Problem with non-commutative functions(quantum mechanics)

[GarethB]

1. I am working through E.Merzbacher quantum mechanics. The problem is;
if G(λ) =e^λAe^λB for two operators A and B, show that
dG/dλ=[A+B+λ[A,B]/1!+λ²[A,[A,B]]/2!+....]G



2. [A,B] is taken to mean AB-BA



3. The only way I can think of proving this is by taylor expanding G(λ) and then differentiang each term in the expansion with respect to λ; this has not worked! Can anyone please help!

 

[GarethB]

Just to expand on my attempt at a solution so far, which I think is barking up the wrong tree even though it seems close is;
1) taylor expand G(λ) at f(0) I get

G=1+λ[A+B]+λ²(A²+2AB+B²)/2!+...
=1+λ[A+B]+λ²(A+B)²/2!+λ³(A+B)³/3!
and then what makes this tempting is that differentiating with respect to λ I get
dG/dλ=[A+B]+λ(A+B)²/1!+λ²(A+B)³/2!

Why I am sure this is the wrong approach is that;
1. the whole expansion is supposed to be multiplied by G, which clearly it is not
2. I have no idea if the (A+B)², (A+B)³ can be expressed it the commutative notation [A,B], [A,[A,B]] etc. probably not so back to square one.

 

[sgd37]

be more careful when expanding quadratics since (A+B)(A+B) = A^2 +AB +BA +B^2 = A^2 +[A,B] +2BA +B^2

 

[GarethB]

Sorry perhaps I was unclear, I know that the quadratic expansions do not equal where I am (was) trying to get.
Anyway I have since cracked it. The procedure is to use the prouct rule to get dG/dλ. You then multiply by e^λbe^-λb. You end up with
dG/dλ=[B+e^λbAe^-λb]G
The term e^λbAe^-λb] is an identity provided in the textbook. Substitution of this identity gives the result. Thanks.