## Dirac ideal Spinors in the Dirac ideal are given by $\begin{aligned} N &= M \frac{1+\gamma_0}{2} \frac{1+\gamma_5\sigma_3}{2} \\ &= (M_+ + M_-)\frac{1+\gamma_0}{2} \frac{1+\gamma_5\sigma_3}{2} \\ &= (M_+ + M_-\gamma_0)\frac{1+\gamma_0}{2} \frac{1+\gamma_5\sigma_3}{2} \\ &= M_+'\frac{1+\gamma_0}{2} \frac{1+\gamma_5\sigma_3}{2} \\ &= (\text{Re}M_+' + \text{Im} M_+' I \sigma_3)\frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2} \\ &= \psi \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2} \\ \end{aligned} $where $M$ is an invertible element of $\mathcal Cl_{1,3}(\mathbb C)$ and $\psi$ an invertible element of $\mathcal Cl_{1,3}^+(\mathbb R)$. #### $\gamma_0$-eigenstates If we are in an eigenstate of $\gamma_0$, then we have $\begin{aligned} N &= \frac{1+\gamma_0}{2}N \\ &= \frac{1+\gamma_0}{2} \psi \frac{1+\gamma_0}{2} \frac{1+\gamma_5\sigma_3}{2}\\ &= \frac{\psi + \gamma_0 \psi \gamma_0}{2} \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2} \\ &= \psi_0 \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2} \end{aligned} $where $\begin{aligned} \psi_0 &= \rho^{1/2} \frac{LUe^{I\beta/2} + \gamma_0 LU e^{I\beta/2} \gamma_0}{2} \\ &= \rho^{1/2} \frac{Le^{I\beta/2} + \widetilde Le^{-I\beta/2}}{2} U \\ &= \rho^{1/2} \frac{(L + \widetilde L) \cos(\beta/2) + (L - \widetilde L)I\sin(\beta/2)}{2} U \\ &= \rho^{1/2} \bigg(\cosh(\alpha/2) \cos(\beta/2) + \hat \alpha \sinh(\alpha/2) I \sin(\beta/2)\bigg) U \\ &= \rho'^{1/2}U' \end{aligned}$ where $L = e^{\vec \alpha/2}$, and so we have that $\gamma_0$ eigenstates are of the form $\psi = \rho^{1/2} U \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2}. $The Lorentz covariant density matrix of a $\gamma_0$ eigenstate takes the form $\begin{aligned} \ket \psi\bra {\bar \psi} &= \rho U \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2} U^\dagger \gamma_0 \\ &= \rho U \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_0}{2} \gamma_0 \widetilde U \\ &= \rho U \frac{1+\gamma_5\sigma_3}{2} \frac{1+\gamma_0}{2} \widetilde U \\ &= \rho U \frac{1+\gamma_5\sigma_3}{2} \widetilde U \frac{1+\gamma_0}{2} \end{aligned} $which is normalized when $\begin{aligned} 1 &= \int d^3x \langle \psi|\psi \rangle \\ &= \int d^3x \text{Tr}(\ket \psi \bra {\bar\psi} \gamma_0) \\ &= 4 \int d^3x \langle \ket \psi \bra {\bar\psi} \rangle_1 \cdot \gamma_0 \\ &= \int d^3x \langle \rho U (1+\gamma_5\sigma_3) \widetilde U (1+\gamma_0)\rangle_1 \cdot \gamma_0 \\ &= \int d^3x \langle (\rho +\gamma_5 \rho U \sigma_3 \widetilde U) (1+\gamma_0)\rangle \\ &= \int d^3x \; \rho(x) \end{aligned} $For a general state in the Dirac ideal, we have $\langle \ket \psi \bra{\overline \psi} \rangle_1 = \frac{1}{4} \psi \gamma_0 \widetilde \psi = \frac{1}{4} J$ . In an eigenstate of $\gamma_0$, we have $J= \rho \gamma_0$. A boost would kick the state out of an eigenstate of $\gamma_0$ and generally, normalization takes the form $\int d^3 x \; J \cdot \gamma_0 = 1$ in the Dirac basis. If we restrict to the ideal of eigenstates of $\gamma_0$ and formally quotient out $\gamma_0$, which we can do because $\gamma_0 \psi = \psi \gamma_0$, then we obtain spinors of the form $\begin{aligned}\hat N = \rho^{1/2} \hat U \frac{1 + \hat \gamma_5 \hat \sigma_3 }{2}\end{aligned}. $The form of a Lorentz covariant density matrix for a $\gamma_0$-eigenstate has the form $\frac{1+\gamma_0}{2} \rho U \frac{1 - i \sigma_1 \sigma_2}{2} \widetilde U = \rho U \frac{1 - i \sigma_1 \sigma_2}{2} \widetilde U \frac{1+\gamma_0}{2}, $so we can quotient out $\gamma_0$, since it commutes with the density matrix, and density matrices take the form $\hat \rho = \rho \frac{1 + \gamma_5 U \sigma_3 U^\dagger}{2}. $ We could also choose $\hat \rho \gamma_0 = \rho \frac{\gamma_0 - i U I \gamma_3 U^\dagger}{2}$ which expresses the density matrix as a complex odd-multivector. The spin degree of freedom is represented as a trivector here. If we represent $\gamma_5 \sigma_3$ as $\hat\sigma_3$ in the Pauli algebra, as we are free to do, then we must remember that $\hat \sigma_3$ implicitly involves the $\gamma_5$ operator in the Dirac basis, as discussed here [[Spacetime Split]]. ## Weyl ideal Consider Dirac spinors of the form $N = M \frac{1+\gamma_5}{2}\frac{1+\gamma_5\sigma_3}{2} = (M_+ + M_-) \frac{1+\gamma_5}{2}\frac{1+\gamma_5\sigma_3}{2}$where $M$ is an invertible element of $\mathcal Cl_{1,3}(\mathbb C)$. Observe that $N_+ = \gamma_5 N_+ = M_+ \frac{1+\gamma_5}{2}\frac{1+\gamma_5\sigma_3}{2} $and $N_- = -\gamma_5 N_- = M_- \frac{1+\gamma_5}{2}\frac{1+\gamma_5\sigma_3}{2} $because $\gamma_5$ commutes with even elements and anticommutes with odd elements. #### $\gamma_5$-eigenstates If $N$ is an eigenstate of $\gamma_5$, then it can be written $\begin{aligned} N_+ &= (\text{Re} M_+ + i \text{Im} M_+)\frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \\ &= (\text{Re} M_+ + I \text{Im} M_+) \frac{1+\gamma_5}{2}\frac{1+\gamma_5\sigma_3}{2} \\ &= \psi \frac{1+\gamma_5\sigma_3}{2} \frac{1+\gamma_5}{2} \end{aligned}$ where $\psi = \rho^{1/2} e^{I \beta/2} R \in \mathcal Cl_{1,3}^+(\mathbb R)$ and $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$. The Lorentz covariant density matrix for $N_+$ is given by $\begin{aligned} \ketbra{\psi}{\bar \psi} &= \psi \frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \gamma_0\widetilde \psi \\ &=\frac{1+\gamma_5}{2} \psi \frac{\gamma_0+\gamma_3}{2} \widetilde \psi \end{aligned} $while normalization is given by $\begin{aligned} 1 &= \int d^3x \bra{\overline\psi}\gamma_0\ket\psi \\ &= \int d^3x \text{Tr}(\ketbra{\psi}{\bar \psi} \gamma_0) \\ &= 4 \int d^3x \langle \frac{1+\gamma_5}{2} \psi \frac{\gamma_0+\gamma_3}{2} \widetilde \psi \rangle_1 \cdot \gamma_0 \\ &= \int d^3x \bigg(\psi (\gamma_0+\gamma_3) \widetilde \psi \bigg) \cdot \gamma_0 \\ \end{aligned} $The transformation $\psi \mapsto \psi \gamma_0$ yields the state $\psi \frac{1+\gamma_5\sigma_3}{2} \frac{1-\gamma_5}{2}\gamma_0, $which is an eigenstate of $-\gamma_5$, and gives us a decomposition of a general state in the Weyl basis into $\psi \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_5}{2} = (\psi_1 + \psi_2 \gamma_0) \frac{1+\gamma_5\sigma_3}{2}\frac{1+\gamma_5}{2} $where $\psi_1, \psi_2 \in \mathcal Cl_{1,3}^+(\mathbb R)$. That is, we have $\psi_2 = (\text{Re} M_- - I\text{Im} M_-)\gamma_0$. Thus, we find that $\text{Tr}(\ket \psi \bra{\overline \psi}\gamma_0) = 4 \langle \ket \psi \bra{\overline \psi} \rangle_1 \cdot \gamma_0 = \langle \psi_1 (\gamma_0 + \gamma_3) \widetilde \psi_1\rangle_1 \cdot \gamma_0 + \langle \psi_2 (\gamma_0 - \gamma_3) \widetilde \psi_2\rangle_1\cdot \gamma_0$ where the vector part $\ket \psi \bra{\overline \psi}$ can be computed from $\begin{aligned} \ket \psi\bra{\overline \psi} &= (\psi_1 + \psi_2 \gamma_0) \frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \gamma_0(\widetilde \psi_1 + \gamma_0 \widetilde \psi_2) \\ &= \psi_1 \frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \gamma_0 \widetilde \psi_1 \\ &+ \psi_2 \gamma_0 \frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \gamma_0 \gamma_0\widetilde \psi_2 \\ &+ \psi_1 \frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \gamma_0 \gamma_0\widetilde \psi_2 \\ &+ \psi_2 \gamma_0 \frac{1+\gamma_5}{2} \frac{1+\gamma_5\sigma_3}{2} \gamma_0 \widetilde \psi_1 \\ &= \frac{1+\gamma_5}{2} \langle \psi_1 \frac{\gamma_0 + \gamma_3}{2} \widetilde \psi_1\rangle_1 \\ &+ \frac{1-\gamma_5}{2} \langle \psi_2 \frac{\gamma_0-\gamma_3}{2}\widetilde \psi_2 \rangle_1 \\ &+ \langle\frac{1+\gamma_5}{2} \psi_1 \frac{1+\sigma_3}{2} \widetilde \psi_2 \rangle_+ \\ &+ \langle\frac{1-\gamma_5}{2} \psi_2 \frac{1+\sigma_3}{2} \widetilde \psi_1 \rangle_+. \end{aligned}$ - [ ] Why don't cross terms contribute? This is not what I would expect. - Count degrees of freedom? When in an eigenstate of $\gamma_5$, we can formally quotient out $\gamma_5$, because $\psi\gamma_5 = \gamma_5 \psi$, which yields an element of an ideal of a quotient algebra isomorphic to $\mathcal Cl_3$: $\psi = e^{I \beta} L U \frac{1+ \sigma_3}{2} = L U e^{I \sigma_3 \beta}\frac{1+ \sigma_3}{2} = L U' \frac{1+ \sigma_3}{2} \in \mathcal Cl_3\frac{1+\sigma_3}{2}$where $\hat L$ is a boost and $\hat U$ is a rotation. In this ideal, all phase transformations are equivalent to a duality transformation, due to the identification $i = I = \gamma_{0123}$, or equivalently, an internal rotation in the $I \hat \sigma_3$ plane. The form of Lorentz covariant density matrices for eigenstates of $\gamma_5$ take the form$\frac{1+\gamma_5}{2} \psi \frac{\gamma_0+\gamma_3}{2} \widetilde \psi = \frac{1 + \gamma_5}{2} \rho R \frac{\gamma_0+\gamma_3}{2}\widetilde R = \rho R \frac{\gamma_0+\gamma_3}{2}\widetilde R \frac{1 - \gamma_5}{2}.$Since this object does not commute with $\gamma_5$, we cannot quotient out $\gamma_5$ in the same way that we did for the spinor above, or as in the case for the spinors in the Dirac ideal, but the vector $\psi \frac{\gamma_0+\gamma_3}{2} \widetilde \psi$ still represents precisely the degrees of freedom of an eigenstate of $\gamma_5$, so in general, a statistical mixture of $\gamma_5$-eigenstates will be represented by a vector. One other significant difference between $\gamma_0$-eigenstates and $\gamma_5$-eigenstates is that $\gamma_5$-eigenstates remain eigenstates under boosts, while $\gamma_0$-eigenstates, by definition, do not remain eigenstates under boosts. Instead, $\gamma_5$-eigenstates possess distinct symmetries arising from the transformations under which the null vector $\gamma_0 + \gamma_3$ is invariant, up to a scalar. Namely,$e^{-\alpha}e^{\sigma_3\alpha/2} (\gamma_0 + \gamma_3)e^{-\sigma_3\alpha/2} = \gamma_0+\gamma_3.$However, in general, a boost yields a distinct $\gamma_5$-eigenstate, and importantly, unlike what one might expect, boosts do not alter the purity of the state. Both these Dirac and Weyl eigenstates possess the same degrees of freedom