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I Polarisation entanglement

  1. Oct 11, 2016 #1
    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.
     
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  3. Oct 11, 2016 #2

    jfizzix

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    The polarization of the state can be a single pure state, which may be a superposition of polarizations, or a mixture of multiple pure states (e.g. partially polarized light)

    In order to figure out the polarization state of the photon, one needs to measure many identical such photons in all three polarization bases (horizontal/vertical diagonal/antidiagonal, and left/right circular). With these measurements, one can reconstruct a strong estimate of the polarization state. This process is called quantum state tomography.
     
  4. Oct 11, 2016 #3

    stevendaryl

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    Definitely not. If you have the situation in which you don't know what the photon's polarization is, then you don't use superpositions, you use mixed states, which are represented by density matrices, not pure states.

    The distinction is the difference between column matrices and square matrices. If you use [itex]\left( \begin{array}\\ 1 \\ 0\end{array} \right)[/itex] to represent a horizontally polarized photon, and [itex]\left( \begin{array}\\ 0 \\ 1\end{array} \right)[/itex] to represent a vertically polarized photon, then a superposition of horizontal and vertical photons would be represented by the general column matrix: [itex]\left( \begin{array}\\ \alpha \\ \beta \end{array} \right)[/itex] where [itex]|\alpha|^2 + |\beta|^2 = 1[/itex]. An observable would be represented in this simple model by a 2x2 matrix [itex]O[/itex]. If the photon is in the state [itex]U[/itex] (a column matrix), then the expected value of a measurement of [itex]O[/itex] would be given by: [itex]\langle O \rangle = U^\dagger O U[/itex].

    In contrast, a mixed state consisting of a probability [itex]p_1[/itex] of being horizontally polarized and [itex]p_2[/itex] of being vertically polarized would be represented by the 2x2 matrix: [itex]\left( \begin{array}\\ p_1 & 0 \\ 0 & p_2 \end{array} \right)[/itex]. If the photon is in the mixed state [itex]D[/itex] (a 2x2 matrix), then the expectation value for a measurement of [itex]O[/itex] would be: [itex]tr(D O)[/itex], where [itex]D O[/itex] means matrix multiplication, and [itex]tr[/itex] means the trace operator (add up the diagonal values of the resulting 2x2 matrix).
     
  5. Oct 11, 2016 #4
    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.
     
  6. Oct 11, 2016 #5

    stevendaryl

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    If you have a pair of photons, then there are three different "systems" that we could look at:
    1. System 1: The first photon.
    2. System 2: The second photon.
    3. System 3: The composite system consisting of two photons.
    Each of these three systems can either be described as a pure state, or as a mixture. If the photons are entangled, then System 3 is in a pure state, but Systems 1 and 2 are mixtures.

    That's an odd fact about quantum mechanics that has no analog in classical physics.
     
  7. Oct 11, 2016 #6

    DrChinese

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    If a photon is polarization entangled, it's polarization is unknown and it's in a superposition of states.
     
  8. Oct 11, 2016 #7

    stevendaryl

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    Don't you mean "mixture", not "superposition"?
     
  9. Oct 11, 2016 #8

    vanhees71

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    Yes, it should be mixture. As I've shown somewhere earlier in this or a recent other thread ;-)), with the usual entangled state
    $$|\psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle),$$
    each of the single photons is in the maximum-entropy state, i.e., it's unpolorized
    $$\hat{\rho}=\frac{1}{2} \hat{1}=\frac{1}{2} (|H \rangle \langle H|+|V \rangle \langle V|).$$
     
  10. Oct 11, 2016 #9

    DrChinese

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    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.
     
  11. Oct 11, 2016 #10
    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.
     
  12. Oct 11, 2016 #11

    stevendaryl

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    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.
     
  13. Oct 11, 2016 #12

    DrChinese

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    Not disputing your or vanhees71 specifically, as I think both of you follow the terminology better than I. I don't understand how the entangled two-photon system is a superposition, but it doesn't make sense to refer to the components as being in a superposition too. There really aren't any components in the first place, it's just a useful way to refer to them.

    Wiki describes a Bell state (what vanhees71 mentions in his post) as an "equal superposition" (under "Qubit"). Of course that is really referring to the system.
     
  14. Oct 11, 2016 #13

    Simon Phoenix

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    I suppose it depends on what people mean by superposition. If by this it is meant that we have a sum of vectors (pure states) then, clearly, the sum is also a vector (pure state). If this is what is meant by 'superposition' then it's absolutely incorrect to say one of the photons in a 2-photon entangled state is in a superposition 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.

    So trying to say that we have an entangled state in which either of the entangled component parts is in a pure state is just a complete non-starter and makes no sense whatsoever.
     
  15. Oct 11, 2016 #14

    DrChinese

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    I already saw you guys have discussed this and other elements around "pure vs mixed" states at length in the other thread. 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?
     
  16. Oct 11, 2016 #15

    Simon Phoenix

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    Yes - very nicely put. Much better than I managed :wideeyed:
     
  17. Oct 12, 2016 #16

    zonde

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    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.
     
  18. Oct 12, 2016 #17

    Simon Phoenix

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    Why do you say this?

    You're not correct, because that's exactly what we have - each of the individual particles in an entangled state is described by a mixed state that is formally identical to a statistical mixture of pure states.

    Look at vanhees' post number 8 above - which bit of that post don't you get?
     
  19. Oct 12, 2016 #18

    zonde

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    Can the second equation describe single photon or it necessarily describes at least two photons?
     
  20. Oct 12, 2016 #19

    stevendaryl

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    A particle can always be in an improper mixed state. You get the corresponding density matrix by tracing out all degrees of freedom not associated with that particle.
     
  21. Oct 12, 2016 #20

    jfizzix

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    If a photon is in a superposition, then there is one way to measure the photon (i.e., a measurement basis) where you will get the same outcome 100 percent of the time.

    For example, if you have a diagonally polarized photon, it can be expressed as a superposition of horizontal and vertical polarization. If you measure in the horizontal/vertical basis, you will have a 50/50 chance of getting either result.
    On the other hand, if you measure in the diagonal/antidiagonal basis, you will measure the diagonal result 100 percent of the time.

    If a photon is in a mixed state, then there is no such basis where you can get the same outcome all the time. This for example corresponds to the statistics of unpolarized light, where you get 50/50 chances of either measurement result in all measurement bases.
     
  22. Oct 12, 2016 #21

    Simon Phoenix

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    The second equation of Vanhees there is obtained by tracing out the degrees of freedom associated with one of the photons - so it is describing the state of just ONE of the photons in the entangled pair.

    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. That's what is meant by the tracing procedure - it's just focusing on part of a bigger picture*. If you can only do measurements on the particle I've given you, you can't even tell whether it's one of an entangled pair. All of the single-particle properties you can measure can be worked out from this statistical mixture.

    *Of course we don't have to actually physically separate things into 'boxes', the properties of just one of the entangled particles are described by a statistical mixture whether or not we actually perform the physical separation. I just think it gives a more vivid description in terms of actually physically separating things.
     
  23. Oct 12, 2016 #22
    Thanks again. 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?
     
  24. Oct 12, 2016 #23

    zonde

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    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?
     
  25. Oct 12, 2016 #24

    stevendaryl

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    No. When you perform a trace, you lose information. The state of the composite system cannot be recovered uniquely from the mixed states of A and B separately. A concrete example: in EPR, an electron/positron pair is created. The electron goes to Alice and the positron goes to Bob.

    The state of Alice's electron, calculated by tracing out Bob's positron, is a mixed state with equal probability of being spin-up or spin-down.

    The state of Bob's positron is also a mixed state with equal probability of spin-up or spin-down.

    Putting those two states together doesn't say anything about whether Alice's electron's spin-state is correlated with Bob's positron's spin-state. That information is lost when you perform the traces.
     
  26. Oct 12, 2016 #25

    zonde

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    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.
     
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