How does the stress tensor transform under a conformal transformation? I'm new to conformal field theory and I expected this question to have a simple answer but I can't find any simple answers online. I'm starting to suspect that there's some subtlety, or that perhaps my question is not well defined.
More precisely, suppose I have a stress tensor $T^{ab}$ defined on a spacetime given by $g_{ab}$, and the theory is a conformal field theory. If I perform a conformal transformation
$$ g_{ab} \rightarrow \tilde{g}_{ab} = \Omega^2 g_{ab}$$
what happens to the stress tensor? I would imagine it behaves like
$$T^{ab} \rightarrow \tilde{T}^{ab} = \Omega^{-\Delta} T^{ab}$$
and for whatever reason I was under the impression that $\Delta$ was equal to the spacetime dimension $d$, which is odd for the case that I'm considering.
Is there more information needed to answer this question? Does the conformal transformation properties of the stress tensor depend on what kind of theory I'm considering? Is my question even well defined?
 A: Properly speaking that isn't a conformal transformal, it's a Weyl transformation. But yes, under a Weyl transformation $T_{\mu \nu}$ transforms precisely like you mention, with $\Delta = d$, regardless of the theory you're considering. You can derive the fact that $\Delta = d$ in several ways, either starting from the definition of $T$ as $\sim \delta S/\delta g^{\mu \nu}$, or in the CFT language you can consider the two-point function
$$\langle T_{\mu \nu}(x) T_{\rho \sigma}(y) \rangle$$
and prove that $\partial_{\mu} T_{\mu \nu} = 0 \leftrightarrow \Delta = d$.
A: Great questions. Well first of all I agree with Marty about the energy dimension of the Energy-Momentum tensor. Conserved currents of spin $l$ have scaling dimension $\Delta = d +l-2$. That makes $\Delta_T = d$.
As for the confusion between Weyl transformation and conformal transformation: a Weyl transformation physically changes the metric space that your theory lives in and that's why the metric changes. A conformal transformation however simply changes your coordinate system. The metric is the same, but it looks different in different coordinates. Just like when you rotate a vector - the object stays the same, but its components look different.
The clearest way I know to express this is the following. You have some physical space with points $P$. Some guy A writes labels on each point and calls them $x(P)$. Some other guy B also labels the space with different labels $\tilde{x}(P)$. Now given one of A's labels we are able to figure out which point it is and therefore we know what $\tilde{x}$ would correspond to it. This gives us a relation $\tilde{x}=f(x)$. If you now ask A and B what their metric at a specific point $P$ looks like they will give different answers. These answers will be related as follows.
\begin{equation}
g_{\mu \nu}(\tilde{x}(P)) = \frac{\partial  x^\alpha}{\partial \tilde{x}^\mu} \frac{\partial x^\beta}{\partial \tilde{x}^\nu} g_{\alpha \beta}(x(P))
\end{equation}
If this takes the specific form
\begin{equation}
g_{\mu \nu}(\tilde{x}(P)) = \Omega^2(x(P))g_{\mu \nu}(x(P))
\end{equation}
we call that a conformal transformation.
Now let's move on to your actual question. As explained before your energy momentum tensor is physically the same as before, just the numbers describing it look different. It is almost true that it transforms the way you suggested. However a tensor also transforms under rotations. Aside from the fact that straight up rotations are indeed conformal transformation, you could also have local twisting going on in more general conformal transformations. Writing this down in all generality seems pretty nasty but I do know how to write it down infinitesimally. Taking the coordinate transform
\begin{equation}
\tilde{x}^\mu = x^\mu + a^\mu +\omega^\mu  _{\; \nu}x^\nu+\lambda x^\mu + 2\, x^\mu (x \cdot b) - b^\mu x^2
\end{equation}
leads to the following field transformation
\begin{equation}
\delta T = -i\left(a^\mu P_\mu +\lambda D + b^\mu K_\mu + \frac{1}{2}\omega^{\mu \nu} L_{\mu \nu}\right) T
\end{equation}
with
\begin{split}
\mathrm{(Scaling) \qquad}  \ D \ \ &= -i (\Delta  - x^\mu \partial_\mu )\\
\mathrm{(Translation) \qquad} \ P_\mu \ &= -i \partial_\mu  \\
\mathrm{(SCT) \qquad}  K_\mu \ &=  2 x_\mu -i\Delta + i( x^2 \partial_\mu  -  2x_\mu x^\nu \partial_\nu ) - 2 x^\nu S_{\mu \nu} \\
\mathrm{(Lorentz) \qquad}  L_{\mu \nu} &= i(x_\mu \partial_\nu - x_\nu \partial_\mu) + S_{\mu \nu}
\end{split}
I really hope I got the minus signs right. Anyway now you see why this is rarely written down explicitly. If you set for example $\omega=0=b$ this can actually be integrated to the finite version and will come out exactly like what you wrote.
