ph135cS1 - Ph135c. Solution set #1, 4/9/10 1. We can...

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Unformatted text preview: 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 at the end): V ( r ) = r < a r > a We look for a solution of Schrodingers equation in the form: km ( r,, ) = u k ( r ) r Y m ( , ) 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 weve 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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This note was uploaded on 12/04/2011 for the course PHY 7070 taught by Professor Smith during the Spring '11 term at Wisconsin.

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ph135cS1 - Ph135c. Solution set #1, 4/9/10 1. We can...

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