How to calculate the commutation relation for generator of $SO(3)$ with finite rotation in the group? I am trying to show that $$e^{-iX_3\theta}X_+e^{iX_3\theta} = e^{-i\theta}X_+$$ where $X_1, X_2, X_3$ are the generators of $SO(3)$, which obey the usual commutation relations of their Lie algebra, and $X_+ := \frac{-1}{\sqrt{2}}(X_1+iX_2)$ is the raising operator.
I assume that the method involves calculating the commutator $[X_+,e^{iX_3\theta}]$, but I'm having a lot of trouble evaluating this. It feels vaguely familiar to something I learnt a few years ago, but if someone could show explicitly what I'm forgetting that would be much appreciated.
 A: You simply apply the standard Hadamard lemma,
$$
\operatorname{Ad}_{e^X}Y = e^{X}Y e^{-X} = e^{\operatorname{ad} _X} Y \\ =Y+\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots,
$$
where $[X_3,X_+]=X_+$, so that the l.h.side reduces to
$$
X_+  -i\theta X_+ +\frac{(-i\theta)^2}{2!} X_+ +... = e^{-i\theta} X_+.
$$
Geometrically, it is straightforward to identify this rotation around $\hat z$ as, anti-clockwise,
$$
X_1 ~~\mapsto \cos\theta ~X_1 +\sin\theta ~X_2,\\
X_2 ~~\mapsto \cos\theta ~X_2 -\sin\theta ~X_1.
$$
A: Hint: call the left-hand side $f(\theta)$ so $f(0)=X_+$. @lux notes we need only prove $f^\prime=-if$. By the product rule, this is equivalent to $[X_+,\,X_3]=-X_+$.
A: You can start with the commutator relation between $X_3$ and $X_+$:
$$[X_3,X_+]=X_+$$
Spell out the commutator, multiply by $(-i\theta)$, add $X_+$,
and factorize a little bit. You get
$$(1-i\theta X_3)X_+(1+i\theta X_3)+O(\theta^2)=(1-i\theta)X_+$$
Here we used the big-$O$ notation because for small $\theta$
the exact form of the $\theta^2$ term is not really important.
We get an equation, which is not restricted to small $\theta$,
by replacing $\theta$ by $\theta/n$
and iterating the equation $n$ times.
$$\left(1-i\frac{\theta}{n}X_3\right)^nX_+\left(1+i\frac{\theta}{n}X_3\right)^n
+O\left(\frac{\theta^2}{n^2}n\right)
=\left(1-i\frac{\theta}{n}\right)^nX_+$$
Now we can go to the limit $n\to\infty$.
The big-$O$ term becomes zero.
And for the other terms we can use the definition of the
exponential function $e^x=\lim_{n\to\infty}(1+\frac{x}{n})^n$.
We finally get:
$$e^{-i\theta X_3}X_+e^{+i\theta X_3}=e^{-i\theta}X_+$$
