Noether's current for dilation transformation

Question

Consider the Lagrangian of $phi^4$ theory

$$
mathcal{L} = frac{1}{2}partial_muphi partial^muphi - frac{lambda}{4!}phi^4.
$$

We define the following dilation transformation

$$
x^mu rightarrow e^alpha x^mu,
phi(x) rightarrow e^{-alpha}phi(x).
$$

How do we find the Noether current associated with this transformation? We can see that the action $S = int d^4x mathcal{L}$ is invariant under dilation because

$$
d^4x rightarrow e^{4alpha} d^4x,
partial_mu rightarrow e^{-alpha} partial_mu,
mathcal{L} rightarrow e^{-4alpha} mathcal{L}.
$$

The problem with $mathcal{L} rightarrow e^{-4alpha} mathcal{L} = mathcal{L} -4alphamathcal{L} + O(alpha^2)$ is that it is not of the form $mathcal{L} rightarrow mathcal{L} + alpha partial_mu J^mu$, so how do we find the conserved current?

Edit: Sorry $m$ is supposed to be $0$

lux · Accepted Answer

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
$$phi'(x) = (1 - alpha)phi(x - alpha x) + mathcal{O}(alpha)= phi(x) - alphaleft(1 + x cdot partialright)phi(x)+ mathcal{O}(alpha)$$
and we have found
$$delta_{alpha}phi(x) = -alphaleft(1 + x cdot partialright)phi(x)$$
Now we will continue to find the variation in the derivative of $phi
$:
$$delta_{alpha}partial_{mu}phi(x) = -alphaleft(2 + x cdot partialright)partial_{mu}phi(x)$$
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 $partial_{mu}J^{mu} = 0$.

Iván Mauricio Burbano · Answer

There is a general formulae for the conserved current due to such translation symmetries. Indeed, assume that under the symmetry $xmapsto x'=x+alpha x$ and $phimapsto phi'$, with $phi'(x')=phi(x)+alpha dphi(x)$ and $alpha$ is infinitesimal, the action is invariant. More precisely if $Omega$ is a region of spacetime and $Omega'$ is the dilation of this region, we assumet that $S_{Omega'}(phi')=S_Omega(phi)$. This fixes the transformation behavior of the Lagrangian. Indeed,
$$S_{Omega'}(phi')-S_Omega(phi)=int_Omega left(delta(d^Dx)mathcal{L}+d^Dxbar{delta}mathcal{L}right).$$
In here $bardelta$ denotes the variation $bar{delta}F(x)=F'(x')-F(x)$. In terms of the functional variation $delta F(x)=F'(x)-F(x)$ we have $bar{delta}=delta x^mupartial_mu+delta$. For example $deltaphi=alpha dphi-alpha x^mupartial_muphi$. Using $bar{delta}mathcal{L}=deltamathcal{L}+delta x^mupartial_mumathcal{L}$ and $delta(d^D x)=d^D xpartial_mudelta x^mu$, we obtain
$$0=S_{Omega'}(phi')-S_Omega(phi)=int_Omega d^Dx left(partial_mu(delta x^mumathcal{L})+deltamathcal{L}right).$$
We conclude that $deltamathcal{L}=-alphapartial_mu(x^mumathcal{L})=partial_mu F^mu$ and thus we have the conserved current
$$j^mu=frac{partialmathcal{L}}{partialpartial_muphi}deltaphi-F^mu=alphaleft(frac{partialmathcal{L}}{partialpartial_muphi}(dphi-x^nupartial_nuphi)+x^mumathcal{L}right).$$
One can specialize this result to your case by taking $d=-1$.

Noether's current for dilation transformation

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