What is the exact relation between $\Theta ^\mu _\mu =0$ and conformal invariance?

Question

This is in part a reiteration of this old phys.SE question, which did not receive much attention.

It is usually stated (see e.g. Ref. 1, §4.2.2) that a traceless energy momentum tensor $\Theta {^\mu} {^\nu}$ implies the invariance of an action $S[\Phi]$ under conformal transformations (here and in what follows the metric $g{_\mu} {_\nu}$ is non-dynamical).

As pointed out in phys.SE/53003, in the usual textbook treatment of the issue, it appears that the underlying assumption that fields $\Phi$ satisfy the EOM, $\frac{\delta S}{\delta \Phi}=0$, is made. Indeed, on the one hand, the total variation of the action should include also a term proportional to $\frac{\delta S}{\delta \Phi}$, if fields carry a non-trivial representation of the conformal group. Moreover, the traceless condition $\Theta ^\mu _\mu =0$ is usually satisfied only on-shell (cf., for instance, the construction of $\Theta {^\mu} {^\nu}$ in §4.2.2 of Ref. 1, where the tracelessnes conditions follows from the conservation of the dilatation current $\partial _\mu j^\mu _D=0$).

This brings me to my question. Consider the three propositions:

1. The theory is conformally invariant.
2. It is possible to define an energy momentum tensor $\Theta ^\mu _\nu$ with identically vanishing trace: $\Theta ^\mu _\mu =0$.
3. It is possible to define an energy momentum tensor $\Theta ^\mu _\nu$ whose trace vanishes identically along the equations of motion: $\Theta ^\mu _\mu \approx 0$.

What are the relations between these three propositions?

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Ref. 1: Conformal field theory, P. Di Francesco, P. Mathieu, D. Senechal.

Answer 1

1. The conformal invariance of the classical action is not an on-shell statement. It is the same as saying that the Lorentz invariance was an on-shell statement. The conservation of the energy-momentum tensor in that case is an on-shell statement, that comes from Noëther theorem. In the case of conformal invariance, the tracelessness of the energy-momentum tensor is an on-shell statement.

Quantum mechanically, Noëther theorem should be changed by Ward Identities. Thus, the trace of the energy-momentum tensor should not be zero as an operator, but its expected value in the vacuum should be zero. This is assuming that the whole quantum theory is conformally invariant, which is the result of the classical action being invariant plus the path integral measure being invariant. If the path integral measure breaks conformal symmetry, then we have a conformal anomaly, and the trace of the energy momentum has a non-zero expected value in the vacuum. If this happens, all the stuff about 2-point and 3-poiny functions is not valid because there you assumed the path integral measure to be invariant.

3. I think that from the on-shell tracelessness condition you could obtain the equations of motion, and from then you could obtain the action and see if it is invariant. Or other way, you could define the variation of the fields to be such that it cancelled the trace of the energy momentum tensor for off-shell fields. This way you would end up discovering the conformal weights of each field as well as that the whole transformation you defined is a conformal transformation (because by definition such a transformation changes the metric that way).

Answer 2

In Ref. 1 of the OP it is proved that $2\implies 1$ (I'm referring to the numbered propositions of the OP). What perhaps should be emphasized is that in such a theory (with $\Theta ^\mu _\mu$ exactly vanishing) conformal invariance holds in a special way: in such a way that all fields are defined to have vanishing conformal weight.

In this answer, conversely, I will prove that $1 \implies 3$, under some technical assumptions which can be found in Ref. 1 of the OP. I will actually prove the statement for a simply scale-invariant theory (under the same technical assumptions, such a theory turns out to be conformally invariant). If all fields have vanishing scaling dimension, the same proof shows that $1\implies 2$.

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When the theory is scale invariant, there exists a conserved dilatation current: $$\partial _\mu j^\mu _D \approx 0,$$ where $\approx$ denotes equality on-shell.

In OP's Ref. 1 it is proved that, given some certain technical assumptions, in a scale invariant theory one can always define an energy-momentum tensor which satisfies: $$\Theta ^{\mu} _{\,\mu} = \partial _\mu j_D^\mu.$$

In particular, this implies:

1. That $\Theta ^{\mu} _{\,\mu} \approx 0$ (trivial).
2. That the theory is actually conformally invariant (not trivial).

Let me sketch the proof of the second point. The variation of a generic matter field $\Phi$ under an infinitesimal conformal transformation $x\to x+\xi(x)$ is given by: $$\delta _c \Phi =\delta _d \Phi + \delta _s \Phi,$$ where $\delta _d\Phi$ is the variation of $\Phi$ as a consequence of the diffeomorphism $x\to x+\xi (x)$, while $\delta _s\Phi$ is a local scale transformation: $$\delta _s \Phi = -\frac{\Delta}{d} (\partial _{\mu} \xi ^{\mu}) \Phi, $$where $\Delta$ is the conformal weight of $\Phi$. Correspondingly, the variation of the action is:$$\delta _c S =\intop \text {d}^d x \frac{\delta S}{\delta \Phi (x)}[\delta _d \Phi (x) + \delta _s \Phi (x)].$$

Now, the two crucial points:

1. The first term is expressed in terms of $\Theta ^{\mu} _{\,\nu}$ by coupling the theory to gravity: $$S[\Phi]\to S[\Phi,g]$$ in such a way that the new action is invariant under diffeomorphisms. In this way, one has: $$\intop \text {d}^d x \frac{\delta S}{\delta \Phi (x)}\delta _d \Phi (x)=-\intop \text {d}^d x \frac{\delta S}{\delta g _{\mu \nu} (x)}\delta g _{\mu \nu} (x)=\intop \text {d}^d x T ^{\mu} _{\,\mu} \partial _{\nu} \xi ^\nu.$$ Here we have defined: $$T ^{\mu \nu}(x)\equiv \frac{\delta S}{\delta g _{\mu \nu} (x)}.$$
2. The second term is, by definition of $j^\mu _D$: $$\intop \text {d}^d x \frac{\delta S}{\delta \Phi (x)}\delta _s \Phi (x)=\intop \text {d}^d x (\partial _\mu j_D ^\mu) (\partial _\nu \xi^\nu) .$$

Now, if we had $T^{\mu \nu}=\Theta ^{\mu \nu}$, then from the condition $\partial _\mu j^\mu _D = \Theta ^\mu _\mu$, summing up the two variations, we would immediately conclude that $\delta _c S=0$, i.e. the theory is conformally invariant. Of course, in general $T\neq \Theta$, but reconsidering the construction of $\Theta$ in Ref. 1 (in particular, Eqs. (4.42) and (4.43) of §4.2), one can see that the substitution $T\to \Theta$ under the integral sign is unconsequential. Conformal invariance then follows.