The stress tensor for a conformal field theory (or any quantum field theory) can be derived from the action $S$ by the functional derivative

$$T^{\mu \nu} ~=~ -\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu \nu}},\tag{2.193}$$

where $g_{\mu \nu}$ is the background metric of signature $(+,-,-,-)$. This formula for the stress tensor appears to be identically symmetric, i.e. $$T^{\mu \nu} = T^{\nu \mu}$$ for any field configuration in the path integral, not just those obeying the classical equations of motion.

On the other hand, I don't see how this is consistent with the Ward identity (e.g. see Di Francesco et al, p. 107)

$$\langle (T^{\mu \nu} - T^{\nu \mu})X \rangle ~=~ -i \sum_i \delta(x - x_i) S_i^{\nu \mu} \langle X \rangle, \tag{4.66}$$

where $X$ is some product of fields $\phi(x_1) \cdots \phi(x_n)$, and the field $\phi$ transforms internally under an infinitesimal rotation $x^{\mu} \to x^{\mu} + \omega^{\mu} {\,}_{\nu} x^\nu$ as $\phi \to \phi + \omega_{\mu \nu} S^{\mu \nu} \phi$.

If $T^{\mu \nu}$ was identically symmetric, then both sides should equal zero.


1 Answer 1

  1. Yes, eq. (2.193) is a classical formula, and the symmetry of the (Hilbert) stress-energy-momentum tensor $T^{\mu\nu}$ is only valid classically.

  2. Quantum mechanically, the symmetry of $T^{\mu\nu}(x)$ is broken by the presence of other fields in positions $x_1,x_2,\ldots$ in the (time-ordered) correlator
    $$\langle T\left\{ (\hat{T}^{\mu \nu}(x) - \hat{T}^{\nu \mu}(x))\hat{X}(x_1,x_2,\ldots)\right\} \rangle $$ $$~=~ -i\hbar \sum_i \delta(x - x_i) ~S_i^{\nu \mu} \langle T\left\{\hat{X}(x_1,x_2,\ldots)\right\} \rangle. \tag{4.66}$$ From the point of view of the path integral, this can be traced back to the time-slicing prescription. Here $T$ denotes time-ordering.


  1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997.
  • 1
    $\begingroup$ Thanks for your answer. If you could elaborate on your statement that the asymmetry at the quantum level is due to the time slicing prescription it would be very useful to me. $\endgroup$ Commented May 2, 2015 at 22:33

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