# Uncertainity in Photon Position (Quantum Physics)

The wavelength of a photon is measured with an uncertainty of 1 ppm (that is $\Delta \lambda / \lambda =10^{-6}$). What is the uncertainty in the position of this photon if it is

a) a visible photon with $\lambda = 500 nm$

b) an X ray photon with $\lambda =0.1 nm$

c) a $\gamma$ photon with $\lambda =1 fm$

The Heisenberg principle for position and momentum is

$\Delta(x) = \hbar/\Delta(p)$

or equivalent

$\lambda*p = h$

By writing the uncertainties we  have

$\Delta(\lambda*p) = \Delta(h)$

Since $h$ is a constant $\Delta(h) =0$ so that

$\lambda*\Delta(p) +\Delta(\lambda)*p =0$

Hence

$\Delta(p) =-p*\frac{\Delta(\lambda)}{\lambda} =-\frac {h}{\lambda}*\frac{\Delta(\lambda)}{\lambda} =-\frac{h*\Delta \lambda }{\lambda ^{2}}$

Back into the expression of $\Delta(x)$ we obtain:

$\Delta x =(\lambda^{2})/(2\pi\Delta\lambda) =\frac{\lambda}{2\pi\left ( \frac{\Delta\lambda}{\lambda} \right )}$

With the given numerical values one gets

a) $\Delta x =0.08 m$

b) $\Delta x =1.59*10^{-5} m$

c) $\Delta x =1.59*10^{-10} m$