angular momentumcommutatoroperatorsquantum mechanicsrotation
Consider for example the rotation of the $x$ component of spin by $\pi$ about the $z$ axis. This flips the spin giving $$e^{i\pi S_z}S_xe^{-i\pi S_z}=-S_x.$$
However, can this be proved directly using the commutation relation $[S_z,S_x]=iS_y$ and expanding the exponential?
As $[S_x,[S_z,S_x]]\neq 0$, I cannot use the usual formula to simplify it, but I am hoping there might be a similar formula which requires something like $[S_z,[S_x,[S_z,S_x]]]=0$.
I think the part you're missing is that $\hat{s}_{z_1}$ only acts on the first spin and has the same properties as if the second spin was not present, i.e. $$ \hat{s}_{z_1} |\uparrow s_2\rangle = \frac{1}{2} |\uparrow s_2\rangle$$ and $$ \hat{s}_{z_1} |\downarrow s_2\rangle = -\frac{1}{2} |\downarrow s_2\rangle$$ with $s_2$ being either $\uparrow$ or $\downarrow$. Analogous properties hold for $\hat{s}_{z_2}$. With this I hope you can solve the problem.
To make it even more clear, here is what is really happening behind the notation given. A state like $|s_1 s_2\rangle$ actually means the tensor product of $s_1$ with $s_2$. $$ |s_1 s_2\rangle := |s_1\rangle \otimes |s_2\rangle$$ The operators perviously only acting on one of the spaces, like $\hat{s}_{z_1}$ and $\hat{s}_{z_2}$ are transferred to this tensor space by tensoring them to a unit operator, like so: $$ \hat{s}_{z_1} \otimes \mathbb{I} \qquad \text{and} \qquad \mathbb{I} \otimes \hat{s}_{z_2}$$ The left operator acts on the first state of the tensor product, the second operator on the right state. Then you get things like $$ (\hat{s}_{z_1} \otimes \mathbb{I}) (|s_1\rangle \otimes |s_2\rangle)\\ = (\hat{s}_{z_1}|s_1\rangle) \otimes (\mathbb{I}|s_2\rangle)$$ which is a tensor product of things we already know (from the first part of the question). We define the operators in tensor space like this, to give them precisely this property of only acting on one of the states, which is reasoned in what measurements are. If we measure just the first spin, we let the second one unperturbed. What is given in your example is simply a shorthand notation for the tensor notation used here.
I am not really sure about the scope of this question and the type of answer OP is looking for but computing those higher spin matrices/representations successively is relatively straight forward:
Lets assume for now that we have understood the spin $\frac{1}{2}$ case: so we know that the spin $\frac{1}{2}$ operators are $$ \begin{align} S_x = \frac{\hbar}{2} \sigma_x= \frac{\hbar}{2}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, \quad S_y = \frac{\hbar}{2} \sigma_y= \frac{\hbar}{2}\begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0 \end{pmatrix}, \quad S_z = \frac{\hbar}{2} \sigma_z= \frac{\hbar}{2}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. \end{align} $$ We can verify that those sattisfy the defining commutator relation $\left[ S_j, S_k\right] = i \hbar \varepsilon_{jkl} S_l$ with $i,j,k\in\{x,y,z\}$. Furthermore we can compute the Casimir operator $S^2$: $$ S^2=S_x^2+S_y^2+S_z^2 =\hbar^2\begin{pmatrix} \frac{3}{4} & 0\\ 0 & \frac{3}{4} \end{pmatrix}. $$ The eigenvalues of $S^2$ characterize the present representation and are $\hbar^2\frac{3}{4}=\hbar^2s(s+1)$ with $s=\frac{1}{2}$. A spin state can be completely characterized by specifying $s$ and one additional spin projection which is conventionally chosen to be along the $z$-direction. $S_z$ has two Eigenvalues $m_\frac{1}{2}=\pm\frac{\hbar}{2}$. The simultaneous Eigenvectors of $S^2$ and $S_z$: $(1,0)^T$ and $(0,1)^T$ form the basis of spin states with well defined quantum numbers $s$ and $m_s$. Eigenvectors of the other spin operators $S_x$ and $S_y$ can be computed and expressed in this basis.
The quoted wikipedia article states that by taking Kronecker products of the spin $\frac{1}{2}$ representation with itself repeatedly, one may construct all higher irreducible representations. What this means in practice is that we can use the Kronecker products to couple spins. The important point here is that when we couple spins the naive basis for states which we get from the direct product of spin states is not an Eigenbasis of the coupled spin operators. Let me illustrate this with an example. We begin with coupling two spin $\frac{1}{2}$: $$ \tilde{S_i}=S_i\otimes \mathrm{Id}_2+\mathrm{Id}_2\otimes S_i $$ which results in $$ \tilde{S}_x=\frac{\hbar}{2} \begin{pmatrix}0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{pmatrix},\quad \tilde{S}_y=\frac{\hbar}{2\mathrm{i}} \begin{pmatrix} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & -1 & 0 \end{pmatrix},\quad \tilde{S}_z=\hbar\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}. $$ $\tilde{S}_z$ has four eigenvalues: $+1$, $0$, $0$ and $-1$. To fully characterize the coupled states we need to compute the Casimir operator $\tilde{S}^2$: $$ \tilde{S}^2 = \tilde{S}_x^2+\tilde{S}_y^2+\tilde{S}_z^2=\hbar^2 \begin{pmatrix}2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}. $$ Using the Eigenvectors of $\tilde{S}^2$ we can construct a matrix $U$ which diagonalizes $\tilde{S}^2$: $$ U= \begin{pmatrix} 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 1 & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \Rightarrow U \tilde{S}^2 U^\dagger=\hbar^2 \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix} $$ We can now classify the coupled Eigenstates of $\tilde{S}^2$ by their eigenvalues: we have three "triplet" states with $s=1$ ($s(s+1)=2$) and one "singlet" state with $s=0$. A representation is classified by the eigenvalue of $\tilde{S}^2$. Using $U$ we can split $\tilde{S}_i$ into block diagonal matrices where each block corresponds to one representation with distinct $s$: $$ U\tilde{S}_xU^\dagger=\hbar \left( \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\ \hline 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ \end{array} \right),\quad U\tilde{S}_yU^\dagger=\hbar \left( \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\\hline 0 & 0 & -\frac{i}{\sqrt{2}} & 0 \\ 0 & \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 & 0 & \frac{i}{\sqrt{2}} & 0 \\ \end{array} \right),\quad U\tilde{S}_zU^\dagger=\hbar\left( \begin{array}{c|ccc} 0 & 0 & 0 & 0 \\\hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right). $$ The $s=1$ blocks of $U\tilde{S}_i U^\dagger$ are the operators of the $s=1$ representation. Due to properties of the Kronecker product and the block diagonal form of $U\tilde{S}_i U^\dagger$ it is clear that the blocks satisfy the defining commutator relation separately thus we succeeded in constructing a $s=1$ representation by coupling two $s=\frac{1}{2}$ spins/representations and computing the Eigenvectors of the Casimir operator of the coupled spins. Higher representations can be computed by coupling more spins/representations: e.g. coupling $s=1$ with $s=\frac{1}{2}$ will yield the $s=\frac{3}{2}$ quartet (and in this case one additional $s=\frac{1}{2}$ doublet).
Best Answer
If we consider the function $$ f(\lambda) = e^{\lambda A} B e^{- \lambda A}, $$ and consider the Taylor expansion of $f(\lambda)$ around $\lambda = 0$, then the observation that $f(0) = B$, and $$ \frac{df}{d\lambda} = [A,f(\lambda)], \\ \frac{d^2f}{d\lambda^2} = \left[ A, \frac{df}{d\lambda} \right] = [A,[A,f(\lambda)]], $$ etc., we find $$ e^{\lambda A} B e^{- \lambda A} = B + \frac{\lambda}{1!} [A,B] + \frac{\lambda^2}{2!} [A,[A,B]] + \frac{\lambda^3}{3!} [A,[A,[A,B]]] + \cdots. $$
Now consider the special case where $$ [A,[A,B]] = \beta B, $$ which is true for the problem you are interested in. This results in a simplification where all terms collapse into terms proportional to either $B$ or $[A,B]$. Explicitly, $$ e^{\lambda A} B e^{- \lambda A} = B + \frac{\lambda}{1!} [A,B] + \frac{\lambda^2}{2!} \beta B + \frac{\lambda^3}{3!} \beta [A,B] + \frac{\lambda^4}{4!} \beta^2 B + \cdots \\ = B \left\{ 1 + \frac{(\lambda \sqrt{\beta})^2}{2!} + \frac{(\lambda \sqrt{\beta})^4}{4!} + \cdots \right\} + \frac{[A,B]}{\sqrt{\beta}} \left\{ \frac{\lambda \sqrt{\beta}}{1!} + \frac{(\lambda \sqrt{\beta})^3}{3!} + \cdots \right\}. $$ Then you can compare this to the Taylor series for the hyperbolic functions, obtaining $$ e^{\lambda A} B e^{- \lambda A} = B \cosh (\lambda \sqrt{\beta}) + \frac{[A,B]}{\sqrt{\beta}}\sinh (\lambda \sqrt{\beta}). $$ In your case, $A = S_z$, $B = S_x$, $\lambda = i \pi$, and $\beta = 1$. By just plugging into the above formula, you quickly find $$ e^{i \pi S_z} S_x e^{-i \pi S_z} = -S_x. $$