# Symmetric stress-energy tensor in CFT

I'm a bit confused reading about the stress-energy tensor and conformal Ward identities in Di Francesco. My question is in a similar spirit to this one from several years ago, but the question was not fully answered there.

Let me first summarize my understanding of the logic in section 2.5.1, pgs 46-49:

1. The stress energy tensor $$T^{\mu \nu}$$, as with any Noether current, is not unique. One is always free to add an identically divergenceless tensor, such as $$\partial_{\rho} B^{\rho \mu \nu}$$ with $$B^{\rho \mu \nu}$$ antisymmetric in $$\mu$$ and $$\nu$$, without destroying the conservation law.

2. If one has Lorentz invariance, then we can always find such a $$B^{\rho \mu \nu}$$ such that the conserved current associated to Lorentz transformations is $$j^{\mu \nu \rho} = T^{\mu \nu} x^{\rho} - T^{\mu \rho} x^{\nu}$$ Then, the conservation law $$\partial_{\mu} j^{\mu \nu \rho} = T^{\rho \nu} - T^{\nu \rho}$$ implies that $$T^{\mu \nu}$$ is classically symmetric. In other words, if the classical equations of motion are satisfied, then $$\partial_{\mu} j^{\mu \nu \rho} = 0$$ and the antisymmetric part of $$T^{\mu \nu}$$ vanishes.

3. Using the equations of motion, we can add another term to $$T^{\mu \nu}$$ which is classically zero, but causes $$T^{\mu \nu}$$ to vanish identically -- this is Eq. (2.181). A priori this modification to $$T^{\mu \nu}$$ affects the Ward identities, but in a trivial way that may be ignored.

All of this is summarized once more in the final paragraph of 2.5.1, in the paragraph on page 49 direclty above 2.5.2.

Now, here comes my confusion: in section 4.3.2, on pgs 106-107, the Ward identity associated with Lorentz invariance is written $$\langle (T^{\rho \nu} - T^{\nu \rho}) X \rangle = -i \sum_i \delta(\vec{x}-\vec{x}_i) S^{\nu \rho}_i \langle X \rangle$$ where $$X$$ is a collection of fields at points $$\vec{x}_i$$. This is derived assuming the stress-energy tensor has been made symmetric and $$j^{\mu \nu \rho} = T^{\mu \nu} x^{\rho} - T^{\mu \rho} x^{\nu}$$. But this Ward identity is manifestly saying that $$T^{\mu \nu}$$ isn't identically symmetric! Just like all the other Ward identities, it appears to be saying that the classical equation of motion only holds away from the locations of other fields. But earlier, we spent some effort to show that $$T^{\mu \nu}$$ can be made identically symmetric without modifying the Ward identities!

So to summarize, my question is: can $$T^{\mu \nu}$$ truly be made identically symmetric, independent of the classical equations of motion? And if so, how should I understand that the Ward identity above seems not to agree?

$$T_{\mu\nu}$$ can be made truly symmetric in a QFT. You can define this by coupling the QFT to a background metric $$g$$. Stress tensor insertions are then defined by $$\langle T_{\mu\nu}(x) O_1(x_1) \cdots O_n(x_n) \rangle_g \equiv - \frac{1}{Z[g]} \frac{2}{\sqrt{-g}} \frac{\delta}{\delta g^{\mu\nu}(x)} \bigg( Z[g] \langle O_1(x_1) \cdots O_n(x_n) \rangle_g \bigg) \tag{1}$$ where $$Z[g]$$ is the partition function of the theory. I have assumed here that the operators $$O_i$$ do not depend on the metric though it is easy to generalize (1) to this case. Diffeomorphism invariance implies the divergence Ward identity $$\langle \nabla^\mu T_{\mu\nu}(x) O_1(x_1) \cdots O_n(x_n) \rangle_g = - \sum_{i=1}^n \langle O_1(x_1) \cdots {\cal L}_{\delta^d(x,x_i) \partial_\nu} O_i(x_i) \cdots O_n(x_n) \rangle_g$$ This stress tensor differs from the one you describe in your question which satisfies a slightly different Ward identities, namely $$\langle \nabla^\mu T^c_{\mu\nu}(x) O_1(x_1) \cdots O_n(x_n) \rangle_g = - \sum_{i=1}^n \delta^d(x,x_i) \langle O_1(x_1) \cdots \nabla_\nu O_i(x_i) \cdots O_n(x_n) \rangle_g$$ $$\langle [ T^c_{\mu\nu}(x) - T^c_{\nu\mu}(x) ] O_1(x_1) \cdots O_n(x_n) \rangle_g = - i \sum_{i=1}^n \delta^d(x,x_i) \langle O_1(x_1) \cdots S^i_{\mu\nu} O_i(x_i) \cdots O_n(x_n) \rangle_g$$ In the "canonical" stress tensor, the divergence Ward identity contains no spin information (all the derivatives are covariant derivatives) and the anti-symmetric part of $$T$$ carries all the spin information.

On the other hand, the stress tensor defined by (1) is identically symmetric but the spin information of the operators $$O_i$$ appears in the divergence Ward identity (the covariant derivative is replaced by the Lie derivative).

Both stress tensors carry the same amount of information, though they are packaged differently. They are related in a very simple way $$T_{\mu\nu} = T^c_{(\mu\nu)}$$ Let us verify this. We will use the Ward identities for $$T^c$$ to derive those for $$T$$. Note that it is identically symmetric by construction. It's divergence is \begin{align} \langle \nabla^\mu T_{\mu\nu}(x) O_1 \cdots O_n \rangle_g &= \langle \nabla^\mu T^c_{\mu\nu}(x) O_1 \cdots O_n \rangle_g + \frac{1}{2} \nabla_x^\mu \langle [ T^c_{\nu\mu}(x) - T^c_{\mu\nu}(x) ] O_1 \cdots O_n \rangle_g \\ &= - \sum_{i=1}^n \delta^d(x,x_i) \langle O_1 \cdots \nabla_\mu O_i \cdots O_n \rangle_g \\ &\qquad - \frac{i}{2} \nabla_x^\mu \sum_{i=1}^n \delta^d(x,x_i) \langle O_1 \cdots S^i_{\mu\nu} O_i \cdots O_n \rangle_g \\ &= - \sum_{i=1}^n \langle O_1 \cdots \bigg( \delta^d(x,x_i) \nabla_\mu - \frac{i}{2} \nabla_x^\mu \delta^d(x,x_i) S^i_{\mu\nu} \bigg) O_i \cdots O_n \rangle_g \end{align} However, the term in the bracket is precisely the Lie derivative so we have \begin{align} \langle \nabla^\mu T_{\mu\nu}(x) O_1 \cdots O_n \rangle_g &= - \sum_{i=1}^n \langle O_1 \cdots {\cal L}_{ \delta^d(x,x_i) \partial_\mu} O_i \cdots O_n \rangle_g \end{align} AS REQUIRED!

• Thank you for your answer, but I think it dodges the question in a similar way to qmechanic in the linked post. The point of the question is as follows: the YB says that $T_{\mu \nu}$ can be made from classically symmetric to truly symmetric without modifying the Ward identities. But then the Ward identity suggests otherwise. Is the YB's statement true or false?
– Zack
Feb 16, 2022 at 16:34
• Also, to complicate matters further: note that the "improvement terms" you mention manifestly cannot modify the Ward identities. So if an equation holds for $T_{\mu \nu}$, it must hold for $T^c_{\mu \nu}$. Indeed, the YB uses exactly your definition of the stress tensor in section 5.2.3, and STILL arrives at Eq. 5.67 suggesting that $T_{\mu \nu}$ is not truly symmetric under correlation functions. Indeed, in the linked post qmechanic states that $T_{\mu \nu}$ as you defined is only classically symmetric. So there still appears to be a paradox.
– Zack
Feb 16, 2022 at 16:36
• You can't make a stress tensor that is not identically symmetric to "truly" symmetric without changing a Ward identity. Improvement terms do change the Ward identities! They do not, however, change the conserved charges of a theory. Feb 16, 2022 at 16:41
• Also, YB does not use the same definition of the stress tensor, despite claiming this. They are defining it as a variation w.r.t. the vielbein $e^\mu_a$, NOT w.r.t. $g_{\mu\nu}$. Symmetry is not inherent in the former definition, but it is in the latter. In particular, their transformations (5.58) is precisely not keeping track of spin information so it is expected that they will recover their original (5.67) but not the ones I mention in this answer. Feb 16, 2022 at 16:42
• In the notation of YB: if you modify $T^{\mu \nu} \to T_B^{\mu \nu} = T^{\mu \nu} + \partial_{\rho} B^{\rho \mu \nu}$ with $B^{\rho \mu \nu} = -B^{\mu \rho \nu}$, then I thought that the Ward identities would remain unchanged since they are all of the form $\langle \partial_{\mu} j^{\mu} X \rangle = \ldots$, and clearly $\partial_{\mu} j^{\mu}$ is unchanged. But this is not actually true: it should really read $\partial_{\mu} \langle j^{\mu} X \rangle = \ldots$. Does the differentiating of the time-ordering symbol mean that the Ward identities are indeed modified? Does YB gloss over this?
– Zack
Feb 16, 2022 at 16:50