Let us assume to have a QFT ($\mathcal{L}$) with translational, Lorentz, scale and conformal invariance.
I ask because we can, for example the free scalar free theory, canonically quantize the system and generate states that don't have scale symmetry (e.g. much like plane waves $a_p^\dagger\left|0\right>$ don't have spherical symmetry).
Are there CFTs that when we quantize can have states that don't have conformal symmetry and can't be written as a sum of states with conformal symmetry?
If the answer is yes, we just disregard them in the state-operator correspondence?
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1$\begingroup$ It is unclear what you mean by a "state not having conformal symmetry". But you may want to read about logarithmic CFTs, e.g. the dilatation operator need not be diagonalizable (and the space of states might be reducible but not completely reducible, etc.) $\endgroup$– AccidentalFourierTransformCommented Dec 22, 2023 at 20:10
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$\begingroup$ @AccidentalFourierTransform I mean that if you have the state $a_p^\dagger\left|0\right>$ when you act on it with the operator $$\phi\left(x\right)=\intop\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{p}}}\left(a_{p}e^{ipx}+a_{p}^{\dagger}e^{-ipx}\right)$$ Meaning $$\left\langle 0\right|\phi\left(x\right)a_{p}^{\dagger}\left|0\right\rangle =\frac{e^{ipx}}{\sqrt{2E_{p}}}$$ It is a plane wave. Now because it spans all functions I can produce a state that doesn't have conformal symmetry. In some sense when you act with an operator with conf. dim. the state would have the same conf. d. $\endgroup$– ssmCommented Dec 23, 2023 at 10:09
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$\begingroup$ *I meant to say scaling dimension $\endgroup$– ssmCommented Dec 23, 2023 at 10:28
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I've got my answer in the comments and answers here https://physics.stackexchange.com/questions/747918/why-is-the-limiting-operator-in-the-cft-state-operator-correspondence-well-defin\ These Dilatation states do not span the entire Hilbert space. So it is just their subset that has the state-operator correspondence.