Past exam of the mathematical methods I course of the University of Oxford / 2013 / 11

Each Pauli matrix has zero diagonal sum and determinant $-1$. Direct multiplication gives ${\sigma }_x^2={\sigma }_y^2={\sigma }_z^2=I$. Thus all requested quantities, for each $j=x,y,z$, areFor example, ${\sigma }_y^2=\mathrm{diag}((-i)i,i(-i))=I$, checking that its imaginary entries do not introduce a minus sign in the square.

The cyclic products obtained by direct matrix multiplication are ${\sigma }_x{\sigma }_y=i{\sigma }_z$, ${\sigma }_y{\sigma }_z=i{\sigma }_x$, and ${\sigma }_z{\sigma }_x=i{\sigma }_y$. Reversing any two distinct indices changes the sign; equal-index products are $I$. Therefore every pair is covered byHere $\{A,B\}$ is the anticommutator and $[A,B]$ the commutator. Explicitly the three diagonal anticommutators are $2I$, all six off-diagonal ones are zero, and all three diagonal commutators vanish. The cyclic commutators are $2i{\sigma }_z,2i{\sigma }_x,2i{\sigma }_y$, respectively, with their reversed partners negative. The convention is ${ϵ}_{xyz}=1$.

Combine the diagonal and cyclic products from b into ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{\sum }_k{ϵ}_{ijk}{\sigma }_k$. Distribute the coefficients in the two Pauli-vector factors:The inner sums are precisely the components of $\mathbf{a}\times \mathbf{b}$, soInterchanging the two factors shows that their commutator is $2i\sigma (\mathbf{a}\times \mathbf{b})$. The three Pauli matrices are linearly independent: the entries of $\sigma (\mathbf{c})$ are $c_z,c_x-i{c}_y,c_x+i{c}_y,-c_z$, which vanish together only when $\mathbf{c}=0$. ThusFor 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.

Setting $\mathbf{b}=\mathbf{a}$ in i removes the cross product term and gives $\sigma (\mathbf{a}{)}^2=(\mathbf{a}\cdot \mathbf{a})I$. Repeated multiplication therefore yields, for nonnegative integers $m$,The zeroth power is $I$. If negative powers are intended, they require $\mathbf{a}\cdot \mathbf{a}\ne 0$, and the same parity formulas then follow from $\sigma (\mathbf{a}{)}^{-1}=\sigma (\mathbf{a})/(\mathbf{a}\cdot \mathbf{a})$. Every nonzero real vector meets that condition; a nonzero complex null vector need not. For example, $\mathbf{a}=(1,i,0)$ has $\sigma (\mathbf{a}{)}^2=0$ and no inverse.

For a real unit vector $\hat{\mathbf{n}}$, let $S=\sigma (\hat{\mathbf{n}})$. Then $S^2=I$, so even powers are $I$ and odd powers are $S$. Split the absolutely convergent matrix exponential into its even and odd series:The result holds for every real $\theta$, not merely small angles. Its two scalar series are the ordinary cosine and sine series.