Uncertainty principle and angular momentum

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Upisoft
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[itex]\Delta{x}\Delta{p}[/itex], [itex]\Delta{E}\Delta{t}[/itex] and particle spin all have units of angular momentum and have ability to be quit uncertain... Any idea if they have something in common (except the units of measurement)?
 
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Upisoft said:
[itex]\Delta{x}\Delta{p}[/itex], [itex]\Delta{E}\Delta{t}[/itex] and particle spin all have units of angular momentum and have ability to be quit uncertain... Any idea if they have something in common (except the units of measurement)?

Before I try to answer the question, one correction is in order. Yes, the uncertainty principle states that [itex]\Delta{x}\Delta{p} <= h/2[/itex], but the same kind of statement does not come from QM (at least not directly) for [itex]\Delta{E}\Delta{t}[/itex]. This is because X and P are are operators in QM, but E and t are not. E and t are real valued parameters.

OK, back to the uncertainty: in QM, any two Hermitian operators which do not commute with each other cannot be simultaneously diagonalized. Translated into plain English, this says that the eigenvalues of the two operators cannot be measured with arbitrary precision without affecting the other's eigenvalues. All uncertainty relations in QM have this feature in common: there is an underlying non-zero commutator for the two operators in question. If on the other hand, you have two commuting operators, then you can measure both of them with arbitrary precision without affecting the other, and there is no corresponding uncertainty relation for them.