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Qm - Hermitian Commutator [A,B]†

  • Thread starter jinksys
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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.
 
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  • #2
George Jones
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What does [itex]\left[ A , B \right] [/itex] equal?
 
  • #3
Dick
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[A,B]=AB-BA. Just use that. You don't need any book references.
 
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What does [itex]\left[ A , B \right] [/itex] equal?
[A,B] = AB - BA

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

and [A†,B†] = A†B† - B†A†.
 
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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?
 
  • #6
George Jones
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So, [A, B]† = (AB - BA)† = (AB)† - (BA)† = B†A† - A†B†
Also,
operators A and B are hermitian
 
  • #7
dextercioby
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[...]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.
 
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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!
 

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