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Quantum erase explained with waves

  1. Jan 7, 2014 #1
    Someone attended me on the article `A double-slit quantum eraser' from Walborn e.a. 2008. This is a much better, clearer and more detailed described quantum eraser then Kim e.a. A basic explanation is in http://grad.physics.sunysb.edu/~amarch/ [Broken]. A simplified diagram:


    A BBO is radiated with a laser. Now and then the BBO emits two entangled photons. The yellow photon goes through a double slit to signal detector Ds, with a quarter wave plate Q1 or Q2 in each path. They are rotated 90° to each other, so produce a polarization which rotates in opposite directions, depending on the polarization of the incoming photon. The green photon goes to detector Dp, with a polarizing cube POL in the path. Only photons in Ds are registered which coincidence with photons in Dp. Dp is situated closer to the BBO then Ds. So photons are first detected by Dp. The intensity of the beam is so low, that photons are detected before a new photon is emitted by the BBO. So it works with single photons. The results were:
    1 Without Q1/2 and POL there was an interference pattern in Ds
    2 With Q1/2 there was no interference.
    3 With Q1/2 and POL, adjusted on Q1 fast axis, there was interference.
    4 With Q1/2 and POL, adjusted on Q2 fast axis, there was interference (but 180º shifted with 3).
    5 1-4 with Dp on a larger distance then Ds (delayed eraser), gives the same result.

    The article explains this with QM math, and:
    2: Q1/Q2 marks the red/blue beams, so one can detect which path the photon went, also when this measurement is not actually done (in 2). This removes the interference pattern.
    3-4: When placing POL at a suitable angle, the polarization information is erased, so it is not possible anymore to know which path the photon went. Therefore the interference pattern is back again.

    Another way to analyse this measurement is with the classical wave property of the photon, which should give the same result. But it doesn't:

    2a The idea is that if the particle goes through Q1, that only Q1 changes its polarization. But polarization is a property of the wave, not of the particle. The wave goes through both slits and should be added at the detector position. The result is not different for particles going through the blue or red slit. Therefore it is still not possible to measure the path of the particle.

    2b Still the interference disappears, so why? The article does remark that the output of 3 + 4 = 2, but it does not draw the obvious conclusion: the interference pattern does not disappear, but there are now two patterns, 180º shifted, which added looks like no interference.

    3-4a The interference pattern of 3 and 4 differ 180º. This will happen when Q1 and Q2 shift the phase 90º in opposite direction. This agrees with the angle of POL and the behaviour of quarter wave plates. If the polarization of the incoming wave is parallel to one of the axes, the output polarization will not rotate, but only shift the phase 90º:


    Above is a view from Ds to the slits and Q1/Q2. If the incoming polarization is vertical then there is no horizontal component, so no rotation of the outgoing polarization. Only S is delayed 90º compared to F. With horizontal incoming polarization it is the other way around. Indeed the article describes that in 3 and 4 the POL is set to the F axis of Q1 or Q2 and it is conformal with the measured interference.

    3-4b This shows that the detection time in Dp is not crucial. This explains that a position of Dp at longer distance then Ds (the so-called delayed erasure) makes no difference. It is the behaviour of POL together with Q1/Q2 which determines the interference pattern.

    So the conclusion should be that there is no marking of the particle, no erasing, so no delayed erasing.
    Last edited by a moderator: May 6, 2017
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  3. Jan 7, 2014 #2


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    Few comments:

    I suppose you mean the right thing, but precise language is important when discussing this experiment. There is an interference pattern in coincidence counts between Ds and Dp, but not on Ds alone.

    That is not correct. Electric fields add vectorially, so orthogonal polarizations do not interfere. At this stage it is possible to measure the path of the particle just by placing a polarizer in front of Ds which only allows one of the two polarizations to pass. This will only let the field which went through one of the slits pass through. Note that it does not matter, whether you actually place the detector there. The mere possibility of doing it at the position where the detection takes places is sufficient.
  4. Jan 7, 2014 #3
    Indeed I suppose so in this measurement. But I think it will not differ, because from all photons detected in Ds the entangled photons will be detected in Dp.

    With "added" I mean taking into account orthogonal. But the point is that the sum, whatever it is, does not differ if the particle goes through one or the other slit. So detector Ds cannot measure which path the photon followed. If you place a polarizer you just block polarization from one slit. But still the particle can goes through each slit. The particle has no polarization.
    Last edited: Jan 7, 2014
  5. Jan 7, 2014 #4


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    No, they will not be and that is one of the most important points of the experiment. The slit in front of Dp is only 300 micrometers wide, so actually just a small part of the photons detected at Ds will see the partner detected at Dp. In fact, if you move Dp around, you are able to change the interference pattern in the coincidence counts.

    Assuming that doing coincidence counting is just some trick to reduce noise or something similar is the main fallacy behind many misconceptions about the entangled photons created from BBOs or similar sources. The two-photon probability amplitudes are absoluteley fundamental and coincidence counting is the single most important thing to do in such measurements. Measurements in one arm alone are absolutely meaningless.

    Put a polarizer there and you can. It is as simple as that.
  6. Jan 7, 2014 #5
    If Q1/Q2 are mounted as above, what will the polarization be at Ds in case the particle goes through one or the other slit?
  7. Jan 7, 2014 #6


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    That depends on the initial polarization of the particle before the double slit. A QWP that has the optical axis oriented at 45 degrees to the polarization axis of the light beam will turn horizontally polarized light into circularly polarized light, either clockwise or counterclockwise - that depends on whether you place it at +45 degrees or -45 degrees. For vertically polarized light, it is the other way round. For any other linear polarization a QWP will produce elliptically polarized light. However, if you write down the Jones vectors for the light fields created by some fixed linear polarization going through
    a) a QWP at some orientation
    b) a QWP oriented at 90 degrees to the first one
    you will always find that the two fields that arise are orthogonal.

    That means that if you now place a polarizer at Ds, you can get different results, depending on how it is oriented: You can get no interference pattern at all (orientation such that only light from one slit passes through), you can get fringes (superposition of both paths) or antifringes (superposition of both paths, but different relative phase). Using the latter orientations also means erasing which-way-information (now in a classical setup). However, without the polarizer the which-way info is always present as the fields are orthogonal.
  8. Jan 7, 2014 #7
    Let us take an easy example: Suppose the photon (not the particle) is polarized as above, parrallel to the fast axis of Q2.
  9. Jan 7, 2014 #8


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    I do not get your point. Is the photon not a particle?

    Anyway, that is a very different example. The previous discussion was aimed at the state given in the paper, where the photon pair creates a Bell state and you ensure that the QWPs are always at +/-45 degrees to the polarization of the incoming photon.

    The case of a well defined polarization parallel to the fast axis of Q2 is pretty boring. Consider the interference pattern you get without having any quarter wave plates inserted. Now by inserting the two quarter wave plates, you just delay the phase of the light field by a fixed amount at Q2 and the same fixed amount +pi/2 at Q1. What you will get is the same polarization as before, but a shifted interference pattern. You will get an interference pattern shifted the other way if you instead have a well defined polarization along the slow axis of Q2. You will get no interference pattern at 45 degrees.
  10. Jan 7, 2014 #9
    No, the photon is a particle and a wave. The particle is a part of the photon properties.

    Right, and is that different if the particle goes through another slit? So how could you measure through which slit the particle went? (the wave goes through both slits).
    Last edited: Jan 7, 2014
  11. Jan 7, 2014 #10


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    Again no! The photon is neither a particle nor a wave (whatever you mean by the latter, I guess a classical wave field). It's a "one-particle Fock state" of the free electromagnetic quantum field.
  12. Jan 7, 2014 #11


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    As has already been pointed out, the notion of particle-wave duality is pretty outdated. All the quantities of interest - no matter whether the layman would consider them wave-like or particle-like from a naive point of view - can be derived from a single, absolutely consistent formulation.

    I do not get your point. Let me try to understand. The difference between a light field with polarization aligned with the fast or slow axis of a QWP (just a phase shift) and a light field polarized at 45 degrees to the fast/slow axis (polarization conversion from linear to circular) is clear? In the latter case some spin angular momentum of +/- hbar is transferred to the photon. This is measurable.
  13. Jan 7, 2014 #12
    What I mean is that above you describe the wave at the detector to be the sum of two waves which differ in phase +pi/2 (suppose the detector is in the middle). Was the photon in that explanation going through Q1 or Q2? Would it be different if the photon went through the other slit? (we are talking about a single photon).
  14. Jan 7, 2014 #13


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    I still do not get what you are aiming at. In the case of alignment along the fast or slow axis you do not get which way information, but that is unrelated to the double slit quantum eraser where you are always at 45 degrees to the optical axis and therefore do not just have a simple phase shift, but also polarization change and transfer of spin orbital momentum. The two cases are fundamentally different.
  15. Jan 8, 2014 #14
    Alright, then at 45 degrees. At the detector Ds now two waves add up (from both slits) which are rotating left and right. How does the result differ if the photon is going through one or the other slit?

    Or in your above argument: the polarizer will only let the field which went through one of the slits pass through. But how does that relate to which slit the particle went through? If the particle passed through slit 1, its wave is going through both slits. If it passed slit 2, again its wave goes through both slits.
  16. Jan 8, 2014 #15


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    As I wrote earlier: Going through the QWP transfers spin angular momentum to the photon. If you perform a measurement of the spin angular momentum of the photon, you can clearly say which way it went.

    I do not get your focus on wave things. QM is based on a wave-like formalism and of course you consider all the paths and probabilities when calculating detection probabilities and averages. Besides calculating probabilities, how is the "wave" relevant for anything that happens in a single run of the experiment?

    As soon as - in an otherwise fixed experimental geometry - you can perform a measurement at the position of the final detector which allows you to identify subsets which can only correspond to slit A or B, you have path information. If you have interference and no path interference such a measurement is not even possible in principle.
  17. Jan 8, 2014 #16
    How do you measure spin?
  18. Jan 8, 2014 #17


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    There are several ways. The boring one consists of another QWP and a polarizer, but I suppose this is not what you want to hear.

    You can put a small, light, frictionless disk there. If it absorbs the photon, the angular momentum gets transferred to the disk and it will start to rotate, either clockwise or counterclockwise, depending on the spin.

    Or you can put a detector based on some spin-sensitive absorption path there. Consider for example the ground state of a quantum dot (artificial atom). The carriers there are excitons, bound electron-hole pairs usually consisting of electrons of spin +/- 1/2 and heavy holes with spin +/- 3/2. Due to Fermi exclusion only two of those pairs fit into the ground state. One with electron spin + 1/2 and hole spin - 3/2 (total spin -1) and one with electron spin -1/2 and hole spin +3/2 (total spin +1). Other combinations are not optically active. Now these pairs with spin +/- 1 can be created by absorption of a photon of matching spin. If you now insert polarized electrons and holes into the system, you can get a quantum dot, where one of the two electron-hole pairs is always present and the other is not. Accordingly only photons with one kind of spin can be absorbed
  19. Jan 8, 2014 #18
    I like the borings ones, because they better show what is happening. QWP and polarizer act on waves, so we are back again to waves and the question: how do the (sum of two) waves look like when the particle is going through Q1 or Q2? In what way do they differ?

    My answer on the question above: Q1 and Q2 cause polarization clockwise and anti-clockwise. So if they add up at the detector, there is no rotation anymore, so no spin.
  20. Jan 8, 2014 #19


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    We are going in circles here. The combination of QWP and polarizers will only let one of the fields pass.

    There is no average spin transfer if you repeat the experiment many times under the same initial conditions. But every single time you do a single experimental run, there will be absorption of one photon and spin will be transferred. The incoherent addition of the "waves" is important for averaging over many runs, for any single repetition it just gives you probabilities for what will happen. That is like having two doors and 50% of the people going through the left one and 50% going through the right one. The average "wave person" would then go through the wall in the middle of the two doors. Averages do not have a 1 to 1-relationship to the single repetition of the experiment.
  21. Jan 8, 2014 #20
    But as mentioned above, when both waves from each QWP combine again (add up) there is no spin anymore. If you add up the clockwise wave from Q1 and the anti-clockwise wave from Q2 at the position of absorption, the result is no-spin.
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