How the factor of $1/2$ is purely conventional?

Question

From Zee's book on group theory, he mentioned that factor $1/2$ is conventional due to historical reasons, but I thought that it was risen to match the Lie algebra of $SO(3)$?

For $N=2$, the hermiticity condition $\begin{pmatrix}u & w \\ z & v\end{pmatrix}^\dagger = \begin{pmatrix}u^* & z^* \\ w^* & v^*\end{pmatrix} = \begin{pmatrix}u & w \\ z & v\end{pmatrix}$ implies that $u$ and $v$ are real and $w = z^*$, while the traceless condition gives $v = - u$. Thus, in general, $$H = \begin{pmatrix}u & z^* \\ z & -u\end{pmatrix} = \frac{1}{2} \begin{pmatrix}\theta_3 & \theta_1 -i \theta_2 \\ \theta_1 + i \theta_2 & - \theta_3\end{pmatrix} \tag{17}$$ where $\theta_1$, $\theta_2$, and $\theta_3$ denote three arbitrary real numbers. (The factor of $\frac{1}{2}$ is conventional, due partly to historical reasons. I explain the reason for its inclusion in chapter IV.5.)

It is standard to define the three traceless hermitean matrices known as Pauli matrices, as $$\sigma_1 = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}, \quad \sigma_2 = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}, \quad \sigma_3 = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} \tag{18}$$ The most general $2$-by-$2$ traceless hermitean matrix $H$ can then be written as a linear combination of the three Pauli matrices: $H = \frac{1}{2}(\theta_1 \sigma_1 + \theta_2 \sigma_2 + \theta_3 \sigma_3) = \sum_{a=1}^{3} \frac{1}{2} \theta_a \sigma_a$. An element of $SU(2)$ can then be written as $U = e^{i \theta_a \sigma_a/2}$ (with the repeated index summation convention).

Answer 1

Lie Algebras are vector spaces, and the commutation relations in then are simply relations between elements of a basis of the vector space. Both in $SU(2)$, $SO(3)$, and in any other group, the normalization chosen is a matter of convention. Notice that if we pick the $SO(3)$ algebra $$[J_i, J_j] = i \epsilon_{ijk} J_k,$$ we could perfectly well define $K_i = \alpha J_i$ and get $$[K_i, K_j] = i \alpha \epsilon_{ijk} K_k,$$ and now describe the algebra in terms of the generators $K_i$. They are just a different basis of the same vector space.

Hence, there is no problem with matching the Lie algebra of $SO(3)$, as that is also defined by convention.

As an example of how the choice of $\alpha$ is merely a convention, notice that, for $\alpha=1$, one has that $-i[K_i,K_j]$ is an element of the Lie algebra. This is the convention used in Physics. It is then interesting to notice that in Mathematics one has that if $X$ and $Y$ are elements of the Lie algebra, $[X,Y]$ is in the Lie algebra instead of $-i[X,Y]$ (see Theorem 3.20 in Brian C. Hall's Lie Groups, Lie Algebras, and Representations). This is because physicists write the group elements as $e^{i t X}$, so the generator $X$ is Hermitian for a unitary group. Mathematicians prefer to write simply $e^{t X}$. Hence, mathematicians use $\alpha = i$. The definitions of Lie algebra are then a little different (and I'm sort of cheating by considering imaginary $\alpha$, since it flips between the definitions), but this shows how arbitrary the convention on even the definition of the Lie algebra is.