will take this opportunity to have so light shed on something that isn´t clear to me, and this context is a good one to me: How do you prepare a mixed state (in this case, a photon that has probability H to be horizontally polarized, and a probability V to be vertically polarized, and how is this physically possible??StevieTNZ said:From an email from Professor Giovanni Vignale sent to me December 2011:
An unpolarized photon is not in a pure state of polarization but in a mixture of different states, e.g. 50% V and 50% H, which is quite different from V+H (For example, V+H will pass with 100% probability through a polarizer at 45 degrees, but an unpolarized photon has 50% probability to pass) Mathematically a pure state is described by a vector in a "Hilbert space", but a mixed state is described by a density matrix.
So, since the trace of the density matrix must be 1, this means that all four polarization modes must be prepared with a contribution ¼?Khashishi said:Unpolarized light is a balanced mixture of polarized light. It could be a mixture of vertical and horizontally polarized light or a mixture of left circularly and right circularly polarized light or something else-- it doesn't matter. There's no way to tell the difference between the different mixtures.
Just take thermal light, the sort that enlightens you, literally. It is already unpolarized, hence prepared as desired.entropy1 said:How do you prepare a mixed state
entropy1 said:I will take this opportunity to have so light shed on something that isn´t clear to me, and this context is a good one to me: How do you prepare a mixed state (in this case, a photon that has probability H to be horizontally polarized, and a probability V to be vertically polarized, and how is this physically possible??
So each photon behaves like it is not polarized, because their polarizations are uniformly randomly distributed?stevendaryl said:Mixed states aren't hard to understand: they can be interpreted as just ignorance as to the true state. Imagine a polarizing filter whose orientation was rapidly changing in an unpredictable way. The light that passes through it would have a random polarization, which would be described as a mixture. That's different from a superposition, because a superposition of H-polarized light and V-polarized is light with a definite polarization at an angle between horizontal and vertical.
entropy1 said:Is thermal light a mixture of polarized photons? Eg, do there even exist unpolarized photons?? Or is a mixed state a physically distinct state per photon?
entropy1 said:So each photon behaves like it is not polarized, because its polarization is uniformly randomly distributed?
This works only in certain cases. If you have a beam of unpolarized light, you cannot know its true pure state - it has none! The preparation procedure decides the state, and it is a mixed state from the very beginning.stevendaryl said:Mixed states aren't hard to understand: they can be interpreted as just ignorance as to the true state.
A. Neumaier said:This works only in certain cases. If you have a beam of unpolarized light, you cannot know its true pure state - it has none! The preparation procedure decides the state, and it is a mixed state from the very beginning.
Being unpolarized is an objective fact about sunlight. If it were polarized (though unknown to an observer with less than maximal knowledge) it could be found out by passing it through various polarization filters and observe how much the intensity decreases. But the intensity decreases by the same amount no matter how you orient it. This proves it is unpolarized.stevendaryl said:Being mixed is a subjective fact about a photon. There is no experiment that can distinguish between a mixed photon and a pure-state photon.
If you make a counting statistics there is a difference.stevendaryl said:Okay, but if you pass that light through a polarizing filter, one photon at a time, there is no experimental difference between unpolarized photons and photons of unknown polarization.
A. Neumaier said:Being unpolarized is an objective fact about sunlight. If it were polarized (though unknown to an observer with less than maximal knowledge) it could be found out by passing it through various polarization filters and observe how much the intensity decreases.
A. Neumaier said:If you make a counting statistics there is a difference.
...and is that the same (also) as a mixed state?stevendaryl said:The other possibility is that the photons all have different unknown polarizations.
entropy1 said:...and is that the same (also) as a mixed state?
The assumption is meaningless.stevendaryl said:That only proves that the photons from the sun don't have the SAME (unknown) polarization. If a process generated randomly polarized photons, the output would be indistinguishable from unpolarized light. Wouldn't it?
You make me curious! What do you mean by ¨usually¨?stevendaryl said:Yes, that's what a mixed state is...well, usually.
To say that all photons have random and unknown (unknowable!) polarizations is saying that they are in the unpolarized state - they have no polarization at all. There is no other possible operational meaning to your statement. Pretending that they have a definite polarization without being able to check it is irrelevant unscientific talk (Ockham's razor).stevendaryl said:The other possibility is that the photons all have different unknown polarizations.
But if that was the case, we would effectively have the same situation, right? (with respect to what we (can) know?)A. Neumaier said:To say that all photons have random and unknown (unknowable!) polarizations is saying that they are in the unpolarized state - they have no polarization at all. There is no other possible operational meaning to your statement. Pretending that they have a definite polarization without being able to check it is irrelevant unscientific talk (Ockham's razor).
A. Neumaier said:To say that all photons have random and unknown (unknowable!) polarizations is saying that they are in the unpolarized state
If it were the case it would look the same. But it would also look the same if it were only left- and right-polarized photons, or only (3|left>+4|right>)/5 and (4|left>-3|right>)/5 polarized photons, or many other possibilities. These are just different ways to prepare unpolarized light if you need to make it from polarized light.entropy1 said:But if that was the case, we would effectively have the same situation, right? (with respect to what we (can) know?)
No. I was arguing against your use of meaningless phrases. I am saying that the only operational sense one can make of your statement is that each of the prepared photons is unpolarized - it makes no sense to assign a random polarization to each single photon, only the prepared non-pure state. Unknowable means ''non-pure'' - this is not the same as ''definitely pure but unknown and different state in each instance''.stevendaryl said:Then you're agreeing with me, since that's what I was saying.
So, in case of thermal light, we can´t know if the individual photons have polarizations or not, for QM does not answer that question, and therefore it is not relevant?A. Neumaier said:But the state is always defined by the preparation procedure, and it makes no operational sense to ascribe a different state to each single photon if in fact a stream of photons is prepared.
entropy1 said:You make me curious! What do you mean by ¨usually¨?
A. Neumaier said:No. I was arguing against your use of meaningless phrases.
Operationally, we cannot collect more information that the coefficients in the density matrix. Thus the shut-up-and-calculate mode in which predictions are made and experiments are interpreted only speaks about that information. Nothing more can be predicted and tested, so everything else is irrelevant (and beyond science).entropy1 said:So, in case of thermal light, we can´t know if the individual photons have polarizations or not, for QM does not answer that question, and therefore it is not relevant?
Any statistical operator can be written in the first way, because it's self-adjoint and thus you can express it in terms of its eigenstates. Then the ##|\psi_j \rangle## are even a complete orthonormal system.stevendaryl said:Well, there are two different ways that mixed states arise in quantum mechanics.
Proper mixed states: You have some process that puts a system into an unknown pure state. In that case, you can describe the system using a mixed state: [itex]\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |[/itex], where [itex]p_j[/itex] is the probability that it is in state [itex]|\psi_j\rangle[/itex]. The "mixed state" reflects our ignorance about the state.
Improper mixed states: If you have two systems that are entangled, then they would be described as a superposition of composite pure states of the form:
[itex]|\Psi\rangle = \sum_{i \alpha} C_{i\alpha} |\psi_i\rangle |\phi_\alpha \rangle[/itex], where [itex]|\psi_i\rangle[/itex] is a complete set of basis states for the first system and [itex]|\phi_\alpha\rangle[/itex] is a complete set of basis states for the second. Now, if the second system is unobservable, then it's inconvenient to include it in the description. In that case, we can get rid of the second system by tracing over its degrees of freedom.
A. Neumaier said:Unknowable means ''non-pure'' - this is not the same as ''definitely pure but unknown and different state in each instance''.
vanhees71 said:Any statistical operator can be written in the first way, because it's self-adjoint and thus you can express it in terms of its eigenstates. Then the ##|\psi_j \rangle## are even a complete orthonormal system.
The difference is described by Ockham's razor - there is no point in postulating something unknowable. Your description is far too detailed and highly ambiguous since one cannot even infer the probability distribution of the different unknown polarizations. Whereas ''unpolarized'' says everything without postulating irrelevant detail.stevendaryl said:There is no difference between unpolarized light and light where the photons have different unknown polarizations.