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.

  • 2
    $\begingroup$ The principle implies that you cannot even fire such a deterministic photon. Firing and detecting it involves E and t. $\endgroup$ – Alchimista Dec 27 '18 at 15:34

The problem is that the photon is also bound by the uncertainty relationship as the comment by Alchimista indicates. A photon of definite wavelength must extend over your entire x axis and cannot be localized. In order to "emit" a photon it will need to be in a localized packet with a distribution of wavelengths and a small, but non-zero, window of time when it was emitted. The uncertainty in exactly when it was emitted and exactly how much momentum it has will make sure your experiment does not disobey the uncertainly relationship.

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.


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