# I have a thought experiment that defies the uncertainty principle - what's wrong with it?

There's a particle somewhere on the x-axis (this whole thing is 1-dimensional) at x0, which we do not yet know. I fire a photon at it from the origin with a known speed (c) and a known wavelength λ1 and so a known momentum h/λ1. It collides with the particle elastically and heads back to the origin. The particle initially had an unknown momentum pi, and a momentum pf after the collision. We measure the time between firing the photon and the photon returning to the origin (tr), and also the wavelength of the photon (λ2) (we thus also know the final momentum of the photon).

The maths:

Because for some reason light travels at the speed of light (how selfish), we have that the light travels a distance of 2 x0, in a time tr, at a speed c. Thus x0 = trc/2.

By conservation of momentum we have that pi + h/λ1 = pf + h/λ2 => pf - pi = h(1/λ1 - 1/λ2)

By conservation of energy we have that pi2/2m + hc/λ1 = pf2/2m + hc/λ2 => pf2 - pi2 = 2 mch(1/λ1 - 1/λ2) = 2 mc (pf - pi)

=> pf + pi = 2 mc

=> pf - pi = h(1/λ1 - 1/λ2)

One can then simply solve for * pf and pi and by p = mv one can determine the velocity, and since we know its original position x0, we can determine the position for all time.

We thus know the exact position and momentum of the particle for all time. This ain't it chief. What part of this is incorrect?

Many thanks to anyone who gives it a go.

• The principle implies that you cannot even fire such a deterministic photon. Firing and detecting it involves E and t. – Alchimista Dec 27 '18 at 15:34

To see why a photon cannot have a definite wavelength and yet be localized consider taking the fourier transform of a light wave localized to some area of space $$c * \Delta t$$ where $$c$$ is the speed of light and $$\Delta t$$ is the window of time during which the photon is emitted from its source.