 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-3/31a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

The

k^{- 1}

μ_{1 x} + i [σ, μ_{2}] = Q μ_{1}

. Substitution in the expression for

B

gives

B = - 2 μ_{1 x}

. Decompose

μ_{1}

into diagonal and off-diagonal parts. Since

Q = i [σ, μ_{1}]

, its off-diagonal part is

i Q σ /2

. Thus

(μ_{1 x})_{off} = \frac{i}{2} Q_{x} σ, (μ_{1 x})_{diag} = (Q μ_{1})_{diag} = \frac{i}{2} Q^{2} σ .

The commutator in the

k^{- 1}

equation has zero diagonal; multiplication of two off-diagonal two-by-two matrices is diagonal. Combining the two parts proves

A = 2 Q, B = - i (Q^{2} + Q_{x}) σ_{3} .

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