$\newcommand{ip}[2]{\left< #1, #2\right>} \newcommand{\d}{\, \mathrm d \lambda^n}$For a Hilbert space $(H,\ip \cdot \cdot)$ and an operator$^1$ $A$ the adjoint of $A$ is defined via the relation:

$$ \ip{x}{ A y} = \ip{A^\dagger x} y \qquad \forall x,y, \in H$$

Supposedly the conformal algebra has a representation on $L^2$ functions for example the dilatation take the form

$$ D = -i x^\mu \partial_\mu $$

However, naively this operator is clearly not Hermitian with the usual scalar product of $L^2$ with Lebesgue measure $\d$

\begin{align}\ip f{Dg} &= \int \bar f (-i x^\mu \partial_\mu g) \d = \int \overline{(-i x^\mu \partial_\mu f)} g \d+ i \int (\bar f g)\, (\partial_\mu x^\mu) \d \\ &= \ip{D f} g + i n \ip fg \quad , \end{align}

where $n$ is the dimension of spacetime and I integrated by parts to get the second equality, lastly I used $\partial_\mu x^\mu = n$.

A second hint that there is a different choice of scalar product comes from the fact that for 2d CFT's the adjoint of an operator is defined (cf. [1] p. 32) via

$$ A(z, \bar z) ^\dagger = A\left( \frac 1 {\bar z} , \frac 1z \right) \frac 1{\bar z^{2h}} \frac1 {z^{2\bar h}} $$

for $(h,\bar h)$ primary fields.

So my question is, what the scalar product one chooses for CFT's $^2$ and how one can derive the "adjoint" rule postulated above by this choice of scalar product.

$^1$ Bounded operator

$^2$ An educated guess would be the Haar measure for the conformal group but I am not sure.

[1] https://arxiv.org/abs/hep-th/9108028v1

Your assumption that the representation you are talking about in the sentence

Supposedly the conformal algebra has a representation on $L^2$ functions

should be Hermitian is incorrect. The space of states of a 2d conformal field theory is not functions on the plane - the functions on the plane are the fields/operators of the theory, not the states. There is no reason to expect the action of the algebra on the operators to be Hermitian w.r.t. to some inner product on the operators.

CFTs are field theories and so their spaces of states are more complicated than the simple $L^2(\mathbb{R}^n)$ spaces of ordinary quantum mechanics. E.g. the free 2d scalar field on a cylinder is a conformal field theory and simply has the Fock space of a free scalar field as its space of states.

  • Ok let me rephrase my question slightly because it could be ambiguos. People claim that there is a representation of the conformal group on $L^2$ with the given $D$ and go on saying that $D$ is hermitian. Furthermore, how would one justify the "adjoint rule" that I described above without specifying the scalar product that one has? – Gonenc Mogol Nov 9 at 22:32
  • 1
    @GonencMogol Which people, where? I've never needed the Hermiticity of this representation for anything in CFT. – ACuriousMind Nov 9 at 22:39
  • Actually, there are unitary representations in $L^2$ of $SL(2,R)$ and they can be also extended to representation of the whole Virasoro algebra passing to Fock representations built upon that $L^2$. I dealt with this stuff several years ago (see e.g. my joint paper V. M. and N. Pinamonti, Aspects of Hidden and manifest SL(2,R) symmetry in 2D near horizon black hole background Nucl.Phys.B 647, 131 (2002) and Bose-Einstein condensate and Spontaneous Breaking of Conformal Symmetry on Killing Horizons J. Math. Phys. 46, 062303 (2005)) – Valter Moretti 2 days ago
  • However, as I said in my partial answer, I am not familiar with CFT and I treated these subjects from the viewpoint of conformal nets of von Neumann algebras instead of standard CFT... – Valter Moretti 2 days ago

The adjoint formula for operators on the radially quantized plane (the one you have written in your question) is derived from a Wick rotation of the Lorentzian definition of adjoint on the cylinder (which is the standard definition).

Let $O(t,x)$ be a Hermitian conformal primary operator with dimension $(h,{\bar h})$ ($t$ is Lorentzian time). It satisfies $$ O^\dagger(t,x) = O(t,x) , \qquad O(t,x) = e^{ i H t} O(0,x) e^{- i H t} $$ In the Wick rotated theory, we define the Euclidean operator $O_E(\tau,x)$ by $$ O_E(\tau,x) = O(-i \tau,x) = e^{H\tau} O(0,x) e^{-H\tau} . $$ The Euclidean opertor has the following adjoint property $$ O_E^\dagger(\tau,x) = O_E(-\tau,x). $$ Now, let us introduce complex coordinates on the plane $w = x - i \tau$ and ${\bar w} = x + i \tau$. The adjoint property then reads $$ O_E^\dagger(w,{\bar w}) = O_E({\bar w},w). $$ Now, consider the theory on a circle with $x \sim x + 2\pi$. We use a conformal transformations to map this to a radially quantized plane. Define $$ z = e^{iw} = e^{\tau} e^{ix} , \qquad {\bar z} = e^{- i {\bar w}} = e^\tau e^{- i x } \quad \implies \quad w = - i \log z , \qquad {\bar w} = i \log {\bar z} $$ Under conformal transformations, primary operators transform as \begin{align} O'_E(z,{\bar z}) &= (\partial_z w)^{h} (\partial_{\bar z} {\bar w})^{{\bar h}} O_E(w,{\bar w}) \\ &= (iz)^{-h} (-i{\bar z})^{-{\bar h}} O_E(- i \log z, i \log {\bar z}) , \end{align} or inversely, $$ O_E(w,{\bar w}) = (ie^{i w})^{h} ( -i e^{-i {\bar w} } )^{{\bar h}} O'_E(e^{iw},e^{-i {\bar w} } ) . $$ Note that $O'_E$ is defined on the radial plane whereas $O_E$ is defined on the cylinder.

We have everything we need! The adjoint of an primary operator on the radially quantized plane is now \begin{align} O'^\dagger_E(z,{\bar z}) &= (-i {\bar z})^{-h} ( i z)^{-{\bar h}} O_E^\dagger (- i \log z, i \log {\bar z}) \\ &= (-i {\bar z})^{-h} ( i z)^{-{\bar h}} O_E(i \log {\bar z},- i \log z) \\ &= \frac{1}{ {\bar z}^{2h} } \frac{1}{ z^{2 \bar h} } O'_E \left( \frac{1}{ {\bar z} } ,\frac{1}{z} \right) . \end{align}

I do not know much on CFT, but I suspect that, dealing with $L^2(\mathbb R_+^n, d^nx)$, there is a trivial adjustment.

The map $\psi(x) \mapsto \psi(\lambda x)$ is not unitary on $L^2(\mathbb R_+^n, d^nx)$: $$\int \overline{\phi(\lambda x)} \psi(\lambda x) d^nx = \lambda^{-n}\int \overline{\phi(\lambda x)} \psi(\lambda x) d^n\lambda x = \lambda^{-n}\int \overline{\phi(x)} \psi(x) d^nx\:.$$ So it cannot define a symmetry in the sense of Wigner. However a unitary action of dilatations is the following $$(U_\lambda \psi)(x):= \lambda^{n/2}\psi(\lambda x)\:.$$ The properly selfadjoint generator of this unitary one-parameter group is not $-ix^\mu\partial_\mu$ but is $$D = -i\left(\frac{n}{2} +x^\mu \partial_\mu\right)\:,$$ and $U_\lambda = e^{i\lambda D}$.

  • 2
    For those wondering where $D$ came from: it is nothing but the symmetrisation of $x\cdot\partial$, that is, $\frac12(x\cdot\partial+\partial\cdot x)$ (Weyl ordering). – AccidentalFourierTransform Nov 9 at 22:07
  • What do you mean by $\mathbb R^n_+$? $\lambda$ must be positive, but $x_i$ need not, right? – AccidentalFourierTransform 2 days ago
  • $[0,+\infty)^n$. But perhaps I am wrong, I had a look at some old paper of mine where I dealt with these topics and it seemed to me that, at least for $n=1$, that is the correct space if you want to construct a unitary rep. of the whole $PSL(2,R)$. – Valter Moretti 2 days ago
  • See arxiv.org/abs/gr-qc/0207072 published as Nucl.Phys. B647 (2002) 131-152 – Valter Moretti 2 days ago

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