Weird condition describing symmetry transformation

[Markus Kahn]

Homework Statement

Consider the real scalar field with interactions
$$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi-\frac{1}{2} m^{2} \phi^{2}-\frac{1}{6} \mu \phi^{3}-\frac{1}{24} \lambda \phi^{4}.$$
The coordinate scaling transformation $x^\prime := a x$ for $a\in\mathbb{R}^+$ can be extended to the scalar field by $\phi^\prime(x^\prime) := a^{-\varepsilon}\phi(x)$ for some $\varepsilon\in \mathbb{R}.$ For which values of the parameters $\{m,\mu,\lambda,\varepsilon\}$ is the action scale-invariant?

Relevant Equations

None.

I'm a bit confused about the condition given in the description of the symmetry transformation of the filed. Usually, given any symmetry transformation $x^\mu \mapsto \bar{x}^\mu$, we require
$$\bar\phi (\bar x) = \phi(x),$$
i.e. we want the transformed field at the transformed coordinates to take the same value as the old field did at the corresponding old coordinates. But in this exercise the corresponding symmetry transformation seems to obey
$$\phi^\prime (x^\prime)= a^{-\varepsilon}\phi(x),$$
which seems to contradict the usual requirement we have for symmetry transformations. I was expecting something like
$$\phi^\prime (x) = \phi(x^\prime)\quad \Longleftrightarrow \quad a^{\varepsilon} \phi (x)=\phi(ax),$$
but I'm not really sure if this makes more sense...

Can somebody explain to me why this makes sense in this case?

 

[Markus Kahn]

I think I've got the answer, feel free to correct me if I'm wrong.

The point that I missed is that we require $\phi^\prime (x^\prime) = \phi(x)$ only for Lorentz transformations, i.e. we want the scalar field to transform like a scalar under a Lorentz transformation, but we don't make any assumptions on the transformation properties of scalar fields for general transformations.

 

[PeroK]

I think I've got the answer, feel free to correct me if I'm wrong.

The point that I missed is that we require $\phi^\prime (x^\prime) = \phi(x)$ only for Lorentz transformations, i.e. we want the scalar field to transform like a scalar under a Lorentz transformation, but we don't make any assumptions on the transformation properties of scalar fields for general transformations.

This is quite fundamental. If we make a coordinate transformation (such as a Lorentz transformation), then a covariant Lagrangian cannot change. You must get the same Lagrangian, but in a different frame/coordinate system. With this variational theory applied to Lagrangians that is not what we are doing.

Instead, we are asking: what if we keep the form of the Lagrangian the same and we keep the same scalar field, but we evaluate that same scalar field at different coordinates, then how does the Lagrangian change? Note that, for example, the derivatives $\partial_{\mu}$ do not change with the coordinates. In the new Lagrangian we will have terms like:
$$\partial_x \phi(x') = \partial_x \phi (x + \delta x)$$
Etc.

Now, at first sight this might seem like an idle question. What does it matter how the Lagrangian changes when we do that? But, we have Noether's theorem. And that relates invariance under such variational processes to symmetry laws. In other words, if we do this somewhat strange variational process on the Lagrangian and we get $L' = L$, then we have a conserved Noether current. This current depends on the parameters associated with the particular coordinate transformation: so the trick is to find the (right) coordinate transformation(s) that lead to invariance of the Lagrangian and hence to a conserved current.

Check out Noether's theorem for the details on this.

Note that in general we also get a conserved current if the Lagrangian varies by a four-divergence. I.e.:
$$\delta L = (\partial_{\mu}W^{\mu})\delta \lambda$$
Where $\lambda$ is the parameter associated with infinitesimal variations. So, we also get a conserved current in this case - of which the $W^{\mu}$ is a component.

Note also that Noether's Theorem extends to variations in the field itself, such as considering $\phi' = e^{i\alpha} \phi$ and asking how that changes the Lagrangian.

The main reference I have for this is Neuenschwander's book:

https://link.springer.com/article/10.1007/s10701-011-9562-3