Griffiths Quantum Question. Unit-less Schrodinger Equation.

Main Question or Discussion Point

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 [itex]\xi \equiv \sqrt{\frac{m\omega}{\hbar}}x[/itex]

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

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

where [itex]K\equiv\frac{2E}{\hbar\omega}[/itex]

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

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

For example, using
[itex]
\frac{d\psi}{dx} = \frac{d\psi}{d\xi} \frac{d\xi}{dx}
[/itex]
then
[itex]
\frac{d^2\psi}{dx^2} = \frac {d}{dx} \Big( \frac{d\psi}{dx} \Big) \\
\frac{d^2\psi}{dx^2} = \frac {d}{dx} \Big( \frac{d\psi}{d\xi} \frac{d\xi}{dx} \Big)\\
\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}\\
\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}\\
\frac{d^2\psi}{dx^2} = \Big(\frac{d\xi}{dx}\Big)^2 \frac{d^2\psi}{d\xi^2}\\
\frac{d^2\psi}{dx^2} = \frac{m\omega}{\hbar} \frac{d^2\psi}{d\xi^2}\\
[/itex]

Does this look legit?
 

Answers and Replies

BvU
Science Advisor
Homework Helper
12,555
2,832
Yes. ##m\omega\over \hbar## is constant.
 

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