How does the Heisenberg uncertainty principle apply to photons?
2 Answers
We literally cannot be certain whatsoever of a photon's position in space from just the facts known about a photon, its interaction with some other body is needed.
Explanation:
More broadly, Heisenberg posits that we cannot be entirely "sure" of a body's position in space. Obviously, as the body's mass increases, there is a better chance at being "certain".
Where
However, a photon is considered to be "massless" with modern theories of physics. I'm not well versed in physics, so I can't elaborate, but if someone can, do so! Let's play with the equation a little:
As you know, anything divided by
There is a different version of the Heisenberg Uncertainty Principle for photons in three dimensions. (In one dimension, they are both
As presented in this paper, in their notation, we have from Eq.
DeltarDeltap >= 4ℏ where:
r is the radial position for a given direction in three dimensions (or the "center of energy").Deltar is then the "spread" in the energy of the light from its center.Deltap = h/(Deltalambda) = (hDeltanu)/(c) is the uncertainty in momentum for photons from the de Broglie relation. That is,lambda = h/p andnu = c/lambda . After realizing thatDeltalambda = h/(Deltap) , the rest follows from there.lambda is the wavelength in"m" andnu is the frequency in"s"^(-1) .c is of course the speed of light in"m/s" .ℏ = h//2pi is the reduced Planck's constant.
The version for electrons in three dimensions is
DeltarDeltap >= (3ℏ)/2 ,where here,
Deltap = mDeltav like usual, since electrons have nonzero rest mass and photons have zero rest mass.
The interesting bit here is that the uncertainty relation for photons only extends "properly" to three dimensions (each dimension multiplied by
In other words, in the limit as