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Homework Help: Normalising schrodinger pdf

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\Psi(x,t) = Ae^{-\lambda |x|}e^{-i \omega t}[/tex]

    2. Relevant equations

    (i.e. it must satisfy [tex]\int_{-\infty}^{\infty} |\Psi (x,t)|^2 dx=1[/tex])

    3. The attempt at a solution

    This is my first attempt at normalisation and I don't know how valid my reasoning is, can someone please check to see if I'm doing it right:

    [tex]\int_{-\infty}^{\infty}|\Psi (x,t)|^2 dx = \frac{A^2}{\lambda} e^{-2 i \omega t}[/tex]

    [tex]\Rightarrow \frac{\lambda}{A^2}e^{2i \omega t} \int_{-\infty}^{\infty}|\Psi (x,t)|^2 dx =1[/tex]

    [tex]\int_{-\infty}^\infty \left(\frac{\sqrt{\lambda}}{A}e^{i\omega t} |\Psi|\right)^2dx =1[/tex]

    [tex]\therefore \Psi (x,t)_{normalised} = \frac{\sqrt{\lambda}}{A}e^{i\omega t} \times \left(Ae^{-\lambda |x|}e^{-i \omega t}\right) = \sqrt{\lambda}e^{-\lambda |x|}[/tex]

  2. jcsd
  3. Nov 22, 2009 #2


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    Staff: Mentor

    First, recall that [itex]|\Psi (x,t)|^2[/itex] means [itex]\Psi^*(x,t) \Psi(x,t)[/itex]. That is, you multiply [itex]\Psi(x,t)[/itex] by its complex conjugate [itex]\Psi^*(x,t)[/itex].

    What is the complex conjugate of the [itex]\Psi(x,t)[/itex] that you were given?
  4. Nov 22, 2009 #3
    Oh, right! So I can't just square it?

    The conjugate would be [tex]Ae^{-\lambda |x|}e^{i \omega t}[/tex]

    So I just times those together to find [tex]|\Psi|^2[/tex]?

    Are my other steps sound though (assuming that I fix the start up)?

    Thanks again
  5. Nov 22, 2009 #4


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    Homework Helper

    Your other steps will be completely different. For one thing, the e^ part becomes 1 and you won't have to worry about it. You should get a value for A, which makes sense because normalization means finding the amplitude A that goes in front of the wavefunction.
  6. Nov 23, 2009 #5
    Thanks for your help! Is this correct now?

    [tex]\begin{array}{cclc}\int_{-\infty}^\infty |\Psi|^2dx &=& 2\int_{0}^\infty (Ae^{-\lambda x}e^{-i \omega t} Ae^{-\lambda x}e^{i \omega t})dx & \ \ \ \ \ \ \ \ \ \int_{-\infty}^\infty f(|x|)dx = 2\int_0^\infty f(x)dx \\\\ &=& 2\int_0^\infty A^2e^{-2\lambda x}dx & \\\\ &=& \dfrac{A^2}{\lambda}& \end{array}[/tex]

    so the answer is [tex]A = \sqrt{\lambda}[/tex]?
    Last edited: Nov 23, 2009
  7. Nov 23, 2009 #6


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    Homework Helper

    Almost, but e^-λx * e^-λx = e^-2λx.
  8. Nov 23, 2009 #7
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