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

[A. Neumaier]

TL;DR Summary

Heisenberg said 1930 that ''the uncertainty relation does not refer to the past''

In post #30 of a now closed thread, vanhees71 wrote:

Do we have the precise paper by Heisenberg, where he makes such a strange and enigmatic statement?

The quote in question is from his 1930 book "The Physical Principles of the Quantum Theory" on p.20 of the English translation of the German original. There Heisenberg writes:

This formulation makes it clear that the uncertainty relation does not refer to the past; if the velocity of the particle is at first known and the position then exactly measured, the position for times previous to the measurement may be calculated. Then for these past times, $\Delta p\Delta q$ is smaller than the usual limiting value, but this knowledge of the past is of a purely speculative character, since it can never (because of the unknown change in momentum caused by the position measurement) be used as an initial condition in any calculation of the future progress of the electron and thus cannot be subjected to experimental verification. It is a matter of personal belief whether such a calculation concerning the past history of the electron can be ascribed any physical reality or not.

 

[AndreiB]

Summary:: Heisenberg said 1930 that ''the uncertainty relation does not refer to the past''

Indeed. For some reason however there is a widespread belief that quantum physics proved that particles can't have trajectories. This is what Nima Arkani-Hamed writes in this paper:

The Future of Fundamental Physics​

Daedalus (2012) 141 (3): 53–66.

"However, it is of fundamental importance to physics that we cannot speak precisely of position and momentum, but only position or momentum."

This is a perfect example of how even the best minds can be misguided.

 

[vanhees71]

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.

 

[A. Neumaier]

For some reason however there is a widespread belief that quantum physics proved that particles can't have trajectories.

There is not the slightest hint that particles could have sharp (point particle) trajectories in standard QM. Even Heisenberg's backward projection is valid only based on the assumption that the prior path of the particle was a straight line. That's why he classifies his observation as belief rather than physics.

 

[CelHolo]

This statement makes no sense to me, because the uncertainty relation is a mathematically deduced property of states

In essence that's what Heisenberg is saying, he's countering a common student error at the time. He's saying that if somebody measured momentum to be within some band $\Delta p$ at $t_{0}$ and then measured position within some error range $\Delta x$ at $t_{1}$, people thought one could deduce an error range at $t_{0}$ for position using basic kinematical arguments. This deduced $\Delta x$ at $t_{0}$ would violate the HUP if combined with $\Delta p$.

Heisenberg is just saying such a retrodicted $\Delta x$ is meaningless physically.

Like a lot of this early stuff it makes little sense unless you learn about general opinions and mistakes at the time. Like the hole debate in GR. It can be made more difficult because word usage was very different at the time as well.

Basically here you have to remember he was talking to people deeply immersed in classical mechanics who didn't know when certain basic tricks of classical mechanics had been rendered meaningless by QM.

 

[AndreiB]

There is not the slightest hint that particles could have sharp (point particle) trajectories in standard QM.

And there is no slightest hint that particles cannot have sharp (point particle) trajectories in standard QM.

Maybe those trajectories exist or maybe not. QM says nothing about them.

Even Heisenberg's backward projection is valid only based on the assumption that the prior path of the particle was a straight line. That's why he classifies his observation as belief rather than physics.

Momentum is a conserved quantity, so, unless there is some field (gravitational, magnetic, electric) along the particle's path, there is no reason for its direction to change.

 

[AndreiB]

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.

Let's assume you measure very accurately the position of an electron to be x0 at t0. The uncertainty relation would imply that the particle's momentum is very uncertain.

You measure again the position, very accurately, to be x1 at t1. You can now calculate the momentum between t0 and t1 as x1-x0/t1-t0. By increasing arbitrarily t1-t0 (delaying the measurement) you can know the momentum between t0 and t1 with any arbitrary accuracy, violating the uncertainty relation.

 

[CelHolo]

Maybe those trajectories exist or maybe not. QM says nothing about them

Well you don't need them for all of chemistry, atomic physics, etc. So these paths seem to be doing very little physically.
Also the theory does contradict the notion of paths fairly clearly. Any state $\rho$ won't give a sharp position trajectory as there will always be fluctuations or "noise" measured by $Tr(\rho X(t)^2 )$ and other statistical moments.

 

[AndreiB]

Well you don't need them for all of chemistry, atomic physics, etc. So these paths seem to be doing very little physically.

I agree, however they might be important when gravity is taken into account (the unification program).

 

[vanhees71]

Let's assume you measure very accurately the position of an electron to be x0 at t0. The uncertainty relation would imply that the particle's momentum is very uncertain.

You measure again the position, very accurately, to be x1 at t1. You can now calculate the momentum between t0 and t1 as x1-x0/t1-t0. By increasing arbitrarily t1-t0 (delaying the measurement) you can know the momentum between t0 and t1 with any arbitrary accuracy, violating the uncertainty relation.

[EDIT: Corrected in view of #13]

I can not, because due to the large uncertainty of the momentum at $t_0$ the uncertainty of the position $x_1$ at $t_1$, $\Delta x_1$, is pretty large at time $t_1$, and this uncertainty gets the larger the larger you choose $t_1-t_0$.

 

[CelHolo]

I agree, however they might be important when gravity is taken into account (the unification program).

Well when we start taking gravity and non-inertial frames into account in quantum theory, or even just special relativistic inertial frames, we don't even have an invariant notion of particle (see the Unruh effect) or a Lorentz covariant position operator or for some particles we have no positon operator at all (e.g. photons). So it doesn't seem like things are going in the direction of trajectories being meaningful at all.

However given the mistake in your post #7, I think this is a case of arguing against quantum theory without knowing the mathematics.

 

[A. Neumaier]

You measure again the position, very accurately, to be x1 at t1.

You don't know where the particle should be measured the second time, since the momentum is very uncertain!

 

[AndreiB]

I can not, because due to the large uncertainty of the momentum at $t_0$ the position $x_1$ at $t_1$ is pretty large at time $t_1$, and this uncertainty gets the larger the larger you choose $t_1-t_0$.

I don't follow you here. Who cares if x1 is large? This does not make X1-X0 less certain.

 

[AndreiB]

You don't know where the particle should be measured the second time, since the momentum is very uncertain!

You use a very large fluorescent screen.

 

[AndreiB]

Well when we start taking gravity and non-inertial frames into account in quantum theory, or even just special relativistic inertial frames, we don't even have an invariant notion of particle (see the Unruh effect) or a Lorentz covariant position operator or for some particles we have no positon operator at all (e.g. photons). So it doesn't seem like things are going in the direction of trajectories being meaningful at all.

All this does not mean that the particles have no trajectory.

However given the mistake in your post #7, I think this is a case of arguing against quantum theory without knowing the mathematics.

What mistake?

 

[CelHolo]

All this does not mean that the particles have no trajectory.

Well if the fact that current physics makes no use of paths, further advances have made the notion increasingly untenable and irrelevant, e.g. in QFT it's not even possible to define in a covariant observer independent way the position of a single particle let alone a whole path, then I think this is just immovable "belief" and I'll stop here.

 

[vanhees71]

I don't follow you here. Who cares if x1 is large? This does not make X1-X0 less certain.

Of course it must read $\Delta x_1$ is very large. I'll correct it in the posting above.

 

[AndreiB]

Of course it must read $\Delta x_1$ is very large. I'll correct it in the posting above.

Why would $\Delta x_1$ be large?

 

[Lord Jestocost]

Maybe those trajectories exist or maybe not. QM says nothing about them.

In case such trajectories exist, could one explain electron diffraction experiments on base of this view?

 

[AndreiB]

In case such trajectories exist, could one explain electron diffraction experiments on base of this view?

Why not? What is the argument here?

 

[A. Neumaier]

You use a very large fluorescent screen.

Then you don't get very accurate position measurements, and the computed momentum will be uncertain enough to satisfy the uncertainty relation.

 

[AndreiB]

Then you don't get very accurate position measurements, and the computed momentum will be uncertain enough to satisfy the uncertainty relation.

Why? The accuracy is given by the resolution of the screen. This has nothing to do with the size of the screen.

 

[vanhees71]

Sure, the resolution of the screen may be way better than the uncertainty of the particle's position, but the particle's position at $t_1$ is large, because the momentum uncertainty was large at $t_0$.

 

[A. Neumaier]

Why? The accuracy is given by the resolution of the screen. This has nothing to do with the size of the screen.

Your arguments are qualitative handwaving only, and convince only yourself - you have no idea how small Planck's constant is. Read the old papers and the book by Heisenberg, where he discussed quite a number of quantitative attempts to get both position and momentum very accurate through a combination of calculation and measurement, none of them succeeding to beat the uncertainty relation.

 

[EPR]

I thought that the double slit experiment with detectors on proved that quantum particles can have a definite trajectory(when being 'watched'). Interference only happens in the absence of which-path information.
If single electrons do not interfere with themselves, they move as if they have a fixed trajectory. And when you tag them in e.g. the quantum eraser, they no longer show superposition effects.

 

[A. Neumaier]

I thought that the double slit experiment with detectors on proved that quantum particles can have a definite trajectory(when being 'watched'). Interference only happens in the absence of which-path information.
If single electrons do not interfere with themselves, they move as if they have a fixed trajectory. And when you tag them in e.g. the quantum eraser, they no longer show superposition effects.

They have approximate trajectories, consistent with the uncertainty relation.

 

[PeterDonis]

You can now calculate the momentum between t0 and t1 as x1-x0/t1-t0.

You can calculate this number, but that doesn't mean it means anything physically.

By increasing arbitrarily t1-t0 (delaying the measurement) you can know the momentum between t0 and t1

No, you can "know" a number that you calculated that has units of momentum, but that doesn't mean that number means anything physically.

 

[AndreiB]

You can calculate this number, but that doesn't mean it means anything physically.

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.

No, you can "know" a number that you calculated that has units of momentum, but that doesn't mean that number means anything physically.

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

 

[AndreiB]

Sure, the resolution of the screen may be way better than the uncertainty of the particle's position, but the particle's position at $t_1$ is large, because the momentum uncertainty was large at $t_0$.

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.

 

[AndreiB]

Your arguments are qualitative handwaving only, and convince only yourself - you have no idea how small Planck's constant is.

Let's calculate!

electron speed: 10^6 m/s
gun opening size: 1µm

The velocity uncertainty calculated using this calculator:

https://www.omnicalculator.com/physics/heisenberg-uncertainty

is ~60 m/s.

Distance is 10^10 Km, so deltaX is 10^13m.
Time measurement error is 1ms. Total travel time would be 10^7 +-10^(-3) s.

The velocity is in the interval 999999.9999 - 1000000.0001 m/s. The uncertainty is about 0.0002 m/s.

0.002 is smaller than 60

Read the old papers and the book by Heisenberg, where he discussed quite a number of quantitative attempts to get both position and momentum very accurate through a combination of calculation and measurement, none of them succeeding to beat the uncertainty relation.

If Heisenberg was logically consistent I guess that those examples are not for the past. His quote in the OP is about 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.