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Why does a quantum field theory invariant under dilations almost always also have to be invariant under proper conformal transformations? To show your favorite dilatation invariant theory is also invariant under proper conformal transformations is seldom straightforward. Integration by parts, introducing Weyl connections and so on and so forth are needed, but yet at the end of the day, it can almost always be done. Why is that?

Qmechanic
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    You can look for Polchinski's paper "Scale and Conformal Invariance in QFT" which discusses the issue in detail. Maybe there have been some developments since then, but that is a good starting point. –  Mar 05 '11 at 18:56
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    For 4d field theory it's still an open question whether scale invariance implies conformal invariance. There's been some recent work on this topic by Slava Rychkov and collaborators, see e.g. 1101.5385. – Matt Reece Mar 06 '11 at 23:06
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    By the way, given that scale invariance does not imply conformal invariance, maybe the question can be rephrased. –  Mar 09 '11 at 18:14
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    I'd like to point to a review by Yu Nakayama, available at http://arxiv.org/abs/1302.0884 – Siva Sep 20 '13 at 07:28

3 Answers3

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As commented in previous answers, conformal invariance implies scale invariance but the converse is not true in general. In fact, you can have a look at Scale Vs. Conformal Invariance in the AdS/CFT Correspondence. In that paper, authors explicitly construct two non trivial field theories which are scale invariant but not conformally invariant. They proceed by placing some conformal field theories in flat space onto curved backgrounds by means of the AdS/CFT correspondence.

xavimol
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    Thanks for that, somehow I missed this one, it is really interesting. –  Mar 09 '11 at 18:14
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    I knew about this paper but reading this question, made it came to my mind again (fortunately, because it's very interesting) – xavimol Mar 09 '11 at 19:23
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A nice article about this: Tutorial on Scale and Conformal Symmetries in Diverse Dimensions.

The rule-of-thumb is that 'conformal ⇒ scale', but the converse is not necessarily true (some condition(s) needs to be satisfied) — but, of course, this varies with the dimensionality of the problem you're dealing.

PS: Polchinski's article: Scale and conformal invariance in quantum field theory.

Daniel
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    It's not a rule-of-thumb that conformal implies scale, it's just a fact. The conditions are mostly locality and low-order of derivatives, which is sometimes imposed by unitarity and renormalizability. – Ron Maimon May 04 '12 at 19:43
  • @RonMaimon: Conformal invariance requires scale invariance in a Poincare invariant theory simply because of the commutator $[K_\mu,P_\nu]=2i(\eta_{\mu\nu}D-M_{\mu\nu})$. The notation should be obvious. – AndyS Jun 15 '12 at 23:10
  • @AndyS: The very existence of D in the conformal group is enough to show conformal implies scale--- it's not a rule of thumb, it's an obvious implication, that's what the comment above was trying to say. You don't need the commutator business to show this, the dilatation is a conformal transformation all by itself. – Ron Maimon Jun 16 '12 at 06:51
  • @RonMaimon: What you're saying is not true; you need the commutator in order to prove what you call "an obvious implication". Also, there is a clear distinction between dilatations and special conformal transformations. – AndyS Jun 19 '12 at 03:23
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    @AndyS: What I am saying is true, and you are saying nonsense. Dilatations are to conformal invariance as rotations about the z-axis are to rotations. They are a special case. If you have rotational invariance, you have rotational invariance around the z-axis. If you have conformal invariance, you have dilatation invariance. This is not an arguable point, it is not a difficult point, and I don't know why you make the comment above. – Ron Maimon Jun 19 '12 at 15:48
  • @RonMaimon: It's pointless to continue this discussion. Just so you know, though, a dilatation is a rescaling of the coordinates, $x^\mu\to\lambda x^\mu$, while a special conformal transformation is an inversion followed by a translation followed by another inversion, $x^\mu\to\frac{x^\mu+b^\mu x^2}{1-2b\cdot x+b^2x^2}$. The global vector $b^\mu$ cannot be chosen to fake a dilatation. (If you allow a local $b^\mu$, then it is possible to create a dilatation by a local conformal transformation, the same way it is possible to create a rotation by a local translation.) – AndyS Jun 19 '12 at 18:35
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    @AndyS: I know what a special conformal transformation is, and the dilatation group is a subgroup of the full conformal group. The special conformal transformations do not close into a group by themselves, and together with rotations and dilatations they make a group. The dilatation subgroup is a special case of the full conformal group and it makes no sense to say you have a conformal theory without dilatation invariance, it is like saying that you have a rotationally invariant theory which is not invariant to rotations about the z axis. – Ron Maimon Jun 19 '12 at 21:12
  • @RonMaimon: The question I'm answering is: "Why is a theory that is invariant under special conformal transformations automatically invariant under dilatations too?" This is very different from the moronic question: "Why is a conformal theory invariant under dilatations?". – AndyS Jun 19 '12 at 22:51
  • @AndyS: I see. Then I agree. But to me this seems backwards, since conformal symmetry is local dilatation symmetry, so dilatation invariance is a more primitive idea, and is assumed in advance before you show conformal invariance. But this is just a point of view. – Ron Maimon Jun 20 '12 at 17:25
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Maybe this does it:
$\begin{array}{rccl} \textrm{Translation:}&P_\mu&=&-i\partial_\mu\\ \textrm{Rotation:}&M_{\mu\nu}&=&i(x_\mu\partial_\nu-x_\nu\partial_\mu)\\ \textrm{Dilation:}&D&=&ix^\mu\partial^\mu\\ \textrm{Special Conformal:}&C_\mu&=&-i(\vec{x}\cdot\vec{x}-2x_\mu\vec{x}\cdot\partial) \end{array}$

Then the commutation relation gives:
$[D,C_\mu] = -iC_\mu$
so $C^\mu$ acts as raising and lowering operators for the eigenvectors of the dilation operator $D$. That is, suppose:
$D|d\rangle = d|d\rangle$
By the commutation relation:
$DC_\mu - C_\mu D = -iC_\mu$
so
$DC_\mu|d\rangle = (C_\mu D -iC_\mu)|d\rangle$
and
$D(C_\mu|d\rangle) = (d-i)(C_\mu|d\rangle)$

But given the dilational eigenvectors, it's possible to define the raising and lowering operators from them alone. And so that defines the $C_\mu$.

P.S. I cribbed this from:

http://web.mit.edu/~mcgreevy/www/fall08/handouts/lecture09.pdf

Carl Brannen
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    Let's say you cited Mr. McGreevy. –  Mar 06 '11 at 13:04
  • The issue is that it just isn't true that scale invariance implies conformal invariance. The simplest counterexample is some self-interacting Levy field theory. – Ron Maimon May 04 '12 at 19:44
  • What are these $|d\rangle$ kets you refer to? Are they fields? And does $d$ correspond to the scaling dimension of the field? I realise this is an old thread, so I would appreciate comments from anyone. – CAF Jul 26 '14 at 16:36