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Schrodinger Eq. in 3D

  1. Nov 27, 2009 #1
    The schrodinger equation in 3D (time independent).

    Letting Phi = X(x).Y(y).Z(z), and solving as a PDE......

    The equation looks pretty much the same except there is a seperate Hamiltonian for each of the Cartesian coordinates x y z. However, the 1/X(x) term etc. really confuses me, I dont know where it comes from. Could someone perhaps explain??

    ie. H_x = [-(h^2)/2m].[1/X(x)].[(d^2)X(x)/d(X(x))^2] + V(x)
    ^^^^

    where h is representing h-bar, and d the partial derivative.

    Cheers guys!!:biggrin:
     
  2. jcsd
  3. Nov 27, 2009 #2
    It occurs because you divide through by 1/XYZ to isolate the equations.

    But note that using a separation in Cartesian coordinates is not always a viable solution, and will only work for some potentials.
     
  4. Nov 27, 2009 #3
    Can you perhaps outline the derivation from the start? It's just clearing it up for me......
     
  5. Nov 27, 2009 #4
    It goes something like assume the potential is an infinite square potential

    [tex]V(x,y,z) = \left(\begin{array}{cc}0 if x,y,z < a \\ \infty else[/tex]

    We can assume a separable solution [tex]\Psi (x,y,z) = X(x)Y(y)Z(z)[/tex]

    [tex]\frac{-\hbar^2}{2m} [Y(y)Z(z) \frac{d^2 X}{dx^2}+X(x)Z(z) \frac{d^2 Y}{dy^2}+X(x)Y(y) \frac{d^2 Z}{dz^2}] + V(x,y,z)XYZ = E(XYZ)[/tex]

    Then just divide everything by 1/XYZ.
     
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