Heisenberg's microscope

The concept was criticized by Heisenberg's mentor Niels Bohr, and theoretical and experimental developments have suggested that Heisenberg's intuitive explanation of his mathematical result might be misleading. While the act of measurement does lead to uncertainty, the loss of precision is less than that predicted by Heisenberg's argument when measured at the level of an individual state. The formal mathematical result remains valid, however, and the original intuitive argument has also been vindicated mathematically when the notion of disturbance is expanded to be independent of any specific state.

Heisenberg's argument

Heisenberg supposes that an electron is like a classical particle, moving in the ${\displaystyle x}$ direction along a line below the microscope. Let the cone of light rays leaving the microscope lens and focusing on the electron make an angle ${\displaystyle \epsilon }$ with the electron. Let ${\displaystyle \lambda }$ be the wavelength of the light rays. Then, according to the laws of classical optics, the microscope can only resolve the position of the electron up to an accuracy of

${\displaystyle \mathrm{\Delta }x=\frac{\lambda }{\sin \epsilon }.}$

An observer perceives an image of the particle because the light rays strike the particle and bounce back through the microscope to the observer's eye. We know from experimental evidence that when a photon strikes an electron, the latter has a Compton recoil with momentum proportional to ${\displaystyle h/\lambda }$, where ${\displaystyle h}$ is the Planck constant. However, the extent of "recoil cannot be exactly known, since the direction of the scattered photon is undetermined within the bundle of rays entering the microscope." In particular, the electron's momentum in the ${\displaystyle x}$ direction is only determined up to

${\displaystyle \mathrm{\Delta }{p}_x\approx \frac{h}{\lambda }\sin \epsilon .}$

Combining the relations for ${\displaystyle \mathrm{\Delta }x}$ and ${\displaystyle \mathrm{\Delta }{p}_x}$, we thus have

${\displaystyle \mathrm{\Delta }x\mathrm{\Delta }{p}_x\approx \left(\frac{\lambda }{\sin \epsilon }\right)\left(\frac{h}{\lambda }\sin \epsilon \right)=h}$,

which is an approximate expression of Heisenberg's uncertainty principle.