 Solution

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

There is no allowed spherical harmonic with

m = l + 1

, so

L_{+} Y_{l}^{l} = 0

- i ℏ \partial_{ϕ} Y = ℏ l Y

first gives

Y = f (θ) e^{i lϕ}

. Combining the differential expressions for

L_{x}, L_{y}

yields

L_{+} = ℏ e^{i ϕ} (\partial_{θ} + i cot θ \partial_{ϕ}) .

Its annihilation of the highest-weight state becomes

f^{'} (θ) - l cot θ f (θ) = 0.

Integrating

f^{'} / f = l cot θ

gives

f = C sin^{l} θ

, with the constant solution included for

l = 0

. Hence

Y_{l}^{l} (θ, ϕ) = C sin^{l} θ e^{i lϕ} .

e^{i lϕ}

is single valued for integral orbital quantum number

l

; normalization fixes only

∣ C ∣

, leaving a conventional constant phase.

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