This is a response to a question regarding the charge of the neutrino in the Geometric Algebra construction of the
Standard Model using C l ( 4 ) ⊗ C Cl(4) \otimes \mathbb{C} Cl ( 4 ) ⊗ C here .
In that model, the neutrino portion of the Clifford object ψ \psi ψ
ψ ν ≡ ( ν 2 + ν 3 q 0 † q 1 † + ν 0 q 0 † q 2 † + ν 1 q 1 † q 2 † ) V 1 + ( ν ^ 2 − ν ^ 3 q 0 q 1 † − ν ^ 0 q 0 q 2 † + ν ^ 1 q 1 † q 2 † ) V 2 + ( ν 2 ∗ − ν 3 ∗ q 0 † q 1 − ν 0 ∗ q 0 † q 2 + ν 1 ∗ q 1 q 2 ) V 15 + ( ν ^ 2 ∗ + ν ^ 3 ∗ q 0 q 1 + ν ^ 0 ∗ q 0 q 2 + ν ^ 1 ∗ q 1 q 2 ) V 16 V 1 ≡ q 0 q 0 † q 1 q 1 † q 2 q 2 † q 3 q 3 † V 2 ≡ q 0 † q 0 q 1 q 1 † q 2 q 2 † q 3 q 3 † V 15 ≡ q 0 q 0 † q 1 † q 1 q 2 † q 2 q 3 † q 3 V 16 ≡ q 0 † q 0 q 1 † q 1 q 2 † q 2 q 3 † q 3 \begin{aligned}
\psi_{\nu} &\equiv (\nu_2 + \nu_3 \mathbf{q}^{\dagger}_0 \mathbf{q}^{\dagger}_1 + \nu_0 \mathbf{q}^{\dagger}_0 \mathbf{q}^{\dagger}_2 + \nu_1 \mathbf{q}^{\dagger}_1 \mathbf{q}^{\dagger}_2) V_1 \\
&+ (\widehat{\nu}_2 - \widehat{\nu}_3 \mathbf{q}_0 \mathbf{q}^{\dagger}_1 - \widehat{\nu}_0 \mathbf{q}_0 \mathbf{q}^{\dagger}_2 + \widehat{\nu}_1 \mathbf{q}^{\dagger}_1 \mathbf{q}^{\dagger}_2) V_2 \\
&+ (\nu^*_2 - \nu^*_3 \mathbf{q}^{\dagger}_0 \mathbf{q}_1 - \nu^*_0 \mathbf{q}^{\dagger}_0 \mathbf{q}_2 + \nu^*_1 \mathbf{q}_1 \mathbf{q}_2) V_{15} \\
&+ (\widehat{\nu}^*_2 + \widehat{\nu}^*_3 \mathbf{q}_0 \mathbf{q}_1 + \widehat{\nu}^*_0 \mathbf{q}_0 \mathbf{q}_2 + \widehat{\nu}^*_1 \mathbf{q}_1 \mathbf{q}_2) V_{16} \\
V_1 &\equiv \mathbf{q}_0 \mathbf{q}^{\dagger}_0 \mathbf{q}_1 \mathbf{q}^{\dagger}_1 \mathbf{q}_2 \mathbf{q}^{\dagger}_2 \mathbf{q}_3 \mathbf{q}^{\dagger}_3 \\
V_2 &\equiv \mathbf{q}^{\dagger}_0 \mathbf{q}_0 \mathbf{q}_1 \mathbf{q}^{\dagger}_1 \mathbf{q}_2 \mathbf{q}^{\dagger}_2 \mathbf{q}_3 \mathbf{q}^{\dagger}_3 \\
V_{15} &\equiv \mathbf{q}_0 \mathbf{q}^{\dagger}_0 \mathbf{q}^{\dagger}_1 \mathbf{q}_1 \mathbf{q}^{\dagger}_2 \mathbf{q}_2 \mathbf{q}^{\dagger}_3 \mathbf{q}_3 \\
V_{16} &\equiv \mathbf{q}^{\dagger}_0 \mathbf{q}_0 \mathbf{q}^{\dagger}_1 \mathbf{q}_1 \mathbf{q}^{\dagger}_2 \mathbf{q}_2 \mathbf{q}^{\dagger}_3 \mathbf{q}_3
\end{aligned} ψ ν V 1 V 2 V 15 V 16 ≡ ( ν 2 + ν 3 q 0 † q 1 † + ν 0 q 0 † q 2 † + ν 1 q 1 † q 2 † ) V 1 + ( ν 2 − ν 3 q 0 q 1 † − ν 0 q 0 q 2 † + ν 1 q 1 † q 2 † ) V 2 + ( ν 2 ∗ − ν 3 ∗ q 0 † q 1 − ν 0 ∗ q 0 † q 2 + ν 1 ∗ q 1 q 2 ) V 15 + ( ν 2 ∗ + ν 3 ∗ q 0 q 1 + ν 0 ∗ q 0 q 2 + ν 1 ∗ q 1 q 2 ) V 16 ≡ q 0 q 0 † q 1 q 1 † q 2 q 2 † q 3 q 3 † ≡ q 0 † q 0 q 1 q 1 † q 2 q 2 † q 3 q 3 † ≡ q 0 q 0 † q 1 † q 1 q 2 † q 2 q 3 † q 3 ≡ q 0 † q 0 q 1 † q 1 q 2 † q 2 q 3 † q 3 ν 2 \nu_2 ν 2 ν 3 \nu_3 ν 3 ν 0 \nu_0 ν 0 ν 1 \nu_1 ν 1 ν ^ \widehat{\nu} ν q 2 / q 3 \mathbf{q}_2 / \mathbf{q}_3 q 2 / q 3 ψ \psi ψ
J ν ≡ ψ ν γ 0 ψ ~ ν = j μ γ μ γ 0 ≡ q 0 † + q 0 γ 1 ≡ q 1 † − q 1 γ 2 ≡ i ( q 1 † + q 1 ) γ 3 ≡ q 0 † − q 0 \begin{aligned}
J_{\nu} \equiv \psi_{\nu} \gamma_0 \tilde{\psi}_{\nu} = j^{\mu} \gamma_{\mu} \\
\gamma_0 \equiv \mathbf{q}^{\dagger}_0 + \mathbf{q}_0 \\
\gamma_1 \equiv \mathbf{q}^{\dagger}_1 - \mathbf{q}_1 \\
\gamma_2 \equiv i(\mathbf{q}^{\dagger}_1 + \mathbf{q}_1) \\
\gamma_3 \equiv \mathbf{q}^{\dagger}_0 - \mathbf{q}_0
\end{aligned} J ν ≡ ψ ν γ 0 ψ ~ ν = j μ γ μ γ 0 ≡ q 0 † + q 0 γ 1 ≡ q 1 † − q 1 γ 2 ≡ i ( q 1 † + q 1 ) γ 3 ≡ q 0 † − q 0 The charge operator Q Q Q ψ \psi ψ
Q ( ψ ) ≡ ( q 3 ∧ q 3 † ) ψ − 1 3 ψ ( ( q 1 ∧ q 1 † ) + ( q 2 ∧ q 2 † ) + ( q 3 ∧ q 3 † ) ) q j ∧ q k † ≡ 1 2 ( q j q k † − q k † q j ) \begin{aligned}
Q(\psi) &\equiv (\mathbf{q}_3 \wedge \mathbf{q}^{\dagger}_3) \psi - \frac{1}{3} \psi ((\mathbf{q}_1 \wedge \mathbf{q}^{\dagger}_1) + (\mathbf{q}_2 \wedge \mathbf{q}^{\dagger}_2) + (\mathbf{q}_3 \wedge \mathbf{q}^{\dagger}_3)) \\
\mathbf{q}_j \wedge \mathbf{q}^{\dagger}_k &\equiv \frac{1}{2}(\mathbf{q}_j \mathbf{q}^{\dagger}_k - \mathbf{q}^{\dagger}_k \mathbf{q}_j)
\end{aligned} Q ( ψ ) q j ∧ q k † ≡ ( q 3 ∧ q 3 † ) ψ − 3 1 ψ (( q 1 ∧ q 1 † ) + ( q 2 ∧ q 2 † ) + ( q 3 ∧ q 3 † )) ≡ 2 1 ( q j q k † − q k † q j ) Note that the charge operator multiplies ψ \psi ψ − 1 2 -\frac{1}{2} − 2 1 + 1 2 +\frac{1}{2} + 2 1 0 0 0
The reason this works this way is detailed in the construction of the model, along with the basic rules of the geometric algebra generators q j \mathbf{q}_j q j q j † \mathbf{q}^{\dagger}_j q j †
q j q k + q k q j = 0 q j q k † + q k † q j = δ j k \begin{aligned}
\mathbf{q}_j \mathbf{q}_k + \mathbf{q}_k \mathbf{q}_j &= 0 \\
\mathbf{q}_j \mathbf{q}^{\dagger}_k + \mathbf{q}^{\dagger}_k \mathbf{q}_j &= \delta_{jk}
\end{aligned} q j q k + q k q j q j q k † + q k † q j = 0 = δ jk