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It is usually said that Dirac got his equation by looking for the square root of the 4-momentum norm (see Dirac’s coop here). The relativistic 4-momentum norm is

$$(E)^2-(\mathbf{p}c)^2=(mc^2)^2 \tag{1}$$

A quick look at this equation, one immediately spots that the left-hand side is similar to the square of a binomial, so, the simplest choice for the square root of this equation is

$$aE+\mathbf{b \cdot p}c=mc^2 \tag{3}$$

As long as the operators $a$ and $\mathbf{b}$ obeys the following rules: (a) $a^2=1, \, b_i^2=-1$ ($i=1,2,3$ is 3D-space index), (b) they all anti-commute and (c) are constant. Replacing $E$ and $\mathbf{p}$ in eq. (4) by their respective operators, one immediately gets the Dirac equation

$$i \hbar a ^ \mu \partial _\mu \Psi - mc^2 \Psi =0. \tag{4}$$

Where we changed the notation for $a$ and $\mathbf{b}$ by the new operators $a ^\mu$ with $\mu=0,1,2,3$ a spacetime index. Using the new notation, the rules above can be summarized by

$$\{a^\mu,a^\nu \}=2\eta_{\mu \nu} \mathbf{1} \tag{2}$$

To my knowledge it’s because of the Dirac's choice of these operators (he choose the $4 \times 4$ Dirac-matrices) that equation (4) is deemed to be valid only for spin-$\frac{1}{2}$ particles. However, from a deeper perspective $a^\mu$ (a.k.a $\gamma ^ \mu$) are just any set of objects as long as they obey conditions (2) which do not include any restrictions on form, nature or dimensions on them, but just demand that cross-terms of eq. (3) to vanish and squared term of $\mathbf{p}$ to be negative.

So, why do we say that Dirac equation is only valid for spin-$\frac{1}{2}$ particles if one could plug a different set of operators (maybe matrix representations of Clifford algebras) representing another spin?. Are the Dirac’s $\gamma$-matrices the only set of operators that when inserted into equation (3) leads to equation (1)?

physicsrev
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2 Answers2

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The way I like to think of spin is not in terms of angular momentum but in terms of representations of the Lorentz group. It can be shown fairly easily, that:

Irreducle finite-dimensional projective representations $\mathcal{D}^{j_1, j_2}$, of $L^\uparrow_+$ are uniquely determined by a pair of half-integer numbers $j_1, j_2 = 0, 1/2, 1, 3/2, \ldots $.

That sounds quite complicated, but it just means that the spin of a particle specifies the representation of the Lorentz transformation it transforms under. For example:

  1. $(0,0)$ are scalar particles,
  2. $(1/2,0)$ and $(0, 1/2)$ are left and right-handed chiral Weyl spinors,
  3. $(1/2,0) \oplus (0, 1/2)$ are Dirac bi-spinors,
  4. $(1/2,1/2)$ are vector bosons.

The equations of motion then follow from covariance (at least for spin-1/2 particles). So I suggest you don't start from the Dirac equation and wonder what kind of fields the equation applies for, but instead, you start from the different representations of the Lorentz group. Then it is obvious that the Dirac equation can only be applied for Dirac bi-spinors because particles of different spin transform under a different representation and thus satisfy a different equation of motion.

As a comment: there are of course also different representations of the $\gamma$-matrices. They are only unique up to unitary transformations, but all of them apply to spin-1/2 particles.

tomtom1-4
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Taff one! Ok, so the answers rightfully pointed out that there are fields of different spin that also satisfy the Dirac equation. But I don't think they can be fully characterized by the Dirac equation, much like any field $\psi$ that satisfies the Dirac-equation also satisfies the Klein-Gordan-equation, yet we still need the Dirac equation. The underlying principle is that particles transform under irreducible representations of the Lorentz group. For instance, one answer suggested taking just two spin-1/2 fields and stacking them on top of each other with block-diagonal $\gamma$-matrices. Of course, this field will satisfy the Dirac-equation but in this form, it is not an irreducible representation of the Lorentz group. And I could be wrong here but I think after decomposing it into irreducible subspaces it no longer satisfies the Dirac equation. I am saying this because it looks pretty similar to Dirac particles: In principle, they transform according to $(1/2,0) \oplus (0,1/2)$, yet their left- and right-handed component mix and they don't satisfy the Weyl equation.

But don't trust me on this one!!! I am not sure about any of this and I am just giving my thoughts here in case no one else answers.

tomtom1-4
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