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Working my way through Griffiths' "Introduction to Quantum Mechanics", I found the following excerpt:

enter image description here

The result is the same as what I'd gotten using the operators' method, and yet in this context, it doesn't make sense since the text clearly states that there must exist a singular n ("some" highest j for which $a_{j+2} = 0$), and since K (and thus E) is related to n by the equation $K = 2n+1$, we should be getting one allowed level of energy E.

I know there obviously must be a flaw in my logic, and yet I am unable to put my hand on it.

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    $\begingroup$ Any given solution must have a highest $j$, and so a fixed $n$. That does not imply that all solutions must have the same $n$. $\endgroup$
    – By Symmetry
    Commented Oct 4, 2021 at 19:18
  • $\begingroup$ @BySymmetry If that is the case, then, depending on the parity of n, why should either of $h_{even}(\varepsilon)$ or $h_{odd}(\varepsilon)$ be eliminated? $\endgroup$
    – Omar El Atyqy
    Commented Oct 5, 2021 at 0:51

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For the solution $\psi_{0}(x)$, the power series must terminate at $j=0$, to satisfy normalization. The corresponding energy will be $$E_{0}=\frac{\hbar\omega}{2}$$ For the solution $\psi_{1}(x)$, the power series must terminate at $j=1$, to satisfy normalization. The corresponding energy will be $$E_{1}=\frac{3\hbar\omega}{2}$$

and so on...

Different solutions will terminate at different values of $j$ and have different energies.

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