8
$\begingroup$

I used to believe the three following statements to be true (at the classical level only):

  1. From scale invariance full conformal invariance follows.

  2. Scale invariance is present if there are no dimensional parameters in the Lagrangian.

  3. The energy-momentum tensor for scale or conformally-invariant theory is traceless.

However, when looking at the particular example of the $\phi^4$ theory in 4d I begin to doubt. The Lagrangian is, of course, $$\mathcal{L}=\frac12(\partial \phi)^2-g \phi^4,\quad S=\int d^4x \mathcal{L}$$

In 4d field $\phi$ is of mass dimension 1 and $g$ is dimensionless. The theory is scale invariant (if under $x'=\lambda x$ field transforms as $\phi'(x')=\lambda^{-1}\phi(x)$), in accordance with statement (2).

However, it seems to me that the theory is not invariant under inversions (I won't bother you with my failed attempts here) and its energy-momentum tensor $$T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-\delta_{\mu\nu}\left(\frac12(\partial\phi)^2-g\phi^4\right)$$ is not traceless $$T^\mu_\mu=(1-d/2)(\partial\phi)^2+dg\phi^4.$$

  • My question is which of the assertions (1,2,3) are in fact true and how all this works at the example of the $\phi^4$ theory?

I stress once again that here I'm interested in the classical aspects only.

$\endgroup$
  • 4
    $\begingroup$ You should use the stress energy tensor obtained by taking the variatianal derivative of the action with a generic non-flat metric $g_{\mu\nu}$ and such that the action also includes the correction term $\phi^2 R/6$ (vanishing in flat spacetime) which makes it conformally invariant in general sense. When you put $g_{ab}=\eta_{ab}$ (flat spacetime) in the final formula for $T_{\mu\nu}$, you see that a term remains however (even if $R=0$) producing a vanishing trace. This improved stress energy tensor is different from the canonical one. $\endgroup$ – Valter Moretti Nov 30 '15 at 11:31
  • $\begingroup$ I thought that what one has to do is to 'covariantize' the action. Hence, given that the term $\phi^2R/6$ is covariant by itself I don't see why it must be included? And why with that specific coefficient etc.? And what about inversion invariance, is it there? How does the field transform under it? Or, maybe, you could just give a reference on this topic?:) Thanks! $\endgroup$ – Weather Report Nov 30 '15 at 11:48
  • 1
    $\begingroup$ The term $\phi^2R/6$ is the one necessary for $d=4$, otherwise there is a complicated coefficient depending on $n$. Actually I am not sure that my idea works in the interacting case $g\neq 0$. You could try to have a look at the (very old) textbook by Birrell and Davies on QFT in curved spacetime... $\endgroup$ – Valter Moretti Nov 30 '15 at 11:57
  • $\begingroup$ ...Also the end of p.448 of Wald's textbook on GR where $T^{\mu}_\mu=0$ is discussed for a symmetric conserved tensor in relation to conformal invariance. $\endgroup$ – Valter Moretti Nov 30 '15 at 12:02
2
$\begingroup$

(2) and (3) are true. (1) is not known in general (for unitary theories).

It is true in 2d (I remember seeing a paper about a similar result for 3d but I can't find it) that scale invariance and unitarity imply conformal invariance. In general dimensions, this is only true when the Virial $$V^\mu=\frac{\delta L}{\delta(\partial^\rho \phi)}(\eta^{\mu\rho}\Delta+i S^{\mu\rho})\phi$$ can be written as a divergence $$V^\mu=\partial_\alpha \sigma^{\alpha\mu}$$ If that is true then you can come up with a modification of the stress-tensor so that $$T^\mu_\mu=\partial_\mu j^\mu_D$$ where $j_D$ is the dilation current. Then scale invariance implies conformal invariance. See section 4.2.2 of DiFrancesco's book to see the full derivation. Basically if you can modify the energy-momentum tensor to make it traceless, in a way that you don't spoil it's conservation nor the Ward identity, then the theory will be conformal by (3).

In your example you can compute the Virial and see whether it can be written as the divergence of something else, if you succeed then it is indeed conformal.

Further reading concerning new developments in the subject of scale invariance $\Rightarrow$ conformal invariance:

http://arxiv.org/abs/1309.2921

http://arxiv.org/abs/1402.6322

http://arxiv.org/abs/1505.01152

$\endgroup$
  • 1
    $\begingroup$ In general, (1) is wrong -- there are counterexamples (theory of elasticity, topologically twisted theories). $\endgroup$ – Hans Moleman Dec 20 '15 at 1:38
  • $\begingroup$ @HansMoleman, I wouldn't say wrong because there's still a large class of theories where scale implies conformal. Counterexamples are more like an exception to the rule and thus it's interesting to understand why they break it. E.g. see the last link where they consider a SFT embedded into a CFT and show that the former has to be trivial. It doesn't say there are no non-trivial SFTs but rather that if there are, they are exotic. (A motivation for this is that RG fixed points are conformal). $\endgroup$ – Prastt Dec 20 '15 at 10:29
  • $\begingroup$ @HansMoleman I might add that I'm considering unitary theories. Otherwise I agree with what you said. $\endgroup$ – Prastt Dec 20 '15 at 10:39
2
+50
$\begingroup$

(1) is not true. Typical counterexamples are Maxwell's theory in dimension $d \neq 4$ or the theory of elasticity in 2 dimensions. See also the other answer.

(3) is not completely true either. The correct statement is that a theory is invariant under scale transformations if $$ T^\mu_\mu = \partial_\mu K^\mu $$ for some operator $K^\mu$ (the virial current mentioned in the other answer), and it is invariant under special conformal transformations if $$ T^\mu_\mu = \partial_\mu \partial_\nu L^{\mu\nu} $$ for some $L^{\mu\nu}$. This is nicely explained in a paper by Polchinski. Equivalently, if $T^{\mu\nu}$ satisfies the above condition, it can be "improved" by adding a term that does not affect its conservation property but cancel the trace.

Explicitly, in your example the canonical energy-momentum tensor is $$ T_c^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial^\nu \phi - \eta^{\mu\nu} \mathcal{L} = \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} \partial_\rho \phi \partial^\rho \phi + g \eta^{\mu\nu} \phi^4 $$ and the improved energy-momentum tensor \begin{align} T^{\mu\nu} &= T_c^{\mu\nu} - \frac{d-2}{4(d-1)} (\partial^\mu \partial^\nu - \eta^{\mu\nu} \square) \phi^2 \\ &= \frac{1}{2(d-1)} \left[ d \partial^\mu \phi \partial^\nu \phi - \eta^{\mu\nu} \partial_\rho \phi \partial^\rho \phi - (d-2) \phi \partial^\mu \partial^\nu \phi + (d-2) \eta^{\mu\nu} \phi \square \phi \right] + g \eta^{\mu\nu} \phi^4 \end{align} Both tensors satisfy the conservation property $\partial_\nu T^{\mu\nu} = \partial_\nu T^{\mu\nu}_c = 0$ when imposing the equation of motion $$ \square \phi + 4 g \phi^3 = 0 $$ But now you can check that the trace of the improved energy-momentum tensor $$ T^\mu_\mu = \frac{d-2}{2} \phi \square \phi + d g \phi^4 $$ also vanishes by the equation of motion in $d = 4$.

(note: this is a slightly edited duplicate of my answer to another question)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.