possible measurement because the operator \u02c6 A \u02c6B is not Hermitian and so cannot

# Possible measurement because the operator ˆ a ˆb is

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possible measurement, because the operator [ ˆ A, ˆ B ] is not Hermitian and so cannot represent an observable. 1 2015-10-08 P424 HW4.tex c 2015 Donald Candela 1
(a) Ananti-HermitianoperatorˆQsatisfiesˆQ=-ˆQ.Using thehermiticity ofˆAandˆB, show that [ˆA,ˆB] is an anti-Hermitianoperator.(b) IfˆQis anti-Hermitian, show thathψ|ˆQ|ψiisimaginary(unlikethe expectation value for a Hermitian operator, which we showedis real).3. (10 pts) Townsend problem 3.15 - For spin 1, find the eigenstates ofˆSxin the usual (Sz) basis. You should normalize the eigenstates after youfind them. Hint: You need to find the eigenvectors of the 3×3 matrix forˆSx, which is given in the text. You already know the eigenvalues, sincethey are the possible results from a measurement of any component ofthe spin for a spin-1 system. Partial answer: The eigenstate of

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• Fall '15
• mechanics, Work, Hilbert space, expectation value, Hermitian