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It is often said that non-commuting observables cannot be simultaneously and precisely measured. The rough idea is that the procedures to measure each observable are different. Is the statement strictly correct? If so, what is the mathematical argument?
Is it possible to use Ozawa's generalized uncertainty relation (Eq 15 of http://arxiv.org/abs/quant-ph/0207121)?
##ε(A)η(B) + ε(A)σ(B) + σ(A)η(B) ≥ \frac{\mid \langle \psi \mid[A,B]\mid \psi \rangle \mid}{2}##
The relation contains both measurement ##ε(A)η(B)## and intrinsic state uncertainties ##σ(A),σ(B)##, and it is impossible to set both measurement uncertainties to zero if ##A## and ##B## do not commute.
If it is correct that non-commuting observables cannot be simultaneously and precisely measured, is there any easier way to prove it?
Is it possible to use Ozawa's generalized uncertainty relation (Eq 15 of http://arxiv.org/abs/quant-ph/0207121)?
##ε(A)η(B) + ε(A)σ(B) + σ(A)η(B) ≥ \frac{\mid \langle \psi \mid[A,B]\mid \psi \rangle \mid}{2}##
The relation contains both measurement ##ε(A)η(B)## and intrinsic state uncertainties ##σ(A),σ(B)##, and it is impossible to set both measurement uncertainties to zero if ##A## and ##B## do not commute.
If it is correct that non-commuting observables cannot be simultaneously and precisely measured, is there any easier way to prove it?