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Schrödinger homework help

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that this solution:

    [tex]\psi(x,y,z)=Ae^{i\mathbf{k}\cdot\mathbf{r}}[/tex]

    is a solution for this equation:

    [tex]\frac{\hbar^{2}}{2m}\Delta \psi(x,y,z)=E\,\psi(x,y,z),[/tex]

    How is K related to E?

    How does the corresponding timedependent function look like?



    3. The attempt at a solution

    My second derivative is:

    [tex] \psi''(x,y,z)=i^{2}k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r} = -k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r}=-k_x^{2} \cdot \psi[/tex]

    [tex]\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) \cdot \psi = E \cdot \psi[/tex]

    [tex]\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) = E[/tex]


    I suppose it is enough to say that

    [tex]\frac{\hbar^{2}}{2m} (k_x^{2} + k_y^{2} + k_z^{2})[/tex] is energy? So it automatically satisfies the S.E.? What is the unit of [tex]\psi[/tex]?

    But the relation between k and E should then be:

    [tex]\frac{\hbar^{2}}{2m} \cdot (k_x^{2} + k_y^{2} + k_z^{2}) = \frac{\hbar^{2}}{2m} \cdot K = E[/tex]

    [tex]\frac{E}{K}= \frac{\hbar^{2}}{2m}[/tex]


    But the last task is a bit more tricky...I will just separate the time-depentent S.E.

    [tex]\frac{\hbar^{2}}{2m}\Delta \psi = i\hbar \frac{d\psi}{dt}[/tex]

    [tex]\frac{\hbar^{2}}{2m}\Delta = i\hbar \frac{d\psi}{dt} \cdot \frac{1}{\psi}[/tex]

    Can someone help me from here?
     
  2. jcsd
  3. Jul 1, 2011 #2

    Redbelly98

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    Science Advisor
    Homework Helper

    Re: Schrödinger

    Welcome to Physics Forums.

    Small correction: it really should be kxx, kyy, etc. here. And it looks like you are using ψ'' to mean ∂²ψ/∂x². But you have the right idea.

    Close, but not quite. You are saying that

    [itex]k_x^2 + k_y^2 + k_z^2 = k^1 [/itex] ???​

    I think you need to modify the k1 term, slightly, to have a true statement here.

    Let's back up 1 step to

    [tex]\frac{\hbar^{2}}{2m}\Delta \psi = i\hbar \frac{d\psi}{dt}[/tex]

    Since ψ is a function of (x,y,z) only, we really need another symbol for the time-dependent part. It's common to assume the full wavefunction is the product of a spatial function (that's ψ(x,y,z) ) and a time-dependent function (call it f(t)). So you want to put ψ(x,y,z)·f(t) into the time-dependent Schrodinger Equation, and work out what f(t) is.

    Hope that helps. By the way, \partial will give you the partial derivative symbol in LaTeX.
     
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