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For the Hamiltonian $H=(\vec{\alpha}\cdot \vec{p}+\vec{\beta}m)$ of the Dirac equation $i\frac{\partial \psi}{\partial t}=H\psi$, it can be shown that $[H,\vec{L}]=-i(\vec{\alpha}\times\vec{p})$. Now, if we define $\vec{S}=\frac{1}{2}\vec{\Sigma}=\frac{1}{2}\begin{pmatrix}\vec{\sigma} & 0\\0 & \vec{\sigma}\end{pmatrix}$, it is clear that $\vec{L}+\vec{S}$ is conserved. So we choose a representation of $\vec{S}$.

Can we derive $\vec{S}$ in a representation independent way? Or this is not a representation at all like those of $\vec{\alpha}$ and $\vec{\beta}$?

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  • $\begingroup$ I do not understand this question. Dirac fermions transform in a very specific spin representation (labeled by (1/2, 1/2)), by their very definition. What do you mean by a "representation-independent" way here? What is your "representation independent" definition of $\vec \Sigma$? $\endgroup$ – ACuriousMind Sep 16 at 19:02
  • $\begingroup$ @ACuriousMind I assume he means independent of the particular matrix representation of the Clifford algebra $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2\eta^{\mu\nu}$ $\endgroup$ – Bence Racskó Sep 16 at 19:09
  • $\begingroup$ @ACuriousMind $\vec{\alpha}$ and $\vec{\beta}$ are representation dependent. There are two representations that I know. The standard Dirac representation and Chiral or Weyl representation in which the $4\times 4$ matrices $\vec{\alpha}$ and $\vec{\beta}$ look different. $\endgroup$ – mithusengupta123 Sep 16 at 19:09
  • $\begingroup$ @ACuriousMind I think, it would be better to say that the $\alpha$ and $\beta$ matrices have two different matrix representation than saying that they are representation dependent. $\endgroup$ – mithusengupta123 Sep 16 at 19:13
  • $\begingroup$ Alright, "independent of the representation as matrices" would probably be unambiguous then. So, what is $\Sigma$ (define your notation!)? $\endgroup$ – ACuriousMind Sep 16 at 19:22

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