I'm not clear as to how the overall interference pattern (or lack thereof) observed in the quantum eraser experiments (http://grad.physics.sunysb.edu/~amarch/ [Broken]) ends up being the same if the eraser is present or not present. My thinking follows. Say N entangled photons are sent through the apparatus without the eraser, and then N with the eraser. Configuration 1: Eraser is NOT present In this configuration some M (<N) photons will end up being detected with appropriate coincidence conditions satisfied. Roughly M/2 of these cases will correspond to the photon knowingly passing through one slit and resulting in a pattern without interference, call it pattern A, on the screen. Similarly, about M/2 cases correspond to the photon knowingly passing through the other slit and resulting in pattern B. In the remaining N-M cases we can't determine which-path info and we get the usual interference pattern, call it pattern C. Configuration 2: Eraser is present In this configuration which-path info is never available and hence pattern C is the result after all N photons pass through the apparatus. So, the OVERALL pattern in configuration 1 should be (M/(2N))A + (M/(2N))B + ((N-M)/N)C, but the overall pattern in configuration 2 should be C. If the overall patterns are the same then: (M/(2N))A + (M/(2N))B + ((N-M)/N)C = C now moving the RHS to the LHS: (M/(2N))A + (M/(2N))B - (2M/(2N))C = 0 now multiplying by 2N/M A + B = 2C This seems impossible to me. Suppose the screen is setup at a distance behind the slits where the maxima in pattern C occurs on the line normal to the screen and midway between the two slits (as appears to be the case in publication by Walborn, Cunha, Padua, ...). I don't think the pattern A+B can have a maxima in that location. I must be missing something here.