 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/15c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

[L_{z}, L_{x}] = i ℏ L_{y}, [L_{z}, L_{y}] = - i ℏ L_{x} .

Consequently

[L_{z}, L_{\pm}] = [L_{z}, L_{x}] \pm i [L_{z}, L_{y}] = i ℏ L_{y} \pm ℏ L_{x} = \pm ℏ L_{\pm} .

The same commutation relations imply

[L^{2}, L_{x}] = [L^{2}, L_{y}] = 0

, and hence

[L^{2}, L_{\pm}] = 0 .

These are the defining identities for the angular momentum ladder operators.

L_{z} ϕ = m ℏ ϕ, L^{2} ϕ = ℓ (ℓ + 1) ℏ^{2} ϕ,

then, whenever

L_{\pm} ϕ \neq = 0

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

Thus $L_{\pm} ϕ$ has quantum numbers $(ℓ, m \pm 1)$ .

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