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Infinite Cubical Well in Cartesian Coordinates

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Use separation of variables in cartesian coordinates to solve the infinite cubical well (or "particle in a box"):

    [tex]V(x,y,z) = \{^{0, if x, y, z are all between 0 and a;}_{\infty , otherwise.}[/tex]


    2. Relevant equations
    Well, I've been trying to use
    [tex]\frac{1}{2}mv^{2} + V = \frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = E\Psi[/tex]


    3. The attempt at a solution
    [tex]\frac{1}{2}mv^{2} + V = \frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = E\Psi[/tex]

    [tex]\frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{x} + \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial y^{2}} + V_{y} + \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial z^{2}} + V_{z}[/tex]

    [tex]\hat{H}_{x} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{x}[/tex]

    [tex]\hat{H}_{y} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial y^{2}} + V_{y}[/tex]

    [tex]\hat{H}_{z} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial z^{2}} + V_{z}[/tex]

    [tex]\Psi = X(x)Y(y)Z(z)[/tex]

    [tex](\hat{H}_{x} + \hat{H}_{y} + \hat{H}_{z})(X(x)Y(y)Z(z)) = E*(X(x)Y(y)Z(z))[/tex]

    So getting it here doesn't really tell me what the energies are. Can I cancel out [tex]\Psi[/tex]?
     
  2. jcsd
  3. Nov 6, 2007 #2

    clem

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    Look in a math book on "separation of variable".
    You can divide each side by XYZ.
    Then use the fact that YZ cancels from H_x(XYZ)/YZ.
     
  4. Nov 6, 2007 #3
    So, if I divide by [tex](X(x)Y(y)Z(z))[/tex] from both sides, then I end up with:

    [tex]\hat{H} = E[/tex]???
     
  5. Nov 6, 2007 #4
    Would it be logical to denote energy in an dimension, like in the x-direction and y-direction and z-direction? (ie. [tex]E = (E_{x} + E_{y} + E_{z})[/tex]
     
  6. Nov 6, 2007 #5

    clem

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    Yes, you have H_x X=E_x X.
    This has a simple (sin) solution since X must be zero at x=0 and a.
     
  7. Nov 6, 2007 #6
    Well, I was a bit rushed to get my homework in on time; but for my energy equation I got:

    [tex]E_{n} = \frac{\hbar^{2} \pi^{2}}{2m}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})[/tex]

    Does that look right? I'd still like to see if what I did was correct or not.
     
    Last edited: Nov 6, 2007
  8. Nov 7, 2007 #7

    Meir Achuz

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    You left out /L^2 of the side.
     
  9. Nov 7, 2007 #8
    [tex]E_{n} = \frac{\hbar^{2} \pi^{2}}{2mL^{2}}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})[/tex]???

    It's important that I know this stuff because we're having a quiz on it tomorrow.
     
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