In the following notes on CFT by Joshua D. Qualls. We're introduced to conformal quantum mechanics with lagrangian:$$L=\frac{1}{2}\dot{Q}^2-\frac{g}{2Q^2}\tag{1.11}$$ It's action is invariant under $SL(2,\mathbb{R})$, and then we proceed to write this group generators $(1.17)$ in infinite-dimensional representation $(1.21)$. This question pertains to scale generator $(D=t\frac{d}{dt})$ commutator with $Q$. I am trying to prove following commutation relation


My calculation using a test function goes like this: $$\bigg[t\frac{d}{dt},Q\bigg]f=t\frac{d}{dt}(Qf)-Qt\frac{d}{dt}(f)$$ $$\require{cancel}\implies tf\frac{d}{dt}Q+\cancel{tQ\frac{d}{dt}f}-\cancel{Qt\frac{d}{dt}(f)}$$ $$\implies\bigg[t\frac{d}{dt},Q\bigg]=t\frac{d}{dt}(Q)$$ Where am I missing the terms? Am I supposed to use equation of motion?


1 Answer 1


It seems Qualls are using the fact that the conformal algebra (1.20) is invariant if we shift the realization (1.21) by $$ D~\longrightarrow~ D + a{\bf 1},\qquad K~\longrightarrow~ K + 2at{\bf 1}, \qquad a~\in~\mathbb{C}.$$ Now pick $a=-\frac{1}{2}$.

Finally rotate all generators by the imaginary unit $i$, cf. remark below eq. (1.20).


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