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Say if we have two energy eingenfunctions and correpsonding eigenvalues, as specified by:

[tex] H \psi_1 =E_1 \psi_1 [/tex]

[tex] H \psi_2 =E_2 \psi_2 [/tex]

I have written in my notes that since the Hamiltonian operator H is linear, then [tex] \phi=C_1 \psi_1 + C_2 \psi_2 [/tex] must also be a solution. But

[tex] H \phi =H(C_1 \psi_1 + C_2 \psi_2) =C_1 E_1 \psi_1 +C_2 E_2 \psi_2 =E_1 C_1 \psi_1 + E_2 C_2\psi_2 != E \phi [/tex], where E is some constant. (!= is supposed to represent NOT equal to)

so how can this combination also be a solution (assuming the eigenfunctions are non degenerate)?

thanks

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# Simple QM question

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