# Does a single photon have a wavelength?

If a single photon has a wavelength, would it violate Heisenberg Uncertainty principle?
Since photon is a particle, that means it can be represented by a wave packet. But wave packet can not have definite wavelength, only pure wave can have wavelength

backward

One form of the uncertainty principle is that the uncertainty in momentum (delta p) multiplied by the uncertainty in position (delta x) is greater than some minimum number. A single photon with a definite wavelength (uncertainty zero) is one limiting case. With definite wavelength, you know it has a definite momentum. The uncertainty principle implies that you have infinite uncertainty in the position. Consider that a single wave stretches across infinite space. So you can't define a position for an infinite wave. In order to try and make the wave start to have a shape at some position, you have to add other wavelengths. So the uncertainty in momentum starts to go up while the uncertainty in the position starts to come down. At the other end of this that you have to add an infinite number of wavelength together in order to put the wave packet at a single location. So you now have definite (zero uncertainty) position but infinite uncertainty in the momentum.

I have diffracted monoenergetic photons through a Bragg diffraction crystal with an angular resolution of a few seconds of arc at a very low counting rate. every photon had the same wavelength within about 1 part in 1000.

No one can answer this question correctly.

For example, when you accelerate one electron to the definite momentum,