The translation changes the <Lagrangian density> by $\delta\mathcal L_0=\epsilon^\nu\partial_\nu\mathcal L_0=\partial_\mu(\epsilon^\mu\mathcal L_0)$. The <Noether current> is therefore
$$
j^\mu=\frac{\partial\mathcal L_0}{\partial(\partial_\mu\psi)}\delta\psi-\epsilon^\mu\mathcal L_0
=\epsilon_\nu T^{\mu\nu},
\qquad
T^{\mu\nu}=i\bar\psi\gamma^\mu\partial^\nu\psi-\eta^{\mu\nu}\mathcal L_0.
$$
This is the <canonical stress-energy tensor>. Directly,
$$
\partial_\mu T^{\mu\nu}
=\left(\frac{\partial\mathcal L_0}{\partial\psi}
-\partial_\mu\frac{\partial\mathcal L_0}{\partial(\partial_\mu\psi)}\right)\partial^\nu\psi
$$
up to the analogous adjoint-field term, so the <Euler-Lagrange field equations> imply $\partial_\mu T^{\mu\nu}=0$ <on shell>. The <Dirac equation> also gives $\mathcal L_0=0$ on shell.

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/a/solution