But if that's the case, then how should I interpret the following 2cases:
$H(\lambda_1 \psi(\vec r)_1 + \lambda_2 \psi(\vec r)_2)= \lambda_1 E_1 \psi(\vec r)_1 + \lambda_2 E_2 \psi(\vec r)_2$

$\psi(\vec r,t)=\phi(\vec r)\chi(t)$

Plugging this in the SE, we get the following two expressions:

$i\hbar \frac 1 {\chi(t)}\frac \partial {\partial t}\chi(t)=B$(a constant.)

Solution: $\chi(t)=C e^{-i\omega t}=e^{-i\omega t}$ (We take C=1 for simplicity)

The 2nd expression is: $H\phi(\vec r)=E\phi(\vec r)$.

And the wave equation would finally be : $\Psi(\vec r,t)=\phi (\vec r)e^{-i\omega t}$.

$\Psi(\vec r,t)=A e^{-i\omega t}\phi (\vec r,t=0)$

$\Psi(\vec r,t)=A e^{-i\frac H \hbar t}\phi (\vec r,t=0)$

$\Psi(\vec r,t)=A e^{-i\frac H \hbar t}\sum_{n=1}c_n\phi_n(\vec r)$

$\Psi(\vec r,t)=\sum_{n=1}A e^{-i\frac {E_n}{\hbar} t}c_n\phi_n(\vec r)$.

Then the solution of the particle in a potential well, which is in my notes must be wrong:
$\Psi(\vec r,t)=A e^{-i\omega t}\phi (\vec r,t=0)$
$\Psi(\vec r,t)=A e^{-i\frac H \hbar t}\phi (\vec r,t=0)$
$\Psi(\vec r,t)=A e^{-i\frac H \hbar t}\sum_{n=1}c_n\phi_n(\vec r)$
$\Psi(\vec r,t)=\sum_{n=1}A e^{-i\frac {E_n}{\hbar} t}c_n\phi_n(\vec r)$.
Because an expression that starts with the ansatz, ends being a superposition.

If we consider a discrete set of square integrable functions {$$u_i(\vec r)}.
If $(u_i,u_j)=\int u_i(\vec r)^*u_j(\vec r)d^3r=\delta_{ij}$ this would be the condition for orthogonality. One can describe this as the inner product of two different vectors, in the Hilbert space where every two functions of the set {$$u_i(\vec r)} are orthogonal. The condition for completeness is: $\sum_i u_i(\vec r)^* u_i(\vec r')=\delta(\vec r- \vec r')$. How do you describe what is going on here?