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?


1 Answer 1


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 $$\mathcal{L} = \frac{1}{2}\dot{Q}^2 - \frac{g}{2Q^2} $$

  • $\begingroup$ Regarding your first two equations: how do we know that a conformal transformation takes this form? For instance why $\lambda^{-\Delta}$ instead of $\lambda^{\Delta}$? ( I apologize if this is a basic question) $\endgroup$
    – Dwagg
    Commented Mar 6, 2018 at 14:05
  • $\begingroup$ Think of it in terms of numbers in different units. I will give an example in 3D because I think it is more visual but the idea in 1D is the same. Say Alice measures things in inches and reports $x=1$ to Bob who prefers centimeters. He now knows $x'=2.54$ in his units. If Alice now tells Bob about some density for example particles per volume and she counts 100 particles per unit volume then in Bob's unit volume he will only find $Q'=2.54^{-3}\cdot100$ particles. It is no coincidence that the energy dimension of this density is 3. $\endgroup$
    – AlexM
    Commented Mar 6, 2018 at 16:04
  • $\begingroup$ However, this theory is not conformal right? I mean, given that any diffeomorphism is a conformal transformation in 1D. $\endgroup$ Commented Jan 25, 2022 at 4:48

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