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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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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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3

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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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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