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In the $7.1.1$ of David Tong's String Theory notes it is said the following about regularization of Polyakov action in a curved target manifold:

$$\tag{7.3} S= \frac{1}{4\pi \alpha'} \int d^2\sigma \ G_{\mu\nu}(X)\partial_a X^\mu \partial^a X^\nu.$$

Classically, the theory defined by (7.3) is conformally invariant. But this is not necessarily true in the quantum theory. To regulate divergences we will have to introduce a UV cut-off and, typically, after renormalization, physical quantities depend on the scale of a given process µ. If this is the case, the theory is no longer conformally invariant.

In this theory the target space metric $G$ takes the role of a set of coupling constants $G_{\mu \nu}$

On an Euclidean worldsheet, by using dimensional regularization the two-point function is now multiplied by a dimensionful scale $\mu^{d-2}$. After renormalization, the renornalized coupling constants $G^R_{\mu \nu}$ actually depend on $\ln(\frac{\mu}{m})$, $m$ being a factor to avoid IR divergencies. The logarithmic derivatives of these coupling constants gives us the beta functional that Tong later refers to, whose vanishing is relates to conformal invariance of the theory.

How this dependence breaks the scale invariance in the mathematical sense of the theory defined by ($7.3$)? I tried to convince myself that the scale $\mu$, if determined experimentally, would tell us that the quantum corrections of the action become measurable at a energy or distance scale associated with $\mu$. However I can't relate this thought with the mathematical notion of scale invariance of the action. The role point of this is to justify the vanishing of the beta functional associated with the couplings.

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  • $\begingroup$ So you're studying string theory before Yang-Mills theory? $\endgroup$ Nov 12, 2023 at 3:45

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The basic idea is that the classical action itself does not fully define the quantum theory. That's actually not that surprising, when you consider there are, eg, ordering ambiguities of operators where different quantum theories could reduce to the same classical theory.

In this context, the point is that you need to define a regularization and renormalization scheme to define the theory. You can already see where there might be some tension. On the one hand, the classical action is scale invariant. Therefore, you might think the quantum theory is scale invariant. If the quantum theory is scale invariant, then there cannot be any special energy scales. On the other hand, introducing renormalization will generically lead to beta functions that run with the coupling constant. That will introduce an energy scale -- for example, the energy scale at which the coupling constants become of order 1 (and therefore the theory transitions from being perturbative to non-perturbative). The only way to avoid this tension, is for the beta functions to be zero. Then there is no special energy scale that appears in the answer to physical questions, so physical quantities will not have a special scale dependence, which is what you would expect for a scale invariant theory.

In other words: the classical action actually is scale invariant. There is no analysis you can apply to it that will tell you whether the quantum theory is scale invariant or not. But for the quantum theory to be scale invariant, it had better be that physical quantities like S matrix elements are scale invariant. A non-zero beta function is a sure sign that physical quantities will have a special scale and therefore not be scale invariant. The reason the quantum theory can end up not being scale invariant, even though the classical theory is, is because to define the quantum theory you need to introduce a regularization and renormalization procedure, which is extra input beyond the classical action, and it may not be possible to introduce these inputs in a way that maintains scale invariance.

This is a specific example of a much more general phenomenon called quantum anomalies, where symmetries of the classical action are not symmetries of the corresponding quantum theory.


To expand on the idea of observables scaling "in a simple way" (as requested in the comments)...

Note that the conformal algebra includes the commutation relation $$ [D, P^\mu] = i P^\mu $$ where $D$ is the dilation operator (the generator of changes in scale), and $P^\mu$ is the translation operator.

To see what this means, consider a finite dilation transformation, $U = e^{i \lambda D}$. Under a dilation, the coordinates transform as $x \rightarrow \lambda x$; this will cause other "simple" operators to transform according to their "scaling dimension". For instance, the derivative $\partial$ will transform with scaling dimension $\Delta=-1$. Now consider a generic operator $\mathcal{O}$ with scaling dimension $\Delta$ \begin{eqnarray} U^\dagger(\lambda) \mathcal{O} U(\lambda) &=& \lambda^\Delta \mathcal{O} \end{eqnarray} For $\lambda = 1 + \epsilon$ and small $\epsilon$... $$(1 - i \epsilon D) \mathcal{O} (1 + i \epsilon D) = (1 + \epsilon)^\Delta \mathcal{O} $$ Keeping terms to first order in $\epsilon$ implies $$ [D, \mathcal{O}] = - i \Delta \mathcal{O} $$ From this, and the commutation relation $[D, P^0]$ given above, we can read off that the conformal scaling dimension of $P^0$ is $\Delta=-1$. Of course, you already knew that, since energy has dimensions of one over distance when $\hbar=c=1$.

The time translation operator $P^0$ is equivalent to the energy. This commutation relation says that if you rescale coordinates $x\rightarrow \lambda x$, then the energy should scale as $P^0=E \rightarrow \lambda^{-1} E$.

We can see this explicitly for a $\phi^4$ theory at the classical level.

In units with $\hbar=c=1$, the energy is

$$ H = \int d^3 x \left(\frac{1}{2} \dot{\phi}^2 + \frac{1}{2}(\nabla \phi)^2 + g\phi^4\right) $$ I'm using $g$ instead of the more conventional $\lambda$ as the coefficient of $\phi^4$ to avoid confusion with the scaling parameter $\lambda$ used above. Each term in the energy density scales as $L^{-4}$, where $L$ is a distance scale, while the measure scales as $L^3$. So the overall energy scales as $L^{-1}$ as expected from the conformal algebra. (Normally physicists look at scaling with energies instead of distances, but here we want to make the connection with the conformal algebra).

Meanwhile, if $g$ is a running coupling that depends on distance scales (or energy scales), then the relative strength of the kinetic/gradient terms and the $g\phi^4$ term in the energy will change at different distances, and the energy will not scale as a simple power law, in contradiction with the conformal algebra. This is a way of seeing how the running coupling will destroy the scaling symmetry of the classical action.

Explicitly, consider transforming $x\rightarrow \lambda x$ inside the integral above. We know how the derivatives and fields transform. Now suppose the coupling $g$ changes as $g\rightarrow \lambda^{\Delta_Q} g$ where $\Delta_Q$ is the "quantum" scaling dimension of $g$ that it picks up from the beta function. If the beta function were 0, then $\Delta_Q=0$. Then $$ H = \lambda^{-1} \int d^3 x \left(\frac{1}{2} \dot{\phi}^2 + \frac{1}{2}(\nabla \phi)^2 + \lambda^{\Delta_Q} g\phi^4\right) $$ The factor of $\lambda^{\Delta_Q}$ prevents $H$ from transforming in the way you'd expect in a scale invariant theory. If the beta function were zero, so $\Delta_Q=0$, then the transformation would be ok.

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  • $\begingroup$ "A non-zero beta function is a sure sign that physical quantities will have a special scale and therefore not be scale invariant". Are you sure about that? Or should we say: A non-zero beta function which diverges/blows up at some specific energy scale is scale invariant, since the blow-up energy point is a fixed energy scale. However, on the other hand, merely a generic non-zero beta function just implies that the coupling constant is running, but it does not necessarily pinpoint any specific energy level that is peculiar, right? $\endgroup$
    – MadMax
    Nov 13, 2023 at 18:00
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    $\begingroup$ @MadMax If the beta function is nonzero, then the physics at one scale is different from the physics at another scale (since the coupling constant runs), so the physics is not scale invariant. $\endgroup$
    – Andrew
    Nov 13, 2023 at 18:07
  • $\begingroup$ Let me take the classical mechanics with a pure kinetic term for example, which you would consider as conformally invariant. However, the total kinetic energy at one velocity scale is different from the one at another scale. Since the kinetic energy "runs", should we regard this as a proof that it is not scale invariant? $\endgroup$
    – MadMax
    Nov 13, 2023 at 18:12
  • $\begingroup$ @MadMax The energy density operator $\sim \dot{\phi}^2$ in that case would be a primary operator with scaling dimension $4$. If you want to think classically, any specific background solution will break scale invariance, but the energy of the solution will scale in a simple way, and you can relate two classical solutions by a scaling transformation. The problem with the running coupling is that the energy will no longer have a scaling dimension because different terms in the energy will scale by different relative amounts due to the running coupling. (...) $\endgroup$
    – Andrew
    Nov 13, 2023 at 21:00
  • $\begingroup$ (...) If you account for the running coupling then classically, the rescaled version of one classical solution will not be a solution. $\endgroup$
    – Andrew
    Nov 13, 2023 at 21:01

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