Heisenberg on ''uncertainty relation does not apply to the past''

[vanhees71]

Usually the momentum of (electrically charged) particles in high-energy particle physics is measured by analyzing the tracks of the particles in a magnetic field. From the curvature of the track you get the momentum (provided you have identified which particle you are measuring).

Why there are indeed "tracks" to be measured here, has been clarified in the late 1920ies in a famous paper by Mott analyzing $\alpha$ particles in a cloud chamber quantum-mechanically.

 

[Morbert]

This statement makes no sense to me, because the uncertainty relation is a mathematically deduced property of states, i.e., for any state in a theory where position and momentum observables with the corresponding commutation relation holds, the standard deviations of these quantities fulfill $\Delta x \Delta p \geq \hbar/2$. Now if you prepare some state at $t=t_0$, then both the time evolution into the past and the future is given by unitary time evolution (for a closed system of course). Thus if your state at $t=t_0$ is a statistical operator also the corresponding time-evolved statistical operators (arguing in the Schrödinger picture for convenience) are again proper statistical operators and for them the uncertainty relation holds. It's clear that an initial uncertainty may also decrease with time evolution (no matter whether you consider the time "evolution" into the future, which is the physical case, of into the past), but it can never violate the uncertainty relation.

What you say is true for anyone state what Heisenberg is likely alluding to was weak values[1] of observables, computed from a two-states framework, formalised by Watanabe, Aharonov Vaidman et al. I.e. If an expectation value for $A$ is $\langle A\rangle_{\rho_1}$ we can compute an expectation value for weak values of A as $\langle A\rangle_{\rho_1,\rho_2}$ which will not have variances subject to the HUP.

[1] https://xqp.physik.uni-muenchen.de/publications/files/theses_master/master_dziewior.pdf

 

[physika]

You do not need to know velocity to establish trajectories, position is enough.