has components $T_{\alpha\beta}$ given by $$T_{\alpha\beta} = \frac{i}{2}\bigg(
\psi^+\gamma_{(\alpha}\nabla_{\beta)}\psi - (\nabla_{(\alpha}\psi^+)\gamma_{\beta)}\psi  
\bigg) \,$$ from which follows the energy density $\rho$: $$\rho = \frac{i}{2}\bigg(
\psi^+\gamma_0\nabla_{0}\psi - (\nabla_{0}\psi^+)\gamma_{0}\psi  
\bigg) \, .$$

Recall from (\[split gamma 0\]), that the spin frame $E_A$ and its Dirac conjugate satisfy $\delta^{AB}E_A\otimes E_B^+ =\gamma_0$. This enables us to write the classical point-split stress energy density $\rho^\mathrm{split}$ of the Dirac field as $$\begin{aligned}
\rho^\mathrm{split}(x,x') &=& \frac{i}{2}\delta^{AB}\bigg( \big( \psi^+E_A  \big)\otimes \big( E^+_B e_{0'}\cdot\nabla\psi\big) {\nonumber \\}&\,& \qquad - \big( [e_0\cdot\nabla\psi^+]E_A \big)\otimes \big( E^+_B\psi \big)
\bigg)(x,x') \, .\end{aligned}$$ One may then use $\rho^\mathrm{split}$ to define the distributional point-split energy density, also denoted $\rho^\mathrm{split}$: Let $f,f' \in C^\infty_0({\mathcal{M}})$ then $$\begin{aligned}
\rho^\mathrm{split}(f',f) &=& \frac{i}{2}\delta^{AB}\int_{{\mathcal{M}}\times{\mathcal{M}}}{\mathrm{dvol}}(x){\mathrm{dvol}}(x') {\nonumber \\}&\,& \quad (f'\otimes f)\bigg( \big( \psi^+E_A  \big)\otimes \big( E^+_B e_{0'}\cdot\nabla\psi\big) {\nonumber \\}&\,& \qquad - \big( [e_0\cdot\nabla\psi^+]E_A \big)\otimes \big( E^+_B\psi \big)
\bigg)(x,x') \\
&=& \frac{i}{2}\delta^{AB} \bigg(
\psi^+(f'E_A)\psi(\nabla\cdot(e_{0'}fE^+_B)) {\nonumber \\}&\,& \qquad - \psi^+(\nabla\cdot(e_0f'E_A))\psi(fE^+_B) \bigg)
\label{class rho split dirac}\end{aligned}$$ where we have expressed the right hand side in more traditional distributional language and $u\nabla\cdot v$ denotes the distributional dual of $(\nabla
u)\cdot v$. It is now clear that the replacement of $\psi^+\otimes\psi$ in (\[class rho split dirac\]) by the two-point function of any Hadamard state $\omega$ defines the expectation point-split energy 