Derivation of Dirac Spinors as the Minimal Left Ideal of a Clifford Algebra

The energy mass shell equation is

(E0c)2=(Ec)2−∣∣p⃗∣∣2\bigg(\dfrac{E_0}{c}\bigg)^2 = \bigg(\dfrac{E}{c}\bigg)^2 - ||\vec{p}||^2

Where E0=mc2E_0 = mc^2 is the rest energy, EE is the total energy and p⃗\vec{p} is the momentum vector.

Replacing energy and momentum with their respective wave equation operators E=iℏ∂∂tE = i\hbar\dfrac{\partial}{\partial t} and p=−iℏ⟨∂∂x,∂∂y,∂∂z⟩p = -i\hbar\bigg\langle\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\bigg\rangle gives the Klein-Gordon equation.

(E0c)2=ℏ2(−1c22∂2∂t2+∂2∂x2+∂2∂y2+∂2∂z2)\bigg(\dfrac{E_0}{c}\bigg)^2 = \hbar^2\bigg(-\dfrac{1}{c^2}^2 \dfrac{\partial^2}{\partial t^2} + \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2}\bigg)
−1ℏ2(E0c)2=(1c22∂2∂t2−∂2∂x2−∂2∂y2−∂2∂z2)-\dfrac{1}{\hbar^2}\bigg(\dfrac{E_0}{c}\bigg)^2 = \bigg(\dfrac{1}{c^2}^2 \dfrac{\partial^2}{\partial t^2} - \dfrac{\partial^2}{\partial x^2} - \dfrac{\partial^2}{\partial y^2} - \dfrac{\partial^2}{\partial z^2}\bigg)

Assume there exists some expression that squares to the operators on the right.

(γt1c∂∂t+γx∂∂x+γy∂∂y+γz∂∂z)2=1c22∂2∂t2−∂2∂x2−∂2∂y2−∂2∂z2\bigg(\gamma_t \dfrac{1}{c} \dfrac{\partial}{\partial t} + \gamma_x \dfrac{\partial}{\partial x} + \gamma_y \dfrac{\partial}{\partial y} + \gamma_z \dfrac{\partial}{\partial z} \bigg)^2 = \dfrac{1}{c^2}^2 \dfrac{\partial^2}{\partial t^2} - \dfrac{\partial^2}{\partial x^2} - \dfrac{\partial^2}{\partial y^2} - \dfrac{\partial^2}{\partial z^2}

Multiplying the left side out gives the below

γt21c2∂2∂t2+γtγx1c∂2∂t∂x+γtγy1c∂2∂t∂y+γtγz1c∂2∂t∂z\gamma_t^2 \dfrac{1}{c^2} \dfrac{\partial^2}{\partial t^2}+ \gamma_t \gamma_x \dfrac{1}{c} \dfrac{\partial^2}{\partial t \partial x}+ \gamma_t \gamma_y \dfrac{1}{c} \dfrac{\partial^2}{\partial t \partial y}+ \gamma_t \gamma_z \dfrac{1}{c} \dfrac{\partial^2}{\partial t \partial z}
+γxγt1c∂2∂x∂t+γx2∂2∂x2+γxγy∂2∂x∂y+γxγz∂2∂x∂z+ \gamma_x \gamma_t \dfrac{1}{c} \dfrac{\partial^2}{\partial x \partial t}+ \gamma_x^2 \dfrac{\partial^2}{\partial x^2}+ \gamma_x \gamma_y \dfrac{\partial^2}{\partial x \partial y}+ \gamma_x \gamma_z \dfrac{\partial^2}{\partial x \partial z}
+γyγt1c∂2∂y∂t+γyγx∂2∂y∂x+γy2∂2∂y2+γyγz∂2∂y∂z+ \gamma_y \gamma_t \dfrac{1}{c} \dfrac{\partial^2}{\partial y \partial t}+ \gamma_y \gamma_x \dfrac{\partial^2}{\partial y \partial x}+ \gamma_y^2 \dfrac{\partial^2}{\partial y^2}+ \gamma_y \gamma_z \dfrac{\partial^2}{\partial y \partial z}
+γzγt1c∂2∂z∂t+γzγx∂2∂z∂x+γzγy∂2∂z∂y+γz2∂2∂z2+ \gamma_z \gamma_t \dfrac{1}{c} \dfrac{\partial^2}{\partial z \partial t}+ \gamma_z \gamma_x \dfrac{\partial^2}{\partial z \partial x}+ \gamma_z \gamma_y \dfrac{\partial^2}{\partial z \partial y}+ \gamma_z^2 \dfrac{\partial^2}{\partial z^2}
=1c22∂2∂t2−∂2∂x2−∂2∂y2−∂2∂z2= \dfrac{1}{c^2}^2 \dfrac{\partial^2}{\partial t^2} - \dfrac{\partial^2}{\partial x^2} - \dfrac{\partial^2}{\partial y^2} - \dfrac{\partial^2}{\partial z^2}

Assuming the derivative operators are commutative then then grouping

γt21c2∂∂t2+γx2∂∂x2+γy2∂∂y2+γz2∂∂z2\gamma_t^2 \dfrac{1}{c^2} \dfrac{\partial}{\partial t^2}+ \gamma_x^2 \dfrac{\partial}{\partial x^2}+ \gamma_y^2 \dfrac{\partial}{\partial y^2}+ \gamma_z^2 \dfrac{\partial}{\partial z^2}
+(γtγx+γxγt)1c∂∂t∂x+ (\gamma_t \gamma_x + \gamma_x \gamma_t) \dfrac{1}{c} \dfrac{\partial}{\partial t \partial x}
+(γtγy+γyγt)1c∂∂t∂y+ (\gamma_t \gamma_y + \gamma_y \gamma_t) \dfrac{1}{c} \dfrac{\partial}{\partial t \partial y}
+(γtγx+γzγt)1c∂∂t∂z+ (\gamma_t \gamma_x + \gamma_z \gamma_t) \dfrac{1}{c} \dfrac{\partial}{\partial t \partial z}
+(γxγy+γyγx)∂∂x∂y+ (\gamma_x \gamma_y + \gamma_y \gamma_x) \dfrac{\partial}{\partial x \partial y}
+(γxγz+γzγx)∂∂x∂z+ (\gamma_x \gamma_z + \gamma_z \gamma_x) \dfrac{\partial}{\partial x \partial z}
+(γyγz+γzγy)∂∂y∂z+ (\gamma_y \gamma_z + \gamma_z \gamma_y) \dfrac{\partial}{\partial y \partial z}
=1c22∂2∂t2−∂2∂x2−∂2∂y2−∂2∂z2= \dfrac{1}{c^2}^2 \dfrac{\partial^2}{\partial t^2} - \dfrac{\partial^2}{\partial x^2} - \dfrac{\partial^2}{\partial y^2} - \dfrac{\partial^2}{\partial z^2}

Assuming the derivatives are independent the only solution can be that

γt2=1\gamma_t^2 = 1
γx2=−1\gamma_x^2 = -1
γy2=−1\gamma_y^2 = -1
γz2=−1\gamma_z^2 = -1
γtγx+γxγt=0\gamma_t \gamma_x + \gamma_x \gamma_t = 0
γtγy+γyγt=0\gamma_t \gamma_y + \gamma_y \gamma_t = 0
γtγx+γzγt=0\gamma_t \gamma_x + \gamma_z \gamma_t = 0
γxγy+γyγx=0\gamma_x \gamma_y + \gamma_y \gamma_x = 0
γxγz+γzγx=0\gamma_x \gamma_z + \gamma_z \gamma_x = 0
γyγz+γzγy=0\gamma_y \gamma_z + \gamma_z \gamma_y = 0

These just so happen to be basis elements of a Clifford algebra Cl(V,Q)\text{Cl}(V, Q) over a vector space with basis elements

tγt+xγx+yγy+zγz∈V=R4t\gamma_t + x\gamma_x + y\gamma_y + z\gamma_z \in V = \mathbb{R}^4

and quadratic form

Q(tγt+xγx+yγy+zγz)=t2−x2−y2−z2Q(t\gamma_t + x\gamma_x + y\gamma_y + z\gamma_z) = t^2 - x^2 - y^2 - z^2

By "square rooting" both sides of the Klein-Gordon equation the resulting equation, the Dirac equation, can then be written as

±iℏE0c=γt1c∂∂t+γx∂∂x+γy∂∂y+γz∂∂z\pm\dfrac{i}{\hbar}\dfrac{E_0}{c} = \gamma_t \dfrac{1}{c} \dfrac{\partial}{\partial t} + \gamma_x \dfrac{\partial}{\partial x} + \gamma_y \dfrac{\partial}{\partial y} + \gamma_z \dfrac{\partial}{\partial z}

(The choice of plus minus will just switch matter and antimatter.) After applying the operator to the wave field ψ(t,x,y,z)\psi(t, x, y, z) we get

±iℏE0cψ(t,x,y,z)=γt1c∂ψ(t,x,y,z)∂t+γx∂ψ(t,x,y,z)∂x+γy∂ψ(t,x,y,z)∂y+γz∂ψ(t,x,y,z)∂z\pm\dfrac{i}{\hbar}\dfrac{E_0}{c}\psi(t, x, y, z) = \gamma_t \dfrac{1}{c} \dfrac{\partial\psi(t, x, y, z)}{\partial t} + \gamma_x \dfrac{\partial\psi(t, x, y, z)}{\partial x} + \gamma_y \dfrac{\partial\psi(t, x, y, z)}{\partial y} + \gamma_z \dfrac{\partial\psi(t, x, y, z)}{\partial z}

Whatever the elements of the field ψ(t,x,y,z)\psi(t, x, y, z) are multiplied on the left by the basis elements of the Clifford algebra they stay in the same group. This behaviour is a left ideal over the Clifford algebra as a ring. A left ideal is a subset S⊂Cl(V,Q)S \subset Cl(V, Q) where ∀(a∈Cl(V,Q)).∀(b∈S).ab∈S\forall (a \in Cl(V, Q)) . \forall (b \in S) . ab \in S it says in the subset when left multiplied by anything.

We can construct a minimal left ideal from any primitive idempotent p=p2p = p^2. The choice of idempotent doesn't matter as all minimal left ideals are isomorphic. We will arbitrarily choose

p=12(1+γt)p = \frac{1}{2}(1 + \gamma_t)
and
p2=14(1+2γt+γt2)=14(2+2γt)=12(1+γt)=pp^2 = \frac{1}{4}(1 + 2\gamma_t + \gamma_t^2) = \frac{1}{4}(2 + 2\gamma_t) = \frac{1}{2}(1 + \gamma_t) = p

Then to generate the minimal left ideal we project all elements a∈CL(V,Q)a \in \text{CL}(V, Q) to apap

Any element xx can be written in its 16 dimensional basis:

a=a =
XX
+Xtγt+ X_t\gamma_t
+Xxγx+ X_x\gamma_x
+Xyγy+ X_y\gamma_y
+Xzγz+ X_z\gamma_z
+Xtxγtγx+ X_{tx}\gamma_t\gamma_x
+Xtyγtγy+ X_{ty}\gamma_t\gamma_y
+Xtzγtγz+ X_{tz}\gamma_t\gamma_z
+Xxyγxγy+ X_{xy}\gamma_x\gamma_y
+Xxzγxγz+ X_{xz}\gamma_x\gamma_z
+Xyzγyγz+ X_{yz}\gamma_y\gamma_z
+Xtxyγtγxγy+ X_{txy}\gamma_t\gamma_x\gamma_y
+Xtxzγtγxγz+ X_{txz}\gamma_t\gamma_x\gamma_z
+Xtyzγtγyγz+ X_{tyz}\gamma_t\gamma_y\gamma_z
+Xtyzγxγyγz+ X_{tyz}\gamma_x\gamma_y\gamma_z
+Xtxyzγtγxγyγz+ X_{txyz}\gamma_t\gamma_x\gamma_y\gamma_z

After multiplying by pp

ap=a12(1+γt)=12(ap = a\frac{1}{2}(1 + \gamma_t) = \frac{1}{2}\bigg(
X(1+γt)X(1 + \gamma_t)
+Xt(γt+1)+ X_t(\gamma_t+1)
+Xx(γx−γtγx)+ X_x(\gamma_x - \gamma_t\gamma_x)
+Xy(γy−γtγy)+ X_y(\gamma_y - \gamma_t\gamma_y)
+Xz(γz−γtγz)+ X_z(\gamma_z - \gamma_t\gamma_z)
+Xtx(γtγx−γx)+ X_{tx}(\gamma_t\gamma_x - \gamma_x)
+Xty(γtγy−γy)+ X_{ty}(\gamma_t\gamma_y - \gamma_y)
+Xtz(γtγz−γz)+ X_{tz}(\gamma_t\gamma_z - \gamma_z)
+Xxy(γxγy+γtγxγy)+ X_{xy}(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y)
+Xxz(γxγz+γtγxγz)+ X_{xz}(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z)
+Xyz(γyγz+γtγyγz)+ X_{yz}(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
+Xtxy(γtγxγy+γxγy)+ X_{txy}(\gamma_t\gamma_x\gamma_y + \gamma_x\gamma_y)
+Xtxz(γtγxγy+γxγz)+ X_{txz}(\gamma_t\gamma_x\gamma_y + \gamma_x\gamma_z)
+Xtyz(γtγyγz+γyγz)+ X_{tyz}(\gamma_t\gamma_y\gamma_z + \gamma_y\gamma_z)
+Xxyz(γxγyγz−γtγxγyγz)+ X_{xyz}(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z)
+Xtxyz(γtγxγyγz−γxγyγz))+ X_{txyz}(\gamma_t\gamma_x\gamma_y\gamma_z - \gamma_x\gamma_y\gamma_z) \bigg)

Then grouping them makes an 8 dimensional basis:

ap=12(ap = \frac{1}{2}\bigg(
(X+Xt)(1+γt)(X + X_t)(1 + \gamma_t)
+(Xx−Xtx)(γx−γtγx)+ (X_x - X_{tx})(\gamma_x - \gamma_t\gamma_x)
+(Xy−Xty)(γy−γtγy)+ (X_y - X_{ty})(\gamma_y - \gamma_t\gamma_y)
+(Xz−Xtz)(γz−γtγz)+ (X_z - X_{tz})(\gamma_z - \gamma_t\gamma_z)
+(Xxy+Xtxy)(γxγy+γtγxγy)+ (X_{xy} + X_{txy})(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y)
+(Xxz+Xtxz)(γxγz+γtγxγz)+ (X_{xz} + X_{txz})(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z)
+(Xyz+Xtyz)(γyγz+γtγyγz)+ (X_{yz} + X_{tyz})(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
+(Xxyz−Xtxyz)(γxγyγz−γtγxγyγz))+ (X_{xyz} - X_{txyz})(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z) \bigg)

Then if we represent this as a vector ⟨A,B,C,D,E,F,G,H⟩\langle A, B, C, D, E, F, G, H \rangle. These elements of a minimal left ideal are the elements of the field ψ(t,x,y,z)\psi(t, x, y, z).

ψ(t,x,y,z)=12(\psi(t, x, y, z) = \frac{1}{2}\bigg(
A(1+γt)A(1 + \gamma_t)
+B(γx−γtγx)+ B(\gamma_x - \gamma_t\gamma_x)
+C(γy−γtγy)+ C(\gamma_y - \gamma_t\gamma_y)
+D(γz−γtγz)+ D(\gamma_z - \gamma_t\gamma_z)
+E(γxγy+γtγxγy)+ E(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y)
+F(γxγz+γtγxγz)+ F(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z)
+G(γyγz+γtγyγz)+ G(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
+H(γxγyγz−γtγxγyγz))+ H(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z) \bigg)

Then substituting into the Dirac equation

±iℏE0cψ(t,x,y,z)=γt1c∂ψ(t,x,y,z)∂t+γx∂ψ(t,x,y,z)∂x+γy∂ψ(t,x,y,z)∂y+γz∂ψ(t,x,y,z)∂z\pm\dfrac{i}{\hbar}\dfrac{E_0}{c}\psi(t, x, y, z) = \gamma_t \dfrac{1}{c} \dfrac{\partial\psi(t, x, y, z)}{\partial t} + \gamma_x \dfrac{\partial\psi(t, x, y, z)}{\partial x} + \gamma_y \dfrac{\partial\psi(t, x, y, z)}{\partial y} + \gamma_z \dfrac{\partial\psi(t, x, y, z)}{\partial z}

with the subsituted in vectors

±iℏE0c[ABCDEFGH]=γt1c∂∂t[ABCDEFGH]+γx∂∂x[ABCDEFGH]+γy∂∂y[ABCDEFGH]+γz∂∂z[ABCDEFGH]\pm\dfrac{i}{\hbar}\dfrac{E_0}{c}\begin{bmatrix}A \\B \\C \\D \\E \\F \\G \\H\end{bmatrix}=\gamma_t \dfrac{1}{c} \dfrac{\partial}{\partial t}\begin{bmatrix}A \\B \\C \\D \\E \\F \\G \\H\end{bmatrix}+ \gamma_x \dfrac{\partial}{\partial x}\begin{bmatrix}A \\B \\C \\D \\E \\F \\G \\H\end{bmatrix}+ \gamma_y \dfrac{\partial}{\partial y}\begin{bmatrix}A \\B \\C \\D \\E \\F \\G \\H\end{bmatrix}+ \gamma_z \dfrac{\partial}{\partial z}\begin{bmatrix}A \\B \\C \\D \\E \\F \\G \\H\end{bmatrix}

We can find how the basis elements γt,γx,γy,γz\gamma_t, \gamma_x, \gamma_y, \gamma_z act on elements of the minimal left ideal, or ψ\psi, when left multiplied..

γt(A(1+γt)\gamma_t(A(1 + \gamma_t)
B(γx−γtγx)+C(γy−γtγy)+D(γz−γtγz)B(\gamma_x - \gamma_t\gamma_x) + C(\gamma_y - \gamma_t\gamma_y) + D(\gamma_z - \gamma_t\gamma_z)
E(γxγy+γtγxγy)+F(γxγz+γtγxγz)+G(γyγz+γtγyγz)E(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y) + F(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z) + G(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
H(γxγyγz−γtγxγyγz))H(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z))
==
A(γt+1)A(\gamma_t + 1)
B(γtγx−γx)+C(γtγy−γy)+D(γtγz−γz)B(\gamma_t\gamma_x - \gamma_x) + C(\gamma_t\gamma_y - \gamma_y) + D(\gamma_t\gamma_z - \gamma_z)
E(γtγxγy+γxγy)+F(γtγxγz+γxγz)+G(γtγyγz+γyγz)E(\gamma_t\gamma_x\gamma_y + \gamma_x\gamma_y) + F(\gamma_t\gamma_x\gamma_z + \gamma_x\gamma_z) + G(\gamma_t\gamma_y\gamma_z + \gamma_y\gamma_z)
H(γtγxγyγz−γxγyγz)H(\gamma_t\gamma_x\gamma_y\gamma_z - \gamma_x\gamma_y\gamma_z)
=[A−B−C−DEFG−H]= \begin{bmatrix}A \\-B \\-C \\-D \\E \\F \\G \\-H\end{bmatrix}

So the matrix representation is

γt↦[100000000−100000000−100000000−100000000100000000100000000100000000−1]\gamma_t \mapsto\begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & -1\end{bmatrix}

Doing it for γx\gamma_x.

γx(A(1+γt)\gamma_x(A(1 + \gamma_t)
B(γx−γtγx)+C(γy−γtγy)+D(γz−γtγz)B(\gamma_x - \gamma_t\gamma_x) + C(\gamma_y - \gamma_t\gamma_y) + D(\gamma_z - \gamma_t\gamma_z)
E(γxγy+γtγxγy)+F(γxγz+γtγxγz)+G(γyγz+γtγyγz)E(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y) + F(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z) + G(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
H(γxγyγz−γtγxγyγz))H(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z))
==
A(γx−γtγx)A(\gamma_x - \gamma_t\gamma_x)
B(−1−γt)+C(γxγy+γtγxγy)+D(γxγz+γtγxγz)B(-1 - \gamma_t) + C(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y) + D(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z)
E(−γy+γtγy)+F(−γz+γtγz)+G(γxγyγz−γtγxγyγz)E(-\gamma_y + \gamma_t\gamma_y) + F(-\gamma_z + \gamma_t\gamma_z) + G(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z)
H(−γyγz−γtγyγz)H(-\gamma_y\gamma_z - \gamma_t\gamma_y\gamma_z)
=[−BA−E−FCD−HG]= \begin{bmatrix}-B \\A \\-E\\-F\\C\\D\\-H\\G\end{bmatrix}

So the matrix representation is

γx↦[0−1000000100000000000−100000000−10000100000000100000000000−100000010]\gamma_x \mapsto\begin{bmatrix}0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\end{bmatrix}

Doing it for γy\gamma_y.

γy(A(1+γt)\gamma_y(A(1 + \gamma_t)
B(γx−γtγx)+C(γy−γtγy)+D(γz−γtγz)B(\gamma_x - \gamma_t\gamma_x) + C(\gamma_y - \gamma_t\gamma_y) + D(\gamma_z - \gamma_t\gamma_z)
E(γxγy+γtγxγy)+F(γxγz+γtγxγz)+G(γyγz+γtγyγz)E(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y) + F(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z) + G(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
H(γxγyγz−γtγxγyγz))H(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z))
==
A(γy−γtγy)A(\gamma_y - \gamma_t\gamma_y)
B(−γxγy−γtγxγy)+C(−1−γt)+D(γyγz+γtγyγz)B(-\gamma_x\gamma_y - \gamma_t\gamma_x\gamma_y) + C(-1 - \gamma_t) + D(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
E(γx−γtγx)+F(−γxγyγz+γtγxγyγz)+G(−γz+γtγz)E(\gamma_x - \gamma_t\gamma_x) + F(-\gamma_x\gamma_y\gamma_z + \gamma_t\gamma_x\gamma_y\gamma_z) + G(-\gamma_z + \gamma_t\gamma_z)
H(γxγz+γtγxγz)H(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z)
=[−CEA−G−BHD−F]= \begin{bmatrix}-C\\E\\A\\-G\\-B\\H\\D\\-F\end{bmatrix}

So the matrix representation is

γy↦[00−1000000000100010000000000000−100−1000000000000010001000000000−100]\gamma_y \mapsto\begin{bmatrix}0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\end{bmatrix}

Doing it for γz\gamma_z.

γz(A(1+γt)\gamma_z(A(1 + \gamma_t)
B(γx−γtγx)+C(γy−γtγy)+D(γz−γtγz)B(\gamma_x - \gamma_t\gamma_x) + C(\gamma_y - \gamma_t\gamma_y) + D(\gamma_z - \gamma_t\gamma_z)
E(γxγy+γtγxγy)+F(γxγz+γtγxγz)+G(γyγz+γtγyγz)E(\gamma_x\gamma_y + \gamma_t\gamma_x\gamma_y) + F(\gamma_x\gamma_z + \gamma_t\gamma_x\gamma_z) + G(\gamma_y\gamma_z + \gamma_t\gamma_y\gamma_z)
H(γxγyγz−γtγxγyγz))H(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z))
==
A(γz−γtγz)A(\gamma_z - \gamma_t\gamma_z)
B(−γxγz−γtγxγz)+C(−γyγz−γtγyγz)+D(−1−γt)B(-\gamma_x\gamma_z - \gamma_t\gamma_x\gamma_z) + C(-\gamma_y\gamma_z - \gamma_t\gamma_y\gamma_z) + D(-1 - \gamma_t)
E(γxγyγz−γtγxγyγz)+F(γx−γtγx)+G(γy−γtγy)E(\gamma_x\gamma_y\gamma_z - \gamma_t\gamma_x\gamma_y\gamma_z) + F(\gamma_x - \gamma_t\gamma_x) + G(\gamma_y - \gamma_t\gamma_y)
H(−γxγy−γtγxγy)H(-\gamma_x\gamma_y - \gamma_t\gamma_x\gamma_y)
=[−DFGA−H−B−CE]= \begin{bmatrix}-D\\F\\G\\A\\-H\\-B\\-C\\E\end{bmatrix}

So the matrix representation is

γz↦[000−100000000010000000010100000000000000−10−100000000−10000000001000]\gamma_z \mapsto\begin{bmatrix}0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\end{bmatrix}

If we then change the vector representation in R8\mathbb{R}^8 to C4\mathbb{C}^4

ψ(t,x,y,z)↦[ABCDEFGH]↦[A+iGF+iEB+iHD+iC]\psi(t, x, y, z) \mapsto \begin{bmatrix}A \\B \\C \\D \\E \\F \\G \\H\end{bmatrix}\mapsto\begin{bmatrix}A + iG \\F + iE \\B + iH \\D + iC\end{bmatrix}

the corresponding matricies map to

γt↦[1000010000−10000−1],γx↦[00−10000110000−100],γy↦[000i00−i00−i00i000],γz↦[000−100−1001001000]\gamma_t \mapsto\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1\end{bmatrix},\gamma_x \mapsto\begin{bmatrix}0 & 0 & -1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 \\0 & -1 & 0 & 0\end{bmatrix},\gamma_y \mapsto\begin{bmatrix}0 & 0 & 0 & i \\0 & 0 & -i & 0 \\0 & -i & 0 & 0 \\i & 0 & 0 & 0\end{bmatrix},\gamma_z \mapsto\begin{bmatrix}0 & 0 & 0 & -1 \\0 & 0 & -1 & 0 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0\end{bmatrix}

and these just so happen to be the Dirac matricies, showing their derivation.