Dirac Spinors as the Minimal Left Ideal of a Clifford Algebra

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The energy mass shell equation is

Where $E_0=m{c}^2$ is the rest energy, $E$ is the total energy and $\overrightarrow{p}$ is the momentum vector.

Replacing energy and momentum with their respective wave equation operators $E=i\hslash {\displaystyle \frac{\partial }{\partial t}}$ and $p=-i\hslash ⟨{\displaystyle \frac{\partial }{\partial x}},{\displaystyle \frac{\partial }{\partial y}},{\displaystyle \frac{\partial }{\partial z}}⟩$ gives the Klein-Gordon equation.

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

Multiplying the left side out gives the below

Assuming the derivative operators are commutative then then grouping

Assuming the derivatives are independent the only solution can be that

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

and quadratic form

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

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

Whatever the elements of the field $\psi (t,x,y,z)$ 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 $S\subset Cl(V,Q)$ where $\forall (a\in Cl(V,Q)).\forall (b\in S).ab\in S$ it says in the subset when left multiplied by anything.

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

and

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

Any element $x$ can be written in its 16 dimensional basis:

After multiplying by $p$

Then grouping them makes an 8 dimensional basis:

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

Then substituting into the Dirac equation

with the subsituted in vectors

We can find how the basis elements ${\gamma }_t,{\gamma }_x,{\gamma }_y,{\gamma }_z$ act on elements of the minimal left ideal, or $\psi$, when left multiplied..

So the matrix representation is

Doing it for ${\gamma }_x$.

So the matrix representation is

Doing it for ${\gamma }_y$.

So the matrix representation is

Doing it for ${\gamma }_z$.

So the matrix representation is

Now we have some choices for how $i^2=-1$ can be represented in the Clifford algebra. The value $i$ can of be ${\gamma }_x{\gamma }_y,{\gamma }_x{\gamma }_z,{\gamma }_y{\gamma }_z$ as all 3 of those bivectors square to $-1$. If we arbitrarily choose $i={\gamma }_y{\gamma }_x$, and be careful to consider that $i$ isn't commutative with everything, we can find its action on the spinor.

So the mapping is

And if we change the vector representation in ${ℝ}^8$ to ${ℂ}^4$ the mapping still holds, and left multiplication with $i$ acts as the complex numbers do.

the corresponding matricies map to

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

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