Griffiths Quantum Question. Unit-less Schrodinger Equation.

[DiogenesTorch]

In Grifftiths Intro to Quantum 2nd edition, page 51, he is re-expressing the Schrodinger equation for a harmonic oscillator in terms of a unit-less quantity \xi \equiv \sqrt{\frac{m\omega}{\hbar}}x

So Griffiths takes the Schrodinger equation in equation [2.70]
-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2x^2\psi =E \psi

and then re-expresses it in equation [2.72] as
\frac{d^2\psi}{d\xi^2} =(\xi^2-K)\psi

where K\equiv\frac{2E}{\hbar\omega}

My question is I am trying to follow what Griffith is doing here and was wondering how to re-express the 2nd derivative of \psi going from \frac{d^2\psi}{dx^2} to \frac{d^2\psi}{d\xi^2}?

Is it just by using a bunch of chain rule stuff?

For example, using
<br /> \frac{d\psi}{dx} = \frac{d\psi}{d\xi} \frac{d\xi}{dx}<br />
then
<br /> \frac{d^2\psi}{dx^2} = \frac {d}{dx} \Big( \frac{d\psi}{dx} \Big) \\<br /> \frac{d^2\psi}{dx^2} = \frac {d}{dx} \Big( \frac{d\psi}{d\xi} \frac{d\xi}{dx} \Big)\\<br /> \frac{d^2\psi}{dx^2} = \frac {d}{d\xi} \Big( \frac{d\psi}{d\xi} \frac{d\xi}{dx} \Big)\frac{d\xi}{dx}\\<br /> \frac{d^2\psi}{dx^2} = \Big( \frac{d\psi}{d\xi} \frac{d}{d\xi}\Big(\frac{d\xi}{dx}\Big) + \frac{d\xi}{dx} \frac{d}{d\xi} \Big(\frac{d\psi}{d\xi}\Big) \Big)\frac{d\xi}{dx}\\<br /> \frac{d^2\psi}{dx^2} = \Big(\frac{d\xi}{dx}\Big)^2 \frac{d^2\psi}{d\xi^2}\\<br /> \frac{d^2\psi}{dx^2} = \frac{m\omega}{\hbar} \frac{d^2\psi}{d\xi^2}\\<br />

Does this look legit?

 

[BvU]

Yes. $m\omega\over \hbar$ is constant.