# Is there a way to make this simple “derivation” of the Trace Anomaly correct?

I think I came up with a simple yet sketchy almost-proof of the trace anomaly (A.K.A. Weyl anomaly) in 2D CFT, but it has the wrong prefactor. I was wondering if anyone could assess whether this "proof" makes any sense at all, and if so, if it can be fixed.

Let's just restrict our attention to the 2D Euclidean free scalar CFT with $$c = 1$$. The two point function is

$$\langle \phi(z_1) \phi(z_2) \rangle = - \ln ( |z_1 - z_2|^2 ).$$

Now imagine that instead of the metric being $$d\bar z dz$$ we introduce a small conformal factor $$\omega$$ so the metric is $$e^{2 \omega} d \bar{z} dz$$. It stands to reason (?) that if the coordinate $$z_1$$ is close to the coordinate $$z_2$$, then we want to use the geodesic distance, which I think should just be

$$\langle \phi(z_1) \phi(z_2) \rangle = - \ln ( e^{2 \omega}|z_1 - z_2|^2 ) = - 2 \omega- \ln (|z_1 - z_2|^2 ).$$

We now want to compute

$$\langle T_{z \bar z} \rangle = \lim_{z_1 \to z_2} \partial_{z_1} \partial_{\bar{z}_2} \langle \phi(z_1) \phi(z_2) \rangle = -2 \partial^2 \omega.$$

When we scale the flat metric by $$e^{2 \omega}$$ we have that $$R = - 2 e^{-2 \omega} \partial^2 \omega$$. For small $$\omega$$ we have $$R = - 2\partial^2 \omega$$. So this gives us

$$\langle T_{z \bar z} \rangle = R.$$

This is of course missing the prefactor $$\tfrac{c}{12} = \tfrac{1}{12}$$. So there is something wrong with this proof. I was wondering if it can be "fixed."

Reasons to be skeptical of this "proof":

• I might not have correctly accounted for normal ordering or computed the geodesic distance to the correct order in $$|z_1 - z_2|$$.
• Seems sketchy and I got the wrong prefactor.
• I didn't seem to have to use that $$\omega$$ was small.

Reasons to be optimistic:

• The reasoning seems intuitively correct, and I almost got the right answer.
• Possible duplicate: physics.stackexchange.com/q/442846/2451 – Qmechanic Sep 13 at 18:51
• You need to explain how you arrived at your equation $\langle T_{z\bar{z}} \rangle = \ldots = -2 \partial^2 \omega$, because if you naively take derivatives of the log, then the limit $z_2 \to z_1$ is divergent. So the first step in getting to a correct proof is to fix this problem. Then with a correct regularization of the 2-point function you should obtain the right result. – M.Jo Sep 23 at 12:47