Can Linear Combinations of Non-Degenerate Eigenstates Form New Eigenstates?

[amir11]

Well, As far as I know the Schrödinger equation for a H atom is linear and real.
Suppose it has two solutions, eigenstates, psi1 and psi2 with non degenerate eigenvalues E1 and E2. It is possible to construct two other states which also are solutions to Schrödinger equation as

psi3= apsi1+bpsi2 and E3=a²E1+b²E2
psi4= apsi2-bpsi1 and E4=a²E2+b²E1

and a²+b²=1

Well you get two eigenstates with two different eigenvalues than those at the beginning.
I know there is a mistake in may approach but where?

 

[jtbell]

psi3= apsi1+bpsi2 and E3=a²E1+b

$\Psi_3$ does not have that energy. In fact it does not have a definite energy.

If you measure the energy of the particle when it is in state $\Psi_3$, you will get either $E_1$, with probability $|a^2|$; or $E_2$, with probability $|b^2|$.

 

[Truecrimson]

Your E3 and E4 are expectation values of the eigenvalues, which generally are not numerically equal to any other eigenvalue. More importantly though, psi3 and psi4 are not eigenvectors, so they don't have an eigenvalue.

Edit: I was too late.