The first way is the one in which we start with the commutator relations $[J_x, J_y]=ihJ_z$, etc. We consider the simultanelus eigenbasis of $J^2$ and $J_z$ : $|j,m\rangle$. We then obtain their eigenvalue sequence with raising and lowering operators, $J_+=J_x+iJ_y, J_-=J_x-iJ_y$. After we have the $J_z$, $J^2$, $J_+$ and $J_-$ matrices, we solve for $J_x$ and $J_y$ matrices.
The spin matrices are then found in the starting eigenvectors of the $J_x$, $J_y$, $J_z$ matrices, expressed in the $|j,m\rangle$ basis.
The second way is the one in which we take the square root of the Klein-Gordon equation to get the Dirac equation.
They both give the same Pauli matrices. As far as I see, the first way doesn't involve any relativity. What's the deeper connection between these two methods?
EDIT I got a hint. The Dirac equation derivation involves taking the square root of the :
$$-\frac{d^2}{dt^2}+\frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2}$$
So basically the minkowski metric signature. Lorentz transformations leave the minkowski metric unchanged. The generators of Lorentz transforms are $J_x$, $J_y$, $J_z$ and whatever $J$ is for boost.
I believe this hint gets us closer to the connection.
Best Answer
The Pauli matrices (more precisely, the first two) and the 4-dimensional $\gamma$-matrices are both manifestations of the general concept of a Clifford algebra. The Pauli matrices are a specific choice of representation for a Clifford algebra in two dimensions, the $\gamma$-matrices form a 4-dimensional Clifford algebra, and in the "chiral basis" these 4d matrices look just like a bunch of Pauli matrices inside a larger matrix.
This is to be expected because you can construct the higher-dimensional Clifford algebras from the 2-dimensional Clifford algebra (the Pauli matrices) by a bootstrapping construction, which constructs the Clifford algebra $\Gamma_i$ of dimension $2n + 2$ from the algebra $\gamma_i$ in dimension $2n$ as \begin{align} \Gamma_i & = \gamma_i \otimes \sigma_3, 0\leq i \leq 2n-1 \\ \Gamma_{2n} &= \mathbf{1}\otimes(\mathrm{i}\sigma_1) \\ \Gamma_{2n+1} &= \mathbf{1}\otimes(\mathrm{i}\sigma_2) \end{align} and if you explicitly perform this for $n=1$ you get precisely the relationship between the Pauli matrices and the $\gamma$-matrices in the chiral basis that you're asking about.
That the Pauli matrices also are the algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ is something of an accident - there just aren't all that many independent 2d complex matrices (the Pauli matrices and the identity form a basis of the 2-by-2 complex matrices), so the Clifford algebra in 2d doesn't have a lot of options except to also involve the Pauli matrices. In general, the isometry group of the metric associated with a Clifford algebra only occurs as the commutators of its generators (i.e. $[\gamma^i,\gamma^j]$ are the generators of the Lorentz algebra in the Minkowski case, but the $\gamma^i$ themselves are not).