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 $0$

  • 1
    $\begingroup$ 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, 2019 at 16:33
  • 2
    $\begingroup$ 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, 2019 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, 2019 at 23:44

2 Answers 2


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

  • $\begingroup$ 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, 2019 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... $\endgroup$
    – nox
    Jan 3, 2020 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 $$j^\mu=\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\delta\phi-F^\mu=\alpha\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}(d\phi-x^\nu\partial_\nu\phi)+x^\mu\mathcal{L}\right).$$ One can specialize this result to your case by taking $d=-1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.