 Solution

ID: past-exam-of-the-quantum-physics-course-of-the-university-of-oxford/2016/8/d/solution

1 = 2 π ∣ C ∣^{2} \int_{0}^{π} sin^{2 l + 1} θ d θ = 4 π ∣ C ∣^{2} \frac{( 2 ^{l} l ! ) ^{2}}{( 2 l + 1 )!} .

Thus

∣ C ∣ = (2 l + 1)! / (4 π) / (2^{l} l!)

. Choosing the usual alternating sign convention,

Y_{l}^{l} = (- 1)^{l} \frac{( 2 l + 1 )!}{2 ^{l} l ! 4 π} sin^{l} θ e^{i lϕ} .

Apply

L_{-}

successively and divide by its ladder coefficient at each step. Explicitly,

Y_{l}^{m} = \frac{( l + m )!}{( 2 l )! ( l - m )!} ℏ^{- (l - m)} (L_{-})^{l - m} Y_{l}^{l} .

The normalization follows because the product of the squared ladder coefficients is

ℏ^{2 (l - m)} (2 l)! (l - m)! / (l + m)!

. This constructs every magnetic sublevel, including negative

m

 New to topics? Read the docs here!