# Noether's current for dilation transformation

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 $$0$$

• 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, $\xi^\mu = x^\mu$. – Prahar Dec 30 '19 at 16:33
• Also, I'm confused about the transformation of your Lagrangian - you have a mass term that breaks the scale symmetry, right? – Prahar Dec 30 '19 at 16:38
• 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 – user4552 Dec 30 '19 at 23:44

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 $$\partial_{\mu}J^{\mu} = 0$$.

• 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 $J^\mu$? – Bernoulli Dec 31 '19 at 5:00
• 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... – lux Jan 3 at 16:25

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 $$d=-1$$.