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Wavefunction and shroedinger equation

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data

    upload_2014-9-27_16-25-56.png

    2. Relevant equations

    upload_2014-9-27_16-26-39.png

    3. The attempt at a solution

    First, I got the wavefunction to look like the one in the question. I think the wavefunction should be n=1 not n=0. So Y(theta,psi) = A constant, that is where the C comes from. But how can I plug this into the shrodinger equation? How can I answer this question?
     
  2. jcsd
  3. Sep 27, 2014 #2

    ShayanJ

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    Gold Member

    In the Schrodinger equation, there is a differentiation operator ([itex]-\frac{\hbar^2}{2m} \nabla^2 [/itex] )and a multiplication-by-a-function([itex]-\frac{kZe^2}{r}[/itex]) operator. You just should apply those operators to [itex] \phi_{0,0,0}(\vec r) [/itex] and add the results and check whether you get a constant times [itex] \phi_{0,0,0}(\vec r) [/itex].
    For [itex]-\frac{\hbar^2}{2m} \nabla^2 [/itex], you should first take the gradient of [itex] \phi_{0,0,0}(\vec r) [/itex] which gives you a vector field which you should get the divergence of. Then multiply by [itex] -\frac{\hbar^2}{2m} [/itex].
     
  4. Sep 28, 2014 #3

    Orodruin

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    Alternatively to the two step approach in first taking the gradient and then the divergence of the gradient, you could apply the Laplace operator in spherical coordinates directly:
    $$
    \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} = \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} ,
    $$
    where I have removed the angular part since your wave function does not depend on the angles.
     
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