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Classically when we solve Newton's equation for $V=\frac{1}{2}m\omega^2x^2$ we get two linearly independent solutions (for $\omega\not=0$): $Ae^{\omega t}$ & $Be^{-\omega t}$, their linear combinations would also be a solution.

In Griffith's QM book (pg 51, 3rd ed.) he says that the time independent Schrodinger equation has two linearly independent solutions for any value of $E$. Does this mean than every $E$ is doubly degenerate? What even are these two solutions? I only know how to write one $\psi(x)$ for normalizable states. This is not as obvious to me as the classical case.

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  • $\begingroup$ Can you take a quote of the main part? $\endgroup$
    – narip
    Commented Oct 1, 2022 at 11:15

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At most one of the two solutions can be normalizable, and therefore an element of the Hibert space. And only a discrete set $E_n=\omega(n+1/2)$ of energies has a normalizable solution.

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  • $\begingroup$ Could you please write down the expression for both the solutions. Are they simply $Ae^{\xi^2/2}$ and $Be^{-\xi^2/2}$? After this the author proceeds to discard the former as it diverges and then he uses power series to derive those normalizeable $\psi(x)$s (the hermite ones). $\endgroup$
    – user292464
    Commented Oct 1, 2022 at 17:17

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