I have several problems with General Definitions of an Operator and Commutator :
- the product of operators is generally not commutative:
$$\hat A \hat B \not= \hat B\hat A .$$
what is this means that generally is not commutative?
and , is the product of two operators $\hat A$ Hermitian Adjoint $\dagger$ then $\hat B$ (because in my book it is just $\hat A \hat B$ )?
$$\hat A^{\dagger} \hat B \not= \hat B^{\dagger}\hat A .$$
Remark
2. Note that the Hermitian adjoint of an operator is not, in general, equal to its complex conjugate:
$$\hat A^{\dagger} \not= \hat A^* .$$
What happens if the Hermitian adjoint of an operator is, in special, equal to its complex conjugate!?
$$\hat A^{\dagger} = \hat A^* .$$
Commutator Algebra
- the commutator of two operators $\hat A$ and $\hat B$, denoted by:
$$[\hat A , \hat B]=\hat A \hat B – \hat B \hat A .$$
again i do not know that, is the product of two operators $\hat A$ Hermitian Adjoint $\dagger$ then $\hat B$?
$$[\hat A , \hat B]=\hat A^{\dagger} \hat B – \hat B^{\dagger} \hat A$$
Best Answer
Unfortunately, "in general" has both a colloquial meaning in English and a mathematical meaning.
The colloquial meaning is the same as "usually". For example, "I generally wake up at 6 am, but sometimes I oversleep."
The mathematical meaning is "true for all instances", as in "the eigenvalue of a Hermitian operator will be real in general", since all eigenvalues of Hermitian operators are real.
When someone says "operators are generally not commutative", it is somewhat ambiguous which meaning they intend. It is most likely the colloquial meaning, so they meant "most of the time, operators don't commute." If they intended the mathematical meaning, they meant "operators do not always commute."
I don't understand your follow-up about adjoints. Certainly, if operators do not commute in general, neither do adjoints of operators (after all, the set of all adjoints is the same as the set of all operators).
If an operator's adjoint is equal to its complex conjugate, it means the operator is symmetric. To see this, simply write the relation in component form. If we represent the components of the original operator by $A_{ij}$, the equation you gave is simply
$$A_{ji}^* = A_{ij}^*$$
which implies
$$A_{ji} = A_{ij}$$
so the operator is symmetric.
Your last question does not make sense to me. The equation $\left[A,B\right] = AB - BA$ is the definition of the commutator. Perhaps you made a typo in your math typesetting? If you are asking whether it is true that
$$\left[A^\dagger, B \right] = A^\dagger B - B^\dagger A$$
then no, that is not correct. Simply by substituting $A = A^\dagger$ in the original expression for the commutator, we get
$$\left[A^\dagger, B \right] = A^\dagger B - B A^\dagger$$