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.