DrChinese said:
Apparently you are aware of the Heisenberg Uncertainty Principle (HUP), which has been around for nearly 100 years. Yet you are imagining that some simple experiment will demonstrate a violation of it. So there are several issues here.
1. The HUP does not itself prevent a simultaneous measurement of non-commuting observables. It prevents such measurement(s) from being performed with unlimited precision. If you measure position quite accurately, you will not get a precise value for momentum (and vice versa).
That's also not entirely true.
The usual Heisenberg uncertainty relation, proved within the first few lectures in QM 1, is about properties of a particle, i.e., it says that it is impossible to prepare a particle such that both position and momentum are arbitrarily accurately determined. The general theorem says that
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i}[\hat{A},\hat{B}] \rangle|,$$
where the expectation value can be taken wrt. an arbitrary state ##\hat{\rho}##,
$$\langle \hat{A} \rangle=\mathrm{Tr}(\hat{\rho} \hat{A}).$$
Only compatible observables can be made simultaneously well determined for all possible values they can take, because this is possible only if there's a common complete set of eigenvectors of the corresponding self-adjoint operators.
There are exceptions in the sense that it can be that there are states, for which two incompatible observables take a determined value. An example is angular momentum. If you are in a state with ##J=0##, then all components of the angular momentum have eigenvalue 0. In general, of course you can only determine one component of angular momenum, for which usually one chooses the 3-component.
Any observable can, at least in principle, be measured at any precision you are able to achieve practically, independent of the state the system is prepared in. The Heisenberg uncertainty relation does not tell you anything about the disturbance of the system due to the measurement. That's a much more complicated question, which also depends on the details of the measurement apparatus used.
DrChinese said:
2. And you wouldn't notice that immediately. It would take repeated measurements of identically prepared systems to notice that you are getting a spread of values.
That's a very important point. At least within the minimal statistical interpretation of QT the only thing we have are the probabilistic laws provided by the formalism, i.e., to, e.g., verify the uncertainty relation for position and momentum of a particle, you have to prepare the particle very often in the same state and measure the position at high accuracy to obtain the statistics of measurement outcomes at the level of significance you want. Then you get an expectation value and the standard deviation. The same you do with momentum for another ensemble of equally prepared particles. Then the QT prediction is that ##\Delta x \Delta p_x \geq \hbar/2##, and this you can verify in the above described way.
Note that we haven't measured position and momentum simultaneously. The possibility to measure incompatible observables on a single quantum system in an arbitrary state is a much more complicated question, and can be described adequately only with a generalized more modern framework of measurement theory, described by socalled positive operator-valued measures.
DrChinese said:
3. And how would you prepare a series of identical systems in the first place? I.e. with known position and momentum? That ends up being a circular problem (since the HUP is standing in your way).
That's the point! You cannot prepare (!!!) the particle in such a way that position and momentum are better determined than the Heisenberg uncertainty relation allows.
DrChinese said:
I could go on, but the bottom line answer to your idea is: you end up with the limits of the HUP in every scenario.