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 :


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.

  • $\begingroup$ It seems like you're confusing the Pauli matrices, which are 2x2 matrices that come out of your first method, and the Dirac gamma matrices, which are 4x4 matrices that come out of your second method. They are related and you are on the right track with the connection, but it helps not to confuse them. $\endgroup$ Commented Mar 15, 2022 at 12:42
  • $\begingroup$ @LukePritchett You are right. When I wrote "Pauli matrices come out in the second method", I implicitly meant that they come out inside those 4x4 matrices. $\endgroup$
    – Ryder Rude
    Commented Mar 15, 2022 at 12:44
  • $\begingroup$ Well the Dirac equation describes spin $1/2$ particles, so it doesn't seem far fetched that the Pauli matricies and spin is involved somehow? $\endgroup$
    – Wihtedeka
    Commented Mar 15, 2022 at 12:46
  • $\begingroup$ @Wihtedeka Yeah, but while we are deriving it purely theoretically (taking the square root), we're not aware that it'll end up describing spin-1/2 particles. $\endgroup$
    – Ryder Rude
    Commented Mar 15, 2022 at 12:49
  • $\begingroup$ I'm don't know the derivation that you have in mind, but the one I know asks for the operator that squares to the Klein Gordon equation essentially. There you won't find the gamma matricies but rather that the matrix valued coefficients have to be part of the Clifford Algebra and the Pauli matricies then come into to form a particular representation of that. $\endgroup$
    – Wihtedeka
    Commented Mar 15, 2022 at 12:52

1 Answer 1


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).

  • $\begingroup$ Good answer but I'll have to learn Clifford algebra before I can comprehend this.. Relativity isn't specifically involved in this, right? For instance, taking the square root of the non-relativistic free particle Hamiltonian should still land you with the similar commutation relations? It all has to do with the metric? That means spin can be naturally "derived" even non-relativistically? $\endgroup$
    – Ryder Rude
    Commented Mar 15, 2022 at 13:39
  • 1
    $\begingroup$ @RyderRude Relativity plays a role because the Clifford algebra involves a metric (the equations in my post are for the Lorentzian, i.e. relativistic case), but nothing stops you from using this for non-relativistic spinors, too. The physical relevance of relativity/QFT with respect to spin is the spin-statistic theorem, which doesn't exist in the non-relativistic case, and this is why you might see people saying that spin is a "relativistic phenomenon" or something like that. $\endgroup$
    – ACuriousMind
    Commented Mar 15, 2022 at 13:46

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