If we transform the massless scalar field Lagrangian $$\mathcal{L}=\frac{1}{2}(\partial_\mu\varphi)^2-\frac{\alpha}{4!}\varphi^4$$ with the simultaneous transformations $$x\mapsto x^\prime= \lambda x,\\ \varphi(x)\mapsto \varphi^\prime(x^\prime)= \lambda^{-\Delta}\varphi(x),$$ we discover that $$\mathcal{L}\mapsto\mathcal{L}^\prime=\lambda^{-2(1+\Delta)}\frac{1}{2}(\partial_\mu\varphi)^2-\lambda^{-4\Delta}\frac{\alpha}{4!}\varphi^4.$$

If we choose $\Delta=1$, we find that $\mathcal{L}^\prime=\lambda^{-4}\mathcal{L}$. But the action $\mathcal{S}=\int d^4x \mathcal{L}$ is unchanged. So this is a symmetry.

  • Sometimes the same transformation is mysteriously defined as $x\mapsto e^\lambda x$ and $\varphi(x)\mapsto e^{-\Delta\lambda}x$. Why?

  • Like Lorentz transformations, do scaling transformations also form a group? Are they examples of symmetry transformations which do not form a group?

  • $\begingroup$ Writing as an exponential is nice because it shows the infinitesimal (Lie algebra) generator - expanding to first order in $\lambda$ also gets your variation $\delta x$ for the Noether current... $\endgroup$
    – nox
    Dec 12, 2019 at 8:58

1 Answer 1


Yes classically massless $\phi^4$ has a scaling symmetry. The reason why people don't mention this is because it is not a quantum symmetry i.e. it doesn't survive the quantization, the technical term for it is that it is anomalous. People write it as $e^\lambda$ because there is an implicit assumption of the transformation having a definite sign. So you can define

$$ x \to f(\lambda) x \qquad \text{for } \quad f(\lambda) >0 \;\forall \lambda$$

instead of writing this we can just say

$$ x \to e^{g(\lambda)} x$$

since any positive function can be written as an exponential by simply writing $g(\lambda):= \log(f(\lambda))$. Also exponentials are easier to keep track of compared to arbitrary functions. Hence, the reason why people write $e^\lambda$ is because it's convenient.

Yes scaling transformations together with Lorentz transformations form the conformal group (technically you need one more generator), which gives rise to beautiful world of Conformal Field Theories (CFTs). Putting CFTs aside for another answer, you can check that scaling transformations forms a group. Recall what this means:

$$ \delta_\lambda (\delta_\kappa \phi) = \delta_{\lambda \circ \kappa} \phi $$

where $\circ$ means the group multiplication, in our case $\circ$ is the usual multiplication (or addition if you use the exponential form).

$$ \delta_\lambda (\delta_\kappa \phi) = e^{- \Delta\lambda } e^{-\Delta \kappa} \phi = e^ {-\Delta(\lambda+\kappa)} \phi = \delta_{\lambda + \kappa} \phi$$

Any symmetry must form a group essentially by definition. We need to see at least 3 defining properties of groups to convince ourselves of this: There exists an identity, an inverse, an a group multiplication.

  • Of course you can choose not to transform your fields; this is the identity transformation.

  • Suppose you have a symmetry, which is given by certain parameter $\lambda$ then you transform your field $\delta_\lambda \phi$ and the transform field is again a solution of equations of motion because it was a symmetry. Suppose you undid the transform (if you couldn't then this couldn't have been a symmetry because "you have lost some information"). Call this undoing transformation $\delta_{-\lambda}$. Then schematically $\delta_\lambda \delta_{-\lambda} = Id$. This shows that you have an inverse transformation.

  • Similarly if you did two different symmetry transformations after another $\delta_\lambda \delta_\kappa$ then the result should again be some symmetry transformation $\delta_\eta$. This shows that you have a "multiplication" schematically given as $\delta_\lambda \delta_\kappa = \delta_{\eta(\lambda,\kappa)}$


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