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Polarization and Bell measurments

  1. Jul 13, 2014 #1
    There have been quite a few Bell threads lately, so I have been looking at them and various other sources. I'm missing something... any guidance appreciated.

    Per Dr. Chinese's, "Once any photon passes through a polarizer lens, its polarization will be aligned exactly with the lens thereafter (even if it wasn't previously).".

    Per Malus' and ubiquitous sites, of the photons passed through a vertical polarizer, some will pass a subsequent polarizer set to 45 degrees. I take this to mean that if a single vertically polarized photon may or may not pass a subsequent 45 degree filter, the same applies of any and all subsequent vertically polarized photons as well. So I am assuming that identically vertically polarized photons that encounter a 45 degree polarizing filter may or may not be passed; for example, of two subsequent such photons (I'll call them a "pair"), their passage or not though the filter does not have to match. If I assign all the possible pairs of a series, I will see some matching and some mismatching.

    With respect to the simple Bell tests, this raises my question...

    The question is, what is the difference between two vertically polarized photons going to a single 45 degree polarizing filter, and two photons identically emitted with vertical polarization, but going opposite directions to their respective but identically aligned 45 degree filters?

    The Bell examples assume that the emitted pairs MUST show a specific and definite relationship between their measurements when the detectors are identically aligned, but two vertically polarized photons encountering a single 45 degree filter DO NOT have to be so... one may pass and the other may not, or vice versa, or both may pass, or neither.

    I'm not seeing why emitted Bell test photons can't mismatch at the detectors even if the detectors' filter alignments match (say both at 45 degrees) and the emitted pair are both polarized vertically. Why can't one pass and the other fail?
  2. jcsd
  3. Jul 13, 2014 #2


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    They can, if they are both polarized vertically. But if they are both known to be polarized vertically (or in any other directions for that matter), then they are no longer entangled, because they've already interacted with (been detected by) a vertical polarization filter.
  4. Jul 13, 2014 #3


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    The difference is between
    • Each photon can be represented as a superposition of states.
    • The pair can be represented as a superposition of states.

    Let me introduce a little bit of notation that I hope will clarify the situation.

    1. [itex]|D\rangle[/itex] A photon in this state definitely passes a filter at 45 degrees, definitely does not pass a filter at 135 degrees.
    2. [itex]|\bar{D}\rangle[/itex] A photon in this state definitely passes 135, definitely does not pass 45.

    I didn't mention probabilities of passing (other than 0% and 100%), because the quantum-mechanical understanding of the probabilities is that they arise from superpositions. If you have a photon in the state

    [itex]|\psi\rangle = \alpha |D\rangle + \beta |\bar{D}\rangle[/itex]

    then it will pass a 45 degree filter with probability [itex]|\alpha|^2[/itex], and will pass a 135 degree filter with probability [itex]|\beta|^2[/itex]

    A vertically polarized photon can be interpreted as a superposition of a 45 degree photon and a 135 degree photon:

    [itex]|V\rangle = \frac{1}{\sqrt{2}}(|D\rangle + |\bar{D}\rangle)[/itex]

    (whether there is a + sign or a - sign or some other phase depends on how the various states are defined, but the above choice illustrates the point)

    If I have a pair of vertically polarized photons, then they are described by the composite state:

    [itex]|V\rangle |V\rangle = \frac{1}{2} |D\rangle |D\rangle + \frac{1}{2} |D\rangle |\bar{D}\rangle + \frac{1}{2} |\bar{D}\rangle |D\rangle + \frac{1}{2} |\bar{D}\rangle |\bar{D}\rangle[/itex]

    Squaring the coefficients tells us, since [itex]|\frac{1}{2}|^2 = \frac{1}{4}[/itex], that:
    1. There is a 25% probability that both photons will pass a 45 degree filter.
    2. There is a 25% probability that the first photon passes, but the second doesn't.
    3. There is a 25% probability that the second photon passes, but the first doesn't.
    4. There is a 25% probability that neither photon passes.

    Now, in an EPR-type experiment, it's not that each photon is in a superposition, it's that the pair is in a superposition. The EPR state could be represented as:

    [itex]\frac{1}{\sqrt{2}} (|D\rangle |D\rangle + |\bar{D}\rangle |\bar{D}\rangle)[/itex]

    In this state, there are no "crossed" states such as [itex]|D\rangle |\bar{D}\rangle[/itex] or [itex]|\bar{D}\rangle |D\rangle[/itex]. So there is 0% probability that one photon will pass the filter and the other will not.
  5. Jul 13, 2014 #4
    OK, but for the emitted pairs in the Bell tests isn't it assumed, thought, expected, or "known" that both will indicate the same polarity if a measurement is applied? Even if the particular polarization attribute is not known until a measurement, is it not the foundation of the experiment that both are the same polarity? Knowing that they are entangled implies this, right?

    I'm not seeing the difference between two that have unknown but identical polarity, and two that have known identical polarity... in both cases the photons of each pair have the same polarity.

    Simply knowing the polarity allowing the pair to interact with a filter differently, but not knowing the polarity making both of the pair interact with the filter identically... seems to suggest an attribute "known or unknown polarity"...


    The notation is new to me, but I think I understand it.
    What I don't understand is the last part...
    You did not rule out that a superposition of two photons could be represented as the superposition of their composites.

    Is that disallowed? That is,

    [itex]\frac{1}{\sqrt{2}} (|V\rangle |V\rangle + |\bar{V}\rangle |\bar{V}\rangle)[/itex]
  6. Jul 13, 2014 #5


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    Clearly, polarization entangled photons have different polarization than a pair vertically polarized. One pair's polarization is described by a single wave function. The other pair is described by 2 separate wave functions.

    You might be interested in this: it is possible to create entangled photon pairs that are vertically (or horizontally) polarized. These pairs are NOT polarization entangled, but they are entangled as to momentum.

    In fact, and somewhat counterintuitively, such pairs CAN be use to create polarization entangled pairs. This in commonly done as a matter of fact, by combining a stream of H> polarized pairs with a stream of V> polarized pairs. The trick is that they must be combined precisely so that it is not possible to determine whether the source was H> or V> polarized. See the following (great) reference, figure 2:

  7. Jul 13, 2014 #6


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    The state [itex]\frac{1}{\sqrt{2}} (|V\rangle |V\rangle + |\bar{V}\rangle |\bar{V}\rangle)[/itex] is a superposition of two mutually exclusive possibilities:

    1. Both photons are vertically polarized.
    2. Both photons are horizontally polarized (if you're using [itex]\bar{V}[/itex] to mean horizontally polarized)

    There is no possibility of one photon passing a vertical filter and the other photon passing a horizontal filter.

    This is the state that would be produced if you generate a pair of entangled photons.

    In contrast, the state [itex]\frac{1}{2} (|V\rangle |V\rangle + |\bar{V}\rangle |{V}\rangle + |V\rangle |\bar{V}\rangle + |\bar{V}\rangle |\bar{V}\rangle)[/itex] is a superposition of four possibilities:

    1. Both photons are vertically polarized.
    2. The first is vertically polarized and the second is horizontally polarized.
    3. The first is horizontally polarized and the second is vertically polarized.
    4. Both are horizontally polarized.

    For this state, there is a nonzero probability that one photon will pass a vertical filter, and the other will pass a horizontal filter.

    This is the state that you would have if you generate two photons in the same single-particle state [itex]\frac{1}{\sqrt{2}} (|V\rangle + |\bar{V}\rangle[/itex]

    Being in the same state, initially, does not mean that the particles are entangle. Just the opposite; if each particle has a definite single-particle state, then the particles are NOT entangled. To be entangled means that the particles are not in definite single-particle states.

    (One point of clarification: Mathematically, two particles are entangled if the composite state [itex]|\Psi\rangle[/itex] CANNOT be represented as a product [itex]|\Psi\rangle = |\psi\rangle |\phi \rangle[/itex])
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