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Does a non-lagrangian field theory have a stress-energy tensor?

Physics Asked on March 23, 2021

In classical field theory, the stress-energy tensor can be defined in terms of the variation of the action with respect to the metric field, or with respect to a frame field if spinors are involved. Of course, this assumes that the field theory is expressed in terms of an arbitrary metric (or frame) field, even if it’s only a background field, so that we can define the variation. This is nicely reviewed in "Currents and the Energy-Momentum Tensor in Classical Field Theory — A fresh look at an Old Problem" (https://arxiv.org/abs/hep-th/0307199).

A similar definition works in quantum field theory, too, ignoring issues of regularization.

We can also have a non-lagrangian field theory whose equations of motion are not necessarily derivable from any lagrangian. We can still express the equations in terms of an arbitrary background metric (or frame) field, but how do we define the stress-energy tensor in this case? Do non-lagrangian field theories still have a stress-energy tensor?

  • I understand that Noether’s theorem assumes a lagrangian, but the stress-energy tensor seems more fundamental, because even non-lagrangian systems couple to gravity, right? Shouldn’t that mean that they have a stress-energy tensor?

  • The existence of a stress-energy tensor seems to be a standard axiom in non-lagrangian formulations of conformal field theory, but I don’t know if these theories could be formulated in terms of a lagrangian, or if they are truly non-lagrangianable. (Can we please make that a word?)


Other posts about non-lagrangian(able) theories:

2 Answers

Most theories do not have a conserved energy-momentum tensor, regardless of whether they are Lagrangian or not. For example you need locality and Lorentz invariance. When you have those, you can define the energy-momentum tensor via the partition function (which always exists, it basically defines the QFT): $$ langle T_{munu}rangle:=frac{delta}{delta g^{munu}}Z[g] $$ You can define arbitrary correlation functions of $T$ by including insertions. And these functions define the operator $T$ itself. So $T$ is defined whenever $Z$ is a (differentiable) function of the metric, i.e., when you have a prescription to probe the dependence of the theory on the background metric. And this prescription is part of the definition of the theory: in order to define a QFT you must specify how the partition function is to be computed, for an arbitrary background. If you cannot or do not want to specify $Z$ for arbitrary $g$, then the derivative cannot be evaluated and $T$ is undefined. (This is not an unreasonable situation, e.g. I may be dealing with a theory that is anomalous and is only defined for a special set of metrics, e.g. Kähler. This theory does not make sense for arbitrary $g$, so I may not be able to evaluate $Z[g]$ for arbitrary $g$, and therefore $T$ may not exist).

If the theory admits an action, then the dependence of $Z$ on $g$ is straightforward: it is given by whatever the path-integral computes. If the theory does not admit an action, then you must give other prescriptions by which to compute $Z$. This prescription may or may not include a definition for arbitrary $g$; if it does not, then $T$ is in principle undefined.

But anyway, for fun consider the following very explicit example: $mathcal N=3$ supersymmetry in $d=4$. This theory is known to be non-lagrangian. Indeed, if you write down the most general Lagrangian consistent with $mathcal N=3$ SUSY, you can actually prove it preserves $mathcal N=4$ SUSY as well. So any putative theory with strictly $mathcal N=3$ symmetry cannot admit a Lagrangian. Such a theory was first constructed in arXiv:1512.06434, obtained almost simultaneously with arXiv:1512.03524. This latter paper analyses the consequences of the anomalous Ward identities for all symmetry currents, in particular the supercurrent and the energy-momentum tensor.

Correct answer by AccidentalFourierTransform on March 23, 2021

Not all non-Lagrangian theories have a stress-energy tensor, an example of this is the critical point of the long-range Ising model, which can be expressed as a "defect" field theory where the action consists of two pieces integrated over spaces of different dimensionality and hence has no single Lagrangian that would describe it.

See chapter 6 of "Conformal Invariance in the Long-Range Ising Model" by Paulos, Rychkov, van Rees, Zan for a discussion of this formulation and what the "missing" stress-energy tensor means for the Ward identities. That paper also refers us in a footnote to "Conformal symmetry in non-local field theories" by Rajabpour, where a "non-local stress tensor" is constructed for a class of theories where the usual kinetic term with a local Laplacian is replaced by the non-local fractional Laplacian, but this object does not behave like one would usually like a stress-energy tensor in a CFT to behave, in particular its operator product expansions are "wrong".

Answered by ACuriousMind on March 23, 2021

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