Show that [A, F(B)] = [A, B]F'(B)

[Jon.G]

Homework Statement

A and B commute with their commutator, C=[A,B]
Show that [A, F(B)] = [A, B]F'(B)
F(B) = ∑_n=0^∞f_nBⁿ

Homework Equations

[A,B] = AB - BA
[A,BC] = [A,B]C + B[A,C]

The Attempt at a Solution

So all I can think to do is:
[A,Bⁿ] = [A,BB^n-1] = [A,B]B^n-1 + B[A,B^n-1]
=ABB^n-1-BAB^n-1+BAB^n-1-BB^n-1A
=ABB^n-1 - BB^n-1A
and this seems to be a dead end? Just loops back around to ABⁿ - BⁿA
If this is along the right track, whereabouts would I go next? If this is not the way to start it, how would I go about beginning this problem?
And even if this is along the right lines, I've ignored the sum here as I'm not quite sure how that affects the whole thing.

Thanks for your time

 

[Dick]

Homework Statement

A and B commute with their commutator, C=[A,B]
Show that [A, F(B)] = [A, B]F'(B)
F(B) = ∑_n=0^∞f_nBⁿ

Homework Equations

[A,B] = AB - BA
[A,BC] = [A,B]C + B[A,C]

The Attempt at a Solution

So all I can think to do is:
[A,Bⁿ] = [A,BB^n-1] = [A,B]B^n-1 + B[A,B^n-1]
=ABB^n-1-BAB^n-1+BAB^n-1-BB^n-1A
=ABB^n-1 - BB^n-1A
and this seems to be a dead end? Just loops back around to ABⁿ - BⁿA
If this is along the right track, whereabouts would I go next? If this is not the way to start it, how would I go about beginning this problem?
And even if this is along the right lines, I've ignored the sum here as I'm not quite sure how that affects the whole thing.

Thanks for your time

Start by showing the case where f(B)=B^2. You should get [A,f(B)]=[A,B]2B. Which is [A,B]f'(B). Then try f(B)=B^3. For the general case use induction to prove if f(B)=B^n, then [A,B^n]=[A,B]nB^(n-1). Which is [A,B]f'(B). Now deal with the sum.

 

[Jon.G]

Ooh ok thanks.
So I went through it with n=2,3 and 4 and [A, F(B)] = [A, B]F'(B) was shown for each of them.
For the sum is it simply that when n is any value, f_n is a constant so that
(for example)
[A,f₂B²] = [A, f₂B]B + f₂B[A, B]
f₂ a constant and can be taken outside of [] as it effect both sides the same so (and as B commutes with [A, B])
[A, f₂B]B + f₂B[A, B] = [A,B]f₂B + [A,B]f₂B
=[A,B]2f₂B = [A, B]F'(B)
?
Is this how it's shown?

 

[Dick]

Ooh ok thanks.
So I went through it with n=2,3 and 4 and [A, F(B)] = [A, B]F'(B) was shown for each of them.
For the sum is it simply that when n is any value, f_n is a constant so that
(for example)
[A,f₂B²] = [A, f₂B]B + f₂B[A, B]
f₂ a constant and can be taken outside of [] as it effect both sides the same so (and as B commutes with [A, B])
[A, f₂B]B + f₂B[A, B] = [A,B]f₂B + [A,B]f₂B
=[A,B]2f₂B = [A, B]F'(B)
?
Is this how it's shown?

Yes, the Lie bracket is linear over scalars, so $[A,cf(B)+dg(B)]=c[A,f(B)]+d[A,g(B)]$. That's how you take care of the sum. It's good that you've shown it works for n=2,3 and 4. To show it's true for general $n$ that $[A,B^n]=[A,B]nB^{n-1}$ there's a general procedure called mathematical induction. You assume it's true for $n$ and then show that it must be true for $n+1$. Since you've already shown that it works for small values of n, that shows it works for all values of n. A less formal argument is to show, for example, that $[A,B^3]=[A,BBB]=[A,B]BB+B[A,B]B+BB[A,B]$ and then say this pattern holds for all n.