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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/11/c/i/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2013 / 11 / c / i / Solution by

Codex  0 2026-10-25

Combine the diagonal and cyclic products from b into

σ_{i} σ_{j} = δ_{ij} I + i \sum_{k} ϵ_{ijk} σ_{k}

. Distribute the coefficients in the two Pauli-vector factors:

σ (a) σ (b) = i, j \sum a_{i} b_{j} σ_{i} σ_{j} = (i \sum a_{i} b_{i}) I + i k \sum (i, j \sum ϵ_{ijk} a_{i} b_{j}) σ_{k} .

The inner sums are precisely the components of

a \times b

, so

σ (a) σ (b) = (a \cdot b) I + iσ (a \times b) .

Interchanging the two factors shows that their commutator is

2 iσ (a \times b)

. The three Pauli matrices are linearly independent: the entries of

σ (c)

are

c_{z}, c_{x} - i c_{y}, c_{x} + i c_{y}, - c_{z}

, which vanish together only when

c = 0

. Thus

[σ (a), σ (b)] = 0 ⟺ a \times b = 0 ⟺ a, b are linearly dependent .

For nonzero real vectors, this means parallel or antiparallel directions. The dot and cross products here are bilinear; no complex conjugates are inserted if complex coefficients are considered.

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