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I Entanglement and photon polarization

  1. Apr 5, 2017 #1
    Hello,
    • A photon can have various types of polarization states (horizontal, vertical, circular, elliptical, linear at an angle ##\theta##). Any valid polarization basis is two-dimensional and can represent any state of polarization. What are the actual eigenvectors of the polarization observable? My understanding is that polarization is an observable, hence an operator.
    • Because the polarization bases are two dimensional, it is convenient to entangle the states of polarization of two photons. I spent some time studying the tensor product and learned that the most general two-photon polarization state is given by
    $$|\Psi>_{12}= a |H>_1 |V>_2+b |V>_1 |V>_2+c|V>_1 |H>_2+d|H>_1 |H>_2$$
    where each term contemplates all possible combinations of polarization H and V for the two photons.
    I know the state below is an entangle state because it cannot be expressed as a product of single photon states:
    $$ (\frac {1}{\sqrt{2}}[ |H>_1 |H>_2+|V>_1 |V>_2]$$
    Because of entanglement, once the polarization of the first photon is measured, the polarization of the second photon is automatically determined. For instance, if one photon turns out H the other turns out H as well and if it was measured as V the other would be V too. But wouldn't that happen also when the the system is in a state like

    $$ |V>_1 |V>_2]$$

    which we know is not entangled? Why not?

    thanks.
     
  2. jcsd
  3. Apr 5, 2017 #2

    DrClaude

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    What do you mean?

    Yes, just like spin.

    For one, you would never measure the first photon in state H. In the later case, the measurements would be correlated, but trivially so. This is not an entangled state because to state of the second photon can be known completely independently of the first photon. There is no particular look between the two photons, and you would get the same results even if the sources of the two photons are completely independent (imagine two classical light sources with polarizing filters).
     
  4. Apr 5, 2017 #3
    Thanks DrClaude.

    1) So polarization is an observable, hence an operator. Are the H and V polarization states the eigenstates of this observable. Does polarization, like spin, not have a specific operator form like other observables (energy, momentum, angular momentum)?

    2) In the case of the state ##|V>_1 |V>_2 ##, if we measure the first photon and its state is V, wouldn't the second photon always be in state V as well? That seems to be a correlation. You mention that state of the second photon can be known completely independently of the first photon. I don't get that which is an important point. Thanks for the patience.
     
  5. Apr 5, 2017 #4

    zonde

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    If rotate your analyzers by 45 deg. and measure entangled photons in this rotated basis you still get perfect correlation (if one is +45 deg. the other one is +45 deg. as well). But this won't happen with ##|V_1\rangle |V_2\rangle## state.
     
  6. Apr 5, 2017 #5

    DrClaude

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    H and V are eigenstates, just as are L and R (left and right circular polarizations). And if I understand correctly the second part of the question, no, you can't write the polarization operator (or the eigenstates) in terms of operators in (or functions of) position.

    See @zonde's post.
     
  7. Apr 5, 2017 #6

    DrChinese

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    Which is a Product state, and therefore not entangled (pretty much by definition).
     
  8. Apr 5, 2017 #7
    Thanks Zonde. your comment helps a lot. What I am getting is that entanglement remains between the two one-photon states even if we vary the experimental setup. What would exactly happen if the analyzers rotated by 45 deg. and the two photons were in the composite and separable state ##|V\rangle_1 |V\rangle_2## ? Wouldn't the two measured polarizations be the same again?

    More general question: a composite system, which is formed by subsystems (two photons in our discussion), is described by a single overall state. However, we keep talking about the individual states of the subsystems that are part of the system. For example, in the case of the hydrogen atom, the proton and electron should not have their own state but be described by their single joint state. What allows us to talk about one-particle states in the context of a composite system and not just stick to a total joint state?
     
  9. Apr 6, 2017 #8

    zonde

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    Coincidence rate for ##|V_1\rangle |V\rangle_2## measurement can be calculated using Malus' law:
    $$C=\cos^2(45^\circ)\cos^2(45^\circ)=0.25$$

    We can talk about substates in composite state only in limited sense of partial trace. I don't know if this answers your question. Maybe you mean that we speak about measurements of individual particles from composite state?
     
  10. Apr 6, 2017 #9
    Thanks Zonde.
    In the case of a single photon, if we say that the photon is in the polarization state ##|V \rangle##, it means that we know its polarization (vertical in this case) with certainty. The photon polarization state is an eigenstate of the polarization observable.
    In the case of a two-photons system, the composite state ##|V\rangle |V\rangle## would indicate that both photons are known to be exactly in a vertical polarization state, correct?
    If we had many identical pairs of photons in the same state ##|V\rangle |V\rangle## and measured them with two analyzers at 45 deg. we would not obtain a high correlation. The output of a 45 deg. analyzer is always a 45 deg. polarized photon (if the photon makes it through otherwise it gets absorbed) but what happens statistically at the output of one analyzer will not be correlated with what happens at the output of the other analyzer. Is this what is going on?
    For example, say we had 4 photons pairs (A,B,C,D) all in the initial state ##|V\rangle |V\rangle## and passed through the two 45 deg. analyzers. For pair A, one photon passes and the other is absorbed. For pair B, both photons pass, for pair C neither photon passes and for pair D we have any of the previous cases.
    What results would we expect if the four photon pairs were entangled?

    Is the coincidence rate the same as correlation?
     
  11. Apr 6, 2017 #10

    DrChinese

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    Entangled pairs would be: both pass, or neither pass. Correlation is 1.

    Keep in mind that correlation varies from -1 to +1, while coincidence varies from 0 to 1.
     
  12. Apr 7, 2017 #11
    I have read a short tutorial article on entanglement written by Frank Wilczek:
    https://www.quantamagazine.org/20160428-entanglement-made-simple/

    Wilczek suggests that the electrons and the protons in a molecule or atom are always entangled due to the interaction. I know that interaction can entangle particles but not all interacting particles are entangled, correct? The description of an atom and its constituents relies on many approximations that lead to consider the subsystems independent from each other but in reality they are not...
     
  13. Apr 7, 2017 #12

    zonde

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    It seems that Wilczek explanation is at the level before Bell's theorem. The part about GHZ state might be ok however.
    I don't think that this article is a good source of information. There is no point calling every classical causal relationship "entanglement".
     
  14. Apr 8, 2017 #13
    Thanks,
    I guess I thought that a system composed of multiple particles would always be described by a single and overall wavefunction and there would not be an individual wavefunction for each particle. If we measure a property for one particle, we surely interact with the wavefuntion of the entire system but we don't affect the other particles. How is that possible?

    In the case of entanglement, if one particle is measured, we immediately know the result of the measurement for the other particle because it is said that the there is one single wavefunction that has already collapsed from the measurement of the first particle...
     
  15. Apr 8, 2017 #14

    DrChinese

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    They (the specific properties associated with the particles) are distinguishable, for one thing. With entangled particles, the specific properties of the individual particles are not distinguishable.
     
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