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.

 

[paranormal]

I can provide some insight into the problem with non-commutative functions in quantum mechanics.

Firstly, let's define what commutativity means in this context. In quantum mechanics, operators are used to represent physical observables, such as position, momentum, and energy. These operators do not always commute with each other, meaning that the order in which they are applied can affect the outcome of a measurement. For example, the position and momentum operators do not commute, so measuring the position and then the momentum of a particle may give different results than measuring the momentum and then the position.

Now, coming to your specific problem, the non-commutativity of operators can lead to complications in calculations, as seen in the expression you have provided. The term [A,B] in your expression represents the commutator of the operators A and B. This means that when these operators are multiplied in different orders, the result is not the same. This can be seen in the example of position and momentum operators mentioned earlier.

To prove the given expression, you can use the Baker-Campbell-Hausdorff (BCH) formula, which is a method commonly used in quantum mechanics to deal with non-commutative operators. The formula states that eXeY = eZ, where Z is a function of X and Y, given by Z = X + Y + [X,Y]/2! + [X,[X,Y]]/3! + [X,[X,[X,Y]]]/4! + ... . Using this, you can show that the term dG/dλ in your expression is equal to the expansion of Z in the BCH formula, which will give you the desired result.

In conclusion, the problem with non-commutative functions in quantum mechanics is that they can complicate calculations and lead to different outcomes depending on the order of operations. However, the use of methods like the BCH formula can help in dealing with these non-commutative operators. I hope this helps in understanding the issue and solving your problem.