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Conformally covariant distributions

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In Conformal Field Theory (in $D$ dimensions) one considers (in particular) correlation functions of the form $$ \langle O(x)O(y)\rangle, $$ where $O$ is a scalar primary field. Scale covariance demands $$ \langle O(\lambda x)O(\lambda y)\rangle=\lambda^{-2\Delta}\langle O(x)O(y)\rangle, $$ where $\Delta\geq\frac{D-2}{2}$ is the scaling dimension of $O$. Translation and rotation invariance require the correlator to be a function of $|x-y|$ only.

More generally for a conformal transformation $x\to x'$ one has $$ \langle O(x)O(y)\rangle=\Omega(x')^{\Delta}\Omega(y')^{\Delta}\langle O(x')O(y')\rangle, $$ where $$ \frac{\partial {x'}^\mu}{\partial x^\nu}=\Omega(x')R_\nu^\mu(x'), $$ and $R$ is an orthogonal matrix. For the two-point function considered above the unique non-zero solution of these constraints is, up to a normalization, $$ \langle O(x)O(y)\rangle=\frac{1}{|x-y|^{2\Delta}}. $$ This is in the class of functions defined for $x\neq y$, and can be seen by using a conformal transformation which brings $x$ and $y$ to some standard positions.

Now, Osterwalder-Schrader axioms (OS) require the correlation functions to be distributions defined on a suitable class of test functions. Clearly, if $\Delta$ is sufficiently large, one cannot naively interpret the above correlation function as a distribution on test functions with compact support due to the singularity at $x=y$. In fact, OS use a class of smooth functions which vanish with all derivatives at $x=y$. This clearly removes the singularity.

If we consider an analogous problem of turning $G(x)=|x|^{-\Delta}$ into a distribution in 1 dimension, one can try something like $$ \langle G_\epsilon,f\rangle=\int_{|x|>\epsilon}G(x)f(x)dx+\int_{|x|<\epsilon}G(x)\left(f(x)-f(0)-f'(0)x-\ldots\right)dx, $$ where in the second integral we subtract sufficiently many terms of Taylor expansion of $f$ at $0$ to make the integral convergent. $G_\epsilon$ defined this way is a distribution on compactly supported test functions without any requirement on behavior at $0$. Note that $G_\epsilon-G_{\epsilon'}$ is supported at $0$.

However, this definition breaks scale covariance, since we explicitly introduce a scale $\epsilon$. If we generalize to higher dimensions, then conformal covariance is also broken. My question is, is it possible to define the above correlation function as a distribution on a class of test functions larger than that of OS, while preserving conformal or scale covariance? I am particularly interested in test functions which would feel distributions supported at $x=y$.

This post imported from StackExchange MathOverflow at 2015-07-07 11:11 (UTC), posted by SE-user Peter Kravchuk
asked Jun 24, 2015 in Theoretical Physics by Peter Kravchuk (40 points) [ no revision ]
retagged Jul 7, 2015
The short answer is you add $i \varepsilon$ to $x-y$. In higher dimensions one has to go to minkowksi space

This post imported from StackExchange MathOverflow at 2015-07-07 11:11 (UTC), posted by SE-user Marcel Bischoff
@MarcelBischoff, can you please expand on the role of $i\epsilon$? I mean, if I wanted to define $1/x$, I could just use the principal value and that would be just fine, or add some delta functions to agree with $i\epsilon$. But I guess I do not quite understand the role of $i\epsilon$ in $1/|x|^\Delta$. Also, how does Minkowski space help?

This post imported from StackExchange MathOverflow at 2015-07-07 11:11 (UTC), posted by SE-user Peter Kravchuk

1 Answer

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Let me extend my comment. The Osterwald-Schrader axioms were invented to make the physical idea of Wick rotating quantum fields rigorous. Namely, the OS axioms are axioms on the correlation function of some classical fields, which ensure that these correlation functions can be analytical continued, such that they fulfill the Wightman axioms, thus give quantum fields on a Hilbert space via the Wightman (which is basically the GNS) reconstruction.

If you assume conformal covariance, the correlation functions are formally the same (off diagonal), but for quantum fields they are well-defined on the diagonal, because they are boundary values of holomorphic function in some tube.

So for 1D example the two-point function gets: $$ W(x,y) = \lim_{\varepsilon\to 0^+} \frac1{(x-y+i\varepsilon)^{2d}} $$ where the limit is in a weak sense. But the fields don't commute anymore, for example if $d=1$ $$\langle [\phi(x),\phi(y)]\rangle=W(x,y)-W(y,x)\sim \delta'(x-y)$$ so the commutator is supported on the diagonal. But it is a perfect conformal covariant distribution, which is conformally covariant. So in your example you could take $$ W(f)=\lim_{\varepsilon\to 0^+} \int_\mathbb{R} \frac{f(x)}{x+i\varepsilon}\,dx $$ as an extension to the Schwartz functions.

This post imported from StackExchange MathOverflow at 2015-07-07 11:11 (UTC), posted by SE-user Marcel Bischoff
answered Jun 25, 2015 by Marcel Bischoff (10 points) [ no revision ]

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