Prove that if the operator A^ is Hermitian, its eigenvalues are real.
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Expert's answer
2010-06-16T08:21:46-0400
Let Ψ be an arbitrary eigenfunction of the operator А^ corresponding to its eigenvalue A. Then, due to the self-adjointness of the operator: ∫Ψ*A^Ψdx = ∫ΨA^*Ψ*dx and A∫Ψ*Ψdx = A*∫ΨΨ*dx, whence A=A*, which is possible only when A is real.
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