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Let us consider a theory defined by an action on a flat space $S[\phi]$ where $\phi$ denotes collectively the fields of the theory. We will study the theory on a general background $g_{\mu\nu}$ and then we will set the metric to be flat.

The Euclidean partition function of the theory in the presence of an external source is

$$ Z[J] = \int [d\phi] e^{-S -\int d^d x\, J \, \mathcal{O}}\,\qquad (1) $$

where $ \mathcal{O}$ can be either an elementary or a composite field (in what follows we will take it to be the trace of the energy-momentum tensor).

Now, a very well-known result is that a traceless energy momentum tensor implies conformal invariance; indeed, under a conformal transformation $g_{\mu\nu} \rightarrow f(x)g_{\mu\nu} $ such that

$$ \partial_{(\mu}\epsilon_{\nu)} = f(x) g_{\mu\nu} $$

the action transforms like

$$ \delta S = \frac{1}{d}\int d^d x T^{\mu}_\mu \partial_\rho \epsilon^\rho $$

Now, another well-known result states that in a generic background metric the expectation value of $T^\mu_\mu$ is not zero but depends on the Weyl-invariant tensors and the Euler density, that is

$$ \langle T^\mu_\mu \rangle = \sum a_i E_d - c_i W_{\mu\nu\rho....}^2 $$

where $\langle T^\mu_\mu \rangle$ is usually defined by the variation of the connected vacuum functional $W = \log Z[J]$ under variations of the metric.

First question. Is $\langle T^\mu_\mu \rangle$ calculable in the usual way using the partition function? That is setting $\mathcal{O} = T^\mu_\mu $ in Eq.(1) we compute

$$ \langle T^\mu_\mu \rangle = \frac{\delta}{\delta\, J} Z[J]\Big|_{J=0}\,\qquad (2) $$

Second question. If the answer of the first question is YES, then I would expect the $\langle T^\mu_\mu \rangle$ computed as the variation of the connected vacuum functional $W[J]$ is the same as the one computed in Eq.(2). Is this true?

Third question.

There are two ways the classical traceless condition can be realized:

  1. on-shell; then, $T^{\mu}_\mu$ is not identically zero but it is so once you apply the equation of motion,e.g. $\lambda \phi^4$ theory in d=4.
  2. $T^\mu_\mu$ is identically zero; that is, you don't need to use the equation of motion (e.g. massless scalar field in d=2 on a curved background)

In the first case, since $T^\mu_\mu$ vanishes on the equation of motion, I agree that it may receive quantum corrections through the coupling of the theory to a curved space; everything is ok.

In the second case instead, namely $T^\mu_\mu$ identically zero, I am unable to compute its expectation value from Eq.(1) and Eq.(2) since $\mathcal{O}=0$ identically; that is, the RHS of Eq.(2) is zero, since Z[J] is actually J-independent. This would imply $\langle T^\mu_\mu \rangle=0$.

Is it still true that the theory enjoys an anomaly on a curved space background? I would say YES, since the anomaly depends only on the central charges. How to resolve this apparent contradiction?

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  • $\begingroup$ You get a energy momentum tensor which can be made traceless once the equation of motion are applied $\endgroup$ – apt45 Nov 22 '17 at 21:04
  • $\begingroup$ EOM has second order derivative on fields whereas $T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}\eta_{\mu\nu}(\partial\phi)^2-\frac{\lambda}{24}\eta_{\mu\nu}\phi^4$ does not contain $\partial^2 \phi$. $\endgroup$ – user110373 Nov 22 '17 at 21:09
  • $\begingroup$ the energy momentum tensor results to be $T^{\mu\nu} = \partial^\mu \phi \partial^\nu \phi -\eta^{\mu\nu} \mathcal{L}$ where $ \mathcal{L}$ is the lagrangian. You can modify it by adding a total derivative such that the conservation law fo $T^{\mu\nu}$ is not spoiled. Then, you can send $T^{\mu\nu} \rightarrow T^{\mu\nu} - 1/6(\partial^\mu\partial^\nu - \eta^{\mu\nu}\square)\phi^2 = \phi(\square\phi + \lambda/3! \phi^3) = 0$ on-shell $\endgroup$ – apt45 Nov 22 '17 at 21:17
  • $\begingroup$ @apt45 Did you eventually find a resolution for this? I'm running into the same problem -- I think for a free massless scalar in $d = 2$, $T^{\mu_\mu}$ vanishes identically. $\endgroup$ – knzhou Mar 13 at 19:31
  • $\begingroup$ Does it perhaps have something to do with how $T^{\mu\nu}$ is defined in the quantum theory? That is, you need something like normal ordering. $\endgroup$ – knzhou Mar 13 at 19:36
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The answer to your first two questions is positive: $\langle T^\mu_\mu \rangle$ can be computed from the partition function, and it is the same as the variation of the connected vacuum functional $W[J]$. I'll let someone else go into the details of a proof, if needed.

Then there is a bit of confusion in your third question: the fact that there is an anomaly tells you exactly that $T^\mu_\mu$ is never identically zero in curved space background (except maybe in very special cases, but then there is no anomaly). The terms that you wrote in your anomaly, i.e. the Euler density and Weyl tensor squared, are curvature tensors that vanish in flat space. So you will always find $\langle T^\mu_\mu \rangle = 0$ with this anomaly.

But this does not yet mean that $T^\mu_\mu$ is identically zero: you could still have $$ \langle T^\mu_\mu \mathcal{O}_1 \cdots \mathcal{O}_n \rangle \neq 0. $$ The question whether all such correlators vanish in a theory in which $\langle T^\mu_\mu \rangle = 0$ is still not completely resolved in dimensions $d > 2$.

If you want to read more on the subject, I suggest looking at https://arxiv.org/abs/1302.0884 and references therein.

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  • $\begingroup$ 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 $\endgroup$ – apt45 Nov 22 '17 at 13:55
  • $\begingroup$ 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? $\endgroup$ – apt45 Nov 23 '17 at 10:41
  • $\begingroup$ @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??? $\endgroup$ – M.Jo Nov 23 '17 at 16:37
  • $\begingroup$ @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! $\endgroup$ – knzhou Mar 13 at 19:44
  • $\begingroup$ @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. $\endgroup$ – octonion Jul 31 at 2:50

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