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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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    I am not going to reply, but I suggest Peskin's and Schroeder's book "An intro to QFT" (chap 3), in which the spin operator is presented and you can see that this operator acting on spin eigenstates yields eigenvalue $\pm\frac{1}{2}$. This should give you some sense of validation, that by writting down the Dirac equation in the current form, we describe spin 1/2 particles. You are right, though, to complain that in order to do so, we have chosen a specific representation. But for you to convince yourself that this is true for every rep, you have to learn more about the Clifford algebra reps. – schris38 Jan 28 '22 at 09:11
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    Does this answer your question? Can a wave function that represents a particle of spin that is not 1/2 be a solution of the Dirac equation? – benrg Jan 28 '22 at 10:05
  • Related: https://physics.stackexchange.com/q/53318/2451 – Qmechanic Jan 28 '22 at 13:28
  • I’m convinced it is $\frac{1}{2}$ when you plug in eq. (4) Dirac’s $\gamma$-matrices. What I want to know is If I can plug a different set of operators there, that obey conditions (2) thus keeping (3) the same, but leading into a different spin. Up to this time I have already being helped a lot, but any answer is welcomed. – physicsrev Jan 28 '22 at 14:08
  • @physicsrev : in fact the Dirac matrices are not the only possible choices. Very popular are Weyl spinors and Majorana spinors but they have the same dimension. To be precise they generate the same Clifford algebra ${\cal Cl}(1,3),.$ The reason why this Clifford algebra is so special is that it is very closely related to the Lorentz group. As we know - Dirac's achievement was to make QM compatible with SR. – Kurt G. Jan 29 '22 at 19:16

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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  • I appreciate the reference to the Lorentz group. It really helps and solves many other questions I had. As I understood, you mention that different spin demand different equations, however, in this very similar question to mine, the accepted answer seems to indicate otherwise. Could you please check if that is the case? Your comment on that will be really appreciated. – physicsrev Jan 31 '22 at 10:12
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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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