Stress-energy Trace of Massless Klein Gordon Field I've calculated the trace of the stress-energy for a massless KG field and I keep getting $T = - (\partial \phi)^2$ in 3+1 dimensions.
I'm using
$$T_{\mu\nu} = \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} (\partial \phi)^2$$
Since $g_{\mu\nu}g^{\mu\nu} = 4$, the second term is double the first.
On the other hand, David Tong's notes suggest that massless KG is a conformal field theory and is thus has trace = 0. He shows this in 2D euclidean space, but then says "This is the key feature of a conformal ﬁeld theory in any dimension. Many theories
have this feature at the classical level"
What am I missing?
 A: The Klein Gordon field as you've written it actually isn't conformally invariant for $D\neq 2$. (It is, classically, scale invariant, but it isn't Weyl invariant. Quantum mechanically even the scale invariance is broken.).
To get a conformal field theory you need to include the so-called "conformal coupling" to gravity. This will change the number of degrees of freedom in your theory--because you are forced to include gravity. So you have two choices: (1) accept that a klein gordon field around a fixed minkowski background is not conformally invariant at the classical level, or (2) look at a different physical theory, the KG field coupled to gravity, which will exhibit the conformal symmetry.
With that caveat. In $D$ dimensions, a conformally coupled scalar field is given by (using a 'mostly plus' signature convention)
\begin{equation}
S = \int d^D x \sqrt{-g} \left(-\frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - \frac{D-2}{8(D-1)}R \phi^2 \right)
\end{equation}
The last term contributes to the stress energy tensor for $\phi$ (computed using $T_{\mu\nu}=\frac{-2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu }}$):
\begin{equation}
T_{\mu\nu} = \partial_\mu\phi \partial_\nu \phi - \frac{1}{2}g_{\mu\nu} (\partial \phi )^2 + \frac{D-2}{4(D-1)} \left(g_{\mu\nu} \square (\phi^2) - \nabla_\mu \nabla_\nu (\phi^2) \right)
\end{equation}
The last term is crucial to get $T=0$, so you see the theory is not conformal without this special coupling.
Now you might naively think that since you are working around Minkowski space, you can ignore the conformal coupling term. This is not so! For one thing, you can see that explicit calculation has shown that the conformal coupling has led to a contribution that does not vanish around Minkowski space.
The basic underlying physics is that a Weyl transformation changes the metric. Thus even if you start with a flat, Minkowski metric with $R=0$, in general after a conformal transformation the metric will become curved. Explicitly, after performing a transformation
\begin{equation}
g_{\mu\nu} \rightarrow e^{2\omega (x) } g_{\mu\nu}
\end{equation}
the Ricci scalar transforms as
\begin{equation}
R \rightarrow R + 2(D-1)\square \omega - (D-1)(D-2) (\partial \omega)^2
\end{equation}
In particular, even if $R=0$ originally, after a Weyl transformation there will generically be a nonzero $R$.
So the scalar field in general must know about the curvature of the metric in order to compensate this transformation. There's a very special, miraculous cancellation that occurs in $D=2$ that allows this fact to be ignored.
Reference: http://folk.uio.no/ingunnkw/art/blackholes.pdf
