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In 2d Conformal Field Theory, the $T_{zz}$ component of energy-momentum tensor is treated as a holomorphic function $T(z)=T_{zz}$ at quantum level such as in OPE involved energy-momentum tensor.

I know, in classical theory, this conclusion comes from the traceless and conservation equation of $T_{\mu\nu}$. The traceless leads to $T_{z\bar{z}}=0$ and then conservation law implies $$\partial_{\bar z} T_{zz}=\partial_z T_{\bar z\bar z}=0$$ However, a quantum field does not need to satisfy the classical equation of motion. (Since in path integral fomalism $\int D\phi\ e^{-S[\phi ]}$, the quantum field $\phi$ can be any smooth field configuration and the energy-momentum tensor is a functional of field $\phi$.) The conservation equation is classical。

Then, why should the $T_{zz}$ still be a holomorphic function at quantum level?

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  • $\begingroup$ Stress-tensor is both conserved and traceless as a quantum operator. Inside path integral this is valid up to contact terms, as all equation of motion are. (Traceless part assumes that at quantum level you still have conformal invariance, i.e. the beta function vanishes.) $\endgroup$
    – Peter Kravchuk
    May 17, 2019 at 5:01

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I will follow Di Francesco's CFT book. By combining the Ward identities associated to conformal invariance (Eq. 5.32), you can get to

$$ \partial_{\bar{z}} \Big[ \langle T(z, \bar{z} ) X \rangle - \sum_{i=1}^n \left( \frac{1}{z-w_i} \partial_{w_i} \langle X \rangle + \frac{h_i}{(z-w_i)^2} \langle X \rangle \right) \Big] = 0 , \qquad \qquad \text{(Eq. 5.39)}$$

where $T(z, \bar{z})$ is the "holomorphic" component of the energy-momentum tensor ( in this convention $T(z, \bar{z}) = - 2 \pi T_{zz}$ ) and $X$ is some string of quasi-primary fields $\Phi_1(w_1, \bar{w}_1) ... \Phi_n(w_n, \bar{w}_n)$ with holomorphic dimensions $h_1, ... , h_n$. This means that the correlation function $\langle T(z, \bar{z} ) X \rangle$ is actually holomorphic up to the sum

$$ \sum_{i=1}^n \left( \frac{1}{z-w_i} \partial_{w_i} \langle X \rangle + \frac{h_i}{(z-w_i)^2} \langle X \rangle \right). $$

Note that this sum is not exactly holomorphic because of the property

$$\partial_{\bar{z}} \frac{1}{z} = \pi \delta (\mathbf{x}), $$

where $\delta (\mathbf{x})$ is the Dirac delta at the point $z$ but in Euclidean coordinates (its normalization may change when changing the coordinates). Therefore, if you apply this property to Eq. 5.39 you can easily see that $\partial_{\bar{z}} \langle T(z, \bar{z} ) X \rangle$ actually vanishes as long as we don't take $z$ to coincide with some $w_i$ (these are called contact terms and are basically always present in the quantum theory). This is I think what is meant by "$T_{zz}$ is holomorphic at the quantum level".

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  • $\begingroup$ +1 – this is also evident in $T_{z \bar{z}} = 0$, which is also only true up to contact terms. $\endgroup$
    – Prof. Legolasov
    May 17, 2019 at 12:30
  • $\begingroup$ So it is the correlation function $\langle ...\rangle$ be holomorphic (up to contact terms) rather than the field $T_{zz}$? Then why we always expand the field $T_{zz}$ as $T_{zz}=\sum \frac{L_n}{z^{n+2}}$ without any $\bar z$ terms? $\endgroup$
    – Jiahao Mao
    May 17, 2019 at 12:54
  • $\begingroup$ @JiahaoMao the operator is nothing more than its action on correlation functions, so T is holomorphic by definition $\endgroup$
    – Prof. Legolasov
    May 17, 2019 at 13:36
  • $\begingroup$ Yes, the operators always appear inside correlation functions at the end of the day. I wouldn't say $T$ is holomorphic by definition though. It's a result from Lorentz and scale invariance $\endgroup$
    – MBolin
    May 17, 2019 at 14:16
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    $\begingroup$ In the operator formalism, all operators satisfy their equation of motion. This is just like in QM where the $\hat q(t)$ heisenber operator in the harmonic oscillator obeys $d^2 \hat q/dt^2+\omega^2 \hat q=0$. $\endgroup$
    – mike stone
    May 17, 2019 at 17:48

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