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Let me rephrase the question in a more general way.
Say we have a classical field theory with the action $$ S[\phi] = \int \!dx\, \mathcal{L}\left(\phi(x),\partial \phi(x) \right) $$ and there is some transformation that does $$ x \rightarrow w; \quad \phi(x) \rightarrow \phi'(w) = \mathsf{F}(\phi(x)). $$ Let me define also the transformed $S'$: $$ S'[\phi] = \int\! dx\, \left\vert\frac{\partial w}{\partial x} \right\vert \mathcal{L}\left(\mathsf{F}(\phi(x)), \frac{\partial x}{\partial w} \frac{\partial}{\partial x} \mathsf{F}(\phi(x)) \right). $$
What I thought is a symmetry: we need $S'[\phi] = S[\phi]$ identically, and only then can we be sure that the field equation is invariant under the said transformation.
What Di Francesco et. al. says about conformal symmetry: $S'[\phi] = S[\phi] + \Delta S[\phi]$ in general, and $\Delta S[\phi] = 0$ if $(\delta S[\phi]/\delta \phi) = 0$ is satisfied. I can't see this as a symmetry. In fact, I think this says the exact opposite: the field equation is not invariant!
Do prove me wrong. Please.
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I know related questions exist, but I don't get my desired answer.
This comes from chapter 4 of the so-called big yellow book by Di Francesco et. al. Specifically concerning eq (4.35) and (4.38). Let's keep everything classical: just the classical action and Euler-Lagrange field equations, no path integral, no correlation function.
Assume that the action $S$ is invariant under uniform translation $$ x^{\mu} \rightarrow x^{\mu} + \epsilon^{\mu} $$ with $\epsilon^\mu$ being a constant. If we now make $\epsilon^{\mu}$ position-dependent, then the change in $S$ is associated with the energy-momentum tensor $$ \delta S = -\int T^{\mu\nu} \, \partial_{\mu} \epsilon_{\nu} + O(\epsilon^2). $$
If $\epsilon^{\mu}$ is an infinitesimal conformal transformation in dimension $d > 2$, and $T^{\mu\nu}$ is symmetric, then we can write $$ \delta S = -\frac1d \int T^\mu_\mu \, \partial^{\nu} \epsilon_{\nu}. $$ Thus the claim that "traceless $T$ implies conformal invariance". (c.f. (4.35))
And then they go on to outline an explicit way to construct a traceless $T$. However, the construction relies on $\partial_\mu j_D^{\mu} = \partial_{\mu} T^{\mu\nu} = 0$ (c.f. (4.38)). These are true only if the equations of motion are enforced.
Now I am confused. If a classical theory is supposed to have a certain symmetry, I would expect its action to just be identically invariant under a symmetry operation. I mean, that's what I do when I'm told to write down a Lorentz-invariant theory, or a Ginzburg-Landau theory for some hexagonal-symmetric unconventional superconductor. I write down terms that are manifestly invariant, whether the equation of motion is satisfied or not. Why is it any different here?
Also $T$ isn't supposed to by identically traceless either. Later in the book we learn that $T^{\mu}_{\mu}$ enters the Ward identity for scaling transformation. If it were identically zero, correlation functions would not transform at all.
