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In double slit experiment, they say they can put a detector after the slit so they know exactly which slit the photon has passed.

But if you successfully detect a photon, isn't the photon also consumed in the detection process?

Because the detection process is basically energy exchange. A photon has a quantum energy. Any energy exchange will consume the photon itself.

So the detected photons are consumed and would never reach the wall. All photons that have reached the wall are the ones not detected.

Am I correct?

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  • $\begingroup$ No, you are not correct. One can reflect the photon on a mirror, which will transfer momentum to the mirror. This, too, will change the experiment, just in a different way (the photon's energy will change, which is a change in wavelength, which also destroys the interference). $\endgroup$ – CuriousOne Jan 2 '16 at 4:23
  • $\begingroup$ If you detect a photon by any method, at least you also change certain property of the photon itself in the process. So the photon is not the same one anyway. Since it is not the same one, you can't expect the experiment to be the same as before. There is no mystery in double slit experiment with detectors then. Is my understanding correct? $\endgroup$ – ayuanx Jan 2 '16 at 4:29
  • $\begingroup$ There never was any mystery in the double slit experiment. It isn't even being used in physics proper to derive quantum mechanics. There is absolutely nothing interesting in it from a physicist's perspective. Unfortunately, just like wave-particle duality and Schroedinger's cat and the EPR phenomenon it has taken on a life of its own among layman and poorly trained teachers who don't actually understand the details of the discipline they are trying to teach. $\endgroup$ – CuriousOne Jan 2 '16 at 4:32
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This statement is wrong:

A photon has a quantum energy.

A photon may, when produced, carry a quantum of energy off a quantized bound state of a system, like an atom, a molecule or a particle resonance, like a pi0. It may also be produced by other processes, as Bremsstrahlung , synchrotron radiaton and in metal antennas, because of the existence of electronic bands, a continuous spectrum comes out.

The correct statement is: a "photon may take away a quantum of energy". A photon can be of any energy whatsoever.

Any energy exchange will consume the photon itself.

Only if the exchange happens with a bound atomic or molecular system it is necessary that a quantum of energy ( +/- the Heisenberg uncertainty) has to be absorbed, and only the correct frequency photons will disappear in the interaction.

Otherwise there exists elastic scattering where only the angle of incidence changes, and Compton scattering and Raleigh scattering. In these processes the photon may lose energy or gain energy and thus just change frequency. ( also red shifts and blue shifts appear in various fields and kinematical frameworks). These interactions happen with various collective fields and there is no quantized spectrum.

So in principle one might design a double slit experiment where the photon will interact minimally (within the dimensions of the problem) so as to see what happens. They have already done that with electrons, demonstrating that it is the change in the boundary conditions that destroys the pattern, the electrons still reach the screen.

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  • $\begingroup$ Yes, you are right. A photon (let's call it P1) can have any amount of energy depending on its frequency. My understanding is that any successful measurement applied on this photon P1 will change its state, which is equivalent to say that P1 is consumed yet in the process anther photon P2 may be produced. However, the produced P2 is not equivalent to the consumed P1. $\endgroup$ – ayuanx Jan 2 '16 at 6:13
  • $\begingroup$ In a feynman diagram of compton scattering the implication is that the scattered photon is a continuation of the incident as much as the scattered electron is the continuation of the incident due to momentum and energy consrvation. This is a consistent framework mathematically . As the model works, i.e. is validated by measurements, we can assume macroscopically the same set up. In any case , photons do not have individual identity cards, nor electrons. you could say the same for the electrons in the diagrams. lyndonashmore.com/feynman_diagrams_for_compton_sca.htm $\endgroup$ – anna v Jan 2 '16 at 6:53
  • $\begingroup$ there are also all the other conservation laws, angular momentum, quantum numbers also $\endgroup$ – anna v Jan 2 '16 at 7:09

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