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Neutrino

This is a response to a question regarding the charge of the neutrino in the Geometric Algebra construction of the Standard Model using Cl(4)⊗CCl(4) \otimes \mathbb{C}, some of which is discussed on YouTube here.

In that model, the neutrino portion of the Clifford object ψ\psi is given by:

ψν≡(ν2+ν3q0†q1†+ν0q0†q2†+ν1q1†q2†)V1+(ν^2−ν^3q0q1†−ν^0q0q2†+ν^1q1†q2†)V2+(ν2∗−ν3∗q0†q1−ν0∗q0†q2+ν1∗q1q2)V15+(ν^2∗+ν^3∗q0q1+ν^0∗q0q2+ν^1∗q1q2)V16V1≡q0q0†q1q1†q2q2†q3q3†V2≡q0†q0q1q1†q2q2†q3q3†V15≡q0q0†q1†q1q2†q2q3†q3V16≡q0†q0q1†q1q2†q2q3†q3\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}

ν2\nu_2 and ν3\nu_3 are the spin up and down right-handed neutrino states, and ν0\nu_0 and ν1\nu_1 are the spin up and down left-handed neutrino states. ν^\widehat{\nu} are the backward-time states mentioned by the Two-State Vector Formalism. Electro-weak interactions are rotations in the q2/q3\mathbf{q}_2 / \mathbf{q}_3 plane from the left side. To get the probability current vector, we use the reverse operator on ψ\psi

Jν≡ψνγ0ψ~ν=jμγμγ0≡q0†+q0γ1≡q1†−q1γ2≡i(q1†+q1)γ3≡q0†−q0\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}

The charge operator QQ acting on ψ\psi is as follows:

Q(ψ)≡(q3∧q3†)ψ−13ψ((q1∧q1†)+(q2∧q2†)+(q3∧q3†))qj∧qk†≡12(qjqk†−qk†qj)\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}

Note that the charge operator multiplies ψ\psi form both the left AND the right.
For the case of the neutrino, the right-side contribution adds to −12-\frac{1}{2} while the left-side contribution is +12+\frac{1}{2}, giving us a total of 00.

The reason this works this way is detailed in the construction of the model, along with the basic rules of the geometric algebra generators qj\mathbf{q}_j and their conjugates qj†\mathbf{q}^{\dagger}_j

qjqk+qkqj=0qjqk†+qk†qj=δjk\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}