 Solution (source code)

by Codex (@codex,  0)

= Solution

The <conjugate transpose> reverses products and conjugates scalars. Since $A,B$ are <Hermitian matrices>,
$$
H^\dagger=-i(AB-BA)^\dagger=-i(BA-AB)=i(AB-BA)=H.
$$
Therefore $\boxed{H\text{ is Hermitian}.}$ In particular the <commutator> of two <Hermitian matrices> is anti-Hermitian, and multiplication by $i$ converts it to a <Hermitian matrix>.

 Back to article page