Polarization of a photon how do we know?

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Main Question or Discussion Point

Initial assumptions :
1. An unpolarised photon has 50% chance of being transmitted through a linear polariser.
2. When a photon meets a polarizer we subject it to an observation.
3. The effect being that we have forced it to collapse from being in the superposition of the vertical and horizontal states to ONE of them.
4. The energy of the transmitted photon is still the same as before it hit the polarizer.


My Question

2 photons are created at some common source , so they have the same energy(frequency).
Let one photon go through a linear polariser . And let another go unpolarised.
Both go to some to some distant place , not linked to our reference frame.

There the two photons arrive and somebody unknown to us wants to determine which of these photons has bee subjected to a polariser .
How can they determine this, without any knowledge of what path they took?
Is it possible , and how?

What does polarization actually do to a photon that we can say , its different from before it went through a linear polariser in this instance?
 

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  • #2
DrChinese
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Initial assumptions :
1. An unpolarised photon has 50% chance of being transmitted through a linear polariser.
2. When a photon meets a polarizer we subject it to an observation.
3. The effect being that we have forced it to collapse from being in the superposition of the vertical and horizontal states to ONE of them.
4. The energy of the transmitted photon is still the same as before it hit the polarizer.


My Question

2 photons are created at some common source , so they have the same energy(frequency).
Let one photon go through a linear polariser . And let another go unpolarised.
Both go to some to some distant place , not linked to our reference frame.

There the two photons arrive and somebody unknown to us wants to determine which of these photons has bee subjected to a polariser .
How can they determine this, without any knowledge of what path they took?
Is it possible , and how?

What does polarization actually do to a photon that we can say , its different from before it went through a linear polariser in this instance?
If one stream is polarized (say as all V*) and the other is not, then it would be possible to identify that stream as such fairly quickly. You simply run it through another polarizer and rotate the setting until you get all Vs. The other stream will be a 50-50 mix of H and V when it is run through a polarizer.

* It is possible to polarize all of a unknown photon stream as V with a combo of a polarizing beam splitter (PBS), and a half wave plate.
 
  • #3
Nugatory
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What does polarization actually do to a photon that we can say , its different from before it went through a linear polariser in this instance?
No matter what the state of the photon is before it encountered the polarizer, its state after it has passed through is such that if it encounters another polarizer: it will pass when the second polarizer is at parallel to the first; not pass when the second polarizer is perpendicular to first; and pass with a probability given by cos2 of the angle between the polarizers when they're in between.

As for what it "actually does" to the photon? There's no answer in the mathematical formalism of quantum mechanics. QM is a theory about the results of measurements, so we can only describe the interaction with the polarizer in terms of its effect on subsequent measurements of the photon.

You could, if you want, say that the polarizer "causes the wave function of the photon to collapse into a particular polarization". That's a very helpful way of thinking about the behavior of photons and polarizers (so helpful that I do it all the time) but not so helpful for other problems - and it's no answer at all to your question about what the polarizer "actually does" to the photon.
 
  • #4
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If one stream is polarized (say as all V*) and the other is not, then it would be possible to identify that stream as such fairly quickly. You simply run it through another polarizer and rotate the setting until you get all Vs. The other stream will be a 50-50 mix of H and V when it is run through a polarizer.
This I understand ,but the fuzzy part for me is when they are only 2 photons , and not a stream.
Because we are using probabilities here , it seems strange to say the 1 photon has 50 % probability of going through, which we can only build up from the results of a stream.
Be that as it may , more specifically if the answer to my question :

What does polarization actually do to a photon that we can say , its different from before it went through a linear polariser in this instance?
If the answer is , there is no measurable property, the person in the distance has no way of saying that one of these photons has been polarized.

Then to his knowledge , they may just as well (for all intensive purposes) be identical unpolarised photons. Yet in "reality" they are not.

In which case if they are passed through a second common randomly orientated polariser he would make similar assumptions as I have previously in my reference frame , which are.

1. An unpolarised photon has 50% chance of being transmitted through a linear polariser.
2. When a photon meets a polarizer we subject it to an observation.
3. The effect being that we have forced it to collapse from being in the superposition of the vertical and horizontal states to ONE of them.
4. The energy of the transmitted photon is still the same as before it hit the polarizer
And clearly this is not the case if you look at the 2 photons common past history.
Will this not lead him to formulate different probability wave functions?
I find this quite confusing.
Is this not a paradox, or again am i missing something again?
 
  • #5
kith
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Will this not lead him to formulate different probability wave functions?
He cannot infer a probability distribution from a single run of the experiment. The only information he gets about two the photons is this:
If a photon is transmitted, its previous polarization wasn't orthogonal to the polarizer orientation.
If a photon is absorbed, its previous polarization wasn't identical to the polarizer orientation.
 
  • #6
Nugatory
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Because we are using probabilities here , it seems strange to say the 1 photon has 50 % probability of going through, which we can only build up from the results of a stream.
You are right about that (and that's one of the arguments for the ensemble interpretation). However, you're asking the practical question "Can we determine which of the two photons is the one that went through the polarizer?" and the answer to that is "No". The best that we can do is set up another polarizer at the same angle; if the photon fails to clear the second polarizer we know it didn't pass through the first, but if it does clear the second polarizer then we don't know.



If the answer is, there is no measurable property, the person in the distance has no way of saying that one of these photons has been polarized.
Yes, see above and my earlier post in my thread. That's the reason why I very carefully avoided saying in that post that the photon "is" polarized.

Then to his knowledge , they may just as well (for all intensive purposes) be identical unpolarised photons. Yet in "reality" they are not.
Good to see that you put the word "reality" in scare-quotes here :smile:. It's a waste of time talking about what the photon really is when the best theory we have only talks about the results of measurements on the photon.

And clearly this is not the case if you look at the 2 photons common past history.
Will this not lead him to formulate different probability wave functions?
I find this quite confusing.
Is this not a paradox, or again am i missing something again?
It is correct to say that the two photons have been prepared differently, and therefore that applying the mathematical formalism of quantum mechanics to their initial states will lead to different statistical predictions for the results of future measurements, and that's close enough to saying that they have different probability wave functions. However, there is no way of experimentally observing this difference without doing the experiment many times and seeing if the statistics validate the calculation.

There's no paradox here, just the inherent limitations of a theory that is statistical and probabilistic at its core.
 
  • #7
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Good to see that you put the word "reality" in scare-quotes here . It's a waste of time talking about what the photon really is when the best theory we have only talks about the results of measurements on the photon.
yes , dare I say what reality is? For me its what I measure/observe. But to say
"Is the moon really there when I don't look" is disturbing to say the least.


It can never be rationalized that it exists if you don't look at it.
Conversely it will never be rationalized that it does not exist if you don't look at it.

So to debate anything rationale , you need to agree that its al least there by some observation.
Who cares if its not there when I don't look.
That's my reality.

Which comes to my next paradox/ or not..


If two entangled particles are separated by a light year and Alice measures the spin of hers to be up.

How can we say that Bob's( who is a light year away ) "instantaneously" becomes down.
After all it should at least take a minimum of one year , to get that information back to Alice
How can you prove or disprove it, its going to take at least 1 year, and who knows what effect that had on Alice and Bobs realities, since they only converge when they exchange their findings.

I mean how is Alice or bob going to prove this phenomena without "Looking", at each others information simultaneously.
We already know that simultaneous events differ relative to an observers frame of reference through relativity.
 
  • #8
Nugatory
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yes , dare I say what reality is? For me its what I measure/observe. But to say
"Is the moon really there when I don't look" is disturbing to say the least.
Decoherence has given us a reasonable answer to this question, and the answer is "yes"... But this was far from obvious a century ago when QM was first being hammered out. There's a book, entitled "Where does the weirdness go", that does a pretty decent non-technical discussion of these issues.

If two entangled particles are separated by a light year and Alice measures the spin of hers to be up.

How can we say that Bob's( who is a light year away ) "instantaneously" becomes down.
After all it should at least take a minimum of one year , to get that information back to Alice
How can you prove or disprove it, its going to take at least 1 year, and who knows what effect that had on Alice and Bobs realities, since they only converge when they exchange their findings.
That's the "spooky action at a distance" thing, and it's one of the situations in which collapse is not an especially illuminating explanation... Not that any of the other explanations are a whole lot more illuminating.... :smile:
QM predicts the results of measurements just fine, but it doesn't tell us anything about the processes by which these results happen.

I mean how is Alice or bob going to prove this phenomena without "Looking", at each others information simultaneously.
We already know that simultaneous events differ relative to an observers frame of reference through relativity.
The correlations only appear when Alice and Bob get together after the fact and compare their observations.
 
  • #9
DrChinese
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I mean how is Alice or bob going to prove this phenomena without "Looking", at each others information simultaneously.
We already know that simultaneous events differ relative to an observers frame of reference through relativity.
We would be hard pressed to say relativity has anything to do with it. Alice and Bob can be in the same inertial frame. And Bell tests have been done to consider the effect if they are not. No impact.
 
  • #10
Nugatory
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We would be hard pressed to say relativity has anything to do with it. Alice and Bob can be in the same inertial frame. And Bell tests have been done to consider the effect if they are not. No impact.
Indeed.

It's worth pointing as well that the predicted measurement results ("The photon passed my filter at angle ##\Theta## - therefore my partner will get a pass if his filter is at ##\Theta\pm\pi/2##" and all the other variations) are symmetrical and unaffected by which measurement happens first. You'll only get into trouble with relativity if you allow yourself to be tempted by the (enticing, but of dubious virtue) assumption that some causal influence is traveling from one measurement to the other.
 
  • #11
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I mean how is Alice or bob going to prove this phenomena without "Looking", at each others information simultaneously.
We already know that simultaneous events differ relative to an observers frame of reference through relativity.

The correlations only appear when Alice and Bob get together after the fact and compare their observations.

This is a crucial statement for me. And correct in all situations. But there can be different ways of interpretation for the phenomena of entanglement.

1. We can say when Alice measure her spin up , Bobs particle "instantaneously" jumps into spin down, and call this spooky action at a distance.
But this is in my reference time frame. We cannot really know anything about Bobs time frame. It may be that he has actually measure his particle before Alice and then destroyed it.
Who knows what is happening to Bob at that instance, if he is 1 light year away from Alice (and perhaps myself , depending where we are measuring from).

So we can say "instantaneous" but we cant prove it until (and this is my point) , Bob somehow relays that information back to Alice.
And when the 2 frames are so to say "united" with a common set of information, that emerged out of 2 particles from a common source.

Then , we may just as well also say , NOW the photons have collapsed into their respective spins, at that instance, when the two sets of information are correlated, 1 year later.

So it depends on your point of view, in my opinion. I agree that the 2 photons have reacted to
ONE of them being measured, and that's a mystery still . But to say it is "instantaneous" depends on your definition.
 
  • #12
Nugatory
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So it depends on your point of view, in my opinion. I agree that the 2 photons have reacted to
ONE of them being measured, and that's a mystery still . But to say it is "instantaneous" depends on your definition.
You do realize that you're the only one who has used the word "instantaneous" in this thread? :smile: If you don't use it, then you don't have to worry about defining it... And as you say, the mystery is still there in the correlation of the measurement results.

The history is somewhat relevant here. The mathematical formalism of quantum mechanics developed from an evolving collection of ad hoc rules to describe experimental results that were not consistent with classical physics. Thus, there was a natural tendency to view the math as describing something "real" (in the same sense that the electrical field described by Coulomb's law is "real") meaning that the wave function is a real physical phenomenon that collapses when something physical happens wave function that really collapses. That's the sense that underlies your original question in this thread: "what does polarization actually do to the photon?"

Only after the formalism was fully developed did it become clear that it only speaks to the results of measurements without explaining how they come about.
 
  • #13
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You do realize that you're the only one who has used the word "instantaneous" in this thread? If you don't use it, then you don't have to worry about defining it... And as you say, the mystery is still there in the correlation of the measurement results
True , I did introduce the word "instantaneous"

According to the formalism of quantum theory, the effect of measurement happens instantly
As did the wiki on entanglement as well. And yes I did make somewhat of a detour trying to explain my view of entanglement , which was not my original question.

.

My original question still is no clearer to me though. That at point a we can polarize a stream of photons, and not polarize the second stream , then measure the probability at a second set of polarisers, giving us consistent results locally.

However if we allow those 2 streams to be sent to a distant scientist.Two him they could be 2 streams of unpolarised photons , there is no way of telling. Yet an observation has been made , collapse has occurred to the polarised photons previously.

So one stream Must show interference and another non-interference if projected from a double slit. Which would be enough for him to make false correlations, since he only knows 2 identical streams.


Don't know if you understand my reasoning?
 
  • #14
atyy
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However if we allow those 2 streams to be sent to a distant scientist.Two him they could be 2 streams of unpolarised photons , there is no way of telling. Yet an observation has been made , collapse has occurred to the polarised photons previously.
Exactly, unless we know Alice's result and Bob's result, the results one sees from making measurements without knowing the results of the other stream will be exactly the same as if Alice never made the measurement.

The technical way to see this is to use something called the "reduced density matrix" for Bob, which predicts the probabilities of Bob's measurements, if he does not know Alice's results. Bob's reduced density matrix is the same, regardless of whether Alice made a measurement and collapsed the wave function, or did nothing at all.

Here when I say wave function and collapse, I do not necessarily mean that wave function or collapse are "real". They are just tools that we use to calculate the probabilities of experimental outcomes. Only experimental outcomes, and their probability distributions are considered real.
 
  • #15
Nugatory
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So one stream Must show interference and another non-interference if projected from a double slit. Which would be enough for him to make false correlations, since he only knows 2 identical streams.
Both the stream that has been through the polarizer and the stream that has not been through the polarizer will produce interference patterns if subsequently passed through a double slit. The state "polarization up" is still a superposition of "left slit" and "right slit".

If you set up an experiment in which two streams of entangled photons are emitted in opposite directions towards double-slit screens, and you interpose a polarizer on one side an interference pattern will form on both sides. The pattern on the side with the polarizer will be only half as bright, and that's because half the light emitted from the source was absorbed at the polarizer, never made it to the screen. You can interpret this as photons being absorbed with 50% probability (most convenient in single-photon experiments) or as the light being attenuated by 50% as it passes through the polarizer.
 
  • #16
DrChinese
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If you set up an experiment in which two streams of entangled photons are emitted in opposite directions towards double-slit screens, and you interpose a polarizer on one side an interference pattern will form on both sides.
Not to contradict you, oh Nugatorial one, but... :smile:

http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

Fig. 2, page 290. Entangled photons do not produce an interference pattern without additional manipulation, which will break the entanglement. And I don't believe a polarizer will do it. We may be talking about different setups though.
 
  • #17
atyy
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I think Nugatory was talking about a different setup than Fig. 2, p290 in the Zeilinger review. In my understanding, Nugatory was talking about polarization entangled photons, whereas Zeilinger was talking about momentum entangled photons (I'm not sure what the right term is, but the labels a, a', b, b' in Zeilinger's Eq 4 are not polarization labels). So I think Nugatory and Zeilinger should both be correct.
 
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  • #18
Nugatory
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Nugatory was talking about polarization entangled photons
Yep... thx.
 
  • #19
DrChinese
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Yep... thx.
I think those babies are both polarization entangled AND momentum entangled. The input is a laser obviously, so a definite direction and energy. The output is a pair of photons where momentum should be conserved, ergo they are entangled on a momentum basis. I don't think I have an exact reference for that however. I will look.
 
  • #20
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"Is the moon really there when I don't look" is disturbing to say the least.
Yea - Einstein and Bohr got stuck on that one. The joke however is they were both wrong - the moon is observed all the time by it's environment, and that's what gives it its classical properties - like being there when not looking

It can never be rationalized that it exists if you don't look at it. Conversely it will never be rationalized that it does not exist if you don't look at it.
You subtly misunderstand what QM says - but don't be too worried - it's rather common :tongue:

Its silent about what's going on when not observed. Its not saying its not there etc etc. Its not saying anything at all - its simply not part of the theory - although various interpretations have their own take.

So to debate anything rationale , you need to agree that its al least there by some observation. Who cares if its not there when I don't look. That's my reality.
Like I said - its bit more subtle than that - it may be there, not there, anything you can think of - its simply silent about it. Its not a matter of caring or not caring - its simply a matter of what the theory is about.

If two entangled particles are separated by a light year and Alice measures the spin of hers to be up. How can we say that Bob's( who is a light year away ) "instantaneously" becomes down. After all it should at least take a minimum of one year , to get that information back to Alice How can you prove or disprove it, its going to take at least 1 year, and who knows what effect that had on Alice and Bobs realities, since they only converge when they exchange their findings.
Since we have zero idea what's going on when not observed how do you know the particles are a light year apart? All we know is if you measure one you know what the other will be. If that entails non local communication is a matter of opinion. I believe it does - but others take a different view:
http://quantum.phys.cmu.edu/CQT/chaps/cqt24.pdf

'Given that the EPR paradox has sometimes been cited to support the claim that there are mysterious nonlocal influences in the quantum world, it is worth emphasizing that the analysis given here does not show any evidence of such influences.'

If you want to delve into foundational issues of QM Griffiths has posted his book on it which examines it via the Consistent History paradigm:
http://quantum.phys.cmu.edu/CQT/index.html

I personally do not hold to that interpretation, but it does introduce the issues.

Thanks
Bill
 
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  • #21
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The history is somewhat relevant here. The mathematical formalism of quantum mechanics developed from an evolving collection of ad hoc rules to describe experimental results that were not consistent with classical physics. Thus, there was a natural tendency to view the math as describing something "real" (in the same sense that the electrical field described by Coulomb's law is "real") meaning that the wave function is a real physical phenomenon that collapses when something physical happens wave function that really collapses. That's the sense that underlies your original question in this thread: "what does polarization actually do to the photon?"
Nicely summarised
 
  • #22
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Yea - Einstein and Bohr got stuck on that one. The joke however is they were both wrong - the moon is observed all the time by it's environment, and that's what gives it its classical properties - like being there when not looking
Well noted - the moon is observed all the time by its environment. There is always something interacting with it, or at least a very high probability(lol). Whether I am involved or not.

Its silent about what's going on when not observed. Its not saying its not there etc etc. Its not saying anything at all - its simply not part of the theory - although various interpretations have their own take.
If that be the case I am in agreement.

The more I read about QM, the more I realize how important "Choice" of words impacts on subtle interpretation. Which inevitably leads to a strange tangent of results.
Only in QM has this become so evident to me. But this is a good sign for QM.

One thing is for sure , recent advancements in technology and the quality of experiments , is definitely converging to a more macroscopic ,local , reality that may be very different from what we are used to.

thx guys ... more food for thought for me.
 
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  • #23
DrChinese
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What Einstein said: "I think that a particle must have a separate reality independent of the measurements. That is: an electron has spin, location and so forth even when it is not being measured. I like to think that the moon is there even if I am not looking at it."

So he wasn't actually talking about the moon, that was just an analogy for objective (as opposed to subjective) quantum properties. He believed in the existence of a more complete specification of a system than QM provides. I.e. that quantum properties did not depend on the act of observation. We now know that for Einstein to be correct, there must be "spooky action at a distance."
 
  • #24
atyy
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Yea - Einstein and Bohr got stuck on that one. The joke however is they were both wrong - the moon is observed all the time by it's environment, and that's what gives it its classical properties - like being there when not looking
Well noted - the moon is observed all the time by its environment. There is always something interacting with it, or at least a very high probability(lol). Whether I am involved or not.
I am going to disagree slightly with bhobba here. Decoherence alone does not mean the moon is there when you are not looking. The moon being "there" means a continuous sequence of particular outcomes, but decoherence does not produce particular outcomes. Quantum mechanics assumes that particular outcomes occur when measurements are made, and calculates the probabilities of those outcomes, but quantum mechanics is silent about what happens when measurements are not made. In order to say that the moon is there when one is not looking, one needs an interpretation such as Bohmian Mechanics. So yes, it is consistent with quantum theory to say that the moon is there when one is not looking, but one has to go beyond quantum theory to say that.
 
  • #25
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So he wasn't actually talking about the moon, that was just an analogy for objective (as opposed to subjective) quantum properties. He believed in the existence of a more complete specification of a system than QM provides. I.e. that quantum properties did not depend on the act of observation. We now know that for Einstein to be correct, there must be "spooky action at a distance."
Absolutely. Its an analogy , in the extreme. Since QM deals predominantly with the microscopic/Sub atomic world.

I think that a particle must have a separate reality independent of the measurements
The act of observation indeed is a fascinating subject in the sub atomic world.
Where does it begin where does it end. How do we interpret it.

Specifically if a photon is created. Will it ever be observed , before it is annihilated.
We say that linear polarization is an act of observation. But what have we actually observed.
There is no change to an independent observer in another far away place as I tried to reason above.

I see nothing there after polarization. We have just normalized the particle to our polarizer , so that we can add a second polarizer to do analysis.
The photon may just as well not be there until something consumes it. It is as if its sole purpose is to transmit information through a vacuum about it's source where it was created.
It tells us nothing about its journey! This is fundemental!
 

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