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 ( θ, ϕ ), Y ll ( θ, ϕ ) = ( − ) l l ! √ 2 π radicalbigg (2 l + 1)! 2 2 l +1 sin l θe ilϕ , (1) normalized so that integraltext d Ω | 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 ± iϕ parenleftbigg ∂ ∂θ ± i cot θ ∂ ∂ϕ parenrightbigg , L z = 1 i ∂ ∂ϕ (3) Solution : Notice that since ∂Y lm /∂ϕ = imY lm , L − applied to Y lm reduces to L − Y lm = − e − iϕ parenleftbigg ∂ ∂θ + m cot θ parenrightbigg Y lm (4) Exploiting this simplification, we evaluate a) Y 11 = − radicalbigg 3 8 π sin θe iϕ (5) Y 10 = 1 √ 2 L − Y 11 = − radicalbigg 3 16 π L − sin θe iϕ = radicalbigg 3 16 π parenleftbigg ∂ ∂θ + cot θ parenrightbigg sin θ = radicalbigg 3 4 π cos θ (6) 1 b) Y 22 = radicalbigg 5! 2 8 π sin 2 θe 2 iϕ = radicalbigg 15 32 π sin 2 θe 2 iϕ Y 21 = 1 √ 4 L − Y 22 = − radicalbigg 15 128 π e iϕ parenleftbigg ∂ ∂θ + 2 cot θ parenrightbigg sin 2 θ = − radicalbigg 15 8 π e iϕ sin θ cos θ Y 20 = 1 √ 6 L − Y 21 = radicalbigg 5 16 π parenleftbigg ∂ ∂θ + cot θ parenrightbigg sin θ cos θ = radicalbigg 5 16 π ( 2 cos 2 θ − sin 2 θ ) = radicalbigg 5 16 π ( 3 cos 2 θ − 1 ) (7) c) Y 33 = − 1 6 radicalbigg 7! 2 8 π sin 3 θe 3 iϕ = − radicalbigg 35 64 π sin 3 θe 3 iϕ Y 32 = 1 √ 6 L − Y 33 = radicalbigg 35 384 π parenleftbigg ∂ ∂θ + 3 cot θ parenrightbigg sin 3 θe 2 iϕ = radicalbigg 105 32 π sin 2 θ cos θe 2 iϕ Y 31 = 1 √ 10 L − Y 32 = − radicalbigg 105 320 π parenleftbigg ∂ ∂θ + 2 cot θ parenrightbigg sin 2 cos θe iϕ = − radicalbigg 21 64 π ( 4 sin θ cos 2 θ − sin 3 θ ) e iϕ = − radicalbigg 21 64 π sin θ ( 5 cos 2 θ − 1 ) e iϕ Y 30 = 1 √ 12 L − Y 31 = radicalbigg 7 256 π parenleftbigg ∂ ∂θ + cot θ parenrightbigg sin θ ( 5 cos 2 θ − 1 ) = radicalbigg 7 256 π ( 5 cos 3 θ − cos θ − 10 sin 2 θ cos θ + 5 cos 3 θ − cos θ ) = radicalbigg 7 256 π ( 20 cos 3 θ − 12 cos θ ) = radicalbigg 7 16 π ( 5 cos 3 θ − 3 cos θ ) (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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This note was uploaded on 01/08/2012 for the course PHY 6346 taught by Professor Staff during the Spring '08 term at University of Florida.

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

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