Condition for the Product of Two Hermitian Operators to be a Hermitian Operator
Theorem
Let two operators $A$, $B$ be Hermitian operators. If $A$, $B$ commute, then $AB$ is also a Hermitian operator. The converse also holds.
Explanation
Considering the contrapositive of the converse, we can see that the product of two operators that do not commute is not a Hermitian operator. That is, all of the following hold.
- Theorem: If two Hermitian operators $A,B$ commute, then $AB$ is also a Hermitian operator.
- Converse: If $AB$ is a Hermitian operator, then $A$ and $B$ commute.
- Inverse: If two Hermitian operators $A, B$ do not commute, then $AB$ is not Hermitian.
Proof
To show that $AB$ is a Hermitian operator, it suffices to show that $(AB)^{\dagger} = AB$. $AB$ is as follows.
$$ \begin{align*} && [A, B] &= AB - BA \\ \implies && AB &= BA + [A, B] \end{align*} $$
Computing $(AB)^{\dagger}$ gives the following.
$$ (AB)^{\dagger} = (BA + [A, B])^{\dagger} = (BA)^{\dagger} + [A,B]^{\dagger} = AB + [A,B]^{\dagger} $$
By assumption $AB = BA$, so if $[A,B]^{\dagger} = 0$ then $(AB)^{\dagger} = AB$. Therefore, if $[A,B]=0$ then $[A,B]^{\dagger} = 0$ and $AB$ is a Hermitian operator.
■
Converse
Since $A,B$ are Hermitian operators, $(AB)^{\dagger}=B^{\dagger} A^{\dagger} = BA$. In this case, if $AB$ is a Hermitian operator, then the following holds.
$$ (AB)^{\dagger} = AB $$
Therefore $AB=BA$.
■