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Jul 31, 2019 at 2:50 comment added octonion @knzhou, I think the anomaly modifies the classical form of the energy momentum tensor. The form you wrote down comes from varying the action with respect to a translation in the position coordinate of the field. Now the idea of the anomaly is that in curved spacetime there is an extra contribution coming from the measure. This isn't usually written down as modifying the form of $T_{\mu\nu}$ but I think that is exactly what it does.
Mar 13, 2019 at 19:44 comment added knzhou @M.Jo It is identically zero and it's not hard to show; the trace is $(\partial_\mu \phi)(\partial^\mu \phi) - \delta^\mu_\mu (1/2)(\partial_\nu \phi)^2 = 0$ even off-shell. I have the same question as the OP and would really appreciate some clarity on this!
Nov 23, 2017 at 16:37 comment added M.Jo @apt45 Are you sure that $T^\mu_\mu$ is identically zero in that case? My guess is that it only vanishes upon the equation of motion, and there is no contradiction. Otherwise you would be telling me that $\langle T^\mu_\mu \rangle \neq 0$ but $T^\mu_\mu$ is exactly zero as an operator???
Nov 23, 2017 at 10:41 comment added apt45 I can give you a counterexample: the massless scalar field action in d=2 enjoys a $T^\mu_\mu=0$ even in curved space background. But this case the anomaly is still present, right?
Nov 22, 2017 at 13:55 comment added apt45 Hi M.Jo. The fact an anomaly is present in curved space tells me that the expectation value of the trace (on a curved bkg) is different from zero. The anomaly I wrote nothing says about its classical expression. The contradiction is easily solved if any theory on a curved space-time has a energy momentum trace whose trace cannot be identically zero classically without using equations of motion. But I don't know if it is true
Nov 22, 2017 at 13:09 history answered M.Jo CC BY-SA 3.0