# Uniqueness of Dirac matrices

I am trying to understand the motivation behind the Dirac equation for a free particle $$i\gamma^\mu\partial_\mu \psi-m\psi=0 \tag{1}$$ I am wondering how to get the concrete form of the matrices $$\gamma^\mu$$.

It is postulated that $$(1)$$ should imply the Klein-Gordon equation $$\partial^2\psi-m^2\psi=0 \tag{2}$$ That implies that $$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu},$$ where $$\eta^{\mu\nu}$$ is the metric tensor.

However the relations $$(2)$$ do not define $$\gamma^\mu$$ uniquely, only up to a conjugacy. (For example, these relations are satisfied by the standard Dirac matrices and by the Majorana matrices.)

However, if I understand correctly, Dirac was looking for an equation such that $$|\psi|^2$$ would be the time component of a conserved 4-current in order to have the standard interpretation of $$\psi$$. I am wondering what kind of restriction this imposes on $$\gamma^\mu$$. Does it exclude the Majorana matrices?

• Have you followed the actual derivation of the dirac matrices? There are further restrictions on them that were part of the original formulation of the dirac equation. Commented Sep 18, 2023 at 16:08
• What restrictions? Do you have a reference for ‘actual’ derivation? In all literature I had Dirac matrices were written down explicitly without much explanations.
– MKO
Commented Sep 18, 2023 at 17:33
• Like when the dirac equation is written with the four matrices $\beta$ and $\vec{\alpha} = (\alpha_1,\alpha_2,\alpha_3)$ so that the Dirac equation became $(\alpha\cdot p + m\beta)\Psi = i \partial_t \Psi$. Squaring this yields several conditions on all the matrices (traceless, hermitian, even dimensional, and with eigenvalues of $\pm1$). Commented Sep 18, 2023 at 18:45
• "I am wondering how to get the concrete form of the matrices..." This is presented in many books. Are you asking for a book recommendation?
– hft
Commented Sep 27, 2023 at 0:07
• @hft: In the known to me book these matrices are just written down without any derivation. I would be happy to see a reference to a derivation.
– MKO
Commented Sep 28, 2023 at 14:19

Matrices that are related by conjugacy are equivalent, as their change can be compensated for by changing $$\psi$$. Indeed such a transformation is required in establishing Lorentz invariance.

• I did not get your point. Dirac and Majorana matrices are conjugate. Nevertheless there is a difference between them. My guess is that for Majorana fermions $|\psi|^2$ is not the time component of a 4-current.
– MKO
Commented Sep 18, 2023 at 17:33
• They are just different representations. Majorana ones have the property that the Majorana condition on $\psi$ is easy to state as it is the condition that $\psi$ be real. In other reps the condition is more complicated, see my review: arxiv.org/abs/2009.00518 Commented Sep 18, 2023 at 17:36
• How the Noether current for Majorana fermions look like? Is it in your survey?
– MKO
Commented Sep 18, 2023 at 17:58
• There is no $U(1)$ for a single Majorana. You will find that if $\psi$ satisfies the Majorana condition $\psi = {\mathcal C}^{-1} \bar \psi^T$ then $\bar \psi \gamma^\mu \psi$ is identically zero for Grassman valued $\psi$'s. There is some discusion of this in my paper, but not using Noether explicitly. Commented Sep 18, 2023 at 18:04

That is Pauli's fundamental theorem. Any two sets of gamma matrices are related by a similarity transformation and the transformation matrix is unique except for a multiplicational constant. If you further postulate that both sets are unitary (or rather all gamma-matrices are unitary in both sets are unitary) which is the case for physically relevant gamma-matrices, you can even find a unitary transformation and the transformation matrix is now unique except for a phase. That preserves (by definition of a unitary matrix) the probability density $$\parallel \psi \parallel^2$$. You can find a proof for example in Messiah, qm2.
And no, that doesn't exclude Majorana-matrices, they are unitary.

• In the case of Majorana matrices is $|\psi|^2$ is still the time component of a four current?
– MKO
Commented Sep 28, 2023 at 14:20
• Yes, also the whole four current doesn't change (that follows directly from the fact that the transformation matrix is unitary) Commented Sep 28, 2023 at 23:36

OP's Eq. (1) is equivalent to Eq. 22-2 in the textbook "Intermediate Quantum Mechanics" by Bethe and Jackiw: $$\frac{1}{c}\frac{\partial \psi}{\partial t} + \vec \alpha\cdot\vec \nabla \psi + \frac{imc}{\hbar}\beta\psi=0\;,$$ i.e., the Dirac equation, where $$\vec \alpha$$ and $$\beta$$ are N-by-N matrices (which will turn out to be four-by-four matrices).

By demanding that the probability density $$\psi^\dagger \psi$$ be postive, we find Bethe's Eq. 22-7: $$\beta^\dagger = \beta\;,$$ $$\alpha^\dagger = \alpha\;.$$

Bethe goes on to show that forcing the Dirac equation to reduce to the Klein-Gordan equation for each component requires the additional properties of his Eq 22-12: $$\frac{1}{2}\left(\alpha^k\alpha^\ell + \alpha^\ell \alpha^k\right)=\delta^{kl}$$ and so on.

Bethe then defines the covariant form of the Dirac matrices: $$\beta = \gamma^0$$ $$\beta\vec \alpha = \vec \gamma\;.$$

Bethe then introduces the 16 "big gamma" matrices $$\Gamma_i$$ and develops some of their properties before returning to the "little gamma" matrices in his section titled "Explicit Form of the Gamma Matrices." In this section he shows that the matrices can be taken to have the explicit form: $$\beta = \left(\begin{matrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{matrix}\right)$$ and $$\vec \alpha = \left(\begin{matrix}0 & \vec \sigma\\ \vec \sigma & 0\\ \end{matrix}\right)$$

There are actually two separate algebras here.

The Dirac algebra is a Clifford algebra generated from four mutually anti-commuting units whose squares have the respective signs $$(+,-,-,-)$$. This is $$C_{1,3}$$.

As a real Clifford algebra, $$C_{1,3}$$ is isomorphic to the $$2×2$$ matrix algebra of quaternions. The quaternions, themselves, are a Clifford algebra as are the real numbers $$ℝ$$ and complex numbers $$ℂ$$.

As complex Clifford algebras, $$C_{0,4}$$, $$C_{1,3}$$, $$C_{2,2}$$, $$C_{3,1}$$ and $$C_{4,0}$$ are all isomorphic to the $$4×4$$ matrix algebra over $$ℂ$$. In turn, they are all isomorphic to the real Clifford algebra $$C_{4,1}$$, that has five mutually anti-commuting units, whose squares have signs $$(+,+,+,+,-)$$.

The Majorana representation pertains to the real $$C_{1,3}$$. The Dirac element $$γ^5 = i γ^0 γ^1 γ^2 γ^3$$ belongs to the complex $$C_{1,3}$$; and the complex $$C_{1,3}$$ - when thought of as a real Clifford algebra - can be obtained from the real $$C_{1,3}$$ by throwing in either $$i$$ as a separate unit, or $$γ^5$$, since it already has the $$i$$ in it. The unit $$i$$ will commute with all the generators, so the process of throwing in an $$i$$ that commutes with everyone is called "complexification". The unit $$γ^5$$ anti-commutes with $$γ^0$$, $$γ^1$$, $$γ^2$$ and $$γ^3$$, so you can consider the complex Dirac algebra as the real Clifford algebra generated by $$\left(γ^1, γ^2, γ^3, γ^4 = γ^0, γ^5\right)$$. Since $$\left(γ^5\right)^2 = 1$$, the complexified Dirac algebra is also equivalent to $$C_{2,3}$$. The name $$γ^5$$ comes from the earlier time when $$γ^0$$ was denoted $$γ^4$$.

The question you're asking is a special case of the more general question: how to find matrix representations of Clifford algebras. This is dealt with in numerous standard references on Clifford algebras; but here's a link I found with a brief search:

Matrix Representations of Clifford Algebras
https://www.cphysics.org/article/86175.pdf

that places emphasis on the matrix representations of the Dirac algebra - both the real and complex representations; and also discusses "complexification" and deals with other Clifford algebras, such as $$C_{2,0}$$ and $$C_{3,0}$$. As real Clifford algebras, the quaternions are $$C_{0,2}$$, the complex numbers $$ℂ$$ are $$C_{0,1}$$ and the real numbers $$ℝ$$ are $$C_{0,0}$$. As a complex Clifford algebra, $$ℂ$$ is just $$C_{0,0}$$ - the complexification of $$ℝ$$; and the complexification of $$C_{0,2}$$ is the algebra of Pauli matrices, which is isomorphic to the $$2×2$$ matrices over $$ℂ$$ and is the same as the real Clifford algebra $$C_{3,0}$$ (since the Pauli matrices are anti-commuting and their squares are $$(+1,+1,+1)$$).

Letting $$ℍ$$ denote the quaternions, and $$M_n(\_)$$ the $$n×n$$ matrix algebra of $$(\_)$$, the spectrum of lower-dimensional real Clifford algebras is: $$C_{0,0} ≅ ℝ, \\ C_{0,1} ≅ ℂ, \quad C_{1,0} ≅ ℝ ⊕ ℝ, \\ C_{0,2} ≅ ℍ, \quad C_{1,1} ≅ M_2 ℝ ≅ C_{2,0}, \\ C_{m,n+2} ≅ ℍ ⊗ C_{n,m}, \quad C_{m+1,n+1} ≅ M_2 C_{m,n}, \quad C_{m+2,n} ≅ M_2 C_{n,m}, \\ C_{3,0} ≅ M_2 ℂ ≅ ℍ ⊗ ℂ, \quad C_{2,2} ≅ M_4 ℝ ≅ ℍ ⊗ ℍ,$$ others you can find by iterating and applying these formulae, as well as $$M_p M_q ≅ M_{pq}$$ for matrix algebras; e.g. $$C_{1,3} ≅ M_2 C_{0,2} ≅ M_2 ℍ$$ for the real Dirac algebra. For complexified Clifford algebras $$ℂ ⊗ C_{m,n} ≅ ℂC_{m+n}$$, the spectrum is: $$ℂC_{0} ≅ ℂ, \quad ℂC_{1} ≅ ℂ ⊕ ℂ, \quad ℂC_{p+2} ≅ M_2 ℂC_{p}.$$ Thus $$ℂ ⊗ C_{1,3} ≅ ℂC_{4} ≅ M_2 ℂC_{2} ≅ M_2 M_2 ℂC_{0} ≅ M_2 M_2 ℂ ≅ M_4 ℂ.$$

The actual reason the Dirac algebra is "complexified" actually has nothing to do with complexification, itself. Instead, it's in order to get $$C_{4,1}$$ - with the fifth generator. You can see the $$(+,+,+,+,-)$$ signature plainly in the mass shell condition: $$(mc)^2 + |𝐩|^2 - \left(\frac{E}{c}\right)^2 = 0.$$ The same applies to the non-relativistic form of this, which is obtained by splitting off $$mc^2$$ from $$E$$ as $$E = H + mc^2$$ and writing: $$|𝐩|^2 - 2mH - \frac{H^2}{c^2} = 0.$$ In the limit as $$1/c^2 → 0$$, this becomes $$|𝐩|^2 - 2mH = 0.$$ These are each special cases of $$|𝐩|^2 - 2mH - αH^2 = 0,$$ with respective values $$α = 1/c^2$$ and $$α = 0$$; and the mass shell equation always reduces to one with signature $$(+,+,+,+,-)$$, independent of what $$α$$ is. (So, there's also a Dirac equation in the non-relativistic case.)

For representations that involve a positive rest mass $$m > 0$$, one uses the real $$C_{4,1}$$ and pretends that it's just $$ℂ ⊗ C_{1,3}$$ that we've been working with all along. For zero-mass representations $$m = 0$$, the mass shell is just $$E^2 - |𝐩c|^2 = 0$$, which has the $$(+,-,-,-)$$ signature of $$C_{1,3}$$. Note, by the way, that $$C_{3,1} ≅ M_2 C_{1,1} ≅ M_2 M_2 ℝ ≅ M_4 ℝ$$, which is different from $$C_{1,3}$$. So, the signs actually matter.