Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 11 / b

Each displayed Pauli matrix has zero trace and determinant $-1$. Direct multiplication gives ${\sigma }_i^2=I$, henceTheir two eigenvalues are therefore $+1$ and $-1$.

Besides their squares, direct multiplication yields ${\sigma }_x{\sigma }_y=i{\sigma }_z$, ${\sigma }_y{\sigma }_z=i{\sigma }_x$ and ${\sigma }_z{\sigma }_x=i{\sigma }_y$. Reversing each pair changes the sign. These identities combine into the Pauli matrix multiplication identityAdding the expression with $i,j$ exchanged cancels the antisymmetric term, proving $\{{\sigma }_i,{\sigma }_j\}=2{\delta }_{ij}I$. Taking the trace instead uses $\mathrm{tr}I=2$ and $\mathrm{tr}{\sigma }_k=0$, proving $\mathrm{tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$.

Expand the two linear combinations and use the Pauli matrix multiplication identity:Interchanging the vectors reverses the cross product, so their commutator is $2i\sigma (\mathbf{a}\times \mathbf{b})$. The three Pauli matrices are linearly independent; alternatively, trace against ${\sigma }_k$ to recover each coefficient. ThusFor 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.