The stress tensor for a conformal field theory (or any quantum field theory) can be derived from the action $S$ by the functional derivative

$$T^{\mu \nu} ~=~ -\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu \nu}},\tag{2.193}$$

where $g_{\mu \nu}$ is the background metric of signature $(+,-,-,-)$. This formula for the stress tensor appears to be *identically* symmetric, i.e. $$T^{\mu \nu} = T^{\nu \mu}$$ for any field configuration in the path integral, not just those obeying the classical equations of motion.

On the other hand, I don't see how this is consistent with the Ward identity (e.g. see Di Francesco et al, p. 107)

$$\langle (T^{\mu \nu} - T^{\nu \mu})X \rangle ~=~ -i \sum_i \delta(x - x_i) S_i^{\nu \mu} \langle X \rangle, \tag{4.66}$$

where $X$ is some product of fields $\phi(x_1) \cdots \phi(x_n)$, and the field $\phi$ transforms internally under an infinitesimal rotation $x^{\mu} \to x^{\mu} + \omega^{\mu} {\,}_{\nu} x^\nu$ as $\phi \to \phi + \omega_{\mu \nu} S^{\mu \nu} \phi$.

If $T^{\mu \nu}$ was identically symmetric, then both sides should equal zero.

This post imported from StackExchange Physics at 2015-05-04 13:53 (UTC), posted by SE-user Dominic Else