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Using the ideas from a previous question here it can be shown that if one takes the boson spin 1 stress-energy tensor of the form

\begin{align} T^{\mu\nu}_{\text{spin one}} = \begin{bmatrix} \frac{1}{8\pi}(E^2+B^2) && S_x/c && S_y/c && S_z/c \\ S_x/c && -\sigma_{xx} && -\sigma_{xy} && -\sigma_{xz} \\ S_y/c && -\sigma_{yx} && -\sigma_{yy} && -\sigma_{yz} \\ S_z/c && -\sigma_{zx} && -\sigma_{zy} && -\sigma_{zz} \end{bmatrix} \end{align}

created from a boson spin 1 electromagnetic field $F=E+iB$ and changes the spin of $F$ from spin one to spin half, $\{F,A\} \rightarrow \{ \phi_+ , \phi_- \} $ using here, then $T^{\mu\nu}_{\text{spin one}}$ changes to

\begin{align} T^{\mu\nu}_{\text{spin half}} = mc^2 \begin{bmatrix} I && 0 && 0 && i\gamma^{0123} \\ \gamma^{01} && 0 && 0 && i\gamma^{23} \\ \gamma^{02} && 0 && 0 && i\gamma^{31} \\ \gamma^{03} && 0 && 0 && i\gamma^{12} \end{bmatrix} , \end{align} which I can only understand $T^{\mu\nu}_{\text{spin half}} $ as a fermion spin half stres-energy tensor.

What is the physical significance of $T^{\mu\nu}_{\text{spin half}} $ compared to $T^{\mu\nu}_{\text{spin one}}$?

I am struggling to understand how to interpret $T^{\mu\nu}_{\text{spin half}} $. This fermi-field stress-energy tensor is asymmetric. The divergence of $T^{\mu\nu}_{\text{spin half}} $ is zero.

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