Does uncertainty mean fixed, but unmeasurable underlying value, or a measurable value that fundamentally fluctuates? WRT specific question

Question

Consider the following question from Eisberg and Resnick's "Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles," pg 84, #4.34

A boy on top of a ladder of height $H$ is dropping marbles of mass $m$ to the floor and trying to hit a crack on the floor. To aim, he is using equipment of the highest possible precision. Show that the marbles will miss the crack by an average distance of the order of $(\hbar/m)^{1/2}(H/g)^{1/4}$

Here is my way to solve this. Assume that the original uncertainty in velocity is $\Delta v_x$, then we have (assuming equality in Heisenberg's principle) that at the point of the marble hitting the floor, its uncertainty in position (by which it misses the crack) is $\Delta x = \frac{\hbar}{2m\Delta v_x}$. If the marble falls for time $t$ then it will deviate from the centre by $\Delta v_xt$ which is also how much it misses the crack by, $\Delta x$. Making this substitution and using $t = \sqrt{\frac{2H}{g}}$, we have that $$\Delta x = \frac{\hbar t}{2m\Delta x}\implies \Delta x = \sqrt{\frac{\hbar t}{2m}} = \sqrt{\frac\hbar{2m}\sqrt{\frac{2H}{g}}}$$which gives us the required order.

The answer relies on the $x$-velocity of the particle being unchanged due to vertical motion, which makes sense. But it also turns that into the uncertainty $\Delta v_x$ remaining the same. I'm a little confused as to why this is possible, because in my understanding this uncertainty reflects a fundamental unknowability in the value of that quantity. My understanding is that we cannot pin the value of the uncertainty in velocity down, even if that pinned-down value is unknown. This is analogous, in my mind, to how the vacuum must have vacuum fluctuation energies because if the energy was zero there would be no uncertainty in its energy, and so it cannot have some fixed value but must fluctuate.

What does the uncertainty principle here mean, exactly?

Answer 1

You write:

My understanding is that we cannot pin the value of the uncertainty in velocity down, even if that pinned-down value is unknown. This is analogous, in my mind, to how the vacuum must have vacuum fluctuation energies because if the energy was zero there would be no uncertainty in its energy, and so it cannot have some fixed value but must fluctuate.

Terms like "fluctuation" and "uncertainty" have some very misleading connotations. The equations of motion of quantum theory describe the evolution of a measurable quantity, such as whether a particle passes through a crack in a floor, by a Hermitian operator called an observable. The eigenvalues of the observable represent the set of possible measurement results.

In general the evolution of an observable depends on what happens to all of the possible measurement results during the experiment: quantum interference. For an example see Section 2 of

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

Quantum theory predicts the expectation values of an observable: a sum or integral over the set of possible values weighted by a probability amplitude. In general the sets of probabilities for different observables constrain one another by the uncertainty principle:

https://arxiv.org/abs/1511.04857

If quantum systems are described by observables, then it is not the case that a particle has a single trajectory. Rather there are several trajectories that interfere with one another to produce the final result. Some physicists don't like this and continue to use those equations but say that they don't describe reality, e.g. - the Copenhagen and statistical interpretations. They don't say specifically what is happening in reality to produce experimental outcomes so in those interpretations there is no answer to your question. Saying that quantum physics describes how reality works is often called the many worlds interpretation because it features multiple interfering versions of the same system:

https://arxiv.org/abs/1111.2189

https://arxiv.org/abs/quant-ph/0104033

In this interpretation quantum theory describes how systems evolve in reality and so that the position and momentum of a particle is described by the relevant observables. If that's true then the uncertainty relations are describing properties of those observables. Measurements are just physical processes that produce records of those observables and this explains why measurement results respect the uncertainty principle:

https://arxiv.org/abs/gr-qc/9304039

https://arxiv.org/abs/gr-qc/9307013

A system doesn't fluctuate around a value. Rather its observables just have a range of different possible values with probabilities constrained by the uncertainty principle and this predicts the probabilities for possible values that deviate from the mean. As for pinning down the uncertainty, that is determined by the observables, the quantum state and the measure of uncertainty you use.

Answer 2

This problem asks You to treat a marble quantum mechanically like you would a photon or electron. This is legitimate. Like an electron, a marble has a wave function. It has an uncertainty in position and momentum. The same principals apply. So you can learn things about how the marble behaves by reading Does the collapse of the wave function happen immediately everywhere?

But size makes a difference. Normally you can follow the trajectory of a marble as closely as you want. You can measure its position and momentum as accurately as you want. If you try the same thing for an electron, it doesn't always work.

You often find electrons bound to a nucleus. You might want to follow the trajectory around the nucleus. You cannot. The uncertainty in position is the size of the orbital.

Another place electrons used to be commonly found was in old cathode ray tube TVs. Electrons were emitted from a hot cathode, steered with magnetic fields, and aimed at a phosphor coated screen. The uncertainty principal does limit how precisely they can be aimed. But they could be aimed precisely enough to make an image.

Cathode ray tubes show a common way of working with quantum mechanics. You prepare an electron in a state. In this state, the electron does not have a position or momentum. The state is a wave function that describes a range of positions and momenta. To predict where the electron could be found next, you have to add up contributions from all of them.

Later, you measure the electron's position by seeing which grain of phosphor lights up. You will find the outcome consistent with the wave function. Now the electron has a new state where the position is in that grain.

You could measure the position or momentum of an electron along the way by bouncing light off it. But an electron is so light that this would significantly change its state.

For a marble, the quantum uncertainties are too small to measure. If you plug numbers into your equation, you will see that $\sqrt{\hbar}$ give you a factor of $10^{-17}$. Never the less, the uncertainties are there.

If you want to get that kind of accuracy, you would have to treat the marble like an electron in a CRT. You would prepare it in a state, and later measure its position. If you measure its position or momentum in between, you would change the state significantly enough that you would not get the required accuracy. So the experiment would have to be in a vacuum, in the dark, and so on.