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I have several problems with General Definitions of an Operator and Commutator :

  1. 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

  1. 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 $\hat B$? $$[\hat A , \hat B]=\hat A^{\dagger} \hat B - \hat B^{\dagger} \hat A$$

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  • $\begingroup$ I'm sure people will have fun answering, but strictly speaking there is not much physical about that question. $\endgroup$ – Nikolaj-K Jun 6 '12 at 10:18
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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

$$\left[A^\dagger, B \right] = A^\dagger B - B A^\dagger$$

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    $\begingroup$ Eichenlau thank you but what is the product of operators that is generally not commutative? i don't know what product means! in this case. $\endgroup$ – user8784 Jun 6 '12 at 10:37
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    $\begingroup$ The product of two operators is the operator obtained by applying the operators in succession. That is, $(AB)|\psi\rangle = A(B(|\psi\rangle))$ $\endgroup$ – Mark Eichenlaub Jun 6 '12 at 10:39
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    $\begingroup$ thanks but i still dont know that what means product of operators because they are complex numbers $\endgroup$ – user8784 Jun 6 '12 at 10:45
  • $\begingroup$ Dear Mr Ares, the ability to compute the product is surely a basic operation one can do with complex numbers, can't he? The only thing one needs to know to multiply complex numbers is the distributive law and $i^2=-1$. Operators aren't really complex numbers, as you mentioned, but matrices - well, operators - constructed out of complex numbers but they may still be multiplied and Mark wrote what it means but you have apparently overlooked the answer even though you have thanked for it. $\endgroup$ – Luboš Motl Jun 6 '12 at 10:54
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    $\begingroup$ @Mark, "operators are not commutative in general" means the same thing in the mathematical meaning of "in general" as in the colloquial one. The sentence says that "it is not true that operators commute with each other in general". In the math jargon interpretation, it says nothing at all about the percentage in which they commute but it's usually the case that when something can be zero as well as nonzero, it's "almost never" zero in the measure-theory as well as colloquial meaning of "almost". ;-) $\endgroup$ – Luboš Motl Jun 6 '12 at 10:57

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