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I propose the following scenario:

At $t=0$, a photon is emitted from a star. At $t=n$, said photon is received and interpreted by some detector.

My question is whether or not it is accurate to say that at $t=0$, the photon was emitted at a correct angle and time such that it had to have landed in that detector. In other words, is photon travel deterministic such that if you knew the emission angle of the photon, you could determine where the photon should eventually end up? It's pretty easy to predict where photons will end up on small scales (otherwise lasers would be fairly useless), but does this ability translate well to cosmic scales (assuming we know the exact emission angle)?

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up vote 7 down vote accepted

Let us be clear about the problem.

A photon is a quantum mechanical entity and follows the laws of quantum mechanics. There is always a probability attached to any possible path it can take so the strict answer is "no, the path of the photon is not deterministic".

BUT the problem changes when speaking of a large ensemble of photons, which is any light we can see. These are described very well by classical electrodynamics in a deterministic way. There exists a smooth transition between the photon quantum mechanical framework and the classical theory of Maxwell,though it needs a lot of physics background to understand it

That is why lasers work so well, they emit huge ensemble of photons correlated and in phase and can practically act like light rays, of classical optics.

One might say then that if one knows the angle of the star the ray path is completely determined, BUT here we hit on General Relativity to clarify what determined means. In General Relativity zero mass objects follow geodesics, so it is not only angles that enter the problem but also massive bodies, before one can ascertain determinism.

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+1 for bending. – Xaqron Apr 7 '13 at 9:29
+1 but shouldn't you mention that a large ensemble of photons behaves ALMOST determinstically but not completely? – centralcharge Jun 22 '13 at 8:46
@dimension10 It is the classical that is deterministic as a wave, yes. – anna v Jun 22 '13 at 10:37

We can satisfy your requirement "the photon was emitted at a correct angle" by "the photon was prepared in a momentum eigenstate". If the photon has definite momentum $\bf{k}$, then its direction of travel is well defined, as you have specified. A photon is a discrete excitation of a "mode", i.e. a solution of Maxwell's equations. For a photon in a momentum eigenstate, this mode will be a plane wave (it will also have a polarization vector).

Now if I've understood it correctly, you also want to be able to say that given such a photon, having detected it at some location you could immediately say where it was produced - its emission point would have to be somewhere along the line traced from its detection point in the direction of its momentum. Now the problem is with "its emission point". Since we've specified that the photon was prepared in a momentum eigenstate, it didn't have a definite emission point - its emission point was completely indeterminate due to the Heisenberg uncertainty principle. (There is also a related problem, namely that photons don't even admit position operators, but this subtlety isn't needed for this discussion).

You may object that surely we know (roughly) the position of emission of a photon which originates from an atomic transition - it must be at the location of the atom (which will be known to some degree of accuracy). This is true, but atomic transitions don't excite photons in plane wave states, so the momentum (in particular the direction) of the photons excited by these transitions is unknown - they're not plane wave momentum eigenstates.

In the laser example, the beam propagation is indeed approximately along classical straight lines, but you cannot trace the activity of a single photon in such a state. Indeed in the coherent state, the number of photons present isn't even definite.

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It is not accurate to say this. Otherwise, the two slit experiment could not work. If every photon had a single direction when it was emitted, it could not go through both slits and interfere with itself; it would have to go through whichever slit it was originally directed towards.

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Don't forget that the photon is a wave as well as a particle. So it spreads out sideways (transversely), and hence travels to its destination by more than one path, which can be verified by the Airy disk in a telescope and other diffraction effects*. The exact landing point is probabilistic, which argues against determinism even if you post-select for a specific starting point and a specific ending point.

*Furthermore, if you believe in Feynman's ideas, the photon traveled throughout all space and time.

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"Furthermore, if you believe in Feynman's ideas, the photon traveled throughout all space and time." "Feynnmans's ideas" are just an interpretation of QM, by the way. – centralcharge Jul 20 '13 at 8:13

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