If momentum is a function of position, then how does the uncertainty principle work?

Question

Update

I found a 2023 article by J.K. Freericks that, I think, answers my question. A couple key quotes:

...any detector that we use is located somewhere. So, when it measures a particle, it is de facto measuring the position. [p. 2]

...one can, in principle, determine the transverse momentum and the position of impact [of a particle] with the screen essentially as accurately as desired...So, where does the uncertainty enter? It enters when we repeat the measurement and obtain data for another shot. Most likely we measure a position and momentum that are different from the first shot. Measure again, and yet a third different result. It is the variance of all of the shots that give us the uncertainty relations. [p. 7]

...the Heisenberg uncertainty principle...does not apply to single shots of a measurement, which can be carried out to as high a precision as desired. Instead, it is the fluctuations between different shots that governs the uncertainty principle. [p. 11]

Original question

If I understand correctly, Heisenberg's Uncertainty Principle holds that position and momentum cannot be known with absolute precision simultaneously. And yet, momentum is a function of position: p=mv, where v=∆x/t.

So how can it be that momentum and position have an “uncertainty relation” when momentum is necessarily inferred from position in this way? The implication of the uncertainty relation, as I understand it, is that zero uncertainty about momentum would mean infinite uncertainty about position. Yet, if position is infinitely uncertain then so is momentum.

The only discussion I have found of this issue is the question "Why isn't momentum a function of position in quantum mechanics?" This simply confuses me more; if momentum isn't a function of position then as far as I am concerned, it is not "momentum" anymore, but something else.

Answer 1

If I understand correctly, Heisenberg's Uncertainty Principle holds that position and momentum cannot be known with absolute precision simultaneously.

That is the consequence of the principle, but it is not the core of the Heisenberg uncertainty principle, or uncertainty relations. Nobody expects absolute precision already in classical physics - there are always errors of measurement or preparation. However in classical physics, it was expected to be possible to decrease these errors to arbitrarily small numbers (but still non-zero).

There are different ideas called "uncertainty principle", and not all of them are equally important (or correct).

One idea that is taught in quantum theory courses is that product of uncertainties about position and momentum of a single particle, induced by complete knowledge of its state, cannot be lower than $\hbar/2 = 5.2\text{e}-35~kg~ms^{-1}$. Thus it cannot be zero (this is the consequence).

This restriction is valid as far as predictions of results of future measurements, based on known quantum state, are concerned. It does not apply to knowledge of past position and momentum; it is possible to assign position and momentum the particle had in the past, with lower product of uncertainties. For example, one can prepare a beam of particles with negligible dispersion of momentum, and then measure position of those particles by detecting mark they left on a screen. For any such mark, we can record the time it appears, and for time right before that, we can consider the particle that left the mark, and retrodict its position and momentum with uncertainties whose product can be smaller than $\hbar/2$. Of course, this retrodiction of past variables cannot be easily verified, because when we measure position, we do not measure momentum, and vice versa, and each measurement affects the particle. So simultaneous existence of those past variables remains hypothetical and is not important in orthodox quantum theory.

The formula $\mathbf p = m\frac{\Delta \mathbf x}{\Delta t}$ does not show momentum is a function of position; it is a formula for calculating average momentum in between two points of detection (notice $\Delta \mathbf x$, which means difference of position between two points). This does not support or invalidate the above uncertainty principle in any way.

Answer 2

In classical physics the position of a particle is described at a a given time by a set of numbers $(x(t),y(t),z(t))$. If you were to measure the $x$ position of a particle at time $t$ you would get the result $x(t)$. The momentum of the particle is $(p_x(t),p_y(t),p_z(t))$ where $$p_x(t)=m\frac{dx}{dt}$$ and similarly for $p_y(t),p_z(t)$. If you measure the $x$ momentum of a particle at time $t$ you would get the result $p_x(t)$. The equations of motion for the particle are given in terms of position and momentum and their derivatives. So the variables that describe the motion of a particle of a motion in classical physics are the same as the possible measurement results.

Quantum physics is very different. The evolution of a physical quantity like position or momentum is described by an operator called an observable. The possible values of an observable are its eigenvalues. Quantum physics predicts the expectation values of those observables. To do that we use the state, which is a description of past records of measurement results in the form of an operator $\rho$. The expectation value of an observable $\hat{A}$ with eigenvalues $a_1,a_2\dots$ is $$\langle\hat{A}\rangle=tr(\rho\hat{A})=\sum_jp_ja_j$$ where $p_1,p_2\dots$ are the probabilities of the measurement outcomes. During quantum experiments what happens to all of the different possible values of the observable can affect the experiment's outcome: this is called quantum interference, see Section 2 of this paper for an example:

https://arxiv.org/abs/math/9911150

Since all of the possible values are required to explain the outcome of the experiment, you can't say that the system has just one of the possible values. As such, a particle in general doesn't have just a single position or momentum. Rather, it has a spread of positions and momenta. So why are these quantities that behave in such strange ways called position and momentum? There are a couple of related reasons. The first is that the equations of motion for quantum systems often have a similar form to those of classical systems with the classical positions and momenta swapped for the relevant operators:

https://arxiv.org/abs/quant-ph/0304202

Another reason is that in the classical limit of quantum theory these observables' expectation values behave a lot like classical positions and momenta. The classical limit arises when information is transferred out of a quantum system, which suppresses interference: this is called decoherence

https://arxiv.org/abs/1911.06282

For the sorts of systems you see around you, like stones and chairs and any dust grain large enough to see, information is copied out of them very quickly compared to the timescales over which they evolve, e.g. air molecules or photons interacting with an object. This results in the positions and momenta of those systems evolving approximately according to classical equations of motion:

https://arxiv.org/abs/0903.1802

Answer 3

There are no trajectories in quantum mechanics. In classical mechanics, you can talk about the position of a particle as a function of time, $x(t)$. Then, from there, you can derive the momentum as a function of time $p(t)=m\dot{x}(t)$. You cannot do this in quantum mechanics. Generally, all you can say at time $t$ is that you have some probability distribution for the position $x$. (More precisely you know the state or wavefunction, which is complex valued, but you can convert the wavefunction to a probability distribution and that is maybe easier to think about at a beginner level).

When passing to quantum mechanics, it's useful to be aware of different formulations of classical mechanics. A particularly useful one is the Hamiltonian formulation. In this way of thinking, the position and momentum are coordinates in a larger phase space. One point in phase space is a particular state of the system; one point in phase space will correspond to a ball of mass $m$ at the north pole moving vertically upward with speed $v$ for example. The fact that $p=m\dot{x}$ is an equation of motion -- particles obeying the laws of classical physics will take a path through phase space that will satisfy the condition that the $p$ coordinate of the particle is related to the time derivative of the $x$ coordinate of the particle in a particular way.

In quantum mechanics, it's useful to think of $x$ and $p$ as independent coordinates, and our knowledge of the two coordinates is generally incomplete. We might have some knowledge of the probability to measure the particle's position at different places at a given time, and knowledge of the probability to measure the particle's momentum at a given time. But we cannot measure the particle's position with certainty for multiple times and build up a trajectory $x(t)$ from which to calculate $\dot{x}$. In general are best thought of as two distinct quantities; we can make a measurement of one without knowing the other. (There are special situations where you can be a bit more direct in the relationship between position and momentum, like if you have a Gaussian "bump" type probability distribution for position centered on some value of $x$ and the center of the distribution is moving with some velocity, then the corresponding probability distribution for momentum will be a bump centered on $m$ times that velocity. But generically the relationship is more complicated in quantum mechanics).

The uncertainty principle then says that there's a lower bound on the product of the uncertainty (technically, standard deviation of the probability distribution) of our knowledge of $x$ and our knowledge of $p$. In the phase space picture, we can never fully localize a system to a single point in phase space; at best we can say a system corresponds to some blob in phase space of area $\hbar$.

Answer 4

Velocity, like any quantity in quantum mechanics, is an operator. We can obtain its form using the standard prescription for obtaining the "operator of time derivative": $$ \hat{v}=\hat{\dot{x}}=\frac{1}{i\hbar}[\hat{x},H]= \frac{p}{m} $$ The last equality assumes the kinetic part of the Hamiltonian as $K=\frac{p^2}{2m}$, but it can be done for more general Hamiltonians. Also note that I am not working here in Heisenberg representation, where visually identical equations govern the time evolution of operators.

This velocity operator does not commute with position. Note also that in presence of a magnetic field the momentum and velocity operators are not simply related by $p=mv$, which brings us to the fact that the Hamiltonian formalism in QM is largely base don the Hamiltonian formalism in classical mechanics (so-called correspondence principles.)

The intuitive relations between velocity, position and momentum are somewhat restored by the Ehrenfest theorem.