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Dadface
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If we don't know the polarisation state of a photon before detection is it reasonable to assume that it's in a superposition of all possible states? Thank you if anyone can clarify.
If we don't know the polarisation state of a photon before detection is it reasonable to assume that it's in a superposition of all possible states? Thank you if anyone can clarify.
Thank you both. I'm not sure if I expressed my question clearly enough but I was referring to entangled photons as in the title of the thread. In the case of two photon polarisation entanglement is it correct to say that before measurement each photon can be considered as being in two polarisation states at once? Thank you.
If we don't know the polarisation state of a photon before detection is it reasonable to assume that it's in a superposition of all possible states? Thank you if anyone can clarify.
If a photon is polarization entangled, it's polarization is unknown and it's in a superposition of states.
Don't you mean "mixture", not "superposition"?
I say superposition is a better description because there is no specific well-defined polarization, not just that it is unknown. But I think vanhees71 has made the point that the entangled state is something of its own animal. He usually has a better knack of labeling these things than I do.
But it's sort of contradicting what others have been saying, that the two-photon state is a superposition, but that there is no one-photon (pure) state. A superposition is a pure state (just a different pure state from any of the basis states), and for entangled photons, neither photon is in a pure state.
This follows from the definition of entanglement. By definition an entangled pure state of 2 objects A and B is one that cannot be written as a product of a pure state of A and a pure state of B.
Your point is that an entangled pair, while itself in a superposition of states, cannot be decomposed into or otherwise considered as 2 particles each in an individual superposition. Else it would be a product state, and it can't be if its entangled too. Does that sum it up?
But if we model mixed state as classical mixture of pure states then entangled particle can't be in mixed state either.
Can the second equation describe single photon or it necessarily describes at least two photons?Look at vanhees' post number 8 above - which bit of that post don't you get?
Entangled particle can't be in pure state (superposition). But if we model mixed state as classical mixture of pure states then entangled particle can't be in mixed state either. So it has to be something third with uncertain (undefined) polarization.
Thanks to all. That's got me looking up the difference between superposition of states and mixture of states as they relate to opposite states. Is it true to say that if a photon is in superposition it will be in opposite states simultaneously whereas if its a mixture it moves between opposite states but at any particular time is in one state only.
Can the second equation describe single photon or it necessarily describes at least two photons?
What you say is that by tracing out A from composite system AB we are left with only the information relevant to B alone. And if we trace out B then we have only the information relevant to A alone. Ok, but it should mean that we can combine A and B back together but we are simply uncertain how to do it properly, right? So if I show that it can't be done then your statement is false, right?That's what is meant by the tracing procedure - it's just focusing on part of a bigger picture*.
What you say is that by tracing out A from composite system AB we are left with only the information relevant to B alone. And if we trace out B then we have only the information relevant to A alone. Ok, but it should mean that we can combine A and B back together but we are simply uncertain how to do it properly, right? So if I show that it can't be done then your statement is false, right?
That is my understanding too but if I got it right you can speak about single particle in a mixed state too if you are using Bayesian interpretation of probabilities.The way I see it so far is that to get a mixed state we need a collection of photons but superposition states can apply to single photons only as well as to collections of photons. Is that true?
But this is exactly what I said, no?No. When you perform a trace, you lose information.
Isn't this (in bold) the same thing as "cannot be recovered uniquely"?Ok, but it should mean that we can combine A and B back together but we are simply uncertain how to do it properly, right?
So if I show that it can't be done then your statement is false, right?
The bold part:which particular statement of mine do you think is false?
Suppose I had two entangled particles. I put one in a box and give it to you. Now if that's all you have access to then what you have is described by a statistical mixture.
ThanksInteresting - have fun trying to prove it is incorrect
So there is no possible way how we could construct full entangled state from statistical subensembles of separate photons for particular case we considered
Let's deal with this first. Where do you see I said there is no unique way how to recover full state? I said there is absolutely no way (unique or not) how you can match prediction worked out from full state to any possible prediction worked out from statistical mixture of separate states in particular case considered.First off, I'm somewhat baffled by your logic. You appear to be saying that because there is no way to uniquely recover (or construct) the full density operator given only the reduced density operators (which is true, there is no unique way) then the reduced density operators cannot be a correct description for the individual components in an entangled pair. I do apologize but your logic here completely escapes me.
I said there is absolutely no way (unique or not) how you can match prediction worked out from full state to any possible prediction worked out from statistical mixture of separate states in particular case considered
But this is exactly what I said, no?
Ok, but it should mean that we can combine A and B back together
So if I show that it can't be done then your statement is false, right?