Deriving Similarity Transformations for SU(2) Algebra I'm working on a project that requires similarity transformations for certain spin operators, but the book I'm working with Mathematical Methods of Quantum Optics doesn't include all the ones I need. From (2.41) the book it has this identity:
$$\exp(-i\theta \textbf{n}\cdot\hat{\textbf{S}})\textbf{a}\cdot\hat{\textbf{S}}\exp(i\theta \textbf{n}\cdot\hat{\textbf{S}}) $$
$$= \cos(\theta)\textbf{a}\cdot\hat{\textbf{S}}\ + (\textbf{n} \times \textbf{a})\cdot\hat{\textbf{S}}\sin(\theta)+[1-\cos(\theta)](\textbf{n}\cdot\hat{\textbf{S}})(\textbf{n} \cdot \textbf{a}) $$
It then goes on to enumerate a few special cases:
$$\hat{S_{\pm}}(\theta) = \exp(-i\theta\hat{S_z})\hat{S_{\pm}}\exp(i\theta\hat{S_z}) = \exp(\mp i\theta)\hat{S_{\pm}}$$
Which is 'derived by noting $d\hat{S_{\pm}}(\theta)/d\theta = \mp i \hat{S_{\pm}}(\theta).$
or:
$$\hat{S_{z}}(\theta) = \exp(-i\theta\hat{S_+})\hat{S_z}\exp(i\theta\hat{S_+}) = \hat{S_{z}} + i\theta\hat{S_{+}}$$
which states 'This is derived by noting that $d\hat{S_{z}}(\theta)/d\theta = i \hat{S_{+}}$. Multiply [the equation] on the left by $\exp(i\theta \hat{S_+})$, differentiate it m times with respect to $\theta$ at $\theta = 0$ and show that $[\hat{S_z},\hat{S_+^m}] = m\hat{S_+^m}$.
I'd like to be able to derive the special cases:
$$\exp(-i\theta\hat{S_-})\hat{S_z}\exp(i\theta\hat{S_-})$$ and $$\exp(-i\theta\hat{S_-})\hat{S_+}\exp(i\theta\hat{S_-}) $$ which aren't present in the book, but I am having difficulty. What would these special cases be? Any assistance would be appreciated.
 A: First rewrite the product using a commutator:
$$\begin{align}
\exp(-i\theta\hat S_+) \, \hat S_z \, \exp(i\theta \hat S_+)
&= \exp(-i\theta\hat S_+) \, \left( [\hat S_z, \exp(i\theta \hat S_+)] + \exp(i\theta \hat S_+) \, \hat S_z \right)
\\
&= \exp(-i\theta\hat S_+) \, [\hat S_z, \exp(i\theta \hat S_+)] + \hat S_z
\end{align}$$
Then calculate the commutator:
$$\begin{align}
[\hat S_z, \exp(i\theta \hat S_+)]
&= [\hat S_z, \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} \hat S_+^n]
= \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} [\hat S_z, \hat S_+^n]
= \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} n \hat S_+^n
\\
&= \sum_{n=1}^{\infty} \frac{(i\theta)^n}{(n-1)!} \hat S_+^n
= \sum_{n=0}^{\infty} \frac{(i\theta)^{n+1}}{n!} \hat S_+^{n+1}
\\
&= \left(\sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} \hat S_+^n\right) ( i\theta \hat S_+)
= \exp(i\theta \hat S_+) ( i\theta \hat S_+ )
\end{align}$$
Inserting this into the first result gives
$$
\exp(-i\theta\hat S_+) \, \hat S_z \, \exp(i\theta \hat S_+)
= i\theta \hat S_+ + \hat S_z.
$$
