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In a two dimensional conformal field theory I have two sets of generators giving a representation of the Virasoro algebra

$$L_n, \bar{L}_n, n \in \mathbb{Z}$$ $$[L_n,L_n] = (m-n) L_{m+n} + c\frac{m(m^2-1)}{12}\delta_{m+n,0},$$ where $L_n$ is the generator associated to the infinitesimal holomorphic diffeomorphism $z^{n+1}\partial_z$.

The canonical Noether charges (under the condition that $z$ and $\bar{z}$ are complex conjugates are given by (eq. 2.7 in https://arxiv.org/abs/hep-th/9108028 after inserting eq. 3.6 )

$$Q_n = L_n +\bar{L}_n.$$

Usually I would expect that these charges can be exponentiated to give unitaries that implement the conformal transformation. I come however to the following question:

Except for $n = 0$, we have $L_n^\dagger = L_{-n} \neq L_n$, thus the noether charge $Q_n$ is not self-adjoint and exponentiating it does not give a unitary operator. This is probably not surprising for $|n|>1$ because the diffeomorphisms are not globally well-defined so I expect there to be some issue that obstructs to compute unitaries implementing the transformation, in particular since the quantum algebra has a central charge and thus should not implement the classical symmetry (I guess the classical symmetry is broken?). However the same issue arises for $n = 1,-1$ so I don't see how I would construct a unitary implementing a $SL(2,C)$ transformation in such a CFT. However according to the review above they exist (its claimed in eq. 3.21). How do I get their generators? I can in principle compute sth like $L_{-1} + L_1$ which is self adjoint but not obviously related to the diffeo $z^{n+1}\partial_z$. I would expect the noether charge itself already to provide me with the correct generator.

This question is similar to Why do we assume local conformal transformations are symmetries in 2D CFT? but not quite the same I believe.

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The statement that $L^\dagger_n = L_{-n}$, or its counterpart in higher dimensional notation which looks like \begin{equation} D^\dagger = D, \quad P_\mu^\dagger = K_\mu, \quad M_{\mu\nu}^\dagger = M_{\mu\nu}, \quad (1) \end{equation} isn't some immutable property of conformal field theory. It's a property of radial quantization which would not hold in equal time quantization for instance. In the former, the conjugate of $\mathcal{O}(x) \left | 0 \right >$ is $\mathcal{O}(x/|x|^2) \left | 0 \right >$ (suppressing contraction with the inversion tensor if $\mathcal{O}$ has Lorentz indices). In the latter, the conjugate of $\mathcal{O}(\tau, y) \left | 0 \right >$ is $\mathcal{O}(-\tau, y) \left | 0 \right >$. So it's not surprising that the Hermiticity of charges will change.

In general, only the charges generating translations along and orthogonal to your chosen foliation direction have to be Hermitian. This is consistent with (1) for radial quantization and $P^\dagger_\mu = P_\mu$ for equal time. So there's no reason to let this stop you from exponentiating a charge. In particular, you can carry out the usual procedure with \begin{align} P_0 = L_{-1} + \bar{L}_{-1}, \quad D = L_0 + \bar{L}_0, \quad K_0 = L_1 + \bar{L}_1 \\ P_1 = L_{-1} - \bar{L}_{-1}, \quad M_{01} = L_0 - \bar{L}_0, \quad K_1 = L_1 - \bar{L}_1 \end{align} to implement finite transformations in $SL(2, \mathbb{C})$. Considering $L_n$ with $|n| > 1$ however won't work because these infinitesimal transformations are not globally defined as you pointed out.

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  • $\begingroup$ Thank you for the answer! My expectation for charges to be hermitian is that if they generates symmetries in the sense of Wigners theorem, they should under exponentiation (In lorentzian signature) generate unitary transformations to be symmetries, which these charges will not do. I think the confusion on my side then is in what sense the SL(2,C) transformations are symmetries. I guess they do map solutions of classical eom to solutions but this seems like a weaker definition of symmetry to what we call symmetries usually in qft. $\endgroup$ Apr 15 at 9:39

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