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Schrodinger equation reduction using substitution

  1. Apr 16, 2009 #1
    SOLVED: Schrodinger equation reduction using substitution

    1. The problem statement, all variables and given/known data

    [tex] \frac{d^2 \psi}{dx^2} - Ax\psi + B\psi = 0[/tex]

    make a substitution using
    [tex]w= A^{1/3} (x - \frac{B}{A})[/tex]

    to get
    [tex] \frac{d^2 \psi}{dw^2} - w\psi = 0[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I use

    [tex] \frac{d\psi}{dx} = \frac{d\psi}{dw} \frac{dw}{dx}[/tex]

    [tex]\frac{d^2\psi}{dx^2} = \frac{d}{dx}[ \frac{d\psi}{dw} \frac{dw}{dx}][/tex]


    [tex]\frac{d^2\psi}{dx^2} = \frac{d\psi}{dwdx} \frac{dw}{dx} + \frac{d^2w}{dx^2} \frac{d\psi}{dw}[/tex]

    but the [tex]\frac{d^2w}{dx^2}[/tex] equals zero, since x is linear in w.

    which implies

    [tex]\frac{d^2\psi}{dx^2} = \frac{d\psi}{dwdx} \frac{dw}{dx}[/tex]

    and [tex]\frac{dw}{dx} = A^{1/3}[/tex]

    but I'm not sure how to evaluate the

    [tex]\frac{d\psi}{dwdx}[/tex] term (I'm not sure If it should even be there.. did I use the chain rule correctly?)

    Then I'll sub in [tex]\frac{d\psi}{dwdx} A^{1/3}[/tex] for [tex] \frac{d^2 \psi}{dx^2}[/tex] and hopefully it all works out.

    Any help would be greatly appreciated.

    Edit: SOLVED
    Last edited: Apr 16, 2009
  2. jcsd
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