High School What is ¨unpolarized¨ light exactly?

[A. Neumaier]

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?

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).

However, different interpretations of quantum mechanics may make additional (uncheckable) assertions beyond that. They are operationally meaningless but can be used as props for the intuition. For example, the Copenhagen interpretation (on which stevendaryl bases his arguments) assigns a pure state to each single state and then has to interpret the density matrix as something subjective due to lack of knowledge. But from what is operationally verifiable it is the opposite: the density matrix contains the full knowable information while the additional information in the assumed pure states - being unobservable - is subjective and spurious.

The situation is slightly different if the light is very dim, so that photons are created so slowly that one can change the polarizer settings before each new photon arrives. Then one can test the state of each photon separately. (This works only if one knows the state from the preparation procedure since one can make only one measurement per photon. Thus one needs ''photons on demand'' or ''heralded photons''.) But even in this case, the state of each prepared photon will typically still be a mixed state. See the photons on demands analysis in the slides mentioned in post #2.

 

[vanhees71]

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: \rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |, where p_j is the probability that it is in state |\psi_j\rangle. 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:
|\Psi\rangle = \sum_{i \alpha} C_{i\alpha} |\psi_i\rangle |\phi_\alpha \rangle, where |\psi_i\rangle is a complete set of basis states for the first system and |\phi_\alpha\rangle 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.

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.

In the second example you simply have a reduced density operator of a pure state of a composite system, i.e., you use the partial trace
$$\hat{\rho}=\mathrm{Tr}_B |\Psi \rangle \langle \Psi|.$$
It's again a self-adjoint trace-class operator with $\mathrm{Tr} \hat{\rho}$ in the Hilbert space of the partial system $A$.

 

[stevendaryl]

Unknowable means ''non-pure'' - this is not the same as ''definitely pure but unknown and different state in each instance''.

I think you're mixing your philosophy up with your physics. There are times when a mixed state is a necessity, as when you are observing one subsystem of an entangled composite system. If that's what you mean, then it would be more useful to say that explicitly.

 

[stevendaryl]

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 isn't in the resulting density matrix, but how that density matrix was computed. Was it computed by statistical averaging over initial states, or was it created by tracing out unobservable degrees of freedom? There is no difference in the resulting density matrix.

 

[A. Neumaier]

There is no difference between unpolarized light and light where the photons have different unknown polarizations.

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.

 

[A. Neumaier]

There are times when a mixed state is a necessity, as when you are observing one subsystem of an entangled composite system.

Every system we look at is such a system!

 

[A. Neumaier]

The difference isn't in the resulting density matrix, but how that density matrix was computed. Was it computed by statistical averaging over initial states, or was it created by tracing out unobservable degrees of freedom? There is no difference in the resulting density matrix.

Nature doesn't compute - so your difference is a human artifact.

 

[stevendaryl]

The difference is described by Ockham's razor

That's philosophy, not physics.

I can set up a situation in which photons are produced with random polarizations. I create a random number generator, and then I use it to randomly pick an orientation for a polarizing filter. When I pass photons through that filter, the photons will have random polarizations. If nobody records the orientations, then they will be random unknown polarizations. This sequence of photons will be indistinguishable (at least if the random number generator is very good) from unpolarized light. Is it REALLY unpolarized light? I can't tell from your posts whether you would consider it unpolarized, or not. But in this scenario, it really is the case that the photons have definite, though unknown, polarizations.

That's a different scenario from the use of entangled photons. In the case of entangled photons, there is a sense in which the individual photons don't have polarizations (rather than having unknown polarizations).[/QUOTE]

 

[stevendaryl]

Nature doesn't compute - so your difference is a human artifact.

It would be nice if you could tag your posts so that I would know which posts have actual information, and which ones are just you being grumpy.

[edit]How about this emoji?

 

[A. Neumaier]

It would be nice if you could tag your posts so that I would know which posts have actual information, and which ones are just you being grumpy.

Then I'd have to tag all my posts with the same tag. That's why I am using a blank as tag.

 

[stevendaryl]

Then I'd have to tag all my posts with the same tag. That's why I am using a blank as tag.

That's too modest. Many of your posts are not grumpy.

 

[A. Neumaier]

When I pass photons through that filter, the photons will have random polarizations. If nobody records the orientations, then they will be random unknown polarizations. This sequence of photons will be indistinguishable (at least if the random number generator is very good) from unpolarized light. Is it REALLY unpolarized light?

Your very artificial preparation procedure - never used in practice - produces approximately unpolarized light.
Approximate only since if your photon rate is much higher than the number of switches and the observer switches its polarizers at a significantly higher rate, the state of each bunch with fixed polarization setting can be determined, and the switching times. At very low intensity it becomes experimentally indistinguishable form unpolarized light.

But having an artificial source of your kind says nothing about sources in nature. You may pretend that one has the same situation in Nature but this is in principle not checkable, hence is no science but subjective belief.

 

[vanhees71]

Everything known about the system is in the statistical operator, i.e., in the state the system is prepared in. If different preparation procedures lead to the same state, you cannot know about the different preparation procedures just observing the so prepared system.

 

[stevendaryl]

Everything known about the system is in the statistical operator, i.e., in the state the system is prepared in. If different preparation procedures lead to the same state, you cannot know about the different preparation procedures just observing the so prepared system.

I wouldn't say that everything known about the system is in the statistical operator. I would say that everything that can be discovered about the system by examining just that system is in the statistical operator. You might know something about the preparation, which is not reflected in the density matrix. But that's not information that is available by just examining the system.

 

[vanhees71]

exactly!

 

[A. Neumaier]

You might know something about the preparation, which is not reflected in the density matrix. But that's not information that is available by just examining the system.

More importantly, it is information from the past that has no consequences at all for the future (unless one couples the system later with the source in which case some correlations might appear). That's why it is generally ignored, with overwhelming success.

 

[Khashishi]

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?

No, because there are only 2 independent polarizations. You can talk about 4 or more types of polarization, but they can all be written as scaled sums of two basis states. So, you can mix 1/2 vertical and 1/2 horizontal. Or you can mix 1/2 CW, 1/2 CCW and get unpolarized light. You can mix all four and still get unpolarized light, but you don't have to.

 

[StevieTNZ]

Eg, do there even exist unpolarized photons??

If, by unpolarized you mean a photon has no polarization at all, then no. Each photon of an ensemble of photons are either |H> or |V> polarized; we just don't know what photon has which, as per the email.

 

[N88]

If, by unpolarized you mean a photon has no polarization at all, then no. Each photon of an ensemble of photons are either |H> or |V> polarized; we just don't know what photon has which, as per the email.

1. Is this wording correct?
2. Is it not meaningless to talk of an UNpolarized photon?
3. Am I correct in thinking that #2 is the point that Prof. Neumaier makes?

 

[entropy1]

2. Is it not meaningless to talk of an UNpolarized photon?
3. Am I correct in thinking that #2 is the point that Prof. Neumaier makes?

Do I understand correctly then that measuring polarisation in general requires a collapse into an eigenstate, so that it is then unknown which polarisation the photon in question had? It becomes polarized in the direction of the eigenvector, so it might have been uniformly polarized or not at all polarized; there is no way to distinquish that?

 

[Khashishi]

It seems like almost every confusion in quantum mechanics boils down into a question about the interpretation of a collapse. We don't really have an answer. We can't pin down a time when the collapse occurs. It seems to occur when the system interacts with a macroscopic environment. So, if you pass a photon into a polarizer and then into a detector, you either measure a count or you don't. You can't say a photon becomes polarized. Reality isn't so simple. We can talk about what we finally measure. What happens in between is kind of fuzzy.

 

[A. Neumaier]

I prefer to avoid using the word photon; then things are much clearer. A beam of light can be unpolarized, partially polarized, or fully polarized, and there is no doubt about what this means. It becomes polarized when passed through a polarization fielter. When a beam falls upon a detector, the detector responds with random clicks whose rate is proportional o the intensity of the incident beam.

Populating the beam of light with photons doesn't add any explanatory value but creats a lot of confusion and puzzlement. Ockham's razor suggests that one better does not do this.

 

[N88]

I prefer to avoid using the word photon; then things are much clearer. A beam of light can be unpolarized, partially polarized, or fully polarized, and there is no doubt about what this means. It becomes polarized when passed through a polarization fielter. When a beam falls upon a detector, the detector responds with random clicks whose rate is proportional o the intensity of the incident beam.

Populating the beam of light with photons doesn't add any explanatory value but creats a lot of confusion and puzzlement. Ockham's razor suggests that one better does not do this.

I understand the need for caution when discussing "photons". But can we clarify this need for caution by moving the discussion to PROTONS?

In his article "Einstein-Podolsky-Rosen experiments", John Bell ('Speakable and unspeakable …', 2004: p.82) writes: "… each particle [PROTON], considered separately, IS UNpolarised here." The italicised IS is Bell's emphasis, yet earlier, on p.81 we have: "… filters that pass only particles [PROTONS] of a given polarization".

How do you represent and clarify these "entangled" matters?

 

[StevieTNZ]

1. Is this wording correct?
2. Is it not meaningless to talk of an UNpolarized photon?
3. Am I correct in thinking that #2 is the point that Prof. Neumaier makes?

An unpolarized photon is defined as per the email I quoted.
A photon which has no polarization, well, I'm not sure the technical term for that (if there is one)

 

[stevendaryl]

1. Is this wording correct?
2. Is it not meaningless to talk of an UNpolarized photon?
3. Am I correct in thinking that #2 is the point that Prof. Neumaier makes?

If a photon is entangled with something else (another photon, for instance), then there is a sense in which it really doesn't have a polarization until measured, as opposed to having an unknown polarization. At least, that's one possible conclusion from Bell's analysis of the EPR experiment.

 

[A. Neumaier]

particles [PROTONS] of a given polarization

Protons have spin, not polarization.

Talking of ''each proton'' in a proton beam as if each were an individual object creates the same sort of problems as talking of photons in a beam of light Talk of one proton makes operational sense only if one can point to this one proton.

 

[N88]

Protons have spin, not polarization.

Talking of ''each proton'' in a proton beam as if each were an individual object creates the same sort of problems as talking of photons in a beam of light Talk of one proton makes operational sense only if one can point to this one proton.

Thank you. I was citing Bell and seeking to understand his terminology in the context of a single particle (considered separately, photon or proton) being UNpolarized.

That is: Under Bell's influence, I was believing that there were such things as UNpolarized photons and protons. I would now be happy to be influenced by your view on this confusing (to me) terminology.

 

[A. Neumaier]

Thank you. I was citing Bell and seeking to understand his terminology in the context of a single particle (considered separately, photon or proton) being UNpolarized.

That is: Under Bell's influence, I was believing that there were such things as UNpolarized photons and protons. I would now be happy to be influenced by your view on this confusing (to me) terminology.

There are unpolarized beams of light (defined by a coherence matrix that is a multiple of the identity), and there are proton beams whose spin coherence matrix is a multiple of the identity. Figuratively, this is interpreted as that the photons or protons in the beam are unpolarized, meaning that their density matrix is half the identity.

 

[N88]

There are unpolarized beams of light (defined by a coherence matrix that is a multiple of the identity), and there are proton beams whose spin coherence matrix is a multiple of the identity. Figuratively, this is interpreted as that the photons or protons in the beam are unpolarized, meaning that their density matrix is half the identity.

Ah, yes! FIGURATIVELY! 'Gold, in the figurative language of the people, was “the tears wept by the sun.”'

 

[A. Neumaier]

Ah, yes! FIGURATIVELY!

I consider all photons with which a beam of light can be populated as figurative talk only.

In quantum field theory, a beam of stationary laser light is an electromagnetic radiation field described by a coherent state. In this state, the expectation value of the photon number operator is given by the beam intensity. It is a finite number. A photodetector at the other end of the laser beam clicks with a rate proportional to the intensity. Each click is figurately interpreted as a photon arriving. If one waits long enough one can extract from the beam as many photons as one likes - in spite of the fact that the expected number of photons can be small (even less than one). Thus the concept of photon number in quantum field theory has very little to do with the concept of photons defined by detector clicks. The only case where the two match is when one uses quantum field theory to calculate a scattering event with a well-defined photon content.

In quantum mechanics, one describes single scattering events in the same manner as in QFT (though usually in simplified models). However, sequences of events are never treated according to a formal prescription but only figuratively by applying the single event formalism in a heuristic way to sequential processes. This is done by
replacing the QFT description of a beam by an imagined sequence of photons created by the source and traveling along the beam to the detector where they produce a click. This is an intuitive picture (hence figurative) that in simple situations gives a quantitatively correct description of the experimental result. This picture is then generalized to entangled photons etc. in a way that allows one to have some intuition about how these behave. This intuition is only of very limited validity, hence it is accompanied by riddles that make everything look very weird.

Thus photons as flying objects causing detector clicks are purely figurative talk. The correct information content can be found only in a quantum field treatment and in approximate formal treatments derived from it - in practice via Lindblad equations.