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Griffiths Quantum Question. Unit-less Schrodinger Equation.

  1. Oct 11, 2015 #1
    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
    \frac{d\psi}{dx} = \frac{d\psi}{d\xi} \frac{d\xi}{dx}
    \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}\\

    Does this look legit?
  2. jcsd
  3. Oct 11, 2015 #2


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