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Unitary vector commuting with Hamiltonian and effect on system
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[QUOTE="strangerep, post: 6854379, member: 70760"] The middle bit looks wrong. You are given that ##H|\psi_n \rangle = E_n|\psi_n\rangle##, but this does not imply ##U|\psi_n \rangle = E_n|\psi_n\rangle##. Not the way you tried to do it. I have already given you the correct method in my post #2. And its eigenvalue is ##E_n##, not 1. This was also shown in my post #2. After you have proven that ##|\phi_n \rangle## is an eigenstate of ##H## with eigenvalue ##E_n##, then... Suppose there was only 1 eigenstate of ##H## (i.e., suppose ##H## were not degenerate). Then you could invoke the theorem that the eigenstates of an Hermitian operator span the Hilbert space, and are mutually orthogonal;-- that would be enough to prove ##|\phi_n\rangle = |\psi_n\rangle##. Even in the degenerate case it is still true that the eigenstates with different eigenvalues are orthogonal. Therefore, since [B]any[/B] state can be expressed as a sum over all ##H##'s eigenstates, we can express ##|\phi_n\rangle## as such a sum. But ##|\phi_n\rangle## must be orthogonal to all ##H##'s eigenstates that [B]don't [/B]have eigenvalue ##E_n##, hence those eigenstates cannot appear in the sum for ##|\phi_n\rangle##. That leaves only the ##|\psi_n^i\rangle## states in the sum. [/QUOTE]
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Unitary vector commuting with Hamiltonian and effect on system
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