Proof that A^B^ is Hermitian only if A^ and B^ commute

Definition of Hermitian operators
Hermitian operators are operators that are equal to their own complex conjugate transpose. That is, if A^ is a Hermitian operator, then A^ = (A^)†.

Proof that A^B^ is Hermitian
To show that A^B^ is Hermitian, we need to show that (A^B^)† = A^B^.

Proof of (A^B^)† = B^A^
Using the definition of the adjoint of a product of operators, we have:

(A^B^)† = (B^)†(A^)†

Since A^ and B^ are Hermitian operators, we know that A^ = A^† and B^ = B^†. Therefore, we can simplify this expression to:

(A^B^)† = B^A^

Proof that A^B^ is Hermitian only if A^ and B^ commute
To show that A^B^ is Hermitian only if A^ and B^ commute, we need to show that if A^B^ = (A^B^)†, then A^ and B^ commute.

Assuming that A^B^ = (A^B^)†, we have:

B^A^ = (A^B^)† = A^B^

Multiplying both sides by A^, we get:

B^A^A^ = A^B^A^

Since A^ is an operator and not a scalar, we cannot simply cancel it out. However, we can use the rule of mutual exclusion to simplify this expression.

Explanation of the rule of mutual exclusion
The rule of mutual exclusion states that if two operators A and B do not commute, then there exists at least one vector v such that the inner product of Av and Bv is not equal to the inner product of Bv and Av.

Using the rule of mutual exclusion to simplify the expression
Since A^ and B^ do not commute, there exists at least one vector v such that the inner product of A^B^v and B^A^v is not equal to the inner product of B^A^v and A^B^v.

However, since A^B^ = B^A^, we have:

A^B^v = B^A^v

Therefore, the inner product of A^B^v and B^A^v is equal to the inner product of B^A^v and A^B^v, which contradicts the rule of mutual exclusion.

Therefore, we must conclude that A^B^ is Hermitian only if A^ and B^ commute.