In what follows, we will work with a cyclic relabeling $\hat \sigma_3 \mapsto \sigma_1$, $\hat \sigma_1 \mapsto \sigma_2$, and $\hat \sigma_2 \mapsto \sigma_3$ of the usual Pauli basis. Let $\ket H$ be an eigenstate of the operator $\sigma_1$, and define $\ket V = \sigma_2 \ket H$. Then for a state $\ket \psi = \alpha \ket H + \beta \ket V = (E_1 + i B_1) \ket H + (E_2 + i B_2) \ket V$ we have $\begin{aligned} \ketbra \psi {\bar\psi} &= \frac{1}{2} \bigg(|\alpha|^2 \ketbra{H}{\bar H} + |\beta|^2 \ketbra{V}{\bar V} + \alpha \beta^* \ketbra{H}{\bar V} + \alpha^* \beta \ketbra{V}{\bar H} \bigg)\\ &= \frac{1}{2} \bigg(|\alpha|^2 (1 + \sigma_1)\gamma_0 + |\beta|^2 (1-\sigma_1)\gamma_0 + \alpha \beta^* (1 + \sigma_1)\sigma_2 \gamma_0 + \alpha^* \beta\sigma_2(1 +\sigma_1)\gamma_0 \bigg)\\ &= \frac{1}{2} \bigg(|\alpha|^2 (\gamma_0 + \gamma_1) + |\beta|^2 (\gamma_0 - \gamma_1) + \alpha \beta^* (\gamma_2 + I \gamma_3) + \alpha^* \beta(\gamma_2 - I\gamma_3) \bigg)\\ &= \frac{1}{2} \big(|\alpha|^2 + |\beta|^2\big)\gamma_0 + \frac{1}{2} \big(|\alpha|^2 - |\beta|^2\big)\gamma_1 + \frac{1}{2}\big( \alpha \beta^* + \beta\alpha^*\big) \gamma_2 + \frac{1}{2}\big( \alpha \beta^* - \beta\alpha^*\big) I \gamma_3 \\ &= \frac{1}{2}\big( \vec E^2 + \vec B^2 \big)\gamma_0 + \frac{1}{2}\big( E_1^2 - E_2^2 + B_1^2 -B_2^2 \big)\gamma_1 + \big( E_1 E_2 + B_1 B_2 \big) \gamma_2 + \big( E_1 B_2 - E_2 B_1 \big) \gamma_5 \gamma_3 \\ \end{aligned}$ Taking $F = (E_1 \sigma_1 + E_2 \sigma_2) + I (B_1 \sigma_1 + B_2 \sigma_2)$, we have $\begin{aligned} T(\gamma_0+\gamma_1) &= \frac{1}{2}F(\gamma_0 + \gamma_1)\widetilde F \\ &= \frac{1}{2}(\vec E^2 + \vec B^2)\gamma_0 + \frac{1}{2}\big( E_1^2 - E_2^2 + B_1^2 - B_2^2 \big)\gamma_1 + (E_1E_2 + B_1 B_2)\gamma_2 + (E_1 B_2 - E_2 B_1) \gamma_3 \end{aligned}$ which are equivalent, component by component, with the sole discrepancy arising from the presence of a $\gamma_5$ associated with the $\gamma_3$ term. When we introduce the Dirac adjoint, we are embedding the Pauli algebra into the Dirac algebra. To do this in a principled manner, we ought not only provide a mapping on the generators of the algebra, along with the "spacetime split" that recovers the original construction, we also wish to identify the Pauli spinor, an element of the minimal Pauli ideal, with an element of a minimal Dirac ideal. For this construction, the most natural identification appears to be $ \begin{gathered} \sigma_j = \gamma_j\gamma_0 \\ \\ (1+\sigma_x)/2 \;\; \mapsto \;\; (1+\sigma_x)/2(1+\gamma_5)/2 \\\\ \bra{\bar \psi} = \bra \psi \gamma_0 \end{gathered} $ so that $ \begin{gathered} \ket \psi \bra{\psi}_\text{Pauli} = \psi \bigg( \frac{1+\sigma_1}{2} \bigg) \psi^\dagger \\ \mapsto \\ \ket \psi \bra{\bar \psi}_\text{Dirac} = \psi \bigg( \frac{1+\sigma_1}{2} \frac{1+\gamma_5}{2} \bigg) \psi^\dagger \gamma_0 \end{gathered} $ along with an embedding $\psi \in \text{Spin}(3) \mapsto \psi \in \text{Spin}(1,3)$. Then we have $ \begin{aligned} \ketbra \psi {\bar\psi} &= \frac{1 + \gamma_5}{2} T_\text{EM}(\gamma_0 + \gamma_1). \end{aligned} $ This recovers the $\gamma_5$ eigenstates of a relativistic qubit, as shown in [[Eigenstates]]. #### Appendix: Calculations $ \begin{aligned} \frac{1}{2} F\gamma_0 \widetilde F &= \frac{1}{2}\bigg( (\vec E + I \vec B) (\vec E - I \vec B) \gamma_0 \bigg) \\ &= \frac{1}{2}(\vec E^2 + \vec B^2)\gamma_0 + (E_1 B_2 - E_2 B_1) \gamma_3 \\ \frac{1}{2} F\gamma_1 \widetilde F &= \frac{1}{2} \bigg( (F_\parallel + F_\perp)(-F_\parallel + F_\perp)\gamma_1 \bigg) \\ &= \frac{1}{2} \bigg( -F_\parallel^2 - F_\perp F_\parallel + F_\parallel F_\perp + F_\perp^2 \bigg)\gamma_1 \\ &= \frac{1}{2}\big( E_1^2 - E_2^2 + B_1^2 - B_2^2 \big)\gamma_1 + (E_1E_2 + B_1 B_2)\gamma_2 \end{aligned} $ $\sigma_1 \gamma_1 = - \gamma_1 \sigma_1$, $\sigma_2 \gamma_1 = \gamma_1 \sigma_2$, $I\sigma_1 \gamma_1 = \gamma_1 I \sigma_1$, and $I \sigma_2 \gamma_1 = -\gamma_1 I \sigma_2$ gives us $F_\parallel = E_2 \sigma_2 + B_1 I \sigma_1$ and $F_\perp = E_1 \sigma_1 + B_2 I \sigma_2$, and thus $\begin{aligned} -F_\parallel^2 &= -E_2^2 +B_1^2 \\ F_\perp^2 &= E_1^2 - B_2^2 \\ -F_\perp F_\parallel &= (E_1 E_2 + B_2 B_1) \gamma_{12} - I \vec E \cdot \vec B \\ F_\parallel F_\perp &= (E_2 E_1 +B_1B_2)\gamma_{12} +I\vec E \cdot \vec B \end{aligned}$ For $\alpha = E_1 + i B_1$ and $\beta = E_2 + i B_2$, $\begin{aligned} \frac{1}{2}\big(|\alpha|^2 + |\beta|^2\big) &= \frac{1}{2}\big( \vec E^2 + \vec B^2 \big) \\ \frac{1}{2}\big(|\alpha|^2 - |\beta|^2\big) &= \frac{1}{2}\big( E_1^2 - E_2^2 + B_1^2 -B_2^2 \big) \\ \frac{1}{2}\big( \alpha \beta^* + \beta\alpha^*\big) &= \big( E_1 E_2 + B_1 B_2 \big) \\ \frac{1}{2}\big( \alpha \beta^* - \beta\alpha^*\big) &= -i\big( E_1 B_2 - E_2 B_1 \big) \end{aligned}$