 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019-resit/11/b/iii/solution

σ (a) σ (b) = ij \sum a_{i} b_{j} (δ_{ij} I + i k \sum ϵ_{ijk} σ_{k}) = (a \cdot b) I + iσ (a \times b) .

Interchanging the vectors reverses the cross product, so their commutator is

2 iσ (a \times b)

. The three Pauli matrices are linearly independent; alternatively, trace against

σ_{k}

to recover each coefficient. Thus

[σ (a), σ (b)] = 0 ⟺ a \times b = 0.

For nonzero real vectors, this means they are parallel or antiparallel, equivalently linearly dependent. The algebra also extends to complex vectors using bilinear dot and cross products, with the same linear-dependence criterion; the dot product there has no complex conjugation.

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