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Well, isn't this very simple? In my opinion you should even introduce the Hilbert-space structure with this example (although I'd prefer the spin of a massive particle since photons shouldn't be treated in QM 1).stevendaryl said:In the case of multiple filters, I don't see how von Neumann filters help explain intuitively what's going on.
You have a sequence of filters, the first oriented at angle 0o, the next at 45o, the next at 90o. Some fraction of the photons, 12.5%, will make it through all 3 filters. You can't understand that in terms of removing photons. Because after the first filter, you've removed all the photons that were polarized at 90o, but at the end, you still have some photons that were polarized at 90o.
I know that you know how to calculate this case, but the words don't actually match up with the facts. At least, not in my mind. The way that you describe von Neumann filters, in terms of throwing out unwanted stuff, sounds like you're selecting based on pre-existing properties, but that's not the case.
Take a single-photon source (e.g., some down-converter crystal and a laser, absorb one of the photons; this gives you perfectly unpolarized single-photons). Then run the corresponding beam through the polarizer oriented at an angle ##\alpha=0##. Then you prepare single photons with the state
$$\hat{\rho}_1=|\alpha=0 \rangle \langle \alpha=0|.$$
According to Born's rule you get on average half the photons through.
Now put another polarizer at angle ##\alpha \neq 0## into the remaining beam of photons. Then the probability for each photon to get through is
$$\langle \alpha|\hat{\rho}_1|\alpha \rangle=|\langle \alpha|\alpha=0 \rangle|^2=\cos^2 \alpha.$$
That's it according to QT. You can't know more about the polarization of the photons. Of course the remaining photons are then in the state
$$\hat{\rho}_2=|\alpha \rangle \langle \alpha|,$$
because by construction you've made a preparation through a von Neumann filter measurement. Of course in each step you get less photons, and then you normalize the state again. So in the experiment with the two polarizers in series you get only about ##(1/2) \cos^2 \alpha## of the original intensity (of the original number of single photons).
The same holds true for a third filter. As long as the relative angle between the two filters is not ##\pi/2##, you have some probability to get photons through, namely ##\cos^2(\alpha_1-\alpha_2)##, when ##\alpha_1## and ##\alpha_2## are the angles of the polarizer orientation (relative to some fixed axis of course).
What's a total enigma to me is, how you can claim that this simple facts are not matched by the theory. Such experiments are done frequently in any quantum-optics lab, and one would have heard about the failure of QED for such a simple and fundamental experiment with single-photon polarization states.
You seem to have fallen in the trap to consider photons just as little (massless) classical billard-ball like particles. But that's not so. They are quanta, and thus if they are polarized in direction ##\alpha## this does not imply that the cannot be found to be polarized in another direction ##\alpha'##, as long as ##\alpha-\alpha' \neq \pm \pi/2##.
By the way, all this holds of course true for classical electromagnetic fields. If you want a classical analgon for photons, it's way better to think about them as ultra-faint electromagnetic radiation than in terms of classical particles. You get almost always the correct results, although there are of course the well-known exceptions, which truly prove the necessity to quantize the em. field, like spontaneous emission, quantum beats, etc.