- #36
A. Neumaier
Science Advisor
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Every system we look at is such a system!stevendaryl said: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!stevendaryl said:There are times when a mixed state is a necessity, as when you are observing one subsystem of an entangled composite system.
Nature doesn't compute - so your difference is a human artifact.stevendaryl said: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 said:The difference is described by Ockham's razor
A. Neumaier said:Nature doesn't compute - so your difference is a human artifact.
Then I'd have to tag all my posts with the same tag. That's why I am using a blank as tag.stevendaryl said: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.
A. Neumaier said:Then I'd have to tag all my posts with the same tag. That's why I am using a blank as tag.
Your very artificial preparation procedure - never used in practice - produces approximately unpolarized light.stevendaryl said: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?
vanhees71 said: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.
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.stevendaryl said: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.
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.entropy1 said: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?
entropy1 said:Eg, do there even exist unpolarized photons??
StevieTNZ said: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.
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?N88 said: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?
A. Neumaier said: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.
An unpolarized photon is defined as per the email I quoted.N88 said: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?
N88 said: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?
Protons have spin, not polarization.N88 said:particles [PROTONS] of a given polarization
A. Neumaier said: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.
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 said: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 said: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.
I consider all photons with which a beam of light can be populated as figurative talk only.N88 said:Ah, yes! FIGURATIVELY!