Lie algebra $su(2)$ consists of the $2\times 2$ skew-hermitian complex matrices with addition and multiplication by real numbers as vector space operations and commutator as Lie bracket. The i-multiples of the Pauli matrices form a basis in $su(2)$ as a real vector space. The Pauli matrices are the following.
$$\begin{align}
\sigma_1 = \sigma_\mathrm{x} &=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix} \\
\sigma_2 = \sigma_\mathrm{y} &=
\begin{pmatrix}
0& -i \\
i&0
\end{pmatrix} \\
\sigma_3 = \sigma_\mathrm{z} &=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix} \\
\end{align}$$
The i-multiple of each of the Pauli matrices (that is, $i\sigma_1,i\sigma_2$ and $i\sigma_3$) is both skew-hermitian and unitary, however, their real linear combinations are generally only skew-hermitian, but not necessarily unitary. As far as I know, $\mathrm{span}({I})\oplus su(2)$ is isomorphic with the algebra of quaternions (where $I$ is the $2\times 2$ identity matrix), and the group formed by the unit quaternions is isomorphic to $SU(2)$. So, the imaginary unit quaternions should be the unit-norm linear combinations of the i-multiples of the Pauli matrices. By "unit-norm linear combination" I mean a linear combination whose squared coefficients sum to 1. Is it really true? Is the unitary subset of $su(2)$ exactly this? How can this subset be determined without using the Pauli-matrices?
Best Answer
Denote $\vec{u}=(u_1,u_2,u_3)$, $\vec{v}=(v_1,v_2,v_3)$ with $u_*, v_*\in \Bbb{R}$. Also let $\vec{\sigma} = (\sigma_1, \sigma_2, \sigma_3)$. Then $$ \begin{align} (\vec{u}\cdot\vec{\sigma})(\vec{v}\cdot\vec{\sigma}) &= \sum_{i, j=1}^3 u_i v_j \sigma_i \sigma_j\\ &=\sum_{i, j, k=1}^3 u_i v_j \sqrt{-1}\, \varepsilon_{ijk}\,\sigma_k +\sum_{i,j=1}^3 u_i v_j\delta_{ij}\mathbf{1}\\ &=\sqrt{-1}\sum_{k=1}^3\left(\sum_{i,j=1}^3 \varepsilon_{ijk} u_i v_j\right)\sigma_k + (\vec{u}\cdot\vec{v})\,\mathbf{1}\\ &=\sqrt{-1}\,(\vec{u}\times\vec{v})\cdot\vec{\sigma} + (\vec{u}\cdot\vec{v})\,\mathbf{1} \end{align} $$
In particular we get that $(\vec{u}\cdot\vec{\sigma})^2 = (\vec{u}\cdot\vec{u})\,\mathbf{1}$ and $(\sqrt{-1}\,\vec{u}\cdot\vec{\sigma})^2 = -(\vec{u}\cdot\vec{u})\,\mathbf{1}$.
Now, if $X\in \mathfrak{su}(2)$ then $\operatorname{trace} X=0$ and $X^\dagger = -X$. So $X$ is unitary if and only if $$\mathbf{1} = X^\dagger X = -X^2$$ or equivalently iff $X^2 = -\mathbf{1}$. Expanding $X=\sqrt{-1}\,\vec{u}\cdot\vec{\sigma}$ in a basis, we see from above that $X$ is unitary if and only if $\vec{u}\cdot\vec{u} = 1$.