Why are Superposition states not solution of TISE?

Question

Form a textbook i am learning for its says:
All valid wave functions (energy eigenfunction with definite energy and superposition states with simultaneously multiple energies) for a given quantum system satisfy the Time dependent Schrodinger equation.
However only the spatial parts $u(x)$ of energy eigenfunction [definite energy states with energy $E$ in $T(t)$] solve the Time-independent Schrodinger equitation. Superposition states are not solution of TISE
Can anyone explain why this is case and show me way to prove this mathematically (if possible)?

JulianDeV · Answer

The TISE is a generic eigenvalue problem:
begin{equation}
hat{H}psi = E psi
end{equation}
where $hat{H}$ is the Hamiltonian and $E$ is the eigenvalue corresponding to the eigenstate $psi$.
Now take $psi_1, psi_2$ to be 2 eigenstates with eigenvalues $E_1, E_2$, look at the superposition $psi_1 + psi_2 = Phi$:
begin{align}
hat{H}(psi_1 + psi_2) &= E_1 psi_1 + E_2psi_2
Rightarrow hat{H} Phi &= E_1 psi_1 + E_2 psi_2 neq E_Phi Phi
end{align}
Clearly $Phi$ is not an eigenstate of $hat{H}$ unless $E_1 = E_2$.

Why are Superposition states not solution of TISE?

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