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Consider the Lagrangian of $\phi^4$ theory

$$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi \partial^\mu\phi - \frac{\lambda}{4!}\phi^4. $$

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

$$ d^4x \rightarrow e^{4\alpha} d^4x,\\ \partial_\mu \rightarrow e^{-\alpha} \partial_\mu,\\ \mathcal{L} \rightarrow e^{-4\alpha} \mathcal{L}. $$

The problem with $\mathcal{L} \rightarrow e^{-4\alpha} \mathcal{L} = \mathcal{L} -4\alpha\mathcal{L} + O(\alpha^2)$ 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 $0$

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    $\begingroup$ The current for a spacetime symmetry generated by Killing vector $\xi^\mu$ is $j_\mu = \xi^\nu T_{\mu\nu}$ (where $T_{\mu\nu}$ is the symmetrized Belinfante stress tensor). For a scale symmetry, $\xi^\mu = x^\mu$. $\endgroup$ – Prahar Dec 30 '19 at 16:33
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    $\begingroup$ Also, I'm confused about the transformation of your Lagrangian - you have a mass term that breaks the scale symmetry, right? $\endgroup$ – Prahar Dec 30 '19 at 16:38
  • $\begingroup$ Please reference the source of this homework question. This is one of the things that we ask you to do in our homework policy: physics.meta.stackexchange.com/q/714 $\endgroup$ – user4552 Dec 30 '19 at 23:44
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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 $\partial_{\mu}J^{\mu} = 0$.

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  • $\begingroup$ Given that $m=0$, the equation of motion is $\partial^2\phi = -\frac{\lambda}{3!}\phi^3$, so $\partial_\mu J^\mu = -4\mathcal{L} = \phi\partial^2\phi - 2\partial_\mu\phi\partial^\mu\phi$. Can we solve for $J^\mu$? $\endgroup$ – Bernoulli Dec 31 '19 at 5:00
  • $\begingroup$ I disagree with your form of $J$ because you have not calculated the infinitesimal variations of the field; moreover, the equations of motion should only be used to show that the continuity equation is satisfied. See my updated answer... $\endgroup$ – lux Jan 3 at 16:25
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There is a general formulae for the conserved current due to such translation symmetries. Indeed, assume that under the symmetry $x\mapsto x'=x+\alpha x$ and $\phi\mapsto \phi'$, with $\phi'(x')=\phi(x)+\alpha d\phi(x)$ 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, $$S_{\Omega'}(\phi')-S_\Omega(\phi)=\int_\Omega \left(\delta(d^Dx)\mathcal{L}+d^Dx\bar{\delta}\mathcal{L}\right).$$ In here $\bar\delta$ 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 $\bar{\delta}=\delta x^\mu\partial_\mu+\delta$. For example $\delta\phi=\alpha d\phi-\alpha x^\mu\partial_\mu\phi$. Using $\bar{\delta}\mathcal{L}=\delta\mathcal{L}+\delta x^\mu\partial_\mu\mathcal{L}$ and $\delta(d^D x)=d^D x\partial_\mu\delta x^\mu$, we obtain $$0=S_{\Omega'}(\phi')-S_\Omega(\phi)=\int_\Omega d^Dx \left(\partial_\mu(\delta x^\mu\mathcal{L})+\delta\mathcal{L}\right).$$ We conclude that $\delta\mathcal{L}=-\alpha\partial_\mu(x^\mu\mathcal{L})=\partial_\mu F^\mu$ and thus we have the conserved current $$j^\mu=\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\delta\phi-F^\mu=\alpha\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}(d\phi-x^\mu\partial_\mu\phi)+x^\mu\mathcal{L}\right).$$ One can specialize this result to your case by taking $d=-1$.

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