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Manipulating the TISE for a SHM

  1. Nov 13, 2005 #1
    Hi guys, hoping someone can help with this manipulation. I need to transform this:
    [tex] \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}u(x) + \frac{1}{2}m\omega^2 x^2 u(x) = Eu(x)[/tex]
    Into its dimensionless form:
    [tex]\frac{d^2}{dy^2}u(y) + (2\epsilon - y^2)u(y) = 0[/tex]
    I have the following info:
    [tex]E = \epsilon\hbar\omega[/tex]
    [tex]x = y\sqrt{\frac{\hbar}{m\omega}}[/tex]
    Heres what ive done so far:
    [tex] \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}u(x) + \frac{1}{2}m\omega^2 y^2 \frac{\hbar}{m\omega} u(x) = \epsilon\hbar\omega u(x)[/tex]
    [tex] \frac{-\hbar}{m}\frac{d^2}{dx^2}u(x) + \omega^2 y^2 u(x) = 2\epsilon\omega u(x)[/tex]
    [tex] \frac{\hbar}{m}\frac{d^2}{dx^2}u(x) + 2\epsilon\omega u(x) - \omega^2 y^2 u(x) = 0[/tex]
    [tex] \frac{\hbar}{m}\frac{d^2}{dx^2}u(x) + (2\epsilon - y^2)\omega u(x) = 0[/tex]
    But i cant see where to go next..i know i must be close to the end though..surely!
  2. jcsd
  3. Nov 13, 2005 #2
    at first sight, there is one thing that remains to be done : the transformation of the derivatives.

    you know that u(x) = u(y)
    Hint : [tex]\frac {du(x)}{dx}= \frac{du(y)}{dx} = \frac {du(y)}{dy} \cdot \frac{dy}{dx}[/tex] . More specifically, can you calculate [tex]\frac {dy}{dx} = constant[/tex] ? What is that constant ? Then, do the same to get the second derivatives.


    ps i did not check the calculations you have done so far but let us first look at the derivatives
    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3
    hmm, the only relationships i have in my notes are:

    [tex]u(x) = Ce^{-\gamma x}[/tex]

    So [tex] \frac{d^2}{dx^2}u(x) = \gamma^2 u(x)[/tex]

    Where [tex]\gamma^2 = \frac{2m}{\hbar^2}(V_0 - E)[/tex]

    edit - just saw your hint..will inspect further!

    edit again - would i be correct to assume that [tex]u(y) = Ce^{-\gamma y \sqrt{\frac{\hbar}{m\omega}}}[/tex] ?
    Last edited: Nov 13, 2005
  5. Nov 13, 2005 #4
    This is all true but you do not need this right now. The problem is quite simple, trust me. Once you get the second derivatives right, the problem is very elementary. just try it, i will keep an eye on it. :approve:

  6. Nov 13, 2005 #5
    meanwhile, your calculations are ok. Once you get my hint (ie the constant), you are done !!!

  7. Nov 13, 2005 #6
    hmm..no im totally lost...sorry :cry:

    I dont see how you can differentiate the function when you say i dont need to know what it is. I can make a wild guess that dy/dx = 2m/hbar^2..but that would just be guessing and not actually understanding.

    Could you give me a further hint? To be honest functions have always confused me slightly.
  8. Nov 13, 2005 #7
    :tongue: your guess is wrong.

    Well it is very easy. if you know that [tex] x = y \sqrt {\frac{\hbar}{m\omega}}[/tex] then [tex] dx = dy \sqrt{\frac{\hbar}{m \omega}}[/tex] and you know the constant.

    But, we need to have second derivatives, this is also easy :
    [tex]\frac {d^2u(x)}{dx^2}= \frac{d^2u(y)}{dx^2} = \frac {d^2u(y)}{dy^2} \cdot ( \frac{dy}{dx} )^2 [/tex]

    So you need to replace the second derivative of u to x by [tex]\frac {d^2u(y)}{dy^2} \frac {m \omega}{\hbar}[/tex]

    Just substitute this into your last equation and all is done
    Last edited: Nov 13, 2005
  9. Nov 13, 2005 #8
    :bugeye: :surprised

    I cannot believe i didnt see that. Whoops! Simple things like that...pretty easy to overlook i guess. Thanks for the help..gonna try it out now! :)
  10. Nov 13, 2005 #9
    no, problem, this happens to all of us

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