Showing $[A,(A^\dagger)^n]=n(A^\dagger)^{n-1}$

Question

Given the operators $$A=\frac{1}{\sqrt{2}}\left(x+\frac{d}{dx}\right)\text{ and } A^\dagger=\frac{1}{\sqrt{2}}\left(x-\frac{d}{dx}\right)$$

and the commutator $[A,B]=AB-BA$ for Operators $A,B$, I want to show that

$$[A,(A^\dagger)^n]=n(A^\dagger)^{n-1}$$

What Ive tried so far is using induction:

$$[A,(A^\dagger)^n]=[A,(A^\dagger)^{n-1}A^\dagger]= A(A^\dagger)^{n-1}A^\dagger-(A^\dagger)^{n-1}A^\dagger A$$

Bot from here on I dont know where to go. Therefore help is very much appreciated!

Answer 1

You are on the right track, just recall the result

$$ [A, BC] = [A,B]C + B[A,C] \tag{1} $$

So that

\begin{eqnarray} [A, (A^\dagger)^n] &=& [A, (A^\dagger)(A^\dagger)^{n-1}] \\ &\stackrel{(1)}{=}& [A, A^\dagger](A^\dagger)^{n-1} + A^\dagger [A, (A^\dagger)^{n-1}] \\ &=& (A^\dagger)^{n-1} + A^\dagger\left\{(n-1)(A^\dagger)^{n-2} \right\} \\ &=& (A^\dagger)^{n-1} + (n-1)(A^\dagger)^{n-1} \\ &=& n(A^\dagger)^{n-1} \end{eqnarray}

Answer 2

This follows from the Leibniz rule. Ie. $$ [A, (A^*)^n] = \sum_{j=0}^{n-1} (A^*)^{j} [A, A^*] (A^*)^{n-1-j} $$ Then the result follows from the case $n=1$.

Answer 3

It's easy to show $[A, A^\dagger]=1$, which gives the identity $AA^\dagger = 1+ A^\dagger A$.

Calculating, we have \begin{align} [A, (A^\dagger)^n]&= A(A^\dagger)^n - (A^\dagger)^nA \\ &= (AA^\dagger)(A^\dagger)^{n-1} - (A^\dagger)^nA \\& =(1+A^\dagger A)(A^\dagger)^{n-1}- (A^\dagger)^nA \\&= (A^\dagger)^{n-1} + (A^\dagger)[A,(A^\dagger)^{n-1}]\end{align} The result then follows by induction.