# Traceless of stress-energy tensor in $d=2$

This is a question regarding Francesco, section 4.3.3. In this section, he considers the two-point function $$S_{\mu\nu\rho\sigma}(x) = \left< T_{\mu\nu}(x) T_{\rho\sigma}(0)\right>$$ He then goes on to claim that symmetry of the stress-energy tensor implies $$S_{\mu\nu\rho\sigma}(x) = S_{\nu\mu\rho\sigma}(x)~~~(1)$$ Though he doesn't mention this, I presume this is true only when $x \neq 0$ since the EM tensor is symmetric in a correlation as long as the other fields in the correlator are not evaluated at the same point.

EDIT: Due to some comments, I'll explain why I think so. If a theory is Poincare invariant, it has conserved currents $T^{\mu\nu}$ for translations and $$j^{\mu\nu\rho} = T^{\mu\nu} x^\rho - T^{\mu\rho} x^\nu$$ for Lorentz transformations. For completeness, we also note that if the theory has scale invariance the dilation current is $$j^\mu_D = T^{\mu\nu} x_\nu$$ In a classical theory, conservation of these currents implies symmetry and tracelessness of the stress-energy tensor. In a quantum theory, we have a Ward Identity, which for each of the currents reads $$\begin{split} \partial_\mu \left< T^\mu{}_\nu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \frac{\partial}{\partial x_i^\nu} \left< X \right> \\ \partial_\mu \left< j^{\mu\nu\rho} X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \left( x_i^\rho\frac{\partial}{\partial x_i^\nu} - x_i^\nu\frac{\partial}{\partial x_i^\rho} - i S_i^{\mu\nu} \right) \left< X \right> \\ \partial_\mu \left< j^\mu_D X \right> &= - \sum\limits_{i=1}^n \delta^d(x-x_i) \left( x_i^\alpha \frac{\partial}{\partial x_i^\alpha} + \Delta_i \right) \left< X \right> \end{split}$$ where $X = \Phi_1(x_1) \cdots \Phi_n(x_n)$, $S^{\mu\nu}_i$ is the representation of the Lorentz algebra under which $\Phi_i(x_i)$ transforms and $\Delta_i$ is the scaling dimension of $\Phi_i(x_i)$. Now plugging in the exact forms of the currents $j^{\mu\nu\rho}$ and $j^\mu_D$, we find $$\begin{split} \partial_\mu \left< T^\mu{}_\nu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \frac{\partial}{\partial x_i^\nu} \left< X \right> \\ \left< \left( T^{\mu\nu} - T^{\nu\mu} \right) X \right> &= i \sum\limits_{i=1}^n \delta^d(x-x_i) S_i^{\mu\nu} \left< X \right> \\ \left< T^\mu{}_\mu X \right> &= \sum\limits_{i=1}^n \delta^d(x-x_i) \Delta_i \left< X \right> \end{split}$$ Clearly, the EM tensor is not symmetric under correlation functions at the points $x = x_i$.

Now, using these symmetry properties and certain other properties under parity, he argues that $$S^\mu{}_\mu{}^\sigma{}_\sigma(x) = \left< T^\mu{}_\mu(x) T^\sigma{}_\sigma(0)\right> = 0$$ Following the above arguments, this should then only be true at $x \neq 0$. However, Francesco claims that this holds everywhere and therefore concludes that $\left< T^\mu{}_\mu(0)^2 \right> = 0$. How does this makes sense?

• Sorry, could you please write a more comprehensible explanation about why you think that any of the symmetries should break for $x=0$? The symmetry in (1) is just coming from the symmetry of the first $T_{\mu\nu}$. It's symmetric as an operator so of course all of its correlators respect the symmetry, too. There can't be any exception for $x=0$ or for Monday evenings. – Luboš Motl Sep 4 '13 at 7:22
• $\left< T^\mu{}_\mu(0)^2 \right>$ is a scalar (relative to Lorentz transformation), translation invariant quantity (there is nothing special with the origin point, and dimensionfull quantity (in 2d it has dimension 4 in energy/mass). So if it were non vanishing, this term could be used to build a characteristic length scale and thus spoil conformal invariance. My guess... – Learning is a mess Sep 4 '13 at 10:53
• @LubošMotl - I have added an explanation as to why I think that the statements should be valid on at $x \neq 0$. What is wrong with my argument? – Prahar Sep 4 '13 at 13:40
• @Learningisamess - I agree with you and I see why it should be true, but it doesn't seem to follow from Francesco's arguments alone (which I have outlined briefly in my question) – Prahar Sep 4 '13 at 13:41
• @Prahar : The variance is always positive, and you have : $V(X) = E(X^2)-(E(X))^2 \geq 0$ – Trimok Sep 5 '13 at 18:28

I think the original source of this claim is the famous unpublished paper of Luescher and Mack. Everyone's citing it. It is more rigorous mathematically and general (they don't assume parity) than Di Francesco. It starts on pages 1-2 of the manuscript. The proof below is basically the same proof, just with added details and a little bit different notation.

Assumptions: $T_{\mu\nu}$ are local covariant fields of dimension 2 and $$T^{\dagger}_{\mu\nu} = T_{\mu\nu}, \quad T_{\mu\nu} = T_{\nu\mu},\quad \partial^{\mu}T_{\mu\nu} = 0.$$ Moreover, one assumes that the correlation functions are well-defined and behaved, and satisfy usual axioms of CFT as stated e.g. in this page of nLab.

Proof: First we define the 2-point functions $$S_{\mu\nu\rho\sigma}(\vec{x}, \vec{y}) := \text{analytic continuation of } \langle \Omega \mid T_{\mu\nu}(\vec{x}) T_{\rho\sigma}(\vec{y}) \mid \Omega \rangle.$$ All $S$'s are translationally invariant, so we may as well set $\vec{y}=0$ so that we are left with $$S_{\mu\nu\rho\sigma}(\vec{x}) := \langle \Omega \mid T_{\mu\nu}(\vec{x}) T_{\rho\sigma}(0) \mid \Omega \rangle.$$ This function is real analytic for $\vec{x}\in\mathbb{R}^2,\, \vec{x}\neq 0$, and because of invariance under dilations, and the assumption that $T_{\mu\nu}$'s are of conformal weight 2, we have

$$S_{\mu\nu\rho\sigma}(\vec{x})=\lambda^4 S_{\mu\nu\rho\sigma}(\lambda \vec{x})\quad\forall\lambda\in\mathbb{R}.\quad\quad (1)$$ Moreover, from $T_{\mu\nu}=T_{\nu\mu}$ we get $$S_{\mu\nu\rho\sigma}(\vec{x})=S_{\nu\mu\rho\sigma}(\vec{x})=S_{\mu\nu\sigma\rho}(\vec{x}).\quad\quad (2)$$ Furthermore, by locality, invariance under translations and Equation 1 with $\lambda=-1$ we get $$S_{\mu\nu\rho\sigma}(\vec{x},0)=S_{\rho\sigma\mu\nu}(0,\vec{x})=S_{\rho\sigma\mu\nu}(-\vec{x},0)=S_{\rho\sigma\mu\nu}(\vec{x},0),$$ or in shorthand notation $$S_{\mu\nu\rho\sigma}(\vec{x}) = S_{\rho\sigma\mu\nu}(\vec{x}) .\quad\quad (3)$$ From Equations 2 and 3 we see that $S_{\mu\nu\rho\sigma}(\vec{x})$ has 6 independent components and also taking into account Equation 1 we can write $$S_{\mu\nu\rho\sigma}(\vec{x}) = (\vec{x}^{\,2})^{-4}\sum_{i=1}^6 \alpha_i f^{\,i} _{\mu\nu\sigma\rho}(\vec{x}),\quad \alpha_i\in\mathbb{C},\quad\quad (4)$$ where \begin{align} &f^{1}_{\mu\nu\rho\sigma}(\vec{x})=(\vec{x}^{\,2})^2 g_{\mu\nu} g_{\rho\sigma},\\\\ &f^{2}_{\mu\nu\rho\sigma}(\vec{x})=(\vec{x}^{\,2})^2 (g_{\mu\rho} g_{\nu\sigma}+g_{\mu\sigma}g_{\nu\rho}),\\\\ &f^{3}_{\mu\nu\rho\sigma}(\vec{x})=\vec{x}^{\,2} (g_{\mu\nu} x_\rho x_\sigma + g_{\rho\sigma} x_\mu x_\nu ),\\\\ &f^{4}_{\mu\nu\rho\sigma}(\vec{x})=x_\mu x_\nu x_\rho x_\sigma,\\\\ &f^{5}_{\mu\nu\rho\sigma}(\vec{x})= \vec{x}^{\,2} \left (g_{\mu\nu} (x_\rho \varepsilon_{\sigma\delta}x^\delta+x_\sigma\varepsilon_{\rho\delta}x^\delta )+ g_{\rho\sigma} (x_\mu\varepsilon_{\nu\delta} x^\delta + x_\nu \varepsilon_{\mu\delta}x^\delta) \right),\\\\ &f^{6}_{\mu\nu\rho\sigma}(\vec{x})= \left( x_\mu \varepsilon_{\nu\delta}x^\delta +x_\nu \varepsilon_{\mu\delta}x^\delta \right) x_\rho x_\sigma + x_\mu x_\nu \left( x_\rho \varepsilon_{\sigma\delta}x^\delta + x_\sigma \varepsilon_{\rho\delta}x^\delta \right), \end{align} and $$g_{\mu\nu}=\delta_{\mu\nu},\quad \varepsilon_{\mu\nu}=-\varepsilon_{\nu\mu},\quad \varepsilon_{01}=+1,\quad \vec{x}^{\,2}=(x^0)^2+(x^1)^2.$$ The continuity equation $\partial^\mu T_{\mu\nu}=0$ implies that $\partial^\mu S_{\mu\nu\rho\sigma}(\vec{x}) = 0$ and thus reduces the number of independent $\alpha_i$'s to 2 arbitrary constants $\alpha_+$ and $\alpha_-$ (checked myself using Mathematica): \begin{alignat*}{6} &\alpha_1 = 3 \alpha_+,\quad &&\alpha_2 = -\alpha_+,\quad \alpha_3 = -4\alpha_+,\quad \alpha_4 = 8\alpha_+,&\\ &\alpha_5 = \alpha_-,\quad &&\alpha_6 = -2\alpha_-. & \end{alignat*} Inserting these values into Equation 4 we obtain: $$S^{\mu}_{\;\mu\rho\sigma} (\vec{x}) = S^{\mu\;\rho}_{\;\mu\;\rho} (\vec{x}) = 0,$$ i.e. $\langle\Omega \mid T^\mu_{\;\mu}(\vec{x})\, T^\rho_{\;\rho} (0)\mid\Omega\rangle = 0$, which by the Reeh-Schlieder Theorem implies that $T^\mu_{\;\mu}(\vec{x})=0$.

Now to be able to apply Reeh-Schlieder, we actually have to go to $\vec{x}=0$ as you have correctly pointed out. And the reason why we can do this is analytic continuation, citing p 110 of R. JOST: The General Theory of Quantized Fields, 1965 (the same step in a related lemma):

"Such a Wigtman distribution thus vanishes in the real regularity points and thus, by analytic continuation identically".

tl;dr analytic continuation