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If I direct a laser pointer north and I put a photodetector eastwards (i.e. at $90^\circ$ ), and I wait for a very very long time (in a perfect vacuum if necessary), will the detector ever be triggered by a photon?

Harish Chandra Rajpoot
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D Deusdat
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Photons traveling from spacetime point A (say the exit of your laser pointer at time $t_1$) to B (somewhere on the line towards the north at time $t_2$) can do this in every way. Via east to the west to the north, via the moon, via Proxima Centauri (with higher than light speed), or whatever. The chance that the photon travels from A to B in a straight line is practically one. There are chances (probability amplitudes) though associated with the other paths taken. The chance that a photon travels from A to B in a different way than a straight line is very small, depending on the path. The more the path differs from a straight line, the smaller the chance that the photon will actually take this path. But if you wait long enough, a photon will hit the detector eastwards.

See for example also this question, or any article on the path integral interpretation of quantum mechanics, which states that a particle (and thus a photon) takes any path in traveling between two spacetime points. All these paths taken together (this taking together is further specified in the approach) give you the wave function of the particle. There is a (very small) chance that a photon is observed light years away from the laser pointer between the two times I specified above. This can give rise to faster than light photons (just as there is a chance that photons travel faster than light in ordinary quantum mechanics), but these photons can't be used for transmitting information faster than light, as these faster than light photons appear only randomly (and, as said, with a very small chance).

Of course, it depends also on the state in which you prepare the photons. Only if you could produce photons with a momentum directed to the detector and with zero momentum uncertainty (and thus infinite position uncertainty along the line pointer-detector while the uncertainty in position perpendicular to this direction is zero) you will not find photons away from the detector. Such a preparation is not realistic though.
If, realistically, you prepare the photons in a state with a small momentum (energy) uncertainty with a relative large corresponding position uncertainty in the direction of momentum like in a laser (the uncertainty in position in the direction perpendicular to the initial momentum being relatively small), there will always develop uncertainties in position in every direction, so photons can always be detected away from the detector. But much less than away from the detector. Of course, I'm assuming the detector to be placed on the line which extends outwards from the laser pointer.

Deschele Schilder
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  • All right. But does the photon actually travel TO point B? Isn't there any kind of spread? If a sniper points his laser sight on a target, a detector NEXT to the target (say 1 meter away) should click once in a while? – D Deusdat May 08 '21 at 12:01
  • @DDeusdat Detectors everywhere around the target should click once in a while (meaning that the photons have actually traveled to these detectors). The more distant the detectors, the fewer clicks, in accordance with the probabilities introduced by quantum theory. – Deschele Schilder May 08 '21 at 12:22
  • This might have some relevance but the problem is that being the answer correct or not, if the photon reaches the detector point B (say the north in OP formulation) isn't relevant anymore. – Alchimista May 08 '21 at 13:06
  • @Alchimista The point is that a photon is subjected to uncertainty in both momentum and position. There is always a probability to find the photon anywhere in the universe. Even if you prepare a photon in a state where it has zero uncertainty in position or momentum, it will always evolve into a state for which there is uncertainty in both. So every photon you send from a laser pointer to a detector will have a chance of not ending up at the detector. Because the associated wavefunction extends all over space (just as all paths extend all over space). – Deschele Schilder May 08 '21 at 16:50
  • @DescheleSchilder yes this sounds more or less what I am convinced of. Perhaps I should have wrote my comment under the OP question also here. It seems to me it would more straight to ask if a photon can move just straight along an hypothetical direction. Than your answer would apply more naturally (again correct or not. I do believe it is correct, but my ground is the basic notion of indeterminacy and a bit of path integral). – Alchimista May 08 '21 at 20:38
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Photons are elementary particles of the standard model. They have mass zero and E=hnu, where nu is the frequency of light the come from, the laser in this case.

As point particles generated by the lasing phenomenon they travel in straight lines because of conservation of energy and momentum. The laser beam will not be perfectly collimated, so there would be a statistical dispersion around the classical ray direction, but 90degrees would mean a very bad laser construction.

(p.s. path integrals cannot be used for paths of real, described by a real four vector , particle. They are a mathematical tool for calculating the probability distributions and the mathematical curves cannot be cut for a calculation that is meant to be going from A to B. )

anna v
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  • For laser photons, the uncertainty in position along directions perpendicular to the direction of initial momentum will grow. It can initially be zero (meaning that there is only position uncertainty in the direction of momentum), but as soon as the photons are created the uncertainty in the other directions will grow. Only if you could create photons with zero energy uncertainty (meaning infinite position uncertainty in the direction of momentum), you cannot find photons away from the target. Conversely, if the target is close enough, this will not be the case also. – Deschele Schilder May 08 '21 at 17:13
  • The last two scenarios are not implied though (the scenario in which you create zero energy uncertainty photons can't even be implied), so there always is a chance to find a photon outside the target. (I didn't downvote, by the way) – Deschele Schilder May 08 '21 at 17:13
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What the other answers do not address is shielding. Electromagnetism has this phenomenon called shielding and we are able to do it pretty well with our current technologies.

EM waves do spread spherically in space always. The only way to go around this is with shielding, in your case, the laser is using very effective mirrors.

The most common type of laser uses feedback from an optical cavity—a pair of mirrors on either end of the gain medium. Light bounces back and forth between the mirrors, passing through the gain medium and being amplified each time. Typically one of the two mirrors, the output coupler, is partially transparent. Some of the light escapes through this mirror.

https://en.wikipedia.org/wiki/Laser

The reason in your example, why you will only detect photons north, is because the shielding (mirrors) are designed so that photons will only escape in that direction. So the answer to your question is, that if this laser is designed to create a narrow beam, then you will only detect photons north because the mirrors only lets them escape that direction.

That being said, no mirror is perfect, and if you wait long enough, you might detect photons that escape in different directions through the mirrors, knowing that our universe is quantum mechanical.

Árpád Szendrei
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