# Measuring the Momentum of a Quantum System using Position Measurements

A basic way to measure the momentum of a charged particle is to know that (classically) it follows a circular trajectory in a uniform magnetic field perpendicular to the plane of its trajectory, with a radius $$R$$ given by $$R=\frac{p}{eB}$$ in the nonrelativistic regimes, where p is the momentum of the particle, e its charge, and B the magnetic field strength.

Reading in my quantum mechanics textbooks, the basic idea thus to measure momentum is by sending a stream of particles through a slit as shown in the figure below, which guarantees (classically) that the momentum is essentially in the $$+\hat{z}$$ direction. Beyond is a region of uniform magnetic field, followed by a set of detectors. If the intensity of the source is very weak, flashes will appear on the detectors one by one; and using the expression for $$R$$ we can associate with each flash a value of momentum in the $$+\hat{z}$$ direction depending on the position of the flash, $$p = ReB$$. My confusion however comes in that in order to measure the momentum, we have measured the position of the particle as well! If we can see the point at which the detector flashed, we have certainly determined the position. Thus again, we know the position, and we also know the momentum. This seems to violate Heisenberg's Uncertainty principle! What am I missing here?

• This seems one of the more clever ways to flout HUP. The exact way it fails is hard to pinpoint since in the quantum reality it doesn't have a well defined trajectory so the initial assumption is wrong. – jacob1729 Mar 18 '19 at 0:42

$$\big [ x_\alpha , p_\beta \big ] = i \hbar \delta_{\alpha,\beta} \, . \ \ \ \ (1)$$
In other words $$\mathbf{p}$$ cannot be a function of $$\mathbf{x}$$ otherwise the variables would commute.