# Use of classical equations of motion inside correlation functions

I am reading this paper by Zamolodchikov about the expectation value of $$T \bar{T}$$ in $$2d$$ QFT and I don't understand how he uses the classical equations of motion. For instance, classically, in any translationally invariant field theory, you have the conservation of the energy-momentum tensor:

$$\partial_{\mu} T^{\mu \nu} = 0$$.

However, this only holds on shell, so in the quantum theory what you have is a Ward identity:

$$\frac{\partial }{\partial x^{\mu}} \langle T^{\mu \nu} (x) X \rangle = - \sum_{i=1}^{n} \delta( \mathbf{x} - \mathbf{x}_i ) \frac{\partial}{\partial x^{\nu}_i} \langle X \rangle$$ ,

where $$X$$ is a string of fields that depend on the points $$\mathbf{x}_1, ..., \mathbf{x}_n$$.

My question is: why does he uses the classical type of equation instead of the quantum type of equation even inside correlation functions?

Comment: I think I have been able to get to Eq. (9) using the Ward identities, but I haven't been able to reproduce Eq. (11). In any case I think my doubt goes beyond the paper. It's just the example I have.

• How is it that T^mu^nu is not a field operator? Or is that not your question?
– user196418
May 9 '19 at 14:20
• Sorry I don't understand what you mean. $T^{\mu \nu}$ is a field operator as far as I know. My question is why it seems that you can forget about its quantum nature and compute as if fields were on shell. May 9 '19 at 14:22
• May 9 '19 at 14:24
• I have a suspicion that Zamolodchikov uses $\partial_{\mu} T^{\mu \nu} = 0$ with a caveat that the domain of correlation functions is $x_i \neq x_j$. The delta functions are by definition zero when their arguments aren't zero. May 9 '19 at 18:56
• Yes, I think it's is simply that. And he can do that because he doesn't take the limit $\lim_{x \rightarrow x'} T(x) \bar{T} (x')$ (at least not naively). May 9 '19 at 19:16

Quantum field theory is a quantum theory. In quantum theory we have operators which act on states. $$T^{\mu\nu}(x)$$ is an operator of this kind. For example, we can write $$|\Psi\rangle =T^{\mu\nu}(x)|0\rangle,$$ and so on. (To be more precise, fields are operator-valued distributions, and for $$|\Psi\rangle$$ to be an honest state we need to smear $$T$$ as $$\int d^dx f(x) T^{\mu\nu}(x)$$ where $$f$$ is a test function. Also, we should interpret all equations below in distributional sense.)

This operator satisfies the equation $$\partial_\mu T^{\mu\nu}(x)=0.$$ There is nothing in the right hand side. It is zero, no contact terms. No, it is not classical.

This means that matrix elements of $$T$$ satisfy the same equations, $$\langle \Phi|\partial_\mu T^{\mu\nu}(x)|\Phi'\rangle=0.$$ The states $$|\Phi\rangle$$ and $$|\Phi'\rangle$$ can be created from vacuum or other states by acting with other operators, $$\langle 0|O_1(x_1)\cdots O_k(x_k)\partial_\mu T^{\mu\nu}(x)O_{k+1}(x_{k+1})\cdots O_n(x_n)|0\rangle=0.$$ Zamolodchikov seems is talking about such expectation values in states, see his discussion below eq. (3).

The contact terms arise only when we start talking about time-ordered correlators. Time-ordered correlators are often denoted by $$\langle O_1(x_1)\cdots O_n(x_n)\rangle$$. (This appears to be not the notation used by Zamolodchikov.) Explicitly, time-ordered correlators are $$\langle 0|\mathcal{T}\{O_1(x_1)\cdots O_n(x_n)\}|0\rangle.$$ Here time-ordering means (for simplicity, in case of two operators) $$\mathcal{T}\{O_1(x_1)O_2(x_2)\}=\theta(x_1^0-x_2^0)O_1(x_1)O_2(x_2)+\theta(x_2^0-x_1^0)O_2(x_2)O_1(x_1).$$ You can see that it contains Heaviside step functions, so if you differentiate wrt, say, $$x_1$$, then you are differentiating not only the operators but also these step functions, which leads to contact terms. Note that if $$x_1$$ is spacelike from $$x_2$$, then there is no difference between the two orderings because $$[O_1(x_1),O_2(x_2)]=0$$ in this case, and time-ordering is trivial. So the contact terms can arise only when $$x_1^0=x_2^0$$ and operators are not spacelike separated. This is only when $$x_1=x_2$$, i.e. at coincident points.

Path integral computes time-ordered correlators, and textbooks which are based around path-integral formulation often do not even discuss expectation values in states, so it is easy to get confused about the origin of contact terms.

• See also equations in section 5 of the paper. May 10 '19 at 3:22

I think it is because this conservation equation for $$T^{\mu\nu}$$ does not come from varying the action, but directly from the definition of the stress-energy tensor as: $$T^{\mu\nu}\propto \frac{\partial S}{\partial g_{\mu\nu}}$$ and the fact that you can set the metric to $$g^{\mu\nu} = \delta^{\mu\nu}$$.

• I don't understand what you mean... Could you elaborate a bit more? Do you mean you can get to $\partial_{\mu} T^{\mu \nu} = 0$ without using the equations of motion? May 9 '19 at 16:31
• Not exactly, in this language conservation of stress-tensor follows from the definition and reparametrization invariance of the action, i.e. if $\delta g_{\mu\nu}=\partial_{(\mu} \xi_{\nu)}$ is an infinitesimal change of coordinates. The variation of the action is $\int d^dx \delta g_{\mu\nu} \frac{\delta S}{\delta g_{\mu\nu}}$. Using the form of $\delta g_{\mu\nu}$ and definition of $T$, one gets conservation after integrating by parts and using the freedom of choosing any $\xi_\mu$. May 10 '19 at 3:16
• If you try to use this derivation in a path integral, you will have to apply reparametrization to other operator insertions, which will lead to the contact terms in the Ward identity. May 10 '19 at 3:17