There is a well defined notion of unitarity of representations in Euclidean Conformal field theories that follows from the requiring unitarity in the Lorentzian space. Under this notion, all states that are created by operator insertions have positive inner product, $$\langle \mathcal O|\mathcal O\rangle = \lim_{z\to\infty} z^{2\Delta} \langle\mathcal O(z)\mathcal O(0)\rangle>0$$ where I am thinking of radial quantization, or equivalently, of a field theory on a cylinder. This also goes by the name of reflection positivity. In arbitrary dimensions this requirement imposes constraints on the dimensions of physical operators:

  • $\Delta\ge\frac{d-2}2$ for non-spinning bosonic operators
  • $\Delta\ge\frac{d-1}2$ for non-spinning fermionic operators
  • $\Delta\ge d+\ell-2$ for operators with spin $\ell$

However, in the study of Harmonic analysis on the Euclidean Conformal groups ($SO(d+1,1)$ for a field theory in d-dimensional Euclidean space), one talks about unitary representations of the conformal group with dimensions $\Delta = \frac d2 + i s, \ s\in\mathbb R$ (the principal series) and, in odd dimensions, additionally $\Delta = \frac d2+\mathbb Z_+$ (the discrete series). Clearly these operators are excluded from the class of 'unitary' operators that follow from the Lorentzian unitarity requirement. From what I understand these class of representations provide a basis of $\mathbb L_2$ normalizable functions on the group manifold.

What is the difference between the two different notions of unitarity and how are they related? Moreover, clearly the operators in the first definition are normalizable in that they have a finite positive inner product ($\langle \mathcal O|\mathcal O\rangle >0$). In what sense are they not normalizable on the group manifold (thereby being excluded from Harmonic analysis basis of functions)?

  • $\begingroup$ By the way, in $d\geq 3$ the spin of an operator is parametrized by more than one number (in fact, by $n$ numbers, where $d=2n+1$ or $d=2n$). Correspondingly, the unitarity bounds are more complicated. The bound $\Delta\geq d+\ell-2$ applies to traceless-symmetric operators. $\endgroup$ May 6, 2019 at 16:53
  • $\begingroup$ @PeterKravchuk: Thanks for mentioning it. I wanted to edit it in question but then I guess I forgot till you mentioned it. $\endgroup$
    – nGlacTOwnS
    May 7, 2019 at 21:20

1 Answer 1


The representations that satisfy the usual unitarity bounds are the only representations that

  1. Are irreps of the Lorentzian conformal group, which is the universal cover $\widetilde{SO}(d,2)$ of $SO(d,2)$.
  2. Are unitary
  3. Have positive spectrum of energies. Note that the Hamiltonian $P^0$ is a part of the conformal algebra and hence its spectrum is fixed by representation theory.

It is in these representations that the physical Hilbert space $\mathcal{H}$ decomposes. By operator-state correspondence you get constraints on operators. See Luscher and Mack "Global Conformal Invariance in Quantum Field theory" and Mack "All unitary ray representations of the conformal group $SU(2,2)$ with positive energy".

The representations used in harmonic analysis are unitary tempered representations of $SO(d+1,1)$. You can see that this is a completely different group. The physical Hilbert space is not in representations of this group, and its unitarity of its representations is completely irrelevant from the quantum-mechanical point of view. Their importance, as you say, is that they furnish a complete set of functions on the group $SO(d+1,1)$ and hence on the homogeneous spaces $SO(d+1,1)/H$, where $H$ are various subgroups. For example, in the analysis of four-point functions the important space is $SO(d+1,1)/(SO(1,1)\times SO(d))$, which is the space of pairs of points on conformal sphere $S^d$.

To summarize, there are just two different stories which happen to both have "unitary" in them.

  • $\begingroup$ Here are some video lectures by David Simmons-Duffin on harmonic analysis in CFT by the way. (There's a bit of light-cone bootstrap in the beginning, but then he goes on to harmonic analysis.) $\endgroup$ May 5, 2019 at 18:46
  • $\begingroup$ Thank you for your answer. In fact, my confusion arose while watching these video lectures. I am still slightly confused about something. I can talk about unitarity in $SO(d,2)$ as reflection positivity in $SO(d+1,1)$ Euclidean CFT. In that sense I don't understand how are the two concepts of unitarity in representations of $SO(d+1,1)$ are related, if they are related. Because IMO, the object of these lectures is to study physical correlation functions but instead of using physical Hilbert space the tempered reps are used. Forgive me if DSD has mentioned it in his lecture and I missed it. $\endgroup$
    – nGlacTOwnS
    May 6, 2019 at 12:57
  • $\begingroup$ @nGlacTOwnS Reflection positivity is really just a different name for $SO(d,2)$ unitarity. For example, if you take Euclidean conformal algebra with the usual conjugation rules (i.e. $P^\dagger = K$ etc), then you can check that the Hermitian generators of this algebra form $so(d,2)$ algebra and not $so(d+1,1)$. So reflection positivity does not give a new concept of unitarity for $SO(d+1,1)$. Perhaps a good exercise is to try to actually construct a representation of $SO(d+1,1)$ from a local operator. By that I mean a vector space with action of $SO(d+1,1)$ and a Hermitian inner product ... $\endgroup$ May 6, 2019 at 16:46
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    $\begingroup$ @nGlacTOwnS, Isn't that what we do in the state-operator correspondence? No, and this is related to what I said about algebra of generators. If you had a unitary irrep of $SO(d+1,1)$, then you would have Hermitian Euclidean generators, i.e. $P^\dagger=P$ and $K^\dagger=K$. You don't, you only have $P^\dagger=K$, and if you build Hermitian linear combinatinos, then these combinations will satisfy $so(d,2)$ algebra (you would have to use $i$ to build Hermitian generators, and this changes the signature in commutation relations). $\endgroup$ May 13, 2019 at 13:28
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    $\begingroup$ @nGlacTOwnS, so by state-operator correspondence you only get unitary irreps of $SO(d,2)$, and these are precisely the irreps classified by Mack: positive-energy unitary irreps. Positive-energy is essential, since $SO(d,2)$ also has principal series irreps as well as others, but these contain negative energies. $\endgroup$ May 13, 2019 at 13:30

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