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

For finite-dimensional operators, use $(AB{)}^{†}=B^{†}{A}^{†}$ and the Hermitian assumptions. The commutator satisfies $[A,B{]}^{†}=-[A,B]$, soThus $i[A,B]$ is Hermitian. These adjoint identities are the intended matrix-level setting; for unbounded operators, domains and self-adjointness require additional analysis.

The commutator is anti-Hermitian: $[A,B{]}^{†}=-[A,B]$. It is also Hermitian precisely when it is zero, so it is not Hermitian in general; it is Hermitian if and only if $A$ and $B$ commute. For example, $[{\sigma }_x,{\sigma }_y]=2i{\sigma }_z\ne 0$ for the Pauli matrices.

Taking the Hermitian adjoint reverses the products and leaves their sum unchanged:Thus the symmetrized product is Hermitian, whether or not $A$ and $B$ commute. This is the Hermitian anticommutator divided by two.

One has $(C^{†}C{)}^{†}=C^{†}C$, with no requirement that $C$ itself be Hermitian. Moreover,Thus $C^{†}C$ is Hermitian and positive semidefinite. The positive semidefinite property is stronger than the Hermitian conclusion alone.

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.