# Derivative of operators/commutators

1. Oct 29, 2013

### d3nat

1. The problem statement, all variables and given/known data
Prove:

$\frac{dG}{d\lambda} = G \Big( A +B + \frac{\lambda}{1!} [A,B] +\frac{\lambda^2}{2!} [A,[A,B]]... \Big)$

2. Relevant equations

$G(\lambda) = e^{\lambda A}e^{\lambda B}$

3. The attempt at a solution

I tried taking the derivative with respect to lambda:
$\frac{dG}{d\lambda} = Ae^{\lambda A}e^{\lambda B} +e^{\lambda A}e^{\lambda B}B$

$= AG + GB$

Doesn't really seem to help me.

I'm assuming from the answer that the identity needs to be used:

$e^{\lambda A}Be^{-\lambda A} = B + \frac{\lambda}{1!} [A,B] +\frac{\lambda^2}{2!} [A,[A,B]]...$

But not sure how to piece it all together.

I'm reviewing for a test, and I'm working through some exercises.
(This is exercise 3.16 in chapter 4 of Merzbacher if anyone wants to reference it)

I think I'm just missing something really simple....

Last edited: Oct 29, 2013
2. Oct 29, 2013

### d3nat

Ah, Okay

$dGdλ = AG+GB = G (A+B) = G (A + e^{A \lambda}Be^{-A \lambda})$

Then just use the identity.

3. Oct 30, 2013

### Staff: Mentor

Typo above. I think you mean dG/dλ. I see where you get AG + GB. Is it reasonable to assume commutivity of multiplication? I'm thinking in terms of matrix multiplication, which is not generally commutative.

In any case, how did you get from G (A + B), which I assume means G(λ) * (A + B), rather than "G of A + B", to what you show on the right side of the equation above?

Also, what do the notations [A, B] and [A, [A, B]] mean?

4. Oct 30, 2013

### Dick

I can do the notation question. That's pretty standard. [A,B] is commutator of A and B. [A,B]=AB-BA.

5. Oct 30, 2013

### Staff: Mentor

Thanks, Dick!