Elementary question:[A,B] = [A-<A>, B-<B>]

[nomadreid]

I know this should be obvious, and I should be ashamed of asking it, but could someone fill in the steps to show that [A,B] = [A-<A>, B-<B>]? Thanks from a non-physicist.

 

[Fredrik]

<A> is a real number (or if A isn't self-adjoint, a complex number), so A-<A> must be interpreted as A-<A>I, where I is the identity operator, which commutes with everything.

The forum's policy on homework and textbook-style questions prevents me from giving you the complete answer, but I think you will find it easy to show that [A+B,C]=[A,C]+[B,C], and [A,B+C]=[A,B]+[A,C] for all A,B,C. Then you can use these formulas to deal with [A+B,C+D]. When you have done that, let B and D be operators that commute with everything ([B,X]=[D,X] for all X) and see what you get.

Edit: Hehe. I cleverly made sure my reply would be the first by posting the first paragraph as soon as I was done with it and then adding the second one in an edit.

 

[dextercioby]

Well, A-<A> means A-<A>1 , where the '1' is the unit operator on the vector space these operators act on. The unit operator commutes with every other operator, including itself, that's why you have the equality of the 2 commutators.

 

[nomadreid]

Thanks, first-past-the-post Fredrik, and also to bigubau. That should do it.

Fredrik: Just as a side note, this was not a homework question: I wish it were, because I miss academia. But you're right, it's textbook style. Alas, I am limited in the number of textbooks I have as reference. Hence my double gratitude for this Forum and knowledgeable people like you on it.