 Solution

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

L^{2} Y_{l}^{m} = ℏ^{2} l (l + 1) Y_{l}^{m}, L_{z} Y_{l}^{m} = ℏ m Y_{l}^{m}, l = 0, 1, \dots, - l \leq m \leq l .

L^{2}

L_{z}

past

L_{\pm}

therefore gives

L^{2} (L_{\pm} Y_{l}^{m}) = ℏ^{2} l (l + 1) (L_{\pm} Y_{l}^{m}), L_{z} (L_{\pm} Y_{l}^{m}) = ℏ (m \pm 1) (L_{\pm} Y_{l}^{m}) .

For a nonzero result these are the eigenvalue equations with unchanged

l

and shifted magnetic quantum number. Taking the squared norm, using

L_{\mp} L_{\pm} = L^{2} - L_{z}^{2} \mp ℏ L_{z}

, fixes the conventional coefficient:

L_{\pm} Y_{l}^{m} = ℏ l (l + 1) - m (m \pm 1) Y_{l}^{m \pm 1} .

At the appropriate endpoint the coefficient is zero. The zero vector is not itself an eigenstate.

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