# Heisenbergs Uncertainty for light

1. Jun 16, 2008

### duffbeerforme

Hi, I was just wondering if heisenbergs uncertainty principal is about position and velocity (or energy and time),.. then how come we can know that a photon travels at light speed (its velocity) exactly and also know that it hit some detector (its position).
If we know the velocity exactly then the light must be everywhere at the same time right?

2. Jun 16, 2008

### pmb_phy

To be precise the uncertainty principle which you're referrring to relates the uncertainty between position and momentum. It seems to me (although I'm not 100% sure) that since its quite possible to measure the momentum p of the photon and its energy E to arbitrary precision (i.e. the energy operator commutes with the position operator) then one can deduce the precise value of the momentum. The relation is p = (E/c^2)c = E/c and therefore c = E/p.

Pete

3. Jun 16, 2008

### Usaf Moji

I'm not an expert, but I think I can answer this one.

As the person above me said, the uncertainty relation doesn't apply to any two observables you can think of. It only applies to observables that don't "commute". So it would apply to position and momentum, but not to position and velocity.

If you consider the position and momentum of a photon, you will see that neither variable is known precisely. The momentum of any given photon is actually a cloud of momenta. So, let's say you think you have photons of a given wavelength (and therefore, a given momentum). It turns out that what you actually have is a bunch of different wavelengths that differ slightly from each other. Think of the photon as a bunch of waves with different wavelengths grouped together in a "wave packet" like here:
http://www.st-andrews.ac.uk/~bds2/ltsn/Edinburgh/wave/index.html

Likewise, the point where the photon contacts the detector is not a precise point but is a small cloud of points.

The smaller the position cloud is, the bigger the momentum cloud will be, and vice-versa.

4. Jun 16, 2008

### pmb_phy

There is no such thing as a velocity operator and as such there is no observable corresponding to such an operator. I guess you could define a velocity operator by using the expression p = mv and rewrite it as v = p/m. Since momentum has an operator and m is a constant then there doesn't seem be be a problem with it. However it is obvious from this definition that the velocity operator commutes with the momentum operator.

As far as measuring position as a function of time and deducing the speed from those measurements then I don't see how that is possible since detecting the position of a photon involves destroying the photon. Two successive measurements of position would then seem to be impossible.

Pete