Observers

Discussion in 'Physics & Math' started by arfa brane, Apr 18, 2017.

  1. iceaura Valued Senior Member

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    We know, if we simply look at the theory and avoid making any assumptions, that they are not correlated but rather entangled. If we say they are "correlated" we become vulnerable to thinking they will behave as other correlated particles behave.
    And this vulnerability is immediately demonstrated:
    That is not true of entangled particles. You learn nothing about the earlier state of an entangled particle by measuring a later state (according to QED, and given no violation of Relativity - in other words, without making assumptions for which we have no evidence whatsoever and must discard current well-established theory to entertain).
     
    Last edited: Jul 17, 2017 at 1:11 AM
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  3. arfa brane call me arf Valued Senior Member

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    And we also know that entangled means correlated and in superposition. When in superposition, correlated states are indistinguishable.

    No, we don't. Coins for instance can be correlated, but they can't be in superposition.
     
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  5. iceaura Valued Senior Member

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    They are not correlated, if one is taking care. That is not a quibble. Entangled and correlated are separate, different, mutually exclusive states in QED. Their relationship is at best metaphorical or analogous, as would be "connected" or "paired".
    As far as I could read, you illustrated the danger by apparently posting exactly that confusion, as quoted.

    You said: "Here's a heuristic I've been using." and then posted an example of correlation, not entanglement, from which one can learn what can be learned from a correlated state but not from an entangled state.

    It's possible I misread that "heuristic", and you were instead referring to an illustration of how entangled particles do not behave, and what one can learn from correlation that cannot be learned from entanglement. If so, I stand corrected - that's not how I read it.
     
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  7. arfa brane call me arf Valued Senior Member

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    Ok, It should have been made clearer that my heuristic is about information and entropy of the classical variety. I suppose what you should say what entanglement is, is formerly correlated states in superposition, or something, but mainly because of our need to involve (the notion of) separated events.

    One other point about all the diagrams of SPDC and the gates, they illustrate maximally entangled states. What about partially (or "weakly") entangled states?
    In the diagram of type I SPDC with a pair of BBO crystals, one crystal is rotated so that the crystal planes of symmetry are orthogonal (one is horizontal the other vertical). What if this rotation angle is arbitrary?

    If you then measure polarization states with counters you can set to detect either horizontal or vertical polarization, what correlations should you expect to see?
     
    Last edited: Jul 17, 2017 at 8:16 AM
  8. arfa brane call me arf Valued Senior Member

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    When you use a heuristic like a coin, you choose what the degrees of freedom are. Exactly as Bell chooses three properties--these can be anything but must be "detectable" in a physical theory.

    So Bell's theorem is about just which degrees of freedom are chosen, but implies a process of detection, or measurement. The entropy of your system is what you don't yet know, but this is bounded by the chosen d.o.f.

    But that's what you do when you flip a coin over without looking at it, you measure something and gain information about the previous state of the coin.
    This previous state doesn't "exist" any more, except that you can remember what you did to the coin (you store information, and erase a state) . . .
     
  9. iceaura Valued Senior Member

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    But that's not what you do.

    When you measure something in an entangled state, you do not gain information about its previous state (you don't even learn, from the measurement, that it used to be entangled. One of the serious technical difficulties in detecting violations of Bell Inequalities is assuring entanglement at the time of measurement - the measurement itself doesn't tell you one way or the other).

    Bell Violations indicate that an entangled particle does not possess the property measured before it is measured. If it did, the measurements could not violate the Inequality (barring superluminal information transfer, or major changes in the logic involved in our reasoning, or the like).
     
  10. arfa brane call me arf Valued Senior Member

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    You mean, that's not what you do with quantum states.
    In quantum algorithms, you need to know where entanglement occurs and what to do with it.
     
  11. iceaura Valued Senior Member

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    What is going on here:
    What is "that", underlined?
     
  12. arfa brane call me arf Valued Senior Member

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    . . . "that" is the choice of d.o.f. you make when you flip a coin without looking at it. It's a measurement, but all measurements are necessarily constrained.

    In the case I'm using, lots of things are ignored (about coins, about human hands and how they function, etc); information is what you decide/choose it to be, constrained by the chosen d.o.f.; these choices are what defines what you can know, given what you do know.

    This does appear at first glance, to be something "special" human observers can do. Except that contravenes physics: no observer is special, or can be.
     
  13. arfa brane call me arf Valued Senior Member

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    But of course, that physics says we're nothing special doesn't mean we aren't complex (systems of particles).

    In perspective, we don't have or are likely to ever have the computing resources that can fully describe a single cell, and we're a cooperating system of billions of cells. But let's not get a head swell going, most of us can't understand quantum mechanics, and it really is about very simple things.
     
  14. arfa brane call me arf Valued Senior Member

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    Here is an interesting paragraph from Wikipedia:

    Isn't there something wrong with electrons and photons having the same rotational degrees of freedom? One's a fermion, one is a boson.

    Or what?

    p.s. when I say QM is really about very simple things, I mean comparatively. A single Feynman diagram of two electrons interacting, by "exchanging momentum", with each other is a simple kind of event, compared to neural cells actively pumping ions against a potential gradient, for instance, which involves a whole lot of extremely complex molecules.
     
    Last edited: Jul 20, 2017 at 3:17 AM
  15. James R Just this guy, you know? Staff Member

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    Electrons has spin 1/2; photons have spin 1, so your fermion/boson comment is valid.

    However, when it comes to eigenvectors, we need to think about what we mean. For electrons, I think, we're talking about something like the component of the electron's magnetic momentum along one particular spatial direction, which can only be "up" or "down". And the components along the other two spatial axes can't be determined, due to the rules of angular momentum in quantum mechanics (or due to the uncertainty principle, if you prefer).

    And for photons, we're talking about the direction of polarisation. A photon can be horizontally or vertically polarised, for example, if we choose those two directions as eigenvectors. But it can't be polarised in the direction of propagation of the photon - that degree of freedom does not exist for photons (unlike the case of some other spin 1 particles).

    So, I don't think there's a problem with saying that electrons and photons both exhibit two degrees of freedom in their "spin".
     
  16. QuarkHead Remedial Math Student Valued Senior Member

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    Don't you mean eigenvalues? In this case you seem to be asserting (reasonably enough) that the quantum state of the electron (a vector), when acted upon by the magnetic momentum operator has a spectrum of eigenvalues of precisely 2.

    Is is not the case that experiments show that an "undetected" photon will be be both horizontally and vertically polarized until this is measured?

    Again, as far as I understood from P.A,M Dirac, the result of a measurement of any quantum state is one of eigenvalues of the Hermitian operators acting upon it. Have I seriously misunderstood?
     
  17. arfa brane call me arf Valued Senior Member

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    Just been reading some stuff about spin. Apparently spin is a property of fields and how they transform under a Lorentz gauge, it isn't an intrinsic property of quantized angular momentum.
    That is, spin captures the transformation properties of a field given certain symmetry contraints. So it's the EM field which is spin-1 (and the gauge 'particle' just represents this).
    --https://physics.stackexchange.com/q...nce-between-spin-and-polarization-of-a-photon
     
  18. James R Just this guy, you know? Staff Member

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    Quarkhead:

    No, I meant eigenvectors. Specifically, I was thinking of state basis vectors in Hilbert space.

    An eigenvalue is just a number that pops out when you operate on an eigenvector.

    Yes.

    Yes, the photon can be in a superposition state - partly horizontally polarised and partly vertically polarised. Classically, this would be described as the polarisation direction of the light being in some intermediate direction between horizontal and vertical. Note: horizontal and vertical polarisation are not the only possible choices for a basis of eigenvectors, but if we choose to detect the photons using that basis (say by passing the light through a polariser), then we'll always measure one of the two directions.

    Similarly, in the electron case the electron can be in a superposition of "spin up" and "spin down" states. And, again, in actual measurements we're free to choose the particular basis that we're using, for example by altering the direction of the magnetic field that we're using to measure the spin direction.

    Your statement about measurements is correct.
     

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