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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.

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Dominic Else
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1 Answers1

3

  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.

References:

  1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997.
Qmechanic
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