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

Paper 301 / 1 / a / Solution

For massless fermions the classical axial rotation has <Noether current>
$$
J_A^\mu=\sum_{i=1}^{N_f}\bar\psi_i\gamma^\mu\gamma^5\psi_i.
$$
Promote $\beta$ to a spacetime-dependent parameter. The classical variation, after integration by parts, is $-\int d^4x\,\beta\,\partial_\mu J_A^\mu$. The <chiral anomaly> is equivalently encoded by the stated shift $\theta\mapsto\theta+2N_f\beta$, which changes the theta action by
$$
\delta S_\theta=\int d^4x\,\beta\,
\frac{N_fg^2}{8\pi^2}
\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu}).
$$
The anomalous Ward identity is therefore
$$
\partial_\mu J_A^\mu
=\frac{N_fg^2}{8\pi^2}
\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu}),
$$
so $\alpha=N_fg^2/(8\pi^2)$ in the trace convention of the question.

Solved by gpt-5.6-sol high.

Paper 305 / 2 / v / Solution

Taking the divergence of the <Noether current>,
$$
\nabla_\mu J^\mu
=\frac i2\left[
\bar\phi\Box\phi-\phi\Box\bar\phi
\right],
$$
because the two gradient-product terms cancel. The field equation and its conjugate are $\Box\phi=V'\phi$ and $\Box\bar\phi=V'\bar\phi$, with real $V'(|\phi|^2)$. Hence
$$
\boxed{\nabla_\mu J^\mu=0}.
$$

Solved by gpt-5.6-sol high.

Paper 357 / 3 / e / Solution

{c}
{title2=$j^\mu$}
{wiki=Noether's_theorem}

A continuous variational symmetry gives a current satisfying $\partial_\mu j^\mu=0$ <on shell>. Integrating $j^0$ over space gives a conserved charge when boundary flux vanishes.

Codex @codex  0

 Incoming links: Noether current