note-math

cf. #link(<massless-spinor-Lagrangian>)[Action of spinor field]

spinor-current-translation-of-spacetime

massless spinor Lagrangian

\begin{matrix} 𝐿 (ϕ, \partial_{𝑥} ϕ) & = & ϕ^{†} {\partial_{spin}}^{◊} ϕ \\ = & ϕ^{†} 𝜎^{𝜇} \partial_{𝜇}^{◊} ϕ \end{matrix}

massive spinor Lagrangian

\begin{matrix} 𝐿 ((\begin{matrix} ϕ \\ 𝜓 \end{matrix}), \partial_{𝑥} (\begin{matrix} ϕ \\ 𝜓 \end{matrix})) & = & {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} (\begin{matrix} i {\partial_{spin}}^{◊} & - 𝑚 𝟙 \\ - 𝑚 𝟙 & i \partial_{spin} \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix}) \\ = & i \cdot (ϕ^{†} 𝜎^{𝜇} \partial_{𝜇}^{◊} ϕ + 𝜓^{†} 𝜎^{𝜇} \partial_{𝜇} 𝜓) - 𝑚 ((ϕ^{†} 𝜓 + 𝜓^{†} ϕ)) \end{matrix}

Since only the $Re$ part of the spinor field action plays a role, a $Re$ type theory can also be used.

Similar to #link(<energy-momentum-tensor-KG>)[the case of scalar field], try to calculate the energy-momentum-tensor

The solution of the spinor eq ${\partial_{spin}}^{◊} ϕ = 0$ or $(\begin{matrix} i {\partial_{spin}}^{◊} & - 𝑚 𝟙 \\ - 𝑚 𝟙 & i \partial_{spin} \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix}) = 0$ makes $𝐿 (ϕ, \partial_{𝑥} ϕ) = 0$

𝑇_{𝜈}^{𝜇} = \frac{\partial 𝐿}{\partial (\partial_{𝜇} ϕ)} \cdot \partial_{𝜈} ϕ

$𝐿 (ϕ, \partial_{𝑥} ϕ) = ϕ^{†} i {\partial_{spin}}^{◊} ϕ$ ==> energy-momentum-tensor-massless-spinor_(tag)

𝑇 = ϕ^{†} (𝜎 i \partial^{◊}) ϕ

or component form

𝑇_{𝜈}^{𝜇} = ϕ^{†} (𝜎^{𝜇} i \partial_{𝜈}^{◊}) ϕ

For massive, the derivative of the mass term with respect to $\partial_{𝜇}$ is zero + the action with solution is still zero $𝐿 = 0$ , so that the energy-momentum tensor is not affected by the mass term.

energy-momentum-tensor-massive-spinor_(tag)

𝑇 = {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} (\begin{matrix} 𝜎 i \partial^{◊} \\ 𝜎 i \partial \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix})

or component form

𝑇_{𝜈}^{𝜇} = {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} (\begin{matrix} i 𝜎^{𝜇} \partial_{𝜈}^{◊} \\ i 𝜎^{𝜇} \partial_{𝜈} \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix})

All are quantities with zero divergence $\partial^{†} 𝑇 = \partial_{𝜇} 𝑇_{𝜈}^{𝜇} = 0$

spinor-energy

Fix $ℝ^{1, 3}$ coordinates and consider $𝑇_{𝜇}^{0} = ϕ^{†} i \partial_{𝜇}^{◊} ϕ$ to be an integrable quantity in $ℝ^{3}$ . Define energy for massless-spinor

\begin{matrix} 𝐸 = \int_{ℝ^{3}} 𝑑 𝑥 (𝑇_{0}^{0}) & = & \int_{ℝ^{3}} 𝑑 𝑥 (ϕ^{†} i \partial_{0} ϕ) \end{matrix}

Similar to #link(<conserved-spatial-integral-energy-KG>)[the case of scalar field], time invariant $\partial_{0} 𝐸 = 0$ by $\partial^{†} 𝑇 = \partial_{𝜇} 𝑇_{𝜈}^{𝜇} = 0$

Example Plane wave, similar to the case of scalar field

massive-spinor energy

\begin{matrix} 𝐸 = \int_{ℝ^{3}} 𝑑 𝑥 (𝑇_{0}^{0}) & = & \int_{ℝ^{3}} 𝑑 𝑥 {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} i \partial_{0} (\begin{matrix} ϕ \\ 𝜓 \end{matrix}) \end{matrix}

spinor-current-rotation-boost-of-spacetime

Using the product rule of differentiation, separate into two parts

The domain differentiation part is still similar to #link(<angular-momentum-KG>)[the case of scalar field]

𝑀 = [𝑥, 𝑇]

𝑀 = ϕ^{†} [𝑥, 𝜎 i \partial^{◊}] ϕ

𝑀_{𝜇 𝜈}^{𝜆} = ϕ^{†} [𝑥_{𝜇}, 𝜎^{𝜆} i \partial_{𝜈}^{◊}] ϕ

The codomain differentiation part

The δ diffeomorphism given by the Lorentz-Lie-algebra $so (1, 3)$ is $\frac{1}{4} {[𝜎_{𝜇}, 𝜎_{𝜈}]}_{◊} ϕ (𝑥)$ . (cf. #link(<square-root-of-Lorentz-Lie-algebra>)[])

0 = \int_{𝑈} 𝑑 𝑥 ({(\frac{1}{4} {[𝜎_{𝜇}, 𝜎_{𝜈}]}_{◊} ϕ)}^{†} i {\partial_{spin}}^{◊} ϕ + ϕ^{†} (𝜎^{𝜆} i \partial_{𝜆}^{◊}) (\frac{1}{4} {[𝜎_{𝜇}, 𝜎_{𝜈}]}_{◊} ϕ))

${\partial_{spin}}^{◊} ϕ = 0$ + product rule + $𝜎^{𝜆} \partial_{𝜆}^{◊} = 𝜎^{𝜆 ◊} \partial_{𝜆}$ gives

0 = \int_{𝑈} 𝑑 𝑥 (\partial_{𝜆} (i ϕ^{†} 𝜎^{𝜆 ◊} \frac{1}{4} {[𝜎_{𝜇}, 𝜎_{𝜈}]}_{◊} ϕ))

forall $𝑈 \subset ℝ^{1, 3}$ ==> integrand is zero ==> divergence-free quantity $\partial^{†} 𝑆 = \partial_{𝜆} 𝑆_{𝜇 𝜈}^{𝜆} = 0$

𝑆_{𝜇 𝜈}^{𝜆} ≔ i ϕ^{†} (𝜎^{◊} \frac{1}{4} {[𝜎, 𝜎]}_{◊}) ϕ

𝑆 = i ϕ^{†} (𝜎^{◊} \frac{1}{4} {[𝜎, 𝜎]}_{◊}) ϕ

spinor-angular-momentum_(tag)

The angular momentum of domain and codomain combined, spinor-angular-momentum is

𝐽 = 𝑀 + 𝑆 = i ϕ^{†} 𝜎^{◊} ([𝑥, \partial] + \frac{1}{4} {[𝜎, 𝜎]}_{◊}) ϕ

𝐽_{𝜇 𝜈}^{𝜆} = i ϕ^{†} 𝜎^{𝜆 ◊} ([𝑥_{𝜇}, \partial_{𝜈}] + \frac{1}{4} {[𝜎_{𝜇}, 𝜎_{𝜈}]}_{◊}) ϕ

The case of massive-spinor is similar. It should be possible to prove by calculation that angular momentum is not affected by the mass term.

\begin{matrix} 𝐽 & = & 𝐿 + 𝑆 = i \cdot {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} (\begin{matrix} 𝜎^{◊} ([𝑥, \partial] + \frac{1}{4} {[𝜎, 𝜎]}_{◊}) \\ 𝜎 ([𝑥, \partial] + \frac{1}{4} {[𝜎, 𝜎]}_{◊}) \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix}) \end{matrix}

current-gauge-spinor

let $ϕ (𝑥)$ be a solution to the spinor eq. Phase change $𝑒^{𝜃 (𝑥)} ϕ (𝑥)$ and its δ change $𝜃 ϕ$ belong to boundary fixed variations near the solution, so

\begin{matrix} 0 & = & \int_{ℝ^{1, 3}} 𝑑 𝑥 (- 𝜃 ϕ^{†} i {\partial_{spin}}^{◊} ϕ + ϕ^{†} i {\partial_{spin}}^{◊} (𝜃 ϕ)) \\ = & \int_{ℝ^{1, 3}} 𝑑 𝑥 ((i ϕ^{†} 𝜎^{◊} ϕ) \cdot \partial 𝜃) \end{matrix}

Using product rule + divergence-free quantity + boundary zero

0 = \int_{ℝ^{1, 3}} 𝑑 𝑥 (\partial^{†} (i ϕ^{†} 𝜎^{◊} ϕ) \cdot 𝜃)

for all $Im (ℂ)$ value function $𝜃 (𝑥)$ , so

\forall 𝑥 \in ℝ^{1, 3}, \partial^{†} (i ϕ^{†} 𝜎^{◊} ϕ) = \partial_{𝜇} (i ϕ^{†} 𝜎^{𝜇 ◊} ϕ) (𝑥) = 0

current-gauge-spinor_(tag) $𝑗 = i ϕ^{†} 𝜎^{◊} ϕ, \partial^{†} 𝑗 = 0$ is called the 4-current of massless-spinor

Similarly, for massive-spinor, the 4-current is $i \cdot {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} (\begin{matrix} 𝜎^{◊} \\ 𝜎 \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix})$

conserved-spatial-integral-charge-spinor_(tag) Fixing $ℝ^{1, 3}$ coordinates, consider $𝑗$ as a quantity integrable over $ℝ^{3}$

time invariant $\partial_{0} \int_{ℝ^{3}} 𝑑 𝑥 (𝑗^{0}) = 0$ by $\partial^{†} 𝑇 = 0$

\begin{matrix} \int_{ℝ^{3}} 𝑑 𝑥 (𝑗^{0}) & = & \int_{ℝ^{3}} 𝑑 𝑥 (i ϕ^{†} 𝜎^{0} ϕ) \\ = & \int_{ℝ^{3}} 𝑑 𝑥 (i ϕ^{†} ϕ) \\ = & \int_{ℝ^{3}} 𝑑 𝑥 | ϕ |^{2} i \end{matrix}

In fact, one can choose a spacetime decomposition coordinate $ℝ^{1, 3}$ and write the spinor eq in the form of unitary evolution of charge $i \partial_{0} ϕ = - i 𝜎^{𝑖 ◊} \partial_{𝑖} ϕ = i 𝜎^{𝑖} \partial_{𝑖} ϕ$ where $- i 𝜎^{𝑖 ◊} \partial_{𝑖}$ is self-adjoint for $ℂ^{2}$ quadratic form + $ℝ^{3}$ integral

(After dropping $i$ ) charge alias probability density or particle number density or electric charge density

The case of massive-spinor is

\begin{matrix} \int_{ℝ^{3}} 𝑑 𝑥 (i \cdot {(\begin{matrix} ϕ \\ 𝜓 \end{matrix})}^{†} (\begin{matrix} 𝜎^{0 ◊} \\ 𝜎^{0} \end{matrix}) (\begin{matrix} ϕ \\ 𝜓 \end{matrix})) \\ = & \int_{ℝ^{3}} 𝑑 𝑥 (| ϕ |^{2} + | 𝜓 |^{2}) i \end{matrix}

Similar to the massless case, the spinor eq can be written in the form of charge unitary evolution.

i \partial_{0} (\begin{matrix} ϕ \\ 𝜓 \end{matrix}) = ((\begin{matrix} i 𝜎^{𝑖} \partial_{𝑖} \\ - i 𝜎^{𝑖} \partial_{𝑖} \end{matrix}) + 𝑚 (\begin{matrix} 𝟙 \\ 𝟙 \end{matrix})) (\begin{matrix} ϕ \\ 𝜓 \end{matrix})

conserved-current on mainfold …