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Deriving Equation

  1. Jan 9, 2012 #1
    I have attached a PDF with some equations.

    (a) is essentially describing a pure state quantum system, with (b) and (c) statistical mixtures.

    I've managed to substitute (d) and (e) into (b) and (c) respectively, however I was wondering how to substitue (d) and (e) into (a).

    Apparently the answer is similiar to 1/square2=[|45,X>+|135,Y>], but was wondering if I could be walked through the steps to get to that.

    I know there are inverses of (b) and (c), but if we apply those inverses to (b) and (c), do we end up with the same result as those in the PDF (d) into (b) and (e) into (c)?

    Attached Files:

  2. jcsd
  3. Jan 9, 2012 #2


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    Only pure states can be written as ket-vectors in Hilbert space, so all three are pure states. You may want to read something about density matrices.

    I don't really get what you want to do. |X> and |Y> just seem to be shorthand notations for |45,A+1> and |135,A-1>. You may also want to read something about direct products / tensor products, because this is the mathematical structure implied by notations like |ab> or |a>|b> or |a>⊗|b>.

    You probably got this from a paper. I don't think you will really understand its contents unless you learn some basic math about quantum mechanical states. Unfortunately, I can't recommend a good beginner's text. Maybe try wikipedia?
  4. Jan 9, 2012 #3
    I've taken the equations out of a book. Essentially what is happening is a 45 degree polarised photon goes through a birefringement crystal, with a detector on the other side. (b) and (c) describe the situation where collapse occurs, and a further test is done on the photon at 45 degrees. (a) describes the situation if the detector goes into superposition, with the photon then undergoing a 45 degree test.

    If we add (c) and (d) to the statistical mixture equations, we get a different result to what would be obtained if we added (c) and (d) to (a).
  5. Jan 9, 2012 #4


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    Please explain what you are trying to accomplish. Your question is still very vague and your notation seems seriously flawed.

    Ok, so |θ> denotes the state of a photon with fixed polarization angle θ wrt to the crystal orientation and |A±1> denotes the state of a detector which has detected a photon in state |45°> respectively |135°>.

    1) How big is the probability for a photon in initial state |45°> to pass the crystal?
    2) How do you get a superposition of |45°> and |135°> from an initial |45°> state?
    3) We have already defined the meaning of |A±1>. Why on earth would you give the same name to the completely different situations (b) and (c)?
    4) Forget about kets for a moment and just look at equations (c) and (e). Both are of the structure x=a-b and x=c-d. If I equate them, I immediately get a=c and b=d. So how can (d) and (e) be more than a renaming of variables?

    Why do you want to add these states in the first place?
  6. Jan 9, 2012 #5
    (a) The photon enters the crystal, and comes out either V or H polarised. V polarised means A+1 on the detector, H - A-1. Strictly speaking the system combined is entangled and in a superposition.
    Equations (b) and (c) describe the situation had collapse occured at the apparatus.

    A further test of 45 degrees is done on the photon. If it were V or H polarised, it'd have 1/2 probability of passing/failing the test. V=|45+135> H=|45-135>|. This information is then substituted into the situation that the system is in a pure state, and the same done for if its a statistical mixture.

    There is a further observable Z that takes on values either X or Y, in accord with the information in the PDF.
  7. Jan 9, 2012 #6


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    Ok. Let's start from the beginning. You have a photon in an initial state |ψ>.
    -First case: You want to pass it through a polarizer and finally onto a detecter.
    -Second case: You want to pass it through two polarizers and finally onto a detector.

    You want to apply this measurement scheme. So please show me how you get to your result (a) (which corresponds to the first case).
  8. Jan 9, 2012 #7
    Equation (a) takes into account both the first and second case.

    EDIT: First case is the photon goes through the crystal, onto a detector (which doesn't absorb it, so allows us to further test it). The next thing we do is send it through a 45 degree filter, and detect whether it goes through or not.
  9. Jan 10, 2012 #8
    Would there be anyone to experimentally differentiate between whether the systems under investigation were in superposition, or whether they were in a mixture? In the original thought experiment, there is an observable Z on the apparatus that takes the value X if the superposition |A+1>+|A-1> exists, otherwise Y for the superposition |A+1>-|A-1>; X being correlated to the photon passing the 45 degree test, Y correlated to the photon failing the 45 degree test.

    In the statistical mixture, X and the failure of the 45 degree photon test can occur together, as well as Y and the passing of the 45 degree test.

    Would there be other ways to experimentally differentiate the two situations (perhaps by only further testing the photon, and without reliance on an observable Z)?

    I take it the equation (a) can still be described the way it is, even without subjecting the photon to the 45 degree polarisation test, as a V or H polarised photon is in superposition of 45 and 135 polarisations, and this is reflected in that equation (except with equation (a), V+H and V-H superpositions exist.)
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