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Normalised eigenspinors and eigenvalues of the spin operator
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[QUOTE="says, post: 5972057, member: 517464"] ## X[SUP]†[/SUP] = (X[SUP]T[/SUP])* where T: transpose of matrix *: complex conjugate Thanks for that. I calculated it and can see that it equals 0. I also found that eigenfunctions of hermitian operators are orthogonal. So I figured if S[SUB]y[/SUB] is hermitian then the eigenfunctions of S[SUB]y[/SUB] will be orthogonal. I essentially did the same thing. If S[SUB]y[/SUB] = S[SUB]y[/SUB][SUP]†[/SUP] then S[SUB]y[/SUB] is hermitian and it's eigenfunctions / eigenspinors are orthogonal. [/QUOTE]
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Normalised eigenspinors and eigenvalues of the spin operator
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