Application of Heisenberg's uncertainty principle on photons

Question

While I was hobby-reading about quantum mechanics I came across Heisenberg's theory. But while I was trying to understand it I thought of this: if the speed of light (as well as the momentum of a photon) is well-defined, then if in an experiment the position of the photon was found, the researchers would know the exact position of the photon, but also its momentum which is constant. Is this an exception from the rule, or am I just ignorant?

PS, 12 grade physics education, please try to explain in simple words

score 3 · Answer 1 · answered Dec 11 '15 at 19:53

3

The momentum of a photon is not a function of its velocity, or else all photons would have identical momentum. Since we know that different photons have different energies, it would be strange if they all had identical momentum. Instead, the momentum of a photon is related to its wavelength: $$ p = \frac{hc}{\lambda} $$ where $h$ is Planck's constant. So our ignorance about the precise momentum state of the photon is in fact a reflection of the fact that we don't know precisely what its wavelength is. (If you ever study Fourier transforms this will make a lot of sense.)

answered Dec 11 '15 at 19:53

zeldredge

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First of all thanks for your correction. Does this mean that all photons, even laser generated ones have an uncertain wavelength? – Anastasis Poutahidis Dec 11 '15 at 19:59
3

Yep. There's always some "line width" to the laser. Experimentalists worry about such things, since they often want to address atomic transitions and the like very precisely. – zeldredge Dec 11 '15 at 20:02
So in a sense photons have the biggest uncertainty... Thanks for all your answers. – Anastasis Poutahidis Dec 11 '15 at 20:10

Application of Heisenberg's uncertainty principle on photons

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