The Exercise

Two operators $A$ and $B$ on the same state space need not commute: $AB$ and $BA$ are in general different operators. The exercise asks for the product $AB$ through the commutator and the anticommutator of $A$ and $B$ .

Rings, Algebras and the Anticommutator

The library states the identity for two elements $a$ and $b$ of a ring $R$ : a set with a sum, a unit $1$ and an associative product that distributes over the sum and need not commute. The commutator of $a$ and $b$ is $[a,b] = ab - ba$ .

The anticommutator $\{a,b\}$ of two elements $a$ and $b$ of any set with a sum and a product is the sum of their products in the two orders, $\{a,b\} = ab + ba.$ (1)^✦

A commutative semiring $\Bbbk$ is a set of scalars with a sum and a commutative product, as the complex numbers have. An algebra $R$ over $\Bbbk$ is a ring whose elements can also be multiplied by scalars $c$ from $\Bbbk$ , with $c(ab) = (ca)b = a(cb)$ .

The Claim

Theorem 1. Let $R$ be a ring in which $2$ has an inverse $\tfrac12$ , and let $a$ and $b$ be elements of $R$ . Then $ab = \tfrac12\,\bigl([a,b] + \{a,b\}\bigr),$ (2)^✦ the product with $\tfrac12$ taken in $R$ .

Theorem 2. Let $\Bbbk$ be a commutative semiring in which $2$ has an inverse $\tfrac12$ , let $R$ be an algebra over $\Bbbk$ , and let $a$ and $b$ be elements of $R$ . Then $ab = \tfrac12 \cdot \bigl([a,b] + \{a,b\}\bigr),$ (3)^✦ with $\tfrac12$ acting on $R$ as a scalar.

With $R$ the operators on a state space, $\Bbbk = \mathbb{C}$ , $a = A$ and $b = B$ , either statement is the exercise.

What Is Checked

Both statements are stated over the commutator and the anticommutator (1) of two elements, with no condition on the elements.