emsol04 - Electromagnetic Theory I Solution Set 4 Due: 19...

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Unformatted text preview: Electromagnetic Theory I Solution Set 4 Due: 19 September 2011 13. We have derived the spherical harmonic Y ll ( , ), Y ll ( , ) = ( ) l l ! 2 radicalbigg (2 l + 1)! 2 2 l +1 sin l e il , (1) normalized so that integraltext d | Y lm | 2 = 1, as a solution of the differential equation L + Y ll = 0 where L = L x iL y are the angular momentum ladder operators. Recall that L raise or lower the m value of a spherical harmonic Y lm by one unit: L Y lm = radicalbig l ( l + 1) m ( m 1) Y lm 1 = radicalbig ( l m )( l m + 1) Y lm 1 . (2) Using these facts, apply L repeatedly to Y ll to derive all the Y lm for 0 m l for the cases a) l = 1, b) l = 2, and c) l = 3. You may use without proof the spherical coordinate form of the L , L z : L = e i parenleftbigg i cot parenrightbigg , L z = 1 i (3) Solution : Notice that since Y lm / = imY lm , L applied to Y lm reduces to L Y lm = e i parenleftbigg + m cot parenrightbigg Y lm (4) Exploiting this simplification, we evaluate a) Y 11 = radicalbigg 3 8 sin e i (5) Y 10 = 1 2 L Y 11 = radicalbigg 3 16 L sin e i = radicalbigg 3 16 parenleftbigg + cot parenrightbigg sin = radicalbigg 3 4 cos (6) 1 b) Y 22 = radicalbigg 5! 2 8 sin 2 e 2 i = radicalbigg 15 32 sin 2 e 2 i Y 21 = 1 4 L Y 22 = radicalbigg 15 128 e i parenleftbigg + 2 cot parenrightbigg sin 2 = radicalbigg 15 8 e i sin cos Y 20 = 1 6 L Y 21 = radicalbigg 5 16 parenleftbigg + cot parenrightbigg sin cos = radicalbigg 5 16 ( 2 cos 2 sin 2 ) = radicalbigg 5 16 ( 3 cos 2 1 ) (7) c) Y 33 = 1 6 radicalbigg 7! 2 8 sin 3 e 3 i = radicalbigg 35 64 sin 3 e 3 i Y 32 = 1 6 L Y 33 = radicalbigg 35 384 parenleftbigg + 3 cot parenrightbigg sin 3 e 2 i = radicalbigg 105 32 sin 2 cos e 2 i Y 31 = 1 10 L Y 32 = radicalbigg 105 320 parenleftbigg + 2 cot parenrightbigg sin 2 cos e i = radicalbigg 21 64 ( 4 sin cos 2 sin 3 ) e i = radicalbigg 21 64 sin ( 5 cos 2 1 ) e i Y 30 = 1 12 L Y 31 = radicalbigg 7 256 parenleftbigg + cot parenrightbigg sin ( 5 cos 2 1 ) = radicalbigg 7 256 ( 5 cos 3 cos 10 sin 2 cos + 5 cos 3 cos ) = radicalbigg 7 256 ( 20 cos 3 12 cos ) = radicalbigg 7 16 ( 5 cos 3 3 cos ) (8) 14. One can sometimes use part of the solution of a simple potential problem as the solution of an apparently more complicated problem. For example, any equipotential surface of the simple problem can be replaced by a conductor of coincident shape....
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emsol04 - Electromagnetic Theory I Solution Set 4 Due: 19...

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