# Conformal Ward Identity (Di Francesco et al)

In the yellow pages (Conformal Field Theory, Di Francesco, Mathieu, Sénéchal), the authors derive the conformal Ward identity in the following way:

1. They show that, for a conformal transformation, $$\partial_\mu (\epsilon_\nu T^{\mu \nu}) = \epsilon_\nu \partial_\mu T^{\mu \nu} + \dfrac12 \partial_\rho \epsilon^\rho \eta_{\mu \nu} T^{\mu \nu} + \dfrac12 \epsilon^{\alpha \beta} \partial_\alpha \epsilon_\beta \epsilon_{\mu \nu} T^{\mu \nu}. \tag{5.42}$$

2. "Integrating both sides of (5.42)", they find $$\delta_\epsilon \langle{X}\rangle= \int_M d^2x \, \partial_\mu \langle{T^{\mu \nu }(x)\epsilon_\nu(x)X}\rangle. \tag{5.44}$$

I don't understand how this last equality is derived from step 1). The domain $$M$$ is taken to include the positions of all the fields contained in $$X$$.

• Divergence aka Stokes' theorem, perhaps? Commented Mar 20, 2023 at 21:26

I encounter the same problem recently. Here is my argument.

For simplicity, let $$X = \phi(y)$$ be a primary field. Start from

$$\int_M d^2x \; \partial_{\mu}\langle T^{\mu\nu}(x)\epsilon_{\nu}(x)\phi(y)\rangle$$ where $$M$$ includes $$y$$. By (5.42), it becomes $$\int_M d^2x \; \epsilon_{\nu} \langle\partial_{\mu}T^{\mu\nu} \phi(y)\rangle + \frac{1}{2}(\partial \cdot \epsilon) \langle {T^{\mu}}_{\mu}\phi(y)\rangle + \frac{1}{2}\varepsilon^{\alpha\beta}\partial_{\alpha}\epsilon_{\beta} \langle\varepsilon_{\mu\nu}T^{\mu\nu}\phi(y)\rangle$$ where $$\varepsilon^{\alpha\beta}$$ is Levi-Civita symbol and $$\epsilon^{\alpha}$$ is the infinitesimal displacement.

By (5.32) in the same book, it becomes $$- \int_{M} d^2x \; \delta(x-y) \bigg( \epsilon_{\mu}\partial^{\mu}\langle\phi(y)\rangle + \frac{1}{2}(\partial\cdot\epsilon)\Delta \langle \phi(y)\rangle + i \frac{s}{2}\varepsilon^{\alpha\beta}\partial_{\alpha}\epsilon_{\beta}\langle\phi(y)\rangle \bigg) \\ = -\epsilon_{\mu}\partial^{\mu}\langle\phi(y)\rangle - \frac{1}{2}(\partial\cdot\epsilon)\Delta \langle \phi(y)\rangle - i \frac{s}{2}\varepsilon^{\alpha\beta}\partial_{\alpha}\epsilon_{\beta}\langle\phi(y)\rangle$$ Now, we switch into the complex coordinate: $$z = x^{1} + ix^{2}$$. Then, $$\varepsilon^{\alpha\beta}\partial_{\alpha}\epsilon_{\beta} = \partial_1\epsilon_2 - \partial_2\epsilon_1 = \frac{1}{i}(\partial_z \epsilon^{z} - \partial_{\bar{z}}\epsilon^{\bar{z}})$$ where we've defined $$\epsilon^{z} = \epsilon^1 + i \epsilon^2$$.

For the sake of convenience, we denote $$\partial_z \to \partial ,\partial_{\bar{z}} \to \bar{\partial}$$ and the same way for $$\epsilon$$. Also, recall that, from (5.21) in the same book: $$\Delta = h + \bar{h}$$ and $$s = h - \bar{h}$$, in which $$h$$ and $$\bar{h}$$ are the conformal weights. As a result,

$$-\bigg(\epsilon\partial + \bar{\epsilon}\bar{\partial} + \frac{1}{2}(\partial\epsilon + \bar{\partial}\bar{\epsilon})(h+\bar{h}) + \frac{1}{2}(\partial\epsilon - \bar{\partial}\bar{\epsilon})(h-\bar{h})\bigg)\langle \phi \rangle \\ = -\bigg[(\epsilon\partial + h\partial\epsilon) - (\bar{\epsilon}\bar{\partial} + \bar{h}\bar{\partial}\bar{\epsilon}) \bigg] \langle \phi \rangle$$

Recall (5.23), it is indeed $$\delta_{\epsilon,\bar{\epsilon}}\langle\phi\rangle$$ as the author claims.