WWCY said:
Also, how would device uncertainty affect the position-momentum uncertainty inherent in a particle?
Finally, are the concepts you mentioned covered in a "standard" particle physics text? Thank you.
I think this common misunderstanding to the uncertainty principle reaches back to Heisenberg's unfortunate misunderstanding of his own discovery. He wrote the first (partially flawed) paper in Copenhagen, but at the time Bohr wasn't around, and that's why there is this confusion. Bohr had to correct Heisenberg, and unfortunately some textbookwriters obviously didn't realize it.
The misunderstanding is that the uncertainty relation refers to the accuracy of measurements and to the perturbation of the measured system by the measurement. This is utterly wrong! The uncertainty relation refers to properties of the state of the system and thus to the preparation of a system in a given state. It's very general since all it needs to be derived is that observables are represented by self-adjoing operators and that the scalar product of a Hilbert space is positive definite.
If ##A## and ##B## are observables, represented by the self-adjoint operator ##\hat{A}## and ##\hat{B}## then for any (pure or mixed) state
$$\Delta A \Delta B \geq \frac{1}{2}|\langle [\hat{A},\hat{B}] \rangle|,$$
where the expectation values are calculated with the state the particle is prepared in, which is represented by a statistical operator ##\hat{\rho}##, cf.
$$\langle f(\hat{A}) \rangle=\mathrm{Tr} (\hat{\rho} \hat{A}).$$
The standard deviations are defined by
$$\Delta A^2=\langle \hat{A}^2 \rangle - \langle \hat{A} \rangle^2.$$
Let's discuss the meaning using the position component ##x## of a particle and the momentum component in the same direction ##p##. Then we have ##[\hat{x},\hat{p}]=\hbar \hat{1}##, and the uncertainty relation reads
$$\Delta x \Delta p \geq \frac{\hbar}{2}.$$
This tells us that if the particle is prepared in a state with a pretty accurate localization, i.e., such that ##\Delta x## is pretty small, then necessarily ##\Delta p## is pretty large, i.e., not very well determined. So the Heisenberg uncertainty relation tells you about certain limits of accurate determination of two non-compatible observables by state preparation.
This doesn't tell you anything about the accuracy with which you might be able to measure the one or the other observable. This is not determined by the state the particle is prepared in but by the construction of the measurement apparatus. You can always, at least in principle measure the particles position or its momentum as accurately as you like. So you can choose to measure very accurately the position to figure out whether its standard deviation is really ##\Delta x## for a given state you prepared the particle in. For that purpose you just have to use a large enough ensemble of equally such prepared particles and make a position measurement that is much more accurate than the uncertainty given by the standard deviation ##\Delta x## due to the particle's preparation in the given state.
Of course there is indeed always some disturbance of the measured system by the measurement since the system must interact with the measurement device to perform this measurement, but this disturbance is much less simple to analyze and it's not such a general property of the QT formalism as the Heisenberg uncertainty relation since it must take into account the individual properties of the measured system and the used measurement device.
Of course, also measurement devices are after all quantum (many-body) systems and thus also are subject to the general laws described by QT, and particularly also to the uncertainty relations. Thus the position measurement of a particle is of course at least limited by the possibility to localize the measurement device itself, and as the uncertainty principle for position and momentum tells you, you can localize the measurement device in principle as accurate as you like, but ##\Delta x=0## is impossible, i.e., there's always some uncertainty in measuring the position of the particle (the same of course also holds for momentum).
Though very difficult to hit this fundamental quantum limit in accuracy is very hard since usually measurement devices, as macrscopic many-body systems, are very hard to control, and there's always thermal motion (fluctuations) which contribute much more to the uncertainty of a position or momentum measurement than the quantum limit, today there are some devices that reach it. One example are the gravitational-wave detectors. Here some limit is the quantum fluctuations of light, and today they even use socalled "squeezed states" (a generalization of the coherent states):
https://arxiv.org/abs/1109.2295
https://www.repo.uni-hannover.de/handle/123456789/3807
