 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015/12/i/solution

σ_{j}^{2} = I

σ_{1} σ_{2} = i σ_{3}

and its cyclic permutations; reversing unequal factors changes the sign. Hence, using the Levi-Civita symbol,

σ_{j} σ_{k} = δ_{jk} I + i ϵ_{jk ℓ} σ_{ℓ}, [σ_{j}, σ_{k}] = 2 i ϵ_{jk ℓ} σ_{ℓ} .

Repeated

ℓ

is summed, while the Kronecker delta handles the equal-index case.

 New to topics? Read the docs here!