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

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.