You know, I think this is the best explanation I've seen yet. I've done plenty of calculations for my classes on which operators commute and which don't. And

You know, I think this is the best explanation I've seen yet. I've done plenty of calculations for my classes on which operators commute and which don't. And if [A, B] ≠ 0, then the two are incompatible observables. And I already knew the mathematics behind the eigenstates/eigenfunctions. But I think the key insight you provided to me is what most people mean when they say "well-defined" -- that the definition for "certainty" implies multiple measurements. In other words, I was thinking: okay, you can measure a particle's position and momentum. You'll get real numbers for both to as much precision as your equipment allows. And I assumed that this was the definition of position and momentum -- the instantaneous values you get for the two measurements are defined to be the particle's position and momentum at time t. So I didn't understand why everyone was saying you can't find both simultaneously -- from my definition, of course you can find them both! But from the definition you've given me, I am guessing most other people define position and momentum as consistent quantities, meaning they have to have the same values upon each new measurement of the system in order to be called "position" or "momentum". Is this what you're saying? If so, that makes a lot of sense. But I would think textbooks would clarify that more, because there are a lot of people who, like me, get confused by this type of wording. You can measure a car's instantaneous speed and that is THE speed at time t. Not "let's measure an identically prepared car multiple times and call the average of its dx/dt measurements 'speed". With regard to your last point about the standard deviations product always being greater than h_bar / 2, I realized my fallacy after I posted it, but didn't think anyone would catch it. Good eye for subtlety!