Choose $C=i/4$. Expanding
$$
S^{\lambda\mu\nu}=\frac12\bar\psi[\gamma^\lambda,\gamma^\mu]\gamma^\nu\psi
$$
with the <Clifford algebra>, applying the product rule and then using the <Dirac equation> and its adjoint gives
$$
\widehat T^{\mu\nu}
=T^{\mu\nu}+\frac i4\left[
\partial_\lambda S^{\lambda\mu\nu}
-\partial^\nu(\bar\psi\gamma^\mu\psi)
\right].
$$
This is the <Belinfante-Rosenfeld stress-energy tensor> written as an improvement of the canonical tensor.

The translation charges are
$$
P^\nu=\int d^3x\,T^{0\nu},
\qquad
\widehat P^\nu=\int d^3x\,\widehat T^{0\nu}.
$$
For spatial $\nu$, their difference is the integral of $\partial_iS^{i0\nu}-\partial^\nu(\bar\psi\gamma^0\psi)$. For $\nu=0$, use conservation of the <Dirac current> to replace $\partial^0(\bar\psi\gamma^0\psi)$ by a spatial divergence; also $S^{00\nu}=0$. Hence $\widehat P^\nu-P^\nu$ is always a spatial boundary integral. Under the usual decay boundary condition it vanishes, so both currents generate the same four-momentum.

Solved by gpt-5.6-sol high.


 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-301/1/c/solution