hermitian operator eigenvalues
,2020年11月30日—Yes.NotonlytheeigenvectorsofaHermitianoperatorconstituteabasis,butitisacompletebasis,i.e.,andfunctioninthespacewhere ...,WewishtoprovethateigenfunctionsofHermitianoperatorsareorthogonal.Infactwewillfirstdothisexceptinthecaseofequa...
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Hermitian.(Prove:T,thekineticenergyoperator,isHermitian).ThenH=T+VisHermitian.PROVE:TheeigenvaluesofaHermitianoperatorarereal.(This.
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