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Unformatted text preview: Midterm Solutions 1) Since we are given that J = 1 and m = 1, we need to use the function Y 1 1 ( θ,φ ) and integrate over the surface of a sphere, using the limits of integration given in the problem: P = Z 2 π Z 2 π/ 3 π/ 3 Y 1 * 1 Y 1 1 sin ( θ ) dθdφ Since Y 1 1 = q 3 8 π sin ( θ ) e iφ , our integral becomes: P = 3 8 π Z 2 π Z 2 π/ 3 π/ 3 sin 2 ( θ ) · 1 · sin ( θ ) dθdφ Since our function has no φ dependence we can carry out this integral immediately to pull out a factor of 2 π , and then we see from the table of integrals that: Z b a sin 3 ( θ ) dθ = 3 cos ( θ ) 4 + cos (3 θ ) 12 b a Using the above results gives us: P = 3 8 π · 2 π · 3 cos ( θ ) 4 + cos (3 θ ) 12 2 π/ 3 π/ 3 = 3 4 3 8 + 1 12 + 3 8 + 1 12 = 11 16 2) Given ˆ A = y d dx and ˆ B = x d dx , we find that: [ ˆ A, ˆ B ] f = y d dx x df dx x d dx y df dx = y df dx + yx d 2 f dx 2 xy d 2 f dx 2 Since the operators for x and y commute, we have xy = yx, so our expression above becomes:...
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 Spring '11
 Linda
 pH, Angular Momentum, Correct choice, dx

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