137A_HW09_solutions.pdf - Physics 137A Spring 2017 Problem...

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Physics 137A, Spring 2017 Problem Set 9 Solutions Problem 1: Angular Quantization and Legendre Polynomials a) We substitute x = cos θ into the theta equation with m = 0 to get 1 Θ sin θ d sin θ d Θ + B sin 2 θ = 0 (1) d sin θ d Θ = - B sin θ Θ (2) - sin θ d dx - sin 2 θ d Θ dx = - B sin θ Θ (3) d dx (1 - cos 2 θ ) d Θ dx = - B Θ (4) d dx (1 - x 2 ) d Θ dx = - B Θ (5) b) We substitute the series expansion Θ( x ) = j =0 c j x j into the theta equation to get d dx (1 - x 2 ) d Θ dx = - B Θ (6) d dx (1 - x 2 ) d dx X j =0 c j x j = - B X j =0 c j x j (7) d dx (1 - x 2 ) X j =0 c j jx j - 1 = X j =0 ( - B ) c j x j (8) - 2 x X j =0 c j jx j - 1 + (1 - x 2 ) X j =0 c j j ( j - 1) x j - 2 = X j =0 ( - B ) c j x j (9) X j =0 ( - 2) c j jx j + X j =0 c j j ( j - 1) x j - 2 - X j =0 c j j ( j - 1) x j = X j =0 ( - B ) c j x j (10) X j =0 c j j ( j - 1) x j - 2 - X j =0 (2 j + j ( j - 1)) c j x j = X j =0 ( - B ) c j x j (11) X j =0 c j j ( j - 1) x j - 2 = X j =0 ( j ( j + 1) - B ) c j x j (12) c) We reindex the sum on the LHS to begin at j = - 2 to get X j = - 2 c j +2 ( j + 2)( j + 1) x j = X j =0 ( j ( j + 1) - B ) c j x j (13) The first two terms in the sum on the LHS are zero, so we can begin the sum at j = 0 X j =0 c j +2 ( j + 2)( j + 1) x j = X j =0 ( j ( j + 1) - B ) c j x j (14) X j =0 [ c j +2 ( j + 2)( j + 1) - ( j ( j + 1) - B ) c j ] x j = 0 (15) 1
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Since j =0 a j x j = 0 implies that each a j = 0, we can conclude that c j +2 ( j + 2)( j + 1) - ( j ( j + 1) - B ) c j = 0 (16) c j +2 = j ( j + 1) - B ( j + 2)( j + 1) c j (17) d) If c j max +2 = 0, then 0 = c j max +2 = j max ( j max + 1) - B ( j max + 2)( j max + 1) c j max (18) = 0 = j max ( j max + 1) - B (19) B = j max (
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