I am reading some lecture notes that unfortunately don't seem to be available online, but that are quite close in spirit in their treatment of the Dirac equation to Sakurai's "Advanced Quantum Mechanics". At some point the $4\times4$ matrices $\Sigma_k$ are introduced as infinitesimal generators of the action of rotations on Dirac spinors; to be precise, a spatial rotation given by a vector $\theta$, whose direction is the axis and whose length the angle of rotation, acts as $\exp(-\frac i2\theta\cdot\Sigma)$. Since these $\Sigma_k$ are block-diagonal with the Pauli matrices $\sigma_k$ on the diagonal, we already see that the components of Dirac spinors pairwise transform just like spin-1/2 states under rotations.
It is clear that the $\Sigma_k$ formally behave like an angular momentum operator (commutation relations), but in these notes the $\Sigma_k$ arose purely as generators or rotations and not as observables (this led to this other question on the dual role of Hermitian operators in quantum mechanics)
It can be seen that the orbital angular momentum $L$ is not a conserved quantity, but $J = L + \frac12\Sigma$ is ($\hbar$ is set to 1). We have $\left|\frac12\Sigma\right|^2 = \frac34$, which corresponds to an (abstract) "orbital" quantum number $s = \frac12$.
According to the notes, this should be interpreted as meaning that a particle described by the Dirac equation must have spin $1/2$.
My question is how exactly this conclusion follows. I could think of three possible reasons, from which it has to follow on physical grounds:
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Is it because for physical reasons we expect the total angular momentum to be conserved, and $\frac12\Sigma$, which formally behaves like an angular momentum is exactly the missing quantity and can therefore be interpreted as an intrinsic angular momentum?
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Is it because the transformation of a Dirac spinor under rotations is similar to that of spin 1/2 particles as identified in an ad-hoc way before the Dirac equation?
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Does it have to do with the fact that $\Sigma$ is the infinitesimal generator of rotations on the finite dimensional "tensor factor" of the state space which can be though of as corresponding to "non-classical degrees of freedom"?
Is any of these facts a compelling reason to draw the conclusion that a particle described by the Dirac equation has spin 1/2? If so, why?
Best Answer
The question puts the cart before the horse. It is not that you derive that particles described by the Dirac equation have spin $\frac 1 2$. Rather, the Dirac equation is found as the equation for spin $\frac 1 2$ particles. A Dirac spinor $\psi$ is an element of the representation $(0,\frac 1 2) \oplus (\frac 1 2, 0)$ of the Lorentz group.1 In both subspaces the generators of rotations are implemented such that $J^2 = \frac 3 4$, i.e., spin $\frac 1 2$. By definition Dirac spinors have spin $\frac 1 2$.
The lowest order Lorentz invariant action we can build with a Dirac spinor is $$\mathcal L = \psi^\dagger \gamma^0 (i \partial_\mu \gamma^\mu + m)\psi$$ for some parameter $m$ with the dimension of mass, which clearly has as field equations the Dirac equation.
1 To understand what this means with all the gory details see How do I construct the $SU(2)$ representation of the Lorentz Group using $SU(2)×SU(2)∼SO(3,1)$? For a less mathematical treatment see Weinberg, vol I, Chapter 2, and also Chapter 5.