 Solution

Introduce a parameter $t$ to keep track of total degree. The power series for the matrix exponential givesSet $S=X+Y$ and $C=\frac{1}{2}[X,Y]$, where the commutator is $[X,Y]=XY-YX$. Expanding a second matrix exponential givesSincethe two expressions agree through total degree two. More explicitly, applying the local matrix logarithm to the first expansion gives $\log (e^{tX}{e}^{tY})=t(X+Y)+\frac{1}{2}{t}^2[X,Y]+O(t^3)$. This proves the displayed order of the Baker--Campbell--Hausdorff formula. The Lie algebra is closed under the commutator, so its displayed exponent lies in the same Lie algebra. The expansion is a formal power series identity, or a convergent identity sufficiently near the identity matrix; this argument does not assert unrestricted global convergence.

The next two terms of the Baker--Campbell--Hausdorff formula are the two cubic commutators:Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\frac{1}{24}[Y,[X,[X,Y]]]$.

For the Pauli matrices, use the Pauli matrix multiplication lawPut $N=\mathbf{n}\cdot \mathbf{\tau }$. The antisymmetric term disappears when contracted with $n_a{n}_b$, so $N^2=|\mathbf{n}{|}^{2}I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the matrix exponential into even and odd terms of its power series yieldsThe Pauli matrices are Hermitian matrices, and the components of $\mathbf{n}$ are real, so $N^{†}=N$. For real $\alpha$, the resulting matrix $U$ satisfiesAlso $\mathrm{tr}N=0$ and $N^2=I$, so its two eigenvalues are $1,-1$. The eigenvalues of $U$ are therefore $e^{i\alpha },e^{-i\alpha }$, whose product is one. Thus $U$ is a unitary matrix with determinant one: it belongs to the SU(2) group.

Multiplying the two exact Pauli matrix exponentials, and using ${\tau }_1{\tau }_2=i{\tau }_3$, givesIn particular, the ${\tau }_3$ term has a minus sign. Its quadratic Taylor expansion isTaking $X=i\alpha {\tau }_1$, $Y=i\beta {\tau }_2$, the Pauli matrix commutator identity gives $[X,Y]=-2i\alpha \beta {\tau }_3$. The quadratic exponent in the Baker--Campbell--Hausdorff formula is therefore $Z=i\alpha {\tau }_1+i\beta {\tau }_2-i\alpha \beta {\tau }_3$. In $e^Z=I+Z+\frac{1}{2}{Z}^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator ${\tau }_1{\tau }_2+{\tau }_2{\tau }_1=0$ gives $Z^2=-({\alpha }^2+{\beta }^2)I+O(3)$, reproducing the exact product's quadratic expansion.