# Photons vs Uncertainty Principle

The uncertainty principle states: $\Delta_x\Delta_p>ħ/4$

We know the photon has a 0 rest mass but we are say that it's momentum is always the same $e=pc$

And more we are certain about the momentum, more we are not certain about the position, however since we are 100% sure about photons momentum then we can't be sure whatever about position! Is this true?

• It is not correct that a photon has a certain exact momentum. Photons have uncertainty in momentum, and uncertainty in frequency, and uncertainty in wavelength, and uncertainty in energy, as well as in position and time! Nov 17, 2013 at 19:22

Given this fact, the HUP can't be applied to light for position-momentum because there are problems defining a position observable for the photon. This has to do with the fact that if $(\vec{E}, \vec{B})$ is a solution of Maxwell's equations, then things like $(x_j \vec{E}, x_j \vec{B})$ (where $x_j$ are the Cartesian co-ordinates) generally aren't (the Gauss laws showing divergencelessness in freespace are violated by such fields). So therefore $(x_j \vec{E}, x_j \vec{B})$ is not in the Hilbert space of valid one-photon states. Contrast this with the scalar quantum electron state in the scalar massive particle nonrelativistic Schrödinger equation where the scalar eigenstates as $\mathbf{L}^2$ complete, so that if $\psi(x)$ is a quantum state in position co-ordinates, then $x \psi(x)$ is also in the Hilbert space of states.
Of course the HUP always applies to noncommuting (conjugate) observables and there are many pairs of those in QED and quantum optics. For example, corresponding Cartesian components of the $\hat{\vec{E}}$ and $c\,\hat{\vec{B}}$ vector valued observables fulfill the canonical commutation relationships. One can of course define an intensity field which yields a probability distribution to (destructively) photodetect a photon, but this is different from asking where (position observable) an electron electron is in an orbital. Electrons can be detected nondestructively - it is very hard to do this for photons. Also, position observables are readily defined only for scalar quantum states in nonrelativistic first quantized descriptions: there is of course no nonrelativitistic first quantised description of the photon. The bispinor valued electron state is also weird and the question of where the electron is cannot be addressed by a simple position observable either. Now you can still define the momentum with the usual observable, because the eigenfunctions of $-i\,\hbar\,\partial_j$ are plane waves, i.e. well defined momentum states. But when you talk of localization of photons and the probability distributions of where to detect them - you are talking about diffraction and you get Heisenberg style reciprocal relationships between wavenumber and photon detection position. This arises by dint of the Fourier transform - the wavevector components are the Fourier transform variables of, say, a classical $\vec{E}$ field. We can also interpret any solution of Maxwell's equation as the quantum state of one photon (I discuss how this arises in this answer to the question "How can we interpret polarization and frequency when we are dealing with one single photon?"). So, if we do so, $\hbar$ times the wavevector is the photon's momentum, so if we handwavingly interpret the value of $|\vec{E}|^2$ as a probability density to destructively detect the photon at a given position, the the HUP will apply: the mathematics of diffraction are the same as the mathematics of a pair of variables fulfilling the HUP: the eigenspaces of two observables fulfilling the CCR are needfully Fourier transforms of one another as I show in this answer to the question "How to prove that $i \hbar \partial_p$ is the operator of $x$ in momentum space?"