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Laplacian in Polar Cooridinates

  1. Apr 12, 2005 #1
    I need to take the [itex]\nabla^2[/itex] of [itex]x^2+y^2+z^2[/itex]. This is how far I got

    [tex]
    \begin{gather*}
    \nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} + \frac{1}{r^2}(\frac{1}{sin^2\theta}\frac{d^2}{d\Phi^2} + \frac{1}{sin\theta}\frac{d}{d\theta} sin\theta\frac{d}{d\theta})\\
    \nabla^2(r^2sin^2\theta cos^2\Phi + r^2sin^2\theta sin^2\Phi + r^2cos^2\theta = \frac{1}{sin\theta} \frac{d}{d\theta}(sin\theta \frac{d}{d\theta}) + \frac{1}{sin^2\theta} \frac{d^2}{d\Phi^2})
    \end{gather*}
    [/tex]


    Also, can degeneracy occur with n not in order? Like for a part. in 3D box can degeneracy occur for [tex]\Psi_{1,3,5}[/tex] [tex]\Psi_{5,3,1}[/tex] or do the n have to be next each other like [tex]\Psi_{1,2,1}[/tex] [tex]\Psi_{2,1,1}[/tex]?
     
  2. jcsd
  3. Apr 12, 2005 #2

    dextercioby

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    Is that a spherical box?And how does the potential look like...?

    Daniel.
     
  4. Apr 13, 2005 #3
    Im not sure. I just need to use the del operator on [tex]x^2+y^2+z^2[/tex].
     
  5. Apr 13, 2005 #4

    dextercioby

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    Del or laplacian....?

    Daniel.
     
  6. Apr 13, 2005 #5
    I mean del squared or laplacian.

    Oh, for the 2nd question: a 3D box with V = infinity outside box.
     
    Last edited: Apr 13, 2005
  7. Apr 14, 2005 #6

    dextercioby

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    It's simple.

    [tex] \nabla^{2}=\Delta=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} [/tex]

    Use it to differentiate what u had to ([itex] x^{2}+y^{2}+z^{2} [/itex]).

    Daniel.
     
  8. Apr 14, 2005 #7
    Oh, but I mean using polar coordinates for [itex]x^2+y^2+z^2 = r [/itex].

    Thanks for the the p chem help dextercioby.
     
  9. Apr 14, 2005 #8

    dextercioby

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    Nope,i think you mean spherical coordinates and

    [tex] x^{2}+y^{2}+z^{2}=r^{2} [/tex]

    Daniel.
     
  10. Apr 14, 2005 #9

    dextercioby

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    And one more thing:please take my advice and compute that in cartesian coordinates...It's easier.Maths should be made as easy as possible,here's an opportunity

    Daniel.
     
  11. Apr 14, 2005 #10
    That's why I was so confused with the first post, why were we straying away from cartesian when the Laplacian operator is so easily used on the described fct?
     
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