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

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).

• Did you read chapter IV.5? Commented Aug 1, 2022 at 0:03
• @Andrew i did read,there is no mention of any historical reason.What was mentioned: Is that in order to show that SO(3) match the same Lie Algebra of Su(2). Commented Aug 1, 2022 at 1:42
• Commented Aug 1, 2022 at 4:24

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.
• @Houssam The historical convention is that $\alpha=1$. Someone in the past picked that as a convention, and now that's the convention everyone uses. If your question is about who the someone is, when they did it, and/or why, this is a history of science question. Commented Aug 1, 2022 at 13:05
• In physics, another common convention is $\alpha=\hbar$. When working in units where $\hbar\neq 1$, this lets you equate the Lie algebra generators with the components of the angular momentum operator. Commented Aug 1, 2022 at 23:06