# Question on Harmonic Oscillator Series Derivation

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• TRB8985

#### TRB8985

TL;DR Summary
What is the mathematical justification for being able to use the solutions of single-order DE's for a second-order DE?
Good afternoon all,

On page 51 of David Griffith's 'Introduction to Quantum Mechanics', 2nd ed., there's a discussion involving the alternate method to getting at the energy levels of the harmonic oscillator. I'm filling in all the steps between the equations on my own, and I have a question between equation [2.74] and [2.75]. In particular, starting with the 2nd order DE:

$$\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi$$​
I found in a similar post elsewhere that we can't attack this expression using the typical characteristic equation method, since we end up with:
$$\lambda^2 = \xi^2$$​
and with ##\xi## playing the role of our variable, that's not allowed. To be honest, I don't quite understand why that's an issue, but I pressed on and found a suggestion to instead factor the first equation in the following manner:
$$\frac{d^2\psi}{d\xi^2} - \xi^2\psi = 0$$
$$(\frac{d^2}{d\xi^2} - \xi^2)\psi = 0$$
$$(\frac{d^2}{d\xi^2}+\xi)(\frac{d^2}{d\xi^2} - \xi)\psi=0$$​
Separating these two factors is easy using separation of variables, and I end up with what I see in the form of Griffiths' answer in 2.75. My two solutions would be:
$$\psi(\xi)_1 = \xi^2/2 + constant$$
$$\psi(\xi)_2 = -\xi^2/2 + constant$$​
What's missing in my understanding is the justification to take these solutions of these single-order differential equations and string them together in exponentials to become the following:
$$\psi(\xi) \approx Ae^{-\xi^2/2} +Be^{\xi^2/2}$$​
I am aware that, generally speaking, when solving 2nd order differential equations the substitution of ##e^{\lambda x}## will be used and the characteristic equation is the main plan of attack. In those circumstances, answers generally take the form of the above equation. But in this situation, we didn't solve the characteristic equation. We just factored the expression, solved it individually, then inserted it in the exponent of an exponential. What allows us to do that?

I assume the formula you propose is
$$(\frac{d}{d\xi}+\xi)(\frac{d}{d\xi}-\xi)\psi=0$$
or
$$(\frac{d}{d\xi}-\xi)(\frac{d}{d\xi}+\xi)\psi=0$$
However they become
$$(\frac{d^2}{d^2\xi}-\xi^2 \pm 1)\psi=0$$
due to commutation relation of ##\frac{d}{d\xi}## and ##\xi##. It differs from the original formula.

• vanhees71