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Why all operators which we consider in CFT have fixed Dilatation value?

As I know in general QFT we haven't such requirement.

What if one will consider questions about operators, which are not Dilatation eingenvalues?

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Actually the dilation operator is not necessarily diagonalizable, see https://en.wikipedia.org/wiki/Logarithmic_conformal_field_theory .

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  • $\begingroup$ Is it possible to consider in theory where dilation operator is diagonalizable operators, which are not eingenvalues? $\endgroup$
    – Nikita
    Jun 22, 2020 at 8:50
  • $\begingroup$ If it is diagonalizable, there is a basis of operators that are eigenvectors. $\endgroup$ Jun 22, 2020 at 18:20
  • $\begingroup$ is it know, in which theories Dilation operator are diagonalizable? Is it true in $Ising_3$ CFT? $\endgroup$
    – Nikita
    Jun 22, 2020 at 20:00
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    $\begingroup$ No, it is not known, and the question is not necessarily very well-defined. For example, in the en.wikipedia.org/wiki/Two-dimensional_critical_Ising_model , the minimal model is not logarithmic, but other observables could be logarithmic. Percolation is thought to be logarithmic. For the 2d O(n) and Potts model, the question is in the process of being settled, see arxiv.org/abs/2005.07708 . $\endgroup$ Jun 23, 2020 at 17:20
  • $\begingroup$ I guess a completely rigorous proof without any unjustified assumptions is still elusive, but for a unitary CFT like Ising 2d or 3d with discrete spectrum, the expectation is that the dilation operator is diagonalizable. For a reasonably convincing argument, see Section 7.4 of arxiv.org/abs/1602.07982 $\endgroup$ Aug 5, 2020 at 15:58
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Also in the article On Scale and Conformal Invariance in Four Dimensions there's appendix about such phenomena.

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