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

The energy mass shell equation is

(\frac{E_{0}}{c})^{2} = (\frac{E}{c})^{2} - | | \vec{p} | |^{2}

Where $E_{0} = m c^{2}$ is the rest energy, $E$ is the total energy and $\vec{p}$ is the momentum vector.

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

(\frac{E_{0}}{c})^{2} = ℏ^{2} (- {\frac{1}{c^{2}}}^{2} \frac{\partial^{2}}{\partial t^{2}} + \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}})

- \frac{1}{ℏ^{2}} (\frac{E_{0}}{c})^{2} = ({\frac{1}{c^{2}}}^{2} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}} - \frac{\partial^{2}}{\partial y^{2}} - \frac{\partial^{2}}{\partial z^{2}})

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

(γ_{t} \frac{1}{c} \frac{\partial}{\partial t} + γ_{x} \frac{\partial}{\partial x} + γ_{y} \frac{\partial}{\partial y} + γ_{z} \frac{\partial}{\partial z})^{2} = {\frac{1}{c^{2}}}^{2} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}} - \frac{\partial^{2}}{\partial y^{2}} - \frac{\partial^{2}}{\partial z^{2}}

Multiplying the left side out gives the below

γ_{t}^{2} \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} + γ_{t} γ_{x} \frac{1}{c} \frac{\partial^{2}}{\partial t \partial x} + γ_{t} γ_{y} \frac{1}{c} \frac{\partial^{2}}{\partial t \partial y} + γ_{t} γ_{z} \frac{1}{c} \frac{\partial^{2}}{\partial t \partial z}

+ γ_{x} γ_{t} \frac{1}{c} \frac{\partial^{2}}{\partial x \partial t} + γ_{x}^{2} \frac{\partial^{2}}{\partial x^{2}} + γ_{x} γ_{y} \frac{\partial^{2}}{\partial x \partial y} + γ_{x} γ_{z} \frac{\partial^{2}}{\partial x \partial z}

+ γ_{y} γ_{t} \frac{1}{c} \frac{\partial^{2}}{\partial y \partial t} + γ_{y} γ_{x} \frac{\partial^{2}}{\partial y \partial x} + γ_{y}^{2} \frac{\partial^{2}}{\partial y^{2}} + γ_{y} γ_{z} \frac{\partial^{2}}{\partial y \partial z}

+ γ_{z} γ_{t} \frac{1}{c} \frac{\partial^{2}}{\partial z \partial t} + γ_{z} γ_{x} \frac{\partial^{2}}{\partial z \partial x} + γ_{z} γ_{y} \frac{\partial^{2}}{\partial z \partial y} + γ_{z}^{2} \frac{\partial^{2}}{\partial z^{2}}

= {\frac{1}{c^{2}}}^{2} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}} - \frac{\partial^{2}}{\partial y^{2}} - \frac{\partial^{2}}{\partial z^{2}}

Assuming the derivative operators are commutative then then grouping

γ_{t}^{2} \frac{1}{c^{2}} \frac{\partial}{\partial t^{2}} + γ_{x}^{2} \frac{\partial}{\partial x^{2}} + γ_{y}^{2} \frac{\partial}{\partial y^{2}} + γ_{z}^{2} \frac{\partial}{\partial z^{2}}

+ (γ_{t} γ_{x} + γ_{x} γ_{t}) \frac{1}{c} \frac{\partial}{\partial t \partial x}

+ (γ_{t} γ_{y} + γ_{y} γ_{t}) \frac{1}{c} \frac{\partial}{\partial t \partial y}

+ (γ_{t} γ_{x} + γ_{z} γ_{t}) \frac{1}{c} \frac{\partial}{\partial t \partial z}

+ (γ_{x} γ_{y} + γ_{y} γ_{x}) \frac{\partial}{\partial x \partial y}

+ (γ_{x} γ_{z} + γ_{z} γ_{x}) \frac{\partial}{\partial x \partial z}

+ (γ_{y} γ_{z} + γ_{z} γ_{y}) \frac{\partial}{\partial y \partial z}

= {\frac{1}{c^{2}}}^{2} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}} - \frac{\partial^{2}}{\partial y^{2}} - \frac{\partial^{2}}{\partial z^{2}}

Assuming the derivatives are independent the only solution can be that

γ_{t}^{2} = 1

γ_{x}^{2} = - 1

γ_{y}^{2} = - 1

γ_{z}^{2} = - 1

γ_{t} γ_{x} + γ_{x} γ_{t} = 0

γ_{t} γ_{y} + γ_{y} γ_{t} = 0

γ_{t} γ_{x} + γ_{z} γ_{t} = 0

γ_{x} γ_{y} + γ_{y} γ_{x} = 0

γ_{x} γ_{z} + γ_{z} γ_{x} = 0

γ_{y} γ_{z} + γ_{z} γ_{y} = 0

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

t γ_{t} + x γ_{x} + y γ_{y} + z γ_{z} \in V = ℝ^{4}

and quadratic form

Q (t γ_{t} + x γ_{x} + y γ_{y} + z γ_{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

\pm \frac{i}{ℏ} \frac{E_{0}}{c} = γ_{t} \frac{1}{c} \frac{\partial}{\partial t} + γ_{x} \frac{\partial}{\partial x} + γ_{y} \frac{\partial}{\partial y} + γ_{z} \frac{\partial}{\partial z}

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

\pm \frac{i}{ℏ} \frac{E_{0}}{c} ψ (t, x, y, z) = γ_{t} \frac{1}{c} \frac{\partial ψ (t, x, y, z)}{\partial t} + γ_{x} \frac{\partial ψ (t, x, y, z)}{\partial x} + γ_{y} \frac{\partial ψ (t, x, y, z)}{\partial y} + γ_{z} \frac{\partial ψ (t, x, y, z)}{\partial z}

Whatever the elements of the field $ψ (t, x, y, z)$ 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 $S \subset C l (V, Q)$ where $\forall (a \in C l (V, Q)) . \forall (b \in S) . a b \in S$ 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

p = \frac{1}{2} (1 + γ_{t})

and

p^{2} = \frac{1}{4} (1 + 2 γ_{t} + γ_{t}^{2}) = \frac{1}{4} (2 + 2 γ_{t}) = \frac{1}{2} (1 + γ_{t}) = p

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

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

a =

X

+ X_{t} γ_{t}

+ X_{x} γ_{x}

+ X_{y} γ_{y}

+ X_{z} γ_{z}

+ X_{t x} γ_{t} γ_{x}

+ X_{t y} γ_{t} γ_{y}

+ X_{t z} γ_{t} γ_{z}

+ X_{x y} γ_{x} γ_{y}

+ X_{x z} γ_{x} γ_{z}

+ X_{y z} γ_{y} γ_{z}

+ X_{t x y} γ_{t} γ_{x} γ_{y}

+ X_{t x z} γ_{t} γ_{x} γ_{z}

+ X_{t y z} γ_{t} γ_{y} γ_{z}

+ X_{t y z} γ_{x} γ_{y} γ_{z}

+ X_{t x y z} γ_{t} γ_{x} γ_{y} γ_{z}

After multiplying by $p$

a p = a \frac{1}{2} (1 + γ_{t}) = \frac{1}{2} (

X (1 + γ_{t})

+ X_{t} (γ_{t} + 1)

+ X_{x} (γ_{x} - γ_{t} γ_{x})

+ X_{y} (γ_{y} - γ_{t} γ_{y})

+ X_{z} (γ_{z} - γ_{t} γ_{z})

+ X_{t x} (γ_{t} γ_{x} - γ_{x})

+ X_{t y} (γ_{t} γ_{y} - γ_{y})

+ X_{t z} (γ_{t} γ_{z} - γ_{z})

+ X_{x y} (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y})

+ X_{x z} (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z})

+ X_{y z} (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

+ X_{t x y} (γ_{t} γ_{x} γ_{y} + γ_{x} γ_{y})

+ X_{t x z} (γ_{t} γ_{x} γ_{y} + γ_{x} γ_{z})

+ X_{t y z} (γ_{t} γ_{y} γ_{z} + γ_{y} γ_{z})

+ X_{x y z} (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z})

+ X_{t x y z} (γ_{t} γ_{x} γ_{y} γ_{z} - γ_{x} γ_{y} γ_{z}))

Then grouping them makes an 8 dimensional basis:

a p = \frac{1}{2} (

(X + X_{t}) (1 + γ_{t})

+ (X_{x} - X_{t x}) (γ_{x} - γ_{t} γ_{x})

+ (X_{y} - X_{t y}) (γ_{y} - γ_{t} γ_{y})

+ (X_{z} - X_{t z}) (γ_{z} - γ_{t} γ_{z})

+ (X_{x y} + X_{t x y}) (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y})

+ (X_{x z} + X_{t x z}) (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z})

+ (X_{y z} + X_{t y z}) (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

+ (X_{x y z} - X_{t x y z}) (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}))

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 $ψ (t, x, y, z)$ .

ψ (t, x, y, z) = \frac{1}{2} (

A (1 + γ_{t})

+ B (γ_{x} - γ_{t} γ_{x})

+ C (γ_{y} - γ_{t} γ_{y})

+ D (γ_{z} - γ_{t} γ_{z})

+ E (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y})

+ F (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z})

+ G (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

+ H (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}))

Then substituting into the Dirac equation

\pm \frac{i}{ℏ} \frac{E_{0}}{c} ψ (t, x, y, z) = γ_{t} \frac{1}{c} \frac{\partial ψ (t, x, y, z)}{\partial t} + γ_{x} \frac{\partial ψ (t, x, y, z)}{\partial x} + γ_{y} \frac{\partial ψ (t, x, y, z)}{\partial y} + γ_{z} \frac{\partial ψ (t, x, y, z)}{\partial z}

with the subsituted in vectors

\pm \frac{i}{ℏ} \frac{E_{0}}{c} [\begin{array}{c} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \end{array}] = γ_{t} \frac{1}{c} \frac{\partial}{\partial t} [\begin{array}{c} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \end{array}] + γ_{x} \frac{\partial}{\partial x} [\begin{array}{c} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \end{array}] + γ_{y} \frac{\partial}{\partial y} [\begin{array}{c} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \end{array}] + γ_{z} \frac{\partial}{\partial z} [\begin{array}{c} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \end{array}]

We can find how the basis elements $γ_{t}, γ_{x}, γ_{y}, γ_{z}$ act on elements of the minimal left ideal, or $ψ$ , when left multiplied..

γ_{t} (A (1 + γ_{t})

B (γ_{x} - γ_{t} γ_{x}) + C (γ_{y} - γ_{t} γ_{y}) + D (γ_{z} - γ_{t} γ_{z})

E (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y}) + F (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z}) + G (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

H (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}))

=

A (γ_{t} + 1)

B (γ_{t} γ_{x} - γ_{x}) + C (γ_{t} γ_{y} - γ_{y}) + D (γ_{t} γ_{z} - γ_{z})

E (γ_{t} γ_{x} γ_{y} + γ_{x} γ_{y}) + F (γ_{t} γ_{x} γ_{z} + γ_{x} γ_{z}) + G (γ_{t} γ_{y} γ_{z} + γ_{y} γ_{z})

H (γ_{t} γ_{x} γ_{y} γ_{z} - γ_{x} γ_{y} γ_{z})

= [\begin{array}{c} A \\ - B \\ - C \\ - D \\ E \\ F \\ G \\ - H \end{array}]

So the matrix representation is

γ_{t} \mapsto [\begin{array}{cccccccc} 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{array}]

Doing it for $γ_{x}$ .

γ_{x} (A (1 + γ_{t})

B (γ_{x} - γ_{t} γ_{x}) + C (γ_{y} - γ_{t} γ_{y}) + D (γ_{z} - γ_{t} γ_{z})

E (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y}) + F (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z}) + G (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

H (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}))

=

A (γ_{x} - γ_{t} γ_{x})

B (- 1 - γ_{t}) + C (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y}) + D (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z})

E (- γ_{y} + γ_{t} γ_{y}) + F (- γ_{z} + γ_{t} γ_{z}) + G (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z})

H (- γ_{y} γ_{z} - γ_{t} γ_{y} γ_{z})

= [\begin{array}{c} - B \\ A \\ - E \\ - F \\ C \\ D \\ - H \\ G \end{array}]

So the matrix representation is

γ_{x} \mapsto [\begin{array}{cccccccc} 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{array}]

Doing it for $γ_{y}$ .

γ_{y} (A (1 + γ_{t})

B (γ_{x} - γ_{t} γ_{x}) + C (γ_{y} - γ_{t} γ_{y}) + D (γ_{z} - γ_{t} γ_{z})

E (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y}) + F (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z}) + G (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

H (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}))

=

A (γ_{y} - γ_{t} γ_{y})

B (- γ_{x} γ_{y} - γ_{t} γ_{x} γ_{y}) + C (- 1 - γ_{t}) + D (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

E (γ_{x} - γ_{t} γ_{x}) + F (- γ_{x} γ_{y} γ_{z} + γ_{t} γ_{x} γ_{y} γ_{z}) + G (- γ_{z} + γ_{t} γ_{z})

H (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z})

= [\begin{array}{c} - C \\ E \\ A \\ - G \\ - B \\ H \\ D \\ - F \end{array}]

So the matrix representation is

γ_{y} \mapsto [\begin{array}{cccccccc} 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{array}]

Doing it for $γ_{z}$ .

γ_{z} (A (1 + γ_{t})

B (γ_{x} - γ_{t} γ_{x}) + C (γ_{y} - γ_{t} γ_{y}) + D (γ_{z} - γ_{t} γ_{z})

E (γ_{x} γ_{y} + γ_{t} γ_{x} γ_{y}) + F (γ_{x} γ_{z} + γ_{t} γ_{x} γ_{z}) + G (γ_{y} γ_{z} + γ_{t} γ_{y} γ_{z})

H (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}))

=

A (γ_{z} - γ_{t} γ_{z})

B (- γ_{x} γ_{z} - γ_{t} γ_{x} γ_{z}) + C (- γ_{y} γ_{z} - γ_{t} γ_{y} γ_{z}) + D (- 1 - γ_{t})

E (γ_{x} γ_{y} γ_{z} - γ_{t} γ_{x} γ_{y} γ_{z}) + F (γ_{x} - γ_{t} γ_{x}) + G (γ_{y} - γ_{t} γ_{y})

H (- γ_{x} γ_{y} - γ_{t} γ_{x} γ_{y})

= [\begin{array}{c} - D \\ F \\ G \\ A \\ - H \\ - B \\ - C \\ E \end{array}]

So the matrix representation is

γ_{z} \mapsto [\begin{array}{cccccccc} 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{array}]

If we then change the vector representation in $ℝ^{8}$ to $ℂ^{4}$

ψ (t, x, y, z) \mapsto [\begin{array}{c} A \\ B \\ C \\ D \\ E \\ F \\ G \\ H \end{array}] \mapsto [\begin{array}{c} A + i G \\ F + i E \\ B + i H \\ D + i C \end{array}]

the corresponding matricies map to

γ_{t} \mapsto [\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{array}], γ_{x} \mapsto [\begin{array}{cccc} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{array}], γ_{y} \mapsto [\begin{array}{cccc} 0 & 0 & 0 & i \\ 0 & 0 & - i & 0 \\ 0 & - i & 0 & 0 \\ i & 0 & 0 & 0 \end{array}], γ_{z} \mapsto [\begin{array}{cccc} 0 & 0 & 0 & - 1 \\ 0 & 0 & - 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}]

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