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A professor made an statement to prove the variational theorem:

Because the Hamiltonian (H operator of quantum physics) is diagonal in its own eigenfunction, the terms in $\left \langle \Phi _{m} \left |\widehat{ H} \right |\Phi _{n}\right \rangle$ reduce to $E_{n}\delta _{mn}$ given $\widehat{H}|\Phi _{n}>=E_{n}|\Phi _{n}>$ . This is not obvious to me, does anyone have a proof that the matrix of the hamiltonian is diagonal? Please assume that I do not even know about properties of a Hermitian matrix (beginner). Thank you for your help.

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It's really a trivial statement. The reasoning is as follows: there exists some basis in which the Hamiltonian is diagonal, i.e. $H |\Phi_n \rangle = E_n |\Phi_n \rangle$ for some energies $E_n.$ We can also suppose that this basis is orthonormal. Once that is done, it's basic: $\langle \Phi_n | H | \Phi_m \rangle = \langle \Phi_n | (E_m |\Phi_m \rangle) = E_m \delta_{mn},$ since $\langle \Phi_i | \Phi_j \rangle = \delta_{ij}$ by assumption. – Vibert Apr 27 '13 at 22:56
@Vibert: That comment could be an answer. – Qmechanic Apr 29 '13 at 18:20

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