Verifying a Conformal Ward Identity for the Free Boson

Question

I've always been very uncomfortable with the following two conformal Ward identities [c.f. Di Francesco, pg 107, Eqs. (4.66) and (4.67)]: $$ \langle (T^{\rho \nu} - T^{\nu \rho})X \rangle = -i \sum_i \delta(\boldsymbol{x} - \boldsymbol{x}_i) S^{\nu \rho}_i \langle X \rangle, \quad \langle T^{\mu}_{\mu} X \rangle = -\sum_i \delta(\boldsymbol{x} - \boldsymbol{x}_i) \Delta_i \langle X \rangle $$ where $X = \phi_1(\boldsymbol{x}_1) \ldots \phi_n(\boldsymbol{x}_n)$ is a product of local fields with spin generators $S^{\nu \rho}_i$ and scaling dimensions $\Delta_i$. In particular, I've always been confused how the identities $T^{\mu \nu} = T^{\nu \mu}$ and $T^{\mu}_{\mu} = 0$ can hold as operator equations, and yet be violated at insertions of other fields. (In the more familiar Ward identities for a conserved current, $\partial_{\mu} \langle j^{\mu} X \rangle = \ldots$, I am much more comfortable: $\partial_0$ should be interpreted as acting on the time-ordering symbol, so $\partial_{\mu} j^{\mu} = 0$ holding as an operator equation isn't inconsistent with $\partial_{\mu} \langle j^{\mu} X \rangle$ being nonvanishing at field insertions.)

To try and alleviate my confusion, I want to verify these equations in the simple case of a 2d free scalar field $\phi$. Here is my attempt. First of all, in 2d complex coordinates, the above two equations can be rewritten as [c.f. Di Francesco, pg 120, Eqs. (5.37) and (5.38)] $$ 2\langle T_{\bar{z} z} X \rangle = -\sum_i \delta(\boldsymbol{x} - \boldsymbol{x}_i) h_i \langle X \rangle, \quad 2\langle T_{z\bar{z}} X \rangle = -\sum_i \delta(\boldsymbol{x} - \boldsymbol{x}_i) \bar{h}_i \langle X \rangle $$ where $h_i = \frac{1}{2}(\Delta_i + s_i)$ and $\bar{h}_i = \frac{1}{2}(\Delta_i - s_i)$ are the conformal dimensions of $\phi_i$. In the case of a scalar field, the stress tensor is $T_{\mu \nu} = \partial_{\mu} \phi \partial_{\nu} \phi - \frac{1}{2} \eta_{\mu \nu} (\partial \phi)^2$, so the off-diagonal components of $T_{\mu \nu}$ in complex coordinates are $$ T_{z \bar{z}} = \partial \phi \bar{\partial} \phi - \frac{1}{2} (\partial \phi \bar{\partial}\phi + \bar{\partial}\phi \partial \phi) = \frac{1}{2} (\partial \phi \bar{\partial}\phi - \bar{\partial} \phi \partial \phi) $$ and an identical calculation seems to give $T_{\bar{z} z} = -T_{z \bar{z}}$. It's very tempting to declare that $T_{z \bar{z}} = T_{\bar{z} z} = 0$, but this is clearly incorrect; I believe the issue has to do with the necessity of regularizing $T_{\mu \nu}$, so let me assume that the above expression is appropriately defined by point-splitting.

Now, to verify a particular case of the above Ward identities, let's take $X = \partial \phi(w_1) \partial \phi(w_2)$, so that $h_1 = h_2 = 1$ and $\bar{h}_1 = \bar{h}_2 = 0$. The result we want to get is $$ \langle 2 T_{\bar{z} z}(\boldsymbol{y}) X \rangle = \left[ \delta(\boldsymbol{y} - \boldsymbol{x}_1) + \delta(\boldsymbol{y} - \boldsymbol{x}_2) \right] \frac{1}{4\pi (w_1 - w_2)^2}, \quad \langle 2 T_{z \bar{z}}(\boldsymbol{x}) X \rangle = 0 \tag{*} $$ where in the above, $\boldsymbol{x}_1 = (x^0_1, x^1_1)$ and $w_1 = x^0_1 + i x^1_1$ (and similarly for $\boldsymbol{x}_2$ and $w_2$). To get the same result from the lefthand side, we need to evaluate correlation functions of the form (ignoring constant prefactors for the moment) $$ \langle T_{\bar{z} z}(\boldsymbol{y}) \partial \phi(w_1) \partial \phi(w_2) \rangle \sim \lim_{z_1, z_2 \to z} \left[ \frac{\delta(\boldsymbol{y}_2 - \boldsymbol{x}_2)}{(z_1 - w_1)^2} + \frac{\delta(\boldsymbol{y}_2 - \boldsymbol{x}_1)}{(z_1 - w_2)^2} - \frac{\delta(\boldsymbol{y}_1 - \boldsymbol{x}_2)}{(z_2 - w_1)^2} - \frac{\delta(\boldsymbol{y}_1 - \boldsymbol{x}_1)}{(z_2 - w_2)^2} \right] $$ where I've used $\langle \partial \phi(z) \bar{\partial} \phi(\bar{w}) \rangle = \frac{1}{4} \delta(\boldsymbol{y} - \boldsymbol{x})$, and $z$ is the complex coordinate for $\boldsymbol{y}$. At this stage, I already feel as though something has gone horribly wrong: if I performed an analogous computation with $T_{z \bar{z}}$ instead of $T_{\bar{z} z}$, I would get exactly minus this result. So, whether this expression is zero or nonzero upon taking the limit $z_1, z_2 \to z$, I get the incorrect result for one of the two Ward identities!

So, here is the question: what is the correct way to evaluate the lefthand side of the two equations $(*)$, in order to get the desired result from the Ward identities?

Answer 1

Ah, the Yellow Pages textbook, a fun one. I learned conformal field theory solely through Polchinski's textbook, so I apologize if something seems arbitrary or to foreign. I will be working in $d=2$ so I don't have any off-diagonal components, but hopefully you'll be able to generalize from there. If my answer does not help at all, please disregard it.

The way I know of computing the expectation values of some product of fields involving the energy-momentum tensor (EMT) is via the operator product expansion (OPE) due to the relation between operators and states. Let me perform 2 examples, one where $T(z) = -\frac{1}{\alpha'}:\partial X^\nu \partial X_\nu:(z)$ where $T(z)\equiv T_{zz}(z)$ since $T^\mu_\mu = 0\implies T_{z\overline{z}} = T_{\overline{z}z} = 0$ in complex coordinates for $d=2$ and $:\cdots:$ represents normal ordering. $X^\mu(z)$ is some scalar field. The general procedure then for computing the OPE is to first perform all possible Wick contractions where I will denote a Wick contraction via $\left[\cdots\right]$. The Wick contraction here is performed by multipling by the number of possible Wick contractions of 2 operators and then inserting for every pair of fields with $\frac{1}{2}\alpha'\eta^{\mu\nu}\ln|z-w|^2$ (which comes from normal ordering). As a first example, consider the OPE between $T(z)X^\mu$ which is really$\langle T(z)X^\mu(0)\rangle$, \begin{align} T(z)X^\mu(0) & =-\frac{1}{\alpha'}:\partial X^\nu \partial X_\nu:(z) X^\mu(0)\\ & = -\frac{2}{\alpha'}\left[\partial X^\nu(z) x^\mu(0)\right]\partial X_\nu(z)\\ & = \frac{2}{\alpha'}\partial\left(-\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z|^2\right)\partial X_\nu(z)\\ & \sim\frac{1}{z}\partial X^\mu(z) \end{align} where $\sim$ denotes up to pole terms only (negating any rational terms). Now as an example similar to yours consider $T(z) = -\frac{1}{\alpha'}:\partial X^\mu\partial X_\mu: + V_\mu\partial^2 X^\mu$ where $V_\mu$ is some vector field (I am saying similar since it is a product of fields). Then the OPE of $T(z)X^\mu(0)$ is \begin{align} T(z)X^\mu(z) & = \left(-\frac{1}{\alpha'}:\partial X^\nu\partial X_\nu: + V_\nu\partial^2 X^\nu\right)X^\mu(0)\\ & = -\frac{2}{\alpha'}\left[\partial X^\mu(0)X^\mu\right]\partial X_\nu(z) + V_\nu\left[\partial^2X^\nu(z)X^\mu(0)\right]\\ & = -\frac{2}{\alpha'}\partial\left(-\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z|^2\right)\partial X_\nu(z) + V_\nu\partial^2\left(-\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z|^2\right)\\ & \sim \frac{\alpha'}{2}\frac{V^\mu}{2} + \frac{\partial X^\mu(0)}{z} \end{align} To make this calculation more general, you can take the position of $X^\mu(0)\rightarrow X^\mu(w)$ then swap all $z$'s with $z-w$ (also you can set $\alpha' = 1$ for convenience.

For your case however, the full OPE will be a bit more hairy since you have a mixing of the left and right (moving) sectors $\partial$ and $\overline{\partial}$, but the method I listed above still holds. The nice part is that since the two sectors do not talk to each other, you are free to the move the product of fields around with no worry of picking up an extra factor due to normal ordering (I think).

So, lets try to apply this to your scenario by looking at the first term of your EMT, $T_{z\overline{z}}\supset \frac{1}{2}\partial\phi\overline{\partial}\phi$. If $X = \partial\phi(w_1)\partial\phi(w_2)$ then we have to compute an OPE at 3 different locations. But lets think for a second, consider a very general expectation value \begin{equation} \langle :X^\nu \partial^k X^\mu(z_2,\overline{z}_2)\mathcal{O}_1(w_1)\cdots:\rangle = f_k(|w_i-z_2|,|w_i-w_j|) \end{equation} where $\cdots$ are insertions of some operators. The functional $f_k$ can be constrained due to the conformal invariance of the system namely $z\rightarrow \lambda z$ implying the operators transform as $\mathcal{O}\rightarrow \lambda^{-h}\mathcal{O}$ for conformal weight $h$ (and similar for $\tilde{h}$). Then the functional is \begin{equation} f_k\propto \prod_{i,j\neq i}|w_i - z_2|^{-(k+h_i)}|w_i - w_j|^{-(h_i-h_j)}g_k(w_i,z_2) \end{equation} where $g$ is some conformally invariant function and $k$ is the scaling dimension. We can then specialize to 3 operators at 3-different points, so $z_2,w_1,w_2$ (in my weird coordinates) from which you can see you will get 2 parts summed together that is the product of the 3-fields/operators present. So, going back to the first part of your EMT, I would expect it to behave as

\begin{align} \frac{1}{2}\partial\phi\overline{\partial}\phi\partial(z)\partial\phi(w_1)\partial\phi(w_2) & = \frac{1}{2}:\partial\phi\overline{\partial}\phi:(z)\partial\phi(w_1)\partial\phi(w_2)\\ & = \frac{\overline{\partial}\phi}{2}\left(\left[\partial\phi(z)\partial\phi(w_1)\right]\partial\phi(w_2) + \left[\partial\phi(z)\partial\phi(w_2)\right]\partial\phi(w_1)\right) \end{align} which from here you can plug in 2-derivatives of $-\frac{1}{2}\eta^{\mu\nu}\ln|z-w_i|$ (might have to re-introduce indices) where the first term will have $\partial_z\partial_{w_1}$ on it and the second term will have $\partial_z\partial_{w_2}$ on it. Notice that if you let $w_1 = w_2\rightarrow w$ then you get twice the answer. The answer to the above will then appear as \begin{equation} \frac{1}{2}\partial\phi\overline{\partial}\phi\partial(z)\partial\phi(w)\partial\phi(w)\sim \overline{\partial}\phi\left(\frac{1}{z-w}\partial\phi\partial\phi(w) + \frac{1}{|z-w|^2}\partial(\partial\phi\partial\phi(w))\right) \end{equation} You might be able to simplify this further using the equations of motion, which I think in your case is $\partial\overline{\partial}\phi = 0$. From here I hope you are able to prove your specific case of the War identity (and this has not just been me rambling). Cheers. (For reference: chapter 2 of Polchinski's string theory, volume 1).