Simple conceptual question conformal field theory

Question

I come up with this conclusion after reading some books and review articles on conformal field theory (CFT).

CFT is a subset of FT such that the action is invariant under conformal transformation of the fields and coordinate but leave the metric unchanged.

Is this correct?

Let me explain further and take the $\phi^4$ theory in $4$-dim as an example (just discuss classical invariance, I know that loops break the invariance). In $4$-dim consider a scalar field with conformal weight $\Delta=1$ such that \begin{align} x \to x' = \lambda x,\\ \phi'(x')=\phi'(\lambda x) = \lambda^{-1}\phi(x). \end{align} Then the action is unchanged \begin{align} S'& = \int d^4 x' \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\partial'_{\mu}\phi'(x')\partial'_{\mu}\phi'(x')-\phi'^4(x')\right\}\\ &= \int d^4 x \lambda^4 \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\lambda^{-4}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\lambda^{-4}\phi^4(x)\right\} \\ &= \int d^4 x \sqrt{g} \left\{\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\phi^4(x)\right\}. \end{align} Note that I did not use $g'^{\mu\nu} = \lambda^2g^{\mu\nu}$, all metrics are unprimed. In this example we see that conformal invariance is realized without changing the metric. I was confused at the beginning since all textbooks and articles derive the conformal group and representations by considering the change of the metric.

If we use $g'_{\mu\nu}x'^{\mu}x'^{\nu} = g_{\mu\nu}x^{\mu}x^{\nu}$, the physical distance does not change at all and we are just choosing a new coordinate chart. My interpretation is, as what we meant by a physical scaling or transformation, we really change the distance between two points. Another reasoning is, the metric in CFTs are just background (not being integrated in the path integral) thus we do not change them. If we consider a theory including the metric as a dynamical field (we path integrate it and perhaps quantize it), the actions has to be invariant including the transformation of the metric.

Is the above correct? Please give me some comments and point out the wrong concepts if there is any. Thank you very much.

If you have time, could you please take a look at my other question.

Answer 1

Imposing that an action should be conformally invariant has a subtle but important difference with respect to imposing that it should be diffeomorphism invariant. Let me try to emphasize the differences between Diffeomorphisms, Conformal transformations and Weyl transformations. I will also clarify how to impose the invariance of the action, namely I'll argue whether the fields or the coordinates must be changed.

Diffeomorphisms

The action on the coordinates is given by a general differentiable function with differentiable inverse: $x^\mu \to {x'}^\mu = {x'}^\mu(x)$. The action on fields with spin 0, 1 and 2 is the following $$ \begin{align} \phi(x) &\to \phi'(x') = \phi(x)\,, \\ \partial_\mu\phi(x) &\to \partial'_\mu\phi'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\,, \\ g_{\mu\nu}(x) &\to g'_{\mu\nu}(x') = \frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta}(x)\,. \end{align} $$ These transformations are just changes of variables, every theory is invariant under them.

Weyl transformations

Weyl transformations do not act on the coordinates but simply rescale the fields $$ \begin{align} \phi(x) &\to \tilde{\phi}(x) = \Omega^{-\Delta}(x)\phi(x)\,, \\ \partial_\mu\phi(x) &\to \partial_\mu\tilde{\phi}(x) = \partial_\mu(\Omega^{-\Delta}(x)\phi(x))\,, \\ g_{\mu\nu}(x) &\to \tilde{g}_{\mu\nu}(x) = \Omega^{2}(x)g_{\mu\nu}(x)\,. \end{align} $$

Conformal transformations

Conformal transformations are coordinate transformations that preserve the metric up to a scale factor, that is $x^\mu \to {x'}^\mu$ and $$ \frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu} g_{\alpha\beta}(x)= \Omega^{-2}(x)g_{\mu\nu}(x)\,.\tag{1} $$ In OP's example ${x'}^\mu = \lambda x^\mu$ and $\Omega(x) = \lambda$. Now, this is the key point: we don't want to impose that the theory is invariant under this change of variables, this is always true because it is a particular case of diffeomorphism invariance. We want instead to compare theories with the same metric. That is, we always imagine a conformal transformation as a diffeomorphism satisfying (1) followed by a Weyl rescaling that cancels the $\Omega$ factor. Therefore the transformation laws are $$ \begin{align} \phi(x) &\to \tilde{\phi}'(x') = \Omega^{-\Delta}(x)\phi(x)\,, \\ \partial_\mu\phi(x) &\to \partial'_\mu\tilde{\phi}'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\,, \\ g_{\mu\nu}(x) &\to \tilde{g}'_{\mu\nu}(x') = g_{\mu\nu}(x)\,. \end{align} $$ With the notation "tilde prime" I denote the composition of the two last transformations.

Invariance of the action

Consider a generic Lagrangian that depends on the field $\phi$, its first derivative and the metric tensor. The action then reads $$ S = \int d^4x \, \mathcal{L}[g_{\mu\nu}(x),\phi(x),\partial_\mu\phi(x)]\,. $$ The transformation acts only on the fields and leaves $x$ unchanged since it is a dummy variable (integrated over). Proving diffeomorphism invariance amounts to requiring $$ \begin{align} S &= \int d^4x\, \mathcal{L}\left[g'_{\mu\nu}(x),\phi'(x),\partial'_\mu\phi'(x)\right] = \\ &=\int d^4x'\, \mathcal{L}\left[g'_{\mu\nu}(x'),\phi'(x'),\partial'_\mu\phi'(x')\right] = \\ &=\int d^4x\,|\Omega(x)|^4 \mathcal{L}\left[\Omega^{-2}(x)g_{\mu\nu}(x),\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\right]\,, \end{align} $$ in the second step I just renamed $x$ to $x'$, it's a trivial operation. In the last step I used the transformation properties of the measure and of the fields (See Di Francesco - Mathieu - Sénéchal for example).

A Weyl rescaling requires simply $$ S = \int d^4x\, \mathcal{L}\left[\Omega^2(x)g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\partial_\mu(\Omega^{-\Delta}(x)\phi(x))\right]\,. $$ Finally, the requirement for conformal invariance can be obtained by composing the two transformations above $$ S = \int d^4x\, |\Omega(x)|^4\mathcal{L}\left[g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\right]\,. $$ This is consistent with what OP says. Notice that since diffeomorphism invariance is always true, just proving Weyl invariance is enough. The converse is not obvious and it is object of current research. We could say that by imposing conformal invariance we are imposing Weyl invariance only by those $\Omega$'s that make $g_{\mu\nu}$ and $\Omega^2 g_{\mu\nu}$ diffeomorphic.

Answer 2

I think in your case it is better to see conformal transformation this way:

1. You don't change coordinates
2. Field with conformal dimension $\Delta$ transforms as $\widetilde \Phi = \lambda^{- \Delta} \Phi$
3. Field $\phi$ has $\Delta=1$ and metrics $g_{\mu\nu}$ has $\Delta = 2$.

So parts of the action change just as you wrote, but for different reasons. Volume form change beacause $\sqrt{\det \widetilde g} = \sqrt{\det \widetilde g_{\mu\nu}}= \lambda^{4} \sqrt {\det g}$. Derivative $\partial_\mu$ doesn't change, however term $g^{\mu\nu}$ changes as $\widetilde g^{\mu\nu}=\lambda^{-2}g^{\mu\nu}$. And the $\phi$ scales the same way.

This gives the same result as you wrote in your question, but here it is more clear that we do actual physical transformation, not just change of coordinates.

In your case, you make conformal transformation as I just described it along with coordinate transformation $x \rightarrow \lambda x$. All together the fields transform as you wrote $dx^\mu \rightarrow \lambda dx^\mu, g_{\mu\nu}\rightarrow g_{\mu\nu}$ e.t.c., except that metric actually changes $\widetilde{ ds}^2 = \lambda^2 g_{\mu\nu} dx^\mu dx^\nu=\lambda^2 ds^2$.