homework and exercises - Most general Lagrangian in CFT in 0+1D

My question is about $CFT_1$. Page 18 of this says that

$$L={\frac{\overset{.}{Q}^2}{2} - \frac{g}{2Q^2}}\tag{1.11}$$ is the most general Lagrangian that preserves time translation and scale invariance in $CFT_1$. How does one prove that?

Answer

Classically a theory is invariant under a transformation if its action is invariant (up to boundary terms). In our case a conformal transformation is given by

$$t'=\lambda t\\ Q'=\lambda ^{-\Delta}Q$$ where $\Delta$ is the scaling dimension of Q, which is just its energy dimension classically.

For now let's assume a Lagrangian with only the kinetic term and infer the dimension. To this end plug the transformed variables into the action

$$\begin{align} S' &=\int \mathrm{d}t' \frac{1}{2}\left(\frac{\mathrm{d}Q'}{\mathrm{d}t'} \right)^2 \\ &=\lambda^{-2\Delta-1}S \end{align}$$ We can read of that $\Delta = -\frac{1}{2}$.

We can now try to add more terms to our Lagrangian but we would not want to include additional kinetic terms so we restrict them to have the form $g_n Q^n$. Plugging these terms into the action we see that they transform as

$$\mathrm{d}t'\ Q'^n = \lambda^{1+\frac{n}{2}} \mathrm{d}t \ Q^n$$ which implies $n=-2$ if we require conformal invariance. We thus showed that the most general Lagrangian is given by $$\mathcal{L} = \frac{1}{2}\dot{Q}^2 - \frac{g}{2Q^2}$$