Scale invariance requires the potential to obey
$$
\phi V'(\phi)=4V(\phi).
$$
Thus, apart from the zero choice,
$$
\boxed{V(\phi)=g\phi^4}
$$
for a dimensionless constant $g$; an additive constant is excluded because it introduces a scale in the action. The canonical <dilatation current> is
$$
\boxed{D^\mu=x_\nu T^{\mu\nu}+\phi\,\partial^\mu\phi}.
$$
Using $\Box\phi+V'(\phi)=0$,
$$
\partial_\mu D^\mu
=4V-\phi V'=0.
$$
Equivalently, after improving the stress tensor to make it traceless, the current is $x_\nu T_{\rm imp}^{\mu\nu}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 301 / 1 / iii / Solution

The free massless <Dirac field> has action
$$
S_{\rm free}=\int d^4x\,i\bar\psi\gamma^\mu\partial_\mu\psi.
$$
Under a finite dilation,
$$
\psi'(x)=\lambda^{-\Delta}\psi(\lambda^{-1}x),
\qquad
\bar\psi'(x)=\lambda^{-\Delta}\bar\psi(\lambda^{-1}x).
$$
The measure, derivative, and fields contribute $\lambda^4$, $\lambda^{-1}$, and $\lambda^{-2\Delta}$, so invariance requires
$$
4-1-2\Delta=0,
\qquad \Delta=\frac32.
$$
For $\lambda=e^\alpha$, the infinitesimal active variations are
$$
\delta\psi=-\alpha\left(x^\nu\partial_\nu+\frac32\right)\psi,
\qquad
\delta\bar\psi=-\alpha\left(x^\nu\partial_\nu+\frac32\right)\bar\psi.
$$
The associated <dilatation current> is
$$
D^\mu=x_\nu T^{\mu\nu},
\qquad
T^{\mu\nu}=i\bar\psi\gamma^\mu\partial^\nu\psi,
$$
up to an improvement term. Its divergence is the on-shell trace $T^\mu{}_{\mu}=0$.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 301 / 2 / i / Solution

{title2=$D^\mu$}

The dilatation current associated with scale transformations is $D^\mu=x_\nu T^{\mu\nu}$ up to improvement terms. Its divergence is the trace of the stress-energy tensor, so classical scale invariance requires a traceless improved tensor on shell.


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