Qm - Hermitian Commutator [A,B]†

[jinksys]

I have this review question: If operators A and B are hermitian, prove that their commutator is "anti-hermitian", ie) [A,B]†=-[A,B]

What has me confused is the placement of the dagger on the commutator. Why [A,B]† and not [A†,B†]? Also, I am using Griffith's Intro to QM as a text. I have failed to find where the author uses this [A,B]† commutator in the book. If anyone has the book and can give me a page number of an example, I'd be grateful.

 

[George Jones]

What does $\left[ A , B \right]$ equal?

 

[Dick]

[A,B]=AB-BA. Just use that. You don't need any book references.

 

[jinksys]

What does $\left[ A , B \right]$ equal?

[A,B] = AB - BA

So, [A, B]† = (AB - BA)† = (AB)† - (BA)† = B†A† - A†B†

and [A†,B†] = A†B† - B†A†.

 

[jinksys]

I think there may be some confusion. I know what [A,B] is; What I'm inquiring on is the placement of the dagger.

Edit: You guys can see my dagger symbol, right?

 

[George Jones]

So, [A, B]† = (AB - BA)† = (AB)† - (BA)† = B†A† - A†B†

Also,

operators A and B are hermitian

 

[dextercioby]

[...]What has me confused is the placement of the dagger on the commutator. Why [A,B]† and not [A†,B†]? [...]

Because they are generally NOT the same operator. You have the adjoint of the commutator ([A,B]†) versus the commutator of the adjoints ([A†,B†]). For bounded operators, which can be extended to the whole Hilbert space, they are equal, provided A and B are self-adjoint and bounded, of course.

 

[jinksys]

Because they are generally NOT the same operator. You have the adjoint of the commutator ([A,B]†) versus the commutator of the adjoints ([A†,B†]). For bounded operators, which can be extended to the whole Hilbert space, they are equal, provided A and B are self-adjoint and bounded, of course.

That makes sense, thanks!