# Noether's current for dilation transformation

Firstly is there a transformation on $m^2$? Otherwise what happens to the second term?

Expand the variation of the Lagrangian to first order in $\alpha$ - Noether's theorem uses the infinitesimal form of the transformation. Then you can look for the solution to the equation $\partial _\mu J^\mu =-\mathcal{L} $.

*Substantial edit:*

In order to examine the conserved current one needs to determine the infinitesimal transformations of the fields. Let's work them out. Firstly the finite transformations are $$\begin{align} x &\longrightarrow x' = \textrm{e}^{\alpha}x \\ \phi(x) & \longrightarrow \phi'(x') = \textrm{e}^{-\alpha}\phi(x) \end{align}$$ where the second line follows from the definition of $\phi$ as a scalar field of conformal weight given in OP's question. Now we define infinitesimal variations in the field as follows: $$ \delta_{\alpha}\phi(x) = \phi'(x) - \phi(x)$$ and we'll be interested in the variation to linear order in $\alpha$.

For us, we will make an active transformation so that with $\phi'(y') = \phi'(\textrm{e}^{\alpha}y) = \textrm{e}^{-\alpha}\phi(y)$ we deduce that $\phi'(x) = \textrm{e}^{-\alpha}\phi(\textrm{e}^{-\alpha}x)$. To find the Noether current we should expand to linear order in $\alpha$ which leads to $$\phi'(x) = (1 - \alpha)\phi(x - \alpha x) + \mathcal{O}(\alpha)= \phi(x) - \alpha\left(1 + x \cdot \partial\right)\phi(x)+ \mathcal{O}(\alpha)$$ and we have found $$\delta_{\alpha}\phi(x) = -\alpha\left(1 + x \cdot \partial\right)\phi(x)$$ Now we will continue to find the variation in the derivative of $\phi $: $$\delta_{\alpha}\partial_{\mu}\phi(x) = -\alpha\left(2 + x \cdot \partial\right)\partial_{\mu}\phi(x)$$ where I assumed that $\alpha$ is constant.

Now we recall that construction of the current: if $\delta_{\alpha}\mathcal{L} = \alpha \partial_{\mu}f^{\mu}$ then the current defined by $\alpha J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\delta_{\alpha}\phi - \alpha f^{\mu}$ satisfies the continuity equation $\partial_{\mu}J^{\mu} = 0$.

In the current case, then I leave it as an exercise to verify $$\delta_{\alpha} \mathcal{L} = -\alpha \partial_{\mu}\left(x^{\mu}\mathcal{L}\right)$$ so that $f^{\mu} = -x^{\mu}\mathcal{L}$ and $$-J^{\mu} = \phi \partial^{\mu}\phi + (\partial^{\mu}\phi) (x \cdot \partial \phi) - x^{\mu}\left(\frac{1}{2}\partial_{\nu}\phi \partial^{\nu}\phi - \frac{\lambda}{4!}\phi^{4}\right)$$ and to check that the equations of motion imply that $\partial_{\mu}J^{\mu} = 0$.