I know that $[H_{D},J]$ = 0 where $H_D$ is the Dirac Hamiltonian and $J = L + S$ where $L = x \times p$ and $S = (1/2)\Sigma$ with $\Sigma = \text{diag}(\sigma,\sigma)$ with $\sigma$ the Pauli matrices. However I wanted to understand better why it is the total angular momentum that is conserved as opposed to either the orbital or the spin. I have a basic background in the Lorentz group as a Lie group, its Lie algebra, the double cover of SL(2,C) of the Lorentz group etc. From the Lie algebra, I know there are 3 generators of rotations. Take $$J_3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ for example.

  1. Is it then true that if I calculated the third component of J = L + S, I would get back this matrix. I tried to do this myself but I got stuck on how to represent the position and momentum operators as matrices in order to combine them with the $\Sigma$ matrix.

  2. Furthermore, assuming that I could do such a thing, is finding that J is the generator of rotations enough to conclude that it must commute with the Hamiltonian?

  3. Is there some other group theoretic way to see why $[H_{D} ,J] = 0$ as opposed to $L$ or $S$ must be true if at all.


To fix the notation, I will write the Hamiltonian as $H_D=\int d^d x \,\psi^{\dagger}(x) \gamma^{0} ( \gamma^i (-i \partial_{x^i})+ m )\psi(x)$, where $m$ is the numerical value for mass and $\gamma_{\mu}$ satisfies Clifford algebra. Please note that although the number of components is 4 (in (3+1)-D), the true size of the first-quantized (one particle) Hamiltonian is actually infinite dimensional, due to the continous space $x_i$.

In this representation, $J_i=L_i+S_i$, $S_{i}=\epsilon_{ijk}\frac{i}{4} [\gamma_i,\gamma_j]$ and $L_i=\epsilon_{ijk}x_j (-i \partial_k)$ (note the second quantized generators are given by $\hat{J}=\int d^d x \, \psi^{\dagger} J \psi$ e.t.c.). One can straightforwardly do the transformation and show the Hamiltonian is invariant. Algebraically it's easy to see why $L_i$ itself is not a symmetry, because for a passive transformation $\mathcal{R}=e^{i \theta \hat{n} \cdot \hat{L}}$ it will give

$H_D=\int d^d x \,\psi^{\dagger}(x) \gamma^{0} ( \gamma^i (-i \partial_{(\mathcal{R}x)^i})+ m )\psi(x)$

which is apparently not equal to the original Hamiltonian: one also need to rotate the spinor part (the space of $\gamma$ matrices).

Another way to "understand" why L and S shouldn't be symmetry separately: Dirac equation is an irreducible spinor representation to accommodate Lorentz symmetry. If L and S are symmetry generators by themselves, it means that the "spin" indices decoupled from the spatial part, they have tensor product structure, and it's not an irreducible representation. This is true for the rotational symmetry of spinful non-relativistic Schrodinger field

$H_S=\int d^d x \,\psi_{\sigma}^{\dagger}(x) ( \frac{1}{2m} (-i \partial_{x^i})^2+ V(x) )\psi_{\sigma}(x)$

where the spinor part $\sigma$ totally decoupled from orbital part $x_i$. In retrospect, the form $\gamma^i (-i \partial_{x^i})$ told us spin and orbital are coupled together already.

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