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

[vanhees71]

I think we speak about different things. I calculate the momentum of a single particle from the measurements at t0 and t1. The uncertainty is only given by the accuracy of those measurements and for a large time and travel distance it becomes insignificant. For this particle, the calculated momentum and position do not obey the uncertainty limit.

For different particles the final position and time of detection would be different so their calculated momenta would be different as well. For an ensemble of such particles the uncertainty relation would hold.

This cannot be. If you accept quantum theory then the uncertainty relation follows. It's a mathematical conclusion.

If you have prepared a particle to be a pretty sharp position at $t=t_0$, then necessarily it must have a pretty large uncertainty in momentum. If you now propagate the particle to a time $t=t_1>t_0$ using the dynamical rules of quantum theory necessarily the position uncertainty (i.e., the standard deviation of the position observable) must get larger. For a free particle the momentum uncertainty stays the same. So you rather get a larger uncertainty product than it was at $t=t_0$. The uncertainty relation thus holds at any later time.

 

[AndreiB]

If you have prepared a particle to be a pretty sharp position at $t=t_0$, then necessarily it must have a pretty large uncertainty in momentum.

Indeed. That uncertainty persists until you finally detect the particle at t1. Before t1 you are uncertain. After t1 you know the momentum, you are not uncertain anymore. This is what Heisenberg points out in that quote. Uncertainty is about future (immediately after t0 - you do not know x1 and t1), not about past (after t1). Uncertainty limits your predictions, not postdictions.

 

[AndreiB]

If you now propagate the particle to a time $t=t_1>t_0$ using the dynamical rules of quantum theory...

Those dynamical rules stop at t1 since the measurement collapses the state.

 

[vanhees71]

But it doesn't stop the meaning of the state as a probabilistic description!

 

[PeterDonis]

Why so? It tells you that the particle traveled the distance X1-X0 in the amount of time given by t1-t0. The physical meaning is crystal clear.

Classically, perhaps. But we are not talking about classical physics. We are talking about quantum physics. You can't apply classical reasoning to quantum physics.

How would you measure momentum so that it means something physically?

By measuring momentum. For example, by measuring the impulse transferred by the object to some measuring device.

 

[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.