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homework_6_solution - (r =[Al rl Bl r(l 1]Pl(cos l=0(r = Al...

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Φ ( r, θ ) = l =0 [ A l r l + B l r ( l +1) ] P l (cos θ ) Φ ( r, θ ) = l =0 A l r l P l (cos θ ) A l = 2 l + 1 2 a l ˆ π 0 V ( θ ) P l (cos θ ) sin θ d θ ˆ π 0 V ( θ ) P l (cos θ ) sin θ d θ = V ˆ 1 0 P l ( x ) dx = V π 2 Γ [1 l 2 ] Γ [ 3+ l 2 ] = π V 2( π / sin( π l/ 2) π ( l 2)!! 2 ( l 1) / 2 ) · ( π ( l +4)!! 2 ( l +5) / 2 ) = 4 sin( π l/ 2) V π ( l + 4)!! / ( l 2)!! G D ( x, x ) = l 4 π 2 l + 1 [ r l r l +1 1 a ( rr a 2 ) l ] Y l 0 ( θ , φ ) Y l 0 ( θ , φ ) = l [ r l r l +1 1 a ( rr a 2 ) l ] P l (cos θ ) P l (cos θ )
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G D r | r = a = l (2 l + 1) r l a l +2 P l (cos θ ) P l (cos θ ) Φ ( x ) = l [ ˆ π 0 Φ ( a, θ ) · 2 π · P l (cos θ ) · sin θ d θ · 2 l + 1 4 π ]( r a ) l P l (cos θ ) = 2 l + 1 2 l [ ˆ π 0 Φ ( a, θ ) · P l (cos θ ) · sin θ d θ ]( r a ) l P l (cos θ ) z = 0 z = L Ψ k ( r, z ) = 1 (2 π ) e ik · r 2 L sin( n π L z ) G ( x, x ) = 4 π ˆ d 2 k n =0 1 4 π 2 2 L e ik ( r r ) sin ( n π L z ) sin( n π L z ) ( k 2 + ( n π L ) 2 ) = 2 π L ˆ d 2 k e ik ( r r ) n =0 1 4 π 2 2 L sin ( n π L z ) sin( n π L z ) ( k 2 + ( n π L ) 2 ) = 1 2 π L ˆ d 2 k e ik ( r r ) + n = −∞ e i n π L ( z + z ) e i n π L ( z z ) ( k 2 + ( n π L ) 2 ) + n = −∞ e i n π L ( z + z ) e i n π L ( z + z ) ( k 2 + ( n π L ) 2 ) = + k = −∞ ˆ dn e i n π L ( z + z ) e i n π L ( z z ) ( k 2 + ( n π L ) 2 ) · e 2 π i · kn
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= L ( + k =0 e k ( z + z ) e k ( z z ) k e 2 kk L + 0 k = −∞ e k ( z + z )
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