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A counterterm is a local term added to a regulated <Lagrangian density> to cancel ultraviolet divergences and impose chosen <renormalization conditions>. Field-strength, mass and coupling counterterms respectively adjust propagator normalization, pole position and interaction strength.

Counterterm

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

Varying the displayed gauge-fixed <Lagrangian density> gives the kinetic operator
$$
D^{\mu\rho}=\eta^{\mu\rho}\Box-\left(1-\frac1\alpha\right)\partial^\mu\partial^\rho.
$$
Its <quantum field theory propagator> should satisfy
$$
D_x^{\mu\rho}\Delta_{\rho\nu}(x-y)=i\delta^\mu_\nu\delta^{(4)}(x-y),
$$
and inversion into transverse and longitudinal projectors gives the numerator
$$
\Pi_{\mu\nu}^{\mathrm{correct}}
=-\eta_{\mu\nu}+(1-\alpha)\frac{k_\mu k_\nu}{k^2}.
$$

The paper instead prints $-\eta_{\mu\nu}+(\alpha-1)k_\mu k_\nu/k^2$. Except at $\alpha=1$, that is not the inverse of the displayed Lagrangian's kinetic operator. Taken literally, for $\alpha\ne2$ it is the Green function of
$$
\left[\eta^{\mu\rho}\Box+\frac{\alpha-1}{2-\alpha}
\partial^\mu\partial^\rho\right]\Delta_{\rho\nu}
=i\delta^\mu_\nu\delta^{(4)},
$$
and at $\alpha=2$ its longitudinal part is noninvertible. Thus the longitudinal sign in the printed propagator is a typographical error; the two forms coincide in <Feynman gauge>.

Solved by gpt-5.6-sol high.

Paper 301 / 3 / a / Solution

{title2=$\mathcal L$}
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A Lagrangian density is integrated over spacetime to give the action $S=\int d^dx\,\mathcal L$.


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 Incoming links: Lagrangian density