I have a thought experiment which in my understanding leads to faster than light messaging, and I had hoped the physics.SE community could help me resolve it. I should say that most of my understanding of quantum mechanics comes from these lectures by Feynman, although I am a mathematics major and can mostly see through the layman's filter he puts on. (He does that marvelously, by the way. I wish I was that good at it.)

Say you have a light source, two semi-transparent mirrors with 4% chance of reflecting a photon and one sensor. If you put the two mirrors parallel to eachother and shine light on it, then the sensor will have between $0\%$ and almost $16\%$ chance of registering a reflected photon, depending on the distance between the mirrors. Set the mirrors up for maximal constructive interference.

If you have full control of the timing of the light source (sending, say, one photon exactly once every millisecond), and the sensor is capable of differing by the time it takes the photon to be reflected from the first mirror and from the second mirror, then as I understand it the interference phenomenon would disappear. You will just get a (bit less than) $8\%$ chance of registering a photon.

Here's the problem: If I have another sensor on the other side of the mirrors, then it would, depending on the time resolution of the first sensor, either register $84\%$ or $92\%$ (i.e. the rest) of the photons sent by the source. If the first sensor is very far away from the mirrors and second sensor, then changing time resolusion (or just swapping the sensor) would have an immediate effect on the number of photons registered at the second sensor.

I don't believe this to be true, but as far as my understanding of quantum mechanics goes, that's the result I get.


First of all, you have to re-check your numbers and your model of the cavity. For one, the square of 4% is 0.16%, but this is not how étalons work: if you set the cavity length up for maximal interference, then (for a monochromatic beam, with an infinite-duration pulse) the transmitted intensity will be limited by the transmittance of the first mirror. (The light accumulates inside the cavity to match the intensity of the incoming beam; 4% of the incoming beam trickles in to replace the 4% that comes out of the second mirror.) On the other hand, if you send in a short pulse (shorter than the cavity round-trip time) the total transmission will be something like $$ \frac{4}{100}\left( \frac{4}{100} +\left(\frac{96}{100}\right)^2\times\frac{4}{100} +\left(\frac{96}{100}\right)^4\times\frac{4}{100} +\cdots \right) =\frac{16}{784}\approx2\% $$ where each term in the series represents a single pulse in a train of reflected pulses.

Thus, there is a difference in the total transmittance depending on the width of the wavepacket, although not the one you state.

Your paradox arises from the fact that it takes time for light to travel from the cavity to the detector. If you change anything in the source, those changes can only propagate at the speed of light, because quantum electrodynamics is a fully local theory. The only circumstance in which the removal of a photon at the far-away detector from the shared beam state is "instantly" measurable at the nearby detector is when the pulse is long enough that it's present in both at the same time - and then it's no big mystery.


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