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# ph135cS1 - Ph135c Solution set#1 1 We can approximate this...

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Ph135c. Solution set #1, 4/9/10 1. We can approximate this potential by an infinite radial square well (taking care to shift the energies by V 0 at the end): V ( r ) = 0 r < a r > a We look for a solution of Schrodinger’s equation in the form: ψ k‘m ( r, θ, φ ) = u k‘ ( r ) r Y m ( θ, φ ) Then we find that u k‘ ( r ) must satisfy (unless otherwise noted, we set ~ = 1): - 1 2 m d 2 u k‘ dr 2 + ( + 1) 2 mr 2 u k‘ = k 2 2 m u k‘ where we’ve written the energy as E = k 2 2 m . The solutions can be expressed in terms of spherical Bessel functions j ( x ): u k‘ ( r ) rj ( kr ) The allowed energies are determined by the condition u k‘ ( a ) = 0. If we denote the n th zero of j ( x ) by x n‘ (that is, j ( x n‘ ) = 0 and 0 < x 1 < x 2 < ... ), then: E n‘ = x n‘ 2 2 ma 2 One can look up the zeros of the spherical Bessel functions in a reference table, and the lowest ten values are: x 1 x 2 x 3 = 0 : 3 . 14 6 . 28 9 . 42 = 1 : 4 . 49 7 . 72 = 2 : 5 . 76 9 . 10 = 3 : 7 . 00 = 4 : 8 . 18 = 5 : 9 . 39 The corresponding energy levels are plotted in figure 1. The degeneracies are determined by the angular momentum via d ( ) = 2 + 1. That is, the s levels have 1 state, the p levels have 3 states, and so on. If we have identical spin- 1 2 fermions, there are 2 total states for each wavefunction, coming from the two independent spins.

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ph135cS1 - Ph135c Solution set#1 1 We can approximate this...

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