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The general HUP has to do with the eigenstates of noncommuting operators. I feel like you've probably seen this before, but one more hand waving try! :) In Q

The general HUP has to do with the eigenstates of noncommuting operators. I feel like you've probably seen this before, but one more hand waving try! :) In QM the state of a system will be a solution to Schroedingers equation for that system. The system is characterized by a Hamiltonian H. If an operator commutes with H, then we can find a basis for the solution space such that every basis state is an eigenstate of that operator. (i.e. it has a definite eigenvalue for that operator; applying the operator to an eigenstate gives back the state times a constant.) If two operators commute, we can even find a basis such that that each basis state is an eigenstates of both operators. But if they don't commute, this is impossible.[1] That means that if a basis state has a definite eigenvalue for one operator, it cannot for the other. What does it mean for a particle to have a well defined momentum? That means it is an eigenstate of the momentum operator. Likewise for position, and the position and momentum operators don't commute. This shows that if a particle is in a state of well-defined momentum, it must be a mixture of states in the basis for position. With more work you can translate this into a well-defined lower bound on the product of the uncertainties of the two operators. >HUP guarantees that the resulting real number will be ≥ ħ/2 (reduced Planck's constant over 2). Well, not quite. You're making an observation of the standard deviation, and it might not accurately measure the actual standard deviation. So we know that if enough measurements are taken, the measured value will converge to a number greater than /hbar / 2. That's different than guaranteeing that any particular such measurement will obey this. 1. If a state |n> is an eigenstate of two operators A and B, such that A|n> = a|n> and B|n> = b|n> (a and b are eigenvalues), then necessarily AB|n> = BA|n> = ab|n>. This implies that [A, B]|n> = 0 which generally contradicts the idea that A and B don't commute; certainly when the commutator is just a constant.

Xcelerate 14y ago
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!

Xcelerate 14y ago

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!