This question has already been asked and answered in Most general Lagrangian in CFT in 0+1D.

However I am just partially convinced with the answer. The idea is to construct the most general Lagrangian that is scale and time invariant in 0+1D (so quantum mechanics). The proposed one is

$$L=\frac{1}{2}\dot{Q}^2-\frac{g}{2Q^2} \, ,\tag{1.11}$$

but why couldn’t we add terms that have $d$ derivatives and $n$ fields with $n = 2(d-1)$ such as $\dot{Q}^3Q$, $\dot{Q}^4Q^2$ and so on $[a]$, because these terms in principle would respect both time and scaling invariance?

In the answer given in the link above it says something that because we don’t want to modify the kinetic term we propose terms of the form $g_nO^n$ but I don’t see why the terms I have mentioned would modify the kinetic term.

$[a]$: with 3 derivatives and 4 fields, obviously I am counting the derivatives and fields separately since each carry dimensions, I could have also said 3 field derivatives and 1 field.


1 Answer 1

  1. A first-order time-translation-invariant Lagrangian $L(Q,\dot{Q},t)$ cannot have any explicit time-dependence.

  2. The claim in Ref. 1 that scale-invariance implies the Lagrangian (1.11) is strictly speaking wrong as written. Scale-invariance (that is assumed compatible with the free kinetic term) only implies that the Lagrangian is of the form $$L(Q,\dot{Q})~=~\frac{f(Q\dot{Q})}{Q^2}\tag{A}$$
    for some function $f$, cf. OP's observation.

  3. However, Ref. 1 presumably implicitly assumes that the Lagrangian is of the form kinetic term minus potential term: $$L(Q,\dot{Q})~=~\frac{1}{2}\dot{Q}^2-V(Q).\tag{B}$$ This indeed leads to the Lagrangian (1.11).

  4. Alternatively, one may appeal to the following proposition.

    Proposition. In order for special conformal transformations (1.12) to be quasi-symmetries of the action $$S~=~\int\! \! \mathrm{d}t~L,\tag{C}$$ [with the Lagrangian (A)], then the function $f$ should be (at most) a 2nd-order polynomial.

    The proposition reproduces the Lagrangian (1.11) up to normalization and total derivative terms.

  5. Sketched proof of proposition: Using the notation of this related Phys.SE post, the quasi-symmetry implies that $$ \frac{f\left(Q\dot{Q}-\frac{cQ^2}{ct+d}\right)-f(Q\dot{Q})}{Q^2} \tag{D}$$ should be a total time derivative, which is only possible if it is a 1st-order polynomial in $\dot{Q}$. This in turn implies that $f$ is a 2nd-order polynomial. $\Box$


  1. J.D. Qualls, Lectures on CFT, arXiv:1511.04074; eq. (1.11).
  • $\begingroup$ FIrst of all many thanks! Okay, with the additional condition of quasi-symmetric under SCT I get it why it has the form of (A). But in prnciple the author does not mention it at all for constructing it (later he points out that it is also SCT symmetric t). Anyway, somehow he must be assuming either the form of (B) or that it is also SCT symmetric. However, in principle, if we allow a potential like $V(\dot{Q}Q)$ with the terms I said ($\dot{Q}³Q$, etc), that would't modify the kinetic term as (implicitly) stated in the question I linked right?. $\endgroup$ Commented Nov 23, 2019 at 8:45
  • $\begingroup$ Essentially the issue becomes whether we allow potential terms to depend on velocity $\dot{Q}$. $\endgroup$
    – Qmechanic
    Commented Nov 23, 2019 at 12:11

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