Diffeomorphism Ward identity

Question

I'm trying to prove: (Exercise 2.1. from "TASI Lectures on the Conformal Bootstrap" by David Simmons-Duffin)

$$\partial_{\mu} \langle T^{\mu \nu} {\cal O}_{1}(x_{1}) \dots {\cal O}(x_{n}) \rangle = -\sum_{i = 1}^{N}\delta(x - x_{i}) \partial^{\nu}_i \langle {\cal O}_{1}(x_{1}) \dots {\cal O}(x_{n}) \rangle\tag{6}$$ under the assumption of (i) flat space with (ii) Euclidean signature, (iii) the action $S$ is diffeomorphism invariant, and (iv) the operators ${\cal O}_i$ have no spin.

The theory has to be invariant under a diffeomorphism of $S[g, \phi]$, thus we have the following equalities:

1. $x^{\prime\mu} = x^{\mu} + \epsilon^{\mu}(x)$
2. $\dfrac{\partial x^{\prime\mu}}{\partial x^{\nu}} = \delta^{\mu}_{\; \nu} + \partial_{\nu}\epsilon^{\mu}$
3. $g^{\prime\mu \nu} = \dfrac{\partial x^{\prime\mu}}{\partial x^{\sigma}}\dfrac{\partial x^{\prime\nu}}{\partial x^{\sigma}}g^{\sigma \sigma} = g^{\mu \nu} + \left( \partial^{\mu}\epsilon^{\nu} + \partial^{\nu}\epsilon^{\mu}\right) + O \left(\epsilon^2 \right)$

As this is a symmetry, I wrote:

$$S[\phi^{\prime}] = S[\phi], \quad D\phi^{\prime} = D\phi, \quad \langle O^{\prime}_{1}(x_{1}) \dots O^{\prime}(x_{n}) \rangle_{g} = \langle O_{1}(x_{1}) \dots O(x_{n}) \rangle_{g}$$

Doing the transformation inside the path integral we modify $$O(x_{1}^{\prime}) = O(x_{1}^{\prime} + \epsilon) = O(x_{1}) +\epsilon^{\nu}\partial_{\mu}O(x_{1})$$

So:

$$\langle O^{\prime}_{1}(x_{1}) \dots O^{\prime}(x_{n}) \rangle_{g} = \int D\phi^{\prime} O^{\prime}_{1}(x_{1}) \dots O^{\prime}(x_{n}) e^{-S[g^{\prime}, \phi^{\prime}]} \\ =\int D\phi O^{\prime}_{1}(x_{1}) \dots O^{\prime}(x_{n}) e^{-S[g, \phi] + \delta g^{\mu \nu} \frac{\delta S[g, \phi] }{\delta g^{\mu \nu}}}$$ Expanding the exponential

$$=\int D\phi O^{\prime}_{1}(x_{1}) \dots O^{\prime}(x_{n})\left[ 1 - \delta g^{\mu \nu} \frac{\delta S[g, \phi] }{\delta g^{\mu \nu}} \right] e^{-S[g, \phi]} \\ = \int D\phi \left[O(x_{1}) +\epsilon^{\nu}\partial_{\mu}O(x_{1}) \right] \dots \left[O(x_{n}) +\epsilon^{\nu}\partial_{\mu}O(x_{n}) \right]\left[ 1 - \delta g^{\mu \nu} \frac{\delta S[g, \phi] }{\delta g^{\mu \nu}} \right] e^{-S[g, \phi]} $$ To mantain $\langle O^{\prime}_{1}(x_{1}) \dots O^{\prime}(x_{n}) \rangle_{g} = \langle O_{1}(x_{1}) \dots O(x_{n}) \rangle_{g}$ and neglecting higher orders, we get the following condition:

$$\langle \frac{\delta S[g, \phi] }{\delta g^{\mu \nu}} O_{1}(x_{1}) \dots O(x_{n}) \rangle = - \sum_{i = 1}^{N} \epsilon^{\nu} \partial_{\mu} \langle O_{1}(x_{1}) \dots O(x_{n}) \rangle $$

That is not the diffeomorphism Ward identity. Am I missing something?

Answer 1

In my opinion, it is more instructive to derive Ward-Takahashi identitities for the general case, $$\partial_{\mu} \left\langle J_a^{\mu}(x) \Phi(x_1) \ldots \Phi(x_n)\right\rangle = -\sum_i \delta(x - x_i) \left\langle \Phi(x_1) \ldots X_a(\Phi)(x_i) \ldots \Phi(x_n)\right\rangle,$$ with $\Phi(x) \to \Phi(x) + \epsilon_a(x) X_a(\Phi)(x)$, and then apply it to specific symmetries, but okay.

First of all, why would $O \to O + \epsilon^{\nu} \partial_{\mu} O$? Let's forget for a second that that's essentially a Taylor expansion. The fact that on the right-hand side you have a spacetime scalar $+$ a spacetime tensor should have alarmed you that something isn't right. Of course, you should have written $\epsilon_{\mu}\partial^{\mu} O$ instead. Furthermore, notice that in each term the argument will be different. So the right-hand side in your last equation shall read $$-\sum_i \epsilon_{\mu}(x_i) \frac{\partial}{\partial x_i^{\mu}} \left\langle O(x_1) \ldots O(x_n) \right\rangle.$$ On the left-hand side, you forgot to multiply by $\delta g^{\mu\nu}$ (as well as the minus sign?). But let's actually not bother with $\delta g$ at all and simply write the change of action as $\delta S$. Putting everything together we then obtain $$\left\langle \delta S\, O(x_1) \ldots O(x_n) \right\rangle = \sum_i \epsilon_{\mu}(x_i) \frac{\partial}{\partial x_i^{\mu}} \left\langle O(x_1) \ldots O(x_n) \right\rangle.$$ Now, you probably remember that for symmetry transformations the change of action takes the form $\delta S = -\int_x \epsilon_k \partial_{\mu} J_k^{\mu} = \int J_k^{\mu} \partial_{\mu} \epsilon_k$, with $J_k^{\mu}$ being conserved currents. For translation symmetry, one has a current for each of the translations $\epsilon_{\nu}$, with $\nu = 0, 1, 2, 3$, and the change in action reads $\delta S = \int_x T^{\mu\nu}(x) \partial_{\mu}\epsilon_{\nu}(x)$ (derive this result if you haven't done this before!). Thus, $$\int_x \partial_{\mu}\epsilon_{\nu}(x) \left\langle T^{\mu\nu}(x) O(x_1) \ldots O(x_n) \right\rangle = \int_x \sum_i \delta(x - x_i) \epsilon_{\nu}(x) \frac{\partial}{\partial x^{\nu}} \left\langle O(x_1) \ldots O(x_n) \right\rangle,$$ where we have made some obvious algebraic manipulations on the right-hand side. Finally, upon integrating the left-hand side by parts, $$-\int_x \epsilon_{\nu}(x) \partial_{\mu} \left\langle T^{\mu\nu}(x) O(x_1) \ldots O(x_n) \right\rangle = \int_x \epsilon_{\nu}(x) \sum_i \delta(x - x_i) \frac{\partial}{\partial x^{\nu}} \left\langle O(x_1) \ldots O(x_n) \right\rangle,$$ and noting that the above identity should hold for any infinitesimal $\epsilon_{\mu}$ we conclude that we may write the following local relation: $$\partial_{\mu} \left\langle T^{\mu\nu}(x) O(x_1) \ldots O(x_n) \right\rangle = -\sum_i \delta(x - x_i) \frac{\partial}{\partial x^{\nu}} \left\langle O(x_1) \ldots O(x_n) \right\rangle,$$ which is exactly the desired Ward identity corresponding to translation symmetry (modulo sign errors and such).

Answer 2

Hints to prove$^1$ eq. (6) in Ref. 1:

1. Hilbert/metric SEM tensor in Euclidean signature: $$T^{\mu\nu}(x)~:=~ -\frac{2}{\sqrt{g(x)}}\frac{\delta S[\phi;g]}{\delta g_{\mu\nu}(x)}.\tag{A}$$

2. Lie derivative of a scalar/zero spin operator: $$ {\cal L}_{\epsilon}\phi(x)~=~\epsilon^{\nu}(x)\partial_{\nu}\phi(x)~=~\epsilon_{\nu}(x)\partial^{\nu}\phi(x).\tag{B}$$

3. Correlation functions$^2$ are defined via a path integral with Euclidean signature: $$ \langle F[\phi]\rangle_g~:=~\frac{1}{Z[g]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{-\frac{1}{\hbar}S[\phi;g]\right\}. \tag{C}$$

4. Use the Schwinger-Dyson (SD) equations $$\begin{align} \hbar\langle {\cal L}_{\epsilon}^{\phi}F[\phi]\rangle_g ~\stackrel{\text{SD}}{=}~~~~&\langle F[\phi]{\cal L}_{\epsilon}^{\phi}S[\phi;g]\rangle_g\cr ~\stackrel{S\text{ diffeo.inv.}}{=}&-\langle F[\phi]{\cal L}_{\epsilon}^{g}S[\phi;g]\rangle_g\cr ~=~~~~& -\int_{V}d^nx~ \langle F[\phi]\frac{\delta S[\phi;g]}{\delta g_{\mu\nu}(x)}\rangle_g{\cal L}_{\epsilon}g_{\mu\nu}(x)\cr ~\stackrel{(A)}{=}~~~~& \int_{V}d^nx~\sqrt{g(x)} \langle F[\phi]T^{\mu\nu}(x)\rangle_g\nabla^{(x)}_{\mu}\epsilon_{\nu}(x)\cr ~\stackrel{\text{IBP}}{=}~~~~& -\int_{V}d^nx~\sqrt{g(x)} \epsilon_{\nu}(x)\nabla^{(x)}_{\mu}\langle T^{\mu\nu}(x)F[\phi]\rangle_g, \end{align}\tag{D}$$ and therefore $$\frac{\hbar}{\sqrt{g(x)}}\frac{\delta}{\delta\epsilon_{\nu}(x)}\langle {\cal L}_{\epsilon}^{\phi}F[\phi]\rangle_g ~\stackrel{(D)}{=}~-\nabla^{(x)}_{\mu}\langle T^{\mu\nu}(x)F[\phi]\rangle_g. \tag{E}$$

5. Now take the flat space limit to derive eq. (6). $\Box$

References:

1. D. Simmons-Duffin, 2015 TASI Lectures on the Conformal Bootstrap; eq. (6).

2. J. Penedones, 2015 TASI lectures on AdS/CFT; eq. (18).

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$^1$ Note that disconnected pieces are missing/implicit in eq. (8) in Ref. 1: $$ \frac{2\hbar}{\sqrt{g(x)}}\frac{\delta \langle F[\phi]\rangle_g}{\delta g_{\mu\nu}(x)} ~\stackrel{(A)+(C)}{=}~ \langle T^{\mu\nu}(x)F[\phi]\rangle_g-\langle T^{\mu\nu}(x)\rangle_g\langle F[\phi]\rangle_g. \tag{8} $$

$^2$ The correlation functions contains an implicit covariant Euclidean time order $T_{\rm cov}$, cf. e.g. eq. (11) in Ref. 1 and this Phys.SE post.