Dirac gauge covariance, action and conserved current

Under $a\mapsto a+\mathrm{\partial }\alpha$ and $\psi \mapsto e^{ig\alpha }\psi$, the chain rule gives $D_a^{\mathrm{\prime }}{\psi }^{\mathrm{\prime }}=e^{ig\alpha }{D}_a\psi$; the conjugate equation gives $D_a^{\mathrm{\prime }}{\bar{\psi }}^{\mathrm{\prime }}=e^{-ig\alpha }{D}_a\bar{\psi }$. The two phases cancel in each term of ${\mathcal{L}}_D$, so the density is gauge invariant. The explicit Countable Choice assumption is [A2].

From [F2], $\beta =\mathrm{diag}(I_2,-I_2)$ is Hermitian and squares to $I_4$; each Pauli matrix is Hermitian, so ${\alpha }_j^{†}={\alpha }_j$, and the block formulas give ${\alpha }_j\beta =-\beta {\alpha }_j$. Therefore $({\gamma }^0{)}^{†}\beta ={\beta }^2=I_4=\beta {\gamma }^0$, while for $j=1,2,3$, $({\gamma }^j{)}^{†}\beta =(\beta {\alpha }_j{)}^{†}\beta ={\alpha }_j{\beta }^{†}\beta ={\alpha }_j={\beta }^2{\alpha }_j=\beta {\gamma }^j\mathrm{.}$ This proves [F3] from the exact matrix definitions. Now vary $\psi$ along a compactly supported complex $\chi$ and $\bar{\psi }$ along $\bar{\chi }$. Expanding the four kinetic terms and integrating the terms containing $D_a\chi$ or $D_a\bar{\chi }$ by parts gives $\delta S_D=2\mathrm{Re}{\int }_U{\chi }^{†}\beta R_D{d}^{4}x,R_D=(i{\gamma }^a{D}_a-\mu )\psi \mathrm{.}$ For example, $\int \bar{\psi }{\gamma }^a{D}_a\chi =-\int (D_a\bar{\psi }){\gamma }^a\chi$ and $-\int (D_a\bar{\chi }){\gamma }^a\psi =\int \bar{\chi }{\gamma }^a{D}_a\psi$; compact support removes both boundary terms. The resulting variation is $i\int (\bar{\chi }{\gamma }^a{D}_a\psi -(D_a\bar{\psi }){\gamma }^a\chi )d^{4}x-\mu \int (\bar{\chi }\psi +\bar{\psi }\chi )d^{4}x=\int (\bar{\chi }{R}_D+\overline{\stackrel{ˉ}{\chi }{R}_D})d^{4}x,$ where the equality uses $({\gamma }^a{)}^{†}\beta =\beta {\gamma }^a$ from [F3]. If $R_D(x_0)\mathrm{\ne }0$, continuity and openness of $U$ give a ball contained in $U$ where $\mathrm{\mid }{R}_D\mathrm{\mid }>0$; choose a bump $\rho$ from [F4] there and take $\chi =\rho \beta R_D$. Since ${\beta }^{†}\beta =I_4$, this gives $\delta S_D=2\int \rho \mathrm{\mid }{R}_D{\mathrm{\mid }}^2{d}^{4}x>0$, contradicting stationarity. Thus stationarity is equivalent to $R_D=0$ pointwise, and taking the adjoint gives $-i(D_a\bar{\psi }){\gamma }^a-\mu \bar{\psi }=0$. This is the charged version of the compact-variation identity in [F1]; the additional $a_a$ terms are smooth multiplication terms.

Contract the first equation in step 1.2 on the left by $\bar{\psi }$ and the adjoint equation on the right by $\psi$. Subtracting cancels the equal mass terms and gives $i(\bar{\psi }{\gamma }^a{D}_a\psi +(D_a\bar{\psi }){\gamma }^a\psi )=0.$ The parenthesis equals ${\mathrm{\partial }}_a(\bar{\psi }{\gamma }^a\psi )$ because the $+ig{a}_a\bar{\psi }$ and $-ig{a}_a\psi$ contributions cancel. Since $i\mathrm{\ne }0$ and $g$ is constant, ${\mathrm{\partial }}_a{j}^a=0$.

From [A1], $g\mathrm{/}{S}_0=q\mathrm{/}\mathrm{\hslash }$, so $g{a}_j=q{A}_j\mathrm{/}\mathrm{\hslash }$ and $g{a}_0=-q\varphi \mathrm{/}(\mathrm{\hslash }c)$; also $x^0=ct$ and $\mu =mc\mathrm{/}\mathrm{\hslash }$. With ${\gamma }^0=\beta$ and ${\gamma }^j=\beta {\alpha }_j$ [F2], multiply the equation in step 1.2 by $\beta$ and then by $\mathrm{\hslash }c$. Collecting the time derivative gives $i\mathrm{\hslash }{\mathrm{\partial }}_t\psi =[c\mathit{\alpha }\cdot (-i\mathrm{\hslash }\mathrm{\nabla }-q\mathbf{A})+\beta m{c}^2+q\varphi ]\psi \mathrm{.}$ This verifies the stated minimal-coupling signs.