Now using x y z makes the problem more difcult since

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Unformatted text preview: , but it can be on one side or the other, and one expects moves between the two sides. It crosses the plane, but it is never found there! Monday, April 1, 2013 A 3D spherical box A complication, not covered in the textbook, is to have a 3D spherical box. Now using x, y, z makes the problem more difficult since the, e.g., x limits depend on the y and z values. Instead we use spherical coordinates, r, θ, φ. The 3D Laplacian operator o perating on ψ makes S.E., slightly rearranged, of the form: ∂￿ ∂2ψ 1 ∂ ￿ 2 ∂ψ ￿ 1 ∂ψ ￿ 1 −2mE 2 ∇ ψ (￿ ) = 2 x r +2 sin θ +2 2 = ψ 2 2 r ∂r ∂r r sin θ ∂θ ∂θ ￿ r sin θ ∂φ It is natural to guess the solution is of the form ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ). Since each term differentiates only one of the three functions, we come up with 3 independent equations, like in the 3D cubical box case. The radial function R needs to go to 0 at the e dge of the box, r = r0, but not at the center, r=0. The angular dependences are usually written as "spherical harmonics" Ylm(θ,φ...
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This document was uploaded on 02/18/2014 for the course PHYS 228 at Rutgers.

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