Exact solutions for scalars and spinors on quantized

(Euclidean) $AdS_{2}$ and the correspondence principle

A<EMAIL_ADDRESS>and A<EMAIL_ADDRESS>

1 Universidade de Brasília, Instituto de Física

70910-900, Brasília, DF, Brasil

and

International Center of Physics

C.P. 04667, Brasília, DF, Brazil

2 Department of Physics, University of Alabama,

Tuscaloosa, Alabama 35487, USA

###### Abstract

We obtain the exact solutions to the field equations for massless scalar and
massless spinor fields on quantized two-dimensional anti-de Sitter space. We
then apply the $AdS/CFT$ correspondence principle to get exact answers for the
two point correlation functions of the associated operators on the boundary.
The results support the conclusion that conformal symmetry on the boundary is
not spoiled by quantization of the bulk. Moreover, quantization of the bulk
has no effect on the spinor two-point correlation function, while it induces
an overall re-scaling in the scalar two-point correlation function.

##  1 Introduction

Research on the quantization of space-time has been carried out with the
intention of probing the quasiclassical regime of quantum gravity. In this
regard, arguments have been given supporting the idea that the quasiclassical
regime can be described in terms of quantum field theory on a non-commutative,
i.e., quantized, background.[1] Meanwhile, the $AdS/CFT$ correspondence
principle offers the possibility of a nonperturbative description of quantum
gravity on asymptotically AdS spaces.[2, 3, 4, 5] (Despite this, many
applications of the correspondence principle rely on classical gravity
calculations.) In an attempt to merge these different approaches, we have
studied the $AdS/CFT$ correspondence principle in a non-commutative
setting.[6, 7, 8] By this we mean that the geometry on the gravity side of the
correspondence was written on a quantized space-time, which we briefly review
below.

In [6, 7] we examined the example where the bulk geometry is the quantized
version of two-dimensional anti-de Sitter space ($AdS_{2}$). In this regard,
$AdS_{2}$ has a unique quantization upon imposing the requirement that all the
isometries of the classical geometry are preserved under the quantization.[9,
10, 11, 12, 13, 14] ‡‡‡Although it may not be possible to make the same demand
for arbitrary $d>2$ dimensional anti-de Sitter spaces, higher-dimensional
generalizations of such an isometry preserving quantization procedure do exist
for other spaces, namely, indefinite complex projective spaces
($\mathbb{CP}^{p,q}$, $q\geq 1$), as was shown in [8]. The $AdS/CFT$
correspondence principle posits a strong/weak duality between quantum gravity
in some $d$ dimensional bulk and a field theory on its $d-1$ boundary.[2, 3,
4, 5] The bulk is traditionally taken to be an asymptotically anti-de Sitter
space, with a corresponding conformal field theory (CFT) residing on its
boundary, although many extensions of the correspondence having been proposed.
We argued in [6, 7] that the quantized version of $AdS_{2}$ is an
asymptotically anti-de Sitter space. This is due to i) the result that the
star product used to realize the noncommutative algebra approaches the trivial
commutative product in the boundary limit, and ii) all quantum corrections to
the Killing vectors vanish in this limit. §§§Analogous results were shown for
the quantization of $\mathbb{CP}^{p,q},\;q\geq 1$, implying that these spaces
are asymptotically commutative. Therefore, according to the correspondence
principle, a dual conformal quantum theory should reside on the boundary of
quantized $AdS_{2}$ (baring known difficulties of the correspondence principle
for two dimensional anti-de Sitter space[15, 16, 17, 18]). An initial study of
the corresponding boundary theory was pursued in [6, 7]. There we examined the
two and three point correlation functions of operators on the boundary which
were sourced by scalar fields on quantized $AdS_{2}$. [More precisely, the
analysis and quantization was carried out for the Euclidean version of
$AdS_{2}$ ($EAdS_{2}$).] Due to rather involved calculations in the approach
followed there, the best we could do was obtain the leading order quantum
(non-commutative) corrections to the usual two and three point correlation
functions, i.e., those associated with a ‘classical’ space-time bulk.

In this article we re-examine the correspondence principle for quantized
$EAdS_{2}$, and obtain exact results for the two point correlation functions
of operators on the boundary. This is done for operators sourced by both
scalar and spinor fields in the bulk. The exact results for the two point
correlators are possible due to the existence of exact solutions to both the
scalar and spinor field equations on quantized $EAdS_{2}$. These equations,
which involve the relevant Laplacian and Dirac operators, can be written
purely algebraically. For simplicity, we shall specialize here to the case of
massless scalar and spinor fields, although exact solutions should be
obtainable for the massive case, as well. Our result shows that the conformal
properties of the boundary two-point correlation functions are not spoiled by
the quantization of the bulk. This is not unexpected since the corresponding
isometries on the gravity side are preserved. Moreover, the boundary two-point
correlation function sourced by spinors on quantized $EAdS_{2}$ is identical
to that of its classical counterpart. Therefore, from such boundary
correlation functions, one cannot distinguish whether the bulk theory is
classical or quantum. On the other hand, the boundary two-point correlation
function sourced by scalars on quantized $EAdS_{2}$ differs from its classical
counterpart by a simple overall factor, which agrees with the leading order
result of [6, 7].

The outline for the rest of this article is as follows: In section 2 we review
the dynamics of massless scalar and spinor fields on $EAdS_{2}$. Two
formulations of spinors are given, which we refer to as the noncovariant and
covariant formulations, and a map between the two is given. While the
noncovariant approach is simpler, the covariant approach has a natural
generalization to quantum space-time. As a preliminary step towards
quantization we introduce a Poisson structure to $EAdS_{2}$. The quantization
of $EAdS_{2}$ is reviewed in section 3. There we write down the Laplacian and
Dirac operators on quantized $EAdS_{2}$, and present exact solutions to the
equations for both massless scalar and spinor fields on this space. From the
asymptotic properties of these solutions, we derive exact results for the
associated two-point correlation functions on the boundary in section 4.
Concluding remarks are presented in section 5. In appendix A we show how to
evaluate boundary terms on quantized $EAdS_{2}$. We find the quantum analogue
of the map between the covariant and noncovariant bases in appendix B, and use
it to construct the Dirac operator in the noncovariant basis. The solutions
for massless spinors in that basis are trivially obtained. They can then be
mapped back to the covariant basis, and the result agrees with the exact
solutions of section 3.

##  2 Scalars and spinors on $EAdS_{2}$

###  2.1 Preliminaries

We begin with a review of the Euclidean $AdS_{2}$ geometry. $EAdS_{2}$ is
conventionally coordinatized using Fefferman-Graham coordinates, $(z,t)$,
spanning a half-plane ${\mathbb{R}}^{2}_{+}$, $z>0$, $-\infty<t<\infty$. The
asymptotic boundary corresponds to $z\rightarrow 0$. The metric tensor in
terms of these coordinates is given by

$ds^{2}=\frac{dz^{2}+dt^{2}}{z^{2}}\;.$ (2.1)

$EAdS_{2}$ is a maximally isometric space, with the isometries generated by
the three Killing vectors

${\cal K}^{-}=-\partial_{t}\quad,\quad{\cal
K}^{1}=-t\partial_{t}-z\partial_{z}\quad,\quad{\cal
K}^{+}=(z^{2}-t^{2})\,\partial_{t}-2zt\,\partial_{z}\;,$ (2.2)

which form a basis for the $su(1,1)$ algebra. Defining ${\cal
K}^{0}=\frac{1}{2}({\cal K}^{+}+{\cal K}^{-})$ and ${\cal
K}^{2}=\frac{1}{2}({\cal K}^{+}-{\cal K}^{-})$, we get the commutator brackets

$[{\cal K}^{0},{\cal K}^{1}]={\cal K}^{2}\;,\qquad\;[{\cal K}^{1},{\cal
K}^{2}]=-{\cal K}^{0}\;,\qquad\;[{\cal K}^{2},{\cal K}^{0}]={\cal
K}^{1}\;.\qquad\;$ (2.3)

The Laplacian on $EAdS_{2}$ is

$\Delta=z^{2}(\partial_{z}^{2}+\partial_{t}^{2})\;.$ (2.4)

It can be written in terms of the three Killing vectors

$\Delta={\cal K}^{a}{\cal K}_{a}\;,$ (2.5)

where the index $a=0,1,2$ is raised and lowered using the metric tensor
$[\eta_{ab}]=$ diag$(-1,+1,+1)$. The equation for a free massless scalar
$\Phi$ on $EAdS_{2}$,

$\Delta\Phi=0\;,$ (2.6)

is easily solved. General solutions are expressed in terms of analytic and
anti-analytic functions of $\zeta=z+it$,

$\Phi(\zeta,\bar{\zeta})={f(\zeta)+g(\bar{\zeta})}\;,$ (2.7)

$f$ and $g$ being complex functions. Upon approaching the boundary $z=0$ the
solution behaves as

$\Phi(\zeta,\bar{\zeta})=\phi_{0}(t)+z\phi_{1}(t)+{\cal O}(z^{2})\;,$
$\phi_{0}(t)={f(it)+g(-it)}\;,\qquad\;\phi_{1}(t)=\Bigl{(}{f^{\prime}(\zeta)+g^{\prime}(\bar{\zeta})}\Bigr{)}|_{z=0}\;,$
(2.8)

the prime denoting a derivative.

The Dirac operator on an arbitrary $d-$dimensional commutative manifold is

$D=\gamma^{\mu}(\partial_{\mu}+\omega_{\mu})\;,$ (2.9)

where $\omega_{\mu}$, $\mu=1,...,d$ are the components of the spin connection
one form and our convention is that the gamma matrices $\gamma^{\mu}$ satisfy
the anti-commutator

$[\gamma^{\mu},\gamma^{\nu}]_{+}=2g^{\mu\nu}{\mathbb{1}}\;.$ (2.10)

Upon assuming that the manifold is locally Euclidean, the gamma matrices are
transformed to a local Euclidean frame using vielbein fields
$e^{i}_{\;\;\mu}$:
$\gamma^{\mu}\rightarrow\hat{\gamma}^{i}=e^{i}_{\;\;\mu}\gamma^{\mu}$. The
transformed gamma matrices $\hat{\gamma}^{i}$ satisfy the anti-commutation
relations: $[\hat{\gamma}^{i},\hat{\gamma}^{j}]_{+}=2\delta^{ij}{\mathbb{1}}$.
We shall assume that the space has zero torsion. Denoting a basis of one forms
by $\\{dx^{\mu}\\}$, we write down the vielbein and spin connection one forms
$e=e_{\mu}dx^{\mu}$ and $\omega=\omega_{\mu}dx^{\mu}$, respectively. $e_{\mu}$
and $\omega_{\mu}$ take values in the Clifford algebra generated by
$\hat{\gamma}^{i}$, $e_{\mu}=e^{i}_{\;\;\mu}\hat{\gamma}_{i}$ and
$\omega_{\mu}=\frac{1}{8}\omega_{ij\mu}[\hat{\gamma}^{i},\hat{\gamma}^{j}]$.
Then the zero torsion condition is

$de+\omega\wedge e+e\wedge\omega=0\;.$ (2.11)

In the next two subsections we shall write down two different realizations of
the Dirac operator, one in what we call the noncovariant basis and another in
the covariant basis. While the former basis is more familiar, the latter is
crucial for the isometry preserving quantization. We also write down the map
between the two realizations. A relevant feature of this map is that it is
singular at the asymptotic boundary. A preliminary step towards quantization
is taken in the third subsection by introducing a natural Poisson structure on
the manifold.

###  2.2 Dirac operator in a noncovariant basis

We denote gamma matrices, the Dirac operator and spinors in the noncovariant
basis with a tilde. (We shall drop the tilde in the next subsection for the
corresponding quantities in the covariant basis.) In writing down the Dirac
operator on $EAdS_{2}$ we can take the two gamma matrices
$\\{\gamma^{\mu}\\}=\\{\tilde{\gamma}^{z},\tilde{\gamma}^{t}\\}$ to be

$\tilde{\gamma}^{z}=z\sigma_{1}\;,\qquad\quad\tilde{\gamma}^{t}=z\sigma_{2}\;,$
(2.12)

$\sigma_{i}$ being Pauli matrices. This choice satisfies (2.10). The vielbein
one-form is then

$e=\sigma_{i}e^{i}_{\;\;\mu}dx^{\mu}=\frac{1}{z}(\sigma_{1}dz+\sigma_{2}dt)\;.$
(2.13)

The torsion free condition (2.11) is satisfied for

$\omega=\frac{i}{2z}\sigma_{3}dt\;.$ (2.14)

The Dirac operator (2.9) then becomes

$\displaystyle{\tilde{D}}$ $\displaystyle=$ $\displaystyle
z(\sigma_{1}\partial_{z}+\sigma_{2}\partial_{t})-\frac{1}{2}\sigma_{1}\;=\;z^{3/2}(\sigma_{1}\partial_{z}+\sigma_{2}\partial_{t})\frac{1}{\sqrt{z}}$
(2.15) $\displaystyle=$ $\displaystyle
z^{3/2}\pmatrix{0&\partial_{z}-i\partial_{t}\cr\partial_{z}+i\partial_{t}&0}\frac{1}{\sqrt{z}}\;.$
(2.17)

The corresponding chirality operator is $\sigma_{3}$ as it anti-commutes with
the Dirac operator $[{\tilde{D}},\sigma_{3}]_{+}=0$, and squares to the
identity. ${\tilde{D}}$ satisfies the Lichnerowicz formula :

$\sqrt{z}\,{\tilde{D}}\frac{1}{z}\,{\tilde{D}}\,\sqrt{z}=\Delta\;.$ (2.18)

The free massless spinor ${\tilde{\Psi}}$ on $EAdS_{2}$ satisfies
${\tilde{D}\tilde{\Psi}}=0$, which is easily solved. General solutions are
again expressed in terms of analytic and anti-analytic functions of
$\zeta=z+it$,

${\tilde{\Psi}}(\zeta,\bar{\zeta})=\sqrt{z}\pmatrix{f(\zeta)\cr
g(\bar{\zeta})}\;,\qquad z=Re\,\zeta\;.$ (2.19)

Here the complex functions $f$ and $g$ associated with solutions of different
chirality. In the boundary limit $z\rightarrow 0$,

${\tilde{\Psi}}(\zeta,\bar{\zeta})\rightarrow\sqrt{z}\,\tilde{\psi}_{0}(t)\;,\qquad\quad\tilde{\psi}_{0}(t)=\pmatrix{f(it)\cr
g(-it)}\;.$ (2.20)

###  2.3 Dirac operator in the covariant basis

In formulating the covariant form of the Dirac operator, it is convenient to
embed $EAdS_{2}$ in ${\mathbb{R}}^{1,2}$. Calling the embedding coordinates
$X^{a}$, $a=0,1,2$, $EAdS_{2}$ corresponds to a single sheeted hyperboloid

$X^{a}X_{a}=-1\;,$ (2.21)

where the metric on ${\mathbb{R}}^{1,2}$ is $[\eta_{ab}]=$diag$(-1,+1,+1)$,
and, for convenience, the $AdS_{2}$ scale is normalized to one.¶¶¶Our index
convention here is different from that in [6, 7]. In the latter, $a=2$
corresponded to the time-like direction. Expressed in terms of Fefferman-
Graham coordinates the embedding is given by

$X^{0}=-\frac{z^{2}+t^{2}+1}{2z}\;,\qquad\;X^{1}=-\frac{t}{z}\;,\qquad\;X^{2}=-\frac{z^{2}+t^{2}-1}{2z}\;,$
(2.22)

putting the single sheeted hyperboloid in the region $X^{0}<0$.

In the covariant formulation of the Dirac operator we choose the chirality
operator to be

$\gamma=X^{a}\tau_{a}\;,$ (2.23)

where $\\{\tau_{a},\,a=0,1,2\\}$ is a basis for $su(1,1)$ in the defining
representation. Our choice is
$(\tau_{0},\tau_{1},\tau_{2})=(\sigma_{3},i\sigma_{1},i\sigma_{2})\,,$
$\sigma_{a}$ denoting Pauli matrices, which satisfies

$\tau_{a}\tau_{b}=-i\epsilon_{abc}\tau^{c}-\eta_{ab}{\mathbb{1}}\;,$ (2.24)

$\epsilon_{abc}$ being the Levi-Civita symbol with $\epsilon_{012}=-1$. We
note that $\gamma$ is not hermitean, and its square is the unit operator.

The chirality operator $\gamma$ can be mapped from the chirality operator in
the noncovariant basis, $\sigma_{3}$, using the similarity transformation

$\gamma=U\sigma_{3}U^{-1}\;,$ (2.25)

where the $2\times 2$ matrix $U$ expressed in complex coordinates $\zeta$ and
$\bar{\zeta}$ is given by

$U=\frac{1}{2\sqrt{z}}\pmatrix{-\zeta+1&-\bar{\zeta}-1\cr\zeta+1&\bar{\zeta}-1}\;.$
(2.26)

$U$ has determinant equal to one and satisfies

$U^{\dagger}\sigma_{3}U=-\sigma_{3}\;.$ (2.27)

While $U$ has unit determinant for all $z$, $U$ is singular in the boundary
limit, $z\rightarrow 0$. More specifically, det$\;\sqrt{z}U\rightarrow 0$ in
the limit.

If we apply the same similarity transformation to the Dirac operator written
in the noncovariant basis (2.17),

${\tilde{D}}\rightarrow D=U{\tilde{D}}U^{-1}\;,$ (2.28)

we get

$D\,=\,i\,\pmatrix{tz\partial_{z}+\frac{1}{2}(1+t^{2}-z^{2})\partial_{t}+i&z(t+i)\partial_{z}-\frac{1}{2}\Bigl{(}z^{2}-(t+i)^{2}\Bigr{)}\partial_{t}\cr-z(t-i)\partial_{z}+\frac{1}{2}\Bigl{(}z^{2}-(t-i)^{2}\Bigr{)}\partial_{t}&-tz\partial_{z}-\frac{1}{2}(1+t^{2}-z^{2})\partial_{t}+i}\;.$
(2.29)

In comparing with (2.9), it corresponds to making the following choice for the
gamma matrices

$\gamma^{z}\,=\,iz\pmatrix{t&t+i\cr-t+i&-t}\qquad\quad\gamma^{t}\,=\,\frac{i}{2}\pmatrix{1+t^{2}-z^{2}&-z^{2}+(t+i)^{2}\cr
z^{2}-(t-i)^{2}&-(1+t^{2}-z^{2})}\;.$ (2.30)

It can be checked that they satisfiy (2.10). The gamma matrices (2.30) are not
obtained with a local Lorentz transformation from
$\hat{\gamma}^{i}=\sigma_{i}$. The transformed Dirac operator $D$ can be re-
expressed in terms of the three Killing vectors (2.2),

$\displaystyle D$ $\displaystyle=$ $\displaystyle i\pmatrix{-\frac{1}{2}({\cal
K}^{-}+{\cal K}^{+})+i&\frac{1}{2}({\cal K}^{-}-{\cal K}^{+})-i{\cal
K}^{1}\cr-\frac{1}{2}({\cal K}^{-}-{\cal K}^{+})-i{\cal
K}^{1}&\frac{1}{2}({\cal K}^{-}+{\cal K}^{+})+i}$ (2.31) $\displaystyle=$
$\displaystyle-i\tau_{a}{\cal K}^{a}-{\mathbb{1}}\;.$ (2.33)

This Dirac operator is preserved under simultaneous $2+1$ Lorentz
transformations on $\tau_{a}$ and ${\cal K}^{a}$. By construction, $D$
anticommutes with the chirality operator $\gamma$. This can also be
independently verified using the properties (2.3), (2.24) and

$[{\cal K}^{a},X^{b}]=\epsilon^{abc}X_{c}\;,\qquad\quad X_{a}{\cal
K}^{a}={\cal K}^{a}X_{a}=0\;.$ (2.34)

The solution $\Psi$ to $D\Psi=0$ is obviously obtained by transforming the
solution (2.19) by $U$

$\Psi=U\tilde{\Psi}=\frac{1}{2}\pmatrix{-\zeta+1&-\bar{\zeta}-1\cr\zeta+1&\bar{\zeta}-1}\pmatrix{f(\zeta)\cr
g(\bar{\zeta})}\;.$ (2.35)

The boundary behavior of $\Psi$ differs from that of $\tilde{\Psi}$ in (2.20)
due to the fact that $U$ is singular at the boundary. Upon Taylor expanding
(2.35) at $z=0$ up to first order, we get

$\displaystyle\Psi(\zeta,\bar{\zeta})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}1-it\\\ 1+it\\\
\end{array}\right)(f(it)-g(-it))$ (2.38) $\displaystyle+$
$\displaystyle\frac{z}{2}\left(\begin{array}[]{c}-f(it)-g(-it)+(1-it)\left(f^{\prime}(it)-g^{\prime}(-it)\right)\\\
\;\;\,f(it)+g(-it)+(1+it)\left(f^{\prime}(it)-g^{\prime}(-it)\right)\\\
\end{array}\right)+O\left(z^{2}\right)\;.$ (2.42)

We can write the result as

$\Psi(\zeta,\bar{\zeta})=(\sigma_{3}-it{\mathbb{1}})\,\psi_{0}(t)-iz\,\partial_{t}\Bigl{(}(\sigma_{3}-it{\mathbb{1}})\,\psi_{1}(t)\Bigr{)}+{\cal
O}(z^{2})\;,$ (2.44)

where

$\psi_{0}(t)=\frac{1}{2}\Bigl{(}f(it)-g(-it)\Bigr{)}\pmatrix{1\cr-1}\;,\quad\quad\psi_{1}(t)=\frac{1}{2}\Bigl{(}f(it)+g(-it)\Bigr{)}\pmatrix{1\cr-1}\;.$
(2.45)

###  2.4 Poisson structure

A preliminary step towards quantization is to attach a Poisson structure to
the manifold. Poisson brackets $\\{\,,\,\\}$ can be uniquely defined on
$EAdS_{2}$ upon demanding that they preserve the isometry. This choice
corresponds to having the embedding coordinates $X^{a}$ define an $su(1,1)$
algebra

$\\{X^{a},X^{b}\\}=\epsilon^{abc}X_{c}\;.$ (2.46)

These relations are consistent with writing down the following Poisson bracket
for the Feffermann-Graham coordinates spanning ${\mathbb{R}}^{2}_{+}$[6, 7]

$\\{t,z\\}=z^{2}\;,$ (2.47)

which can be easily verified using (2.22). It follows that
$z^{-1}=X^{2}-X^{0}$ and $t$ are canonically conjugate coordinates on
${\mathbb{R}}^{2}_{+}$,

$\\{z^{-1},t\\}=1\;.$ (2.48)

Alternatively, one can define a Darboux map to coordinates $(x,y)$ spanning
the entire plane, where

$x=-\ln\,z\,,\qquad y=t/z\;,$ (2.49)

and consequently $\\{x,y\\}=1$.

Using (2.2), the action of the Killing vectors on functions ${\cal F}$ on
$EAdS_{2}$ is implemented from Poisson brackets with the embedding coordinates

${\cal K}^{a}{\cal F}=\\{X^{a},{\cal F}\\}\;.$ (2.50)

It then trivially follows that the Poisson bracket and the defining relation
(2.21) are preserved under the action of the Killing vectors

$\displaystyle{\cal
K}^{d}\\{X^{a},X^{b}\\}=\\{X^{d},\\{X^{a},X^{b}\\}\\}=\epsilon^{abc}\\{X^{d},X_{c}\\}=\epsilon^{abc}{\cal
K}^{d}X_{c}\;,$ (2.51) (2.52) $\displaystyle{\cal
K}^{a}(X^{b}X_{b})=\\{X^{a},X^{b}X_{b}\\}=2\\{X^{a},X^{b}\\}X_{b}=0\;.$ (2.53)

The field equations for the scalar and spinor can be expressed using Poisson
brackets with the embedding coordinates. For the massless scalar field $\Phi$,

$\Delta\Phi=\\{X^{a},\\{X_{a},\Phi\\}\\}=0\;,$ (2.54)

while for the massless spinor $\Psi$,

$D\Psi=-i\\{\gamma,\Psi\\}-\Psi=0\;.$ (2.55)

##  3 Scalars and spinors on quantized $EAdS_{2}$

We begin with the natural isometry preserving quantization of $EAdS_{2}$ in
the first subsection. There we show how the field equations for the scalar and
spinor can be written in an algebraic manner. The exact solutions for massless
fields are presented in the second subsection.

###  3.1 Quantization

The quantization of $EAdS_{2}$ is unique upon demanding that it preseves the
isometry of $EAdS_{2}$.[9, 12] For this one maps the embedding coordinates
$X^{a}$ to operators $\hat{X}^{a}$, satisfying the $su(1,1)$ algebra

$[\hat{X}^{a},\hat{X}^{b}]=i\alpha\epsilon^{abc}\hat{X}_{c}\;,$ (3.1)

where $\alpha$ is the quantization parameter. As with commutative $EAdS_{2}$,
we fix the scale to be one

$\hat{X}^{a}\hat{X}_{a}=-{\mathbb{1}}\;,$ (3.2)

in analogy with (2.21). [Any quantum corrections to the Casimir can be removed
with a rescaling of $\hat{X}_{a}$ and $\alpha$.]

$\hat{X}_{a}$ generate the algebra ${\cal A}$ of quantized $EAdS_{2}$. They
form a basis for $su(1,1)$ in a unitary irreducible representation. It was
shown in [6] that both the commutative limit and boundary limit can be
formulated in terms of these representations. In order for there to be a
consistent commutative limit to $EAdS_{2}$, one has to restrict the allowed
representations to the discrete series $D^{\pm}(k)$, where $k<0$ labels the
representation. Upon choosing $D^{+}(k)$, we can go to an eigenbasis of
$\hat{X}^{0}$, which we denote by $|k,m>$, $m=0,1,2...$. The action of
$\hat{X}^{a}$ on these states is

$\displaystyle\hat{X}_{+}|k,m>$ $\displaystyle=$ $\displaystyle-\alpha
c_{m}|k,m+1>\;,$ (3.3) $\displaystyle\hat{X}_{-}|k,m>$ $\displaystyle=$
$\displaystyle-\alpha c_{m-1}|k,m-1>\;,$ (3.5) $\displaystyle\hat{X}^{0}|k,m>$
$\displaystyle=$ $\displaystyle-\alpha(m-k)|k,m>\;,$ (3.7)

where $\hat{X}_{\pm}=\hat{X}^{2}\pm i\hat{X}^{1}$ and
$c_{m}=\sqrt{(m+1)(m-2k)}$. For the action of the quadratic Casimir one gets

$\displaystyle\hat{X}^{a}\hat{X}_{a}|k,m>$ $\displaystyle=$
$\displaystyle-\alpha^{2}k(k+1)|k,m>\;.$ (3.8)

Upon comparing with (3.2), we get that $k$ is related to noncommutativity
parameter,

$\alpha^{2}k(k+1)=1\;,$ (3.9)

and we should make the restriction $k<-1$. The commutative limit
$\alpha\rightarrow 0$ then corresponds to $k\rightarrow-\infty$. In section
3.2 we shall argue that $\hat{X}^{2}-\hat{X}^{0}$ is the quantum analogue of
$z^{-1}$. From (3.7), its expectation value goes to infinity when
$m\rightarrow\infty$. So it follows that this is boundary limit.

It is straightforward to generalize the notion of Killing vectors to quantized
$EAdS_{2}$. Denoting them by $\hat{\cal K}^{a}$, their action on functions
$\hat{\cal F}$ on ${\cal A}$ is given by

$\hat{\cal K}^{a}\hat{\cal F}=-\frac{i}{\alpha}[\hat{X}^{a},\hat{\cal F}]\;.$
(3.10)

Just as with the Killing vectors ${\cal K}^{a}$, $\hat{\cal K}^{a}$ satisfy
the $su(1,1)$ algebra

$[\hat{\cal K}^{a},\hat{\cal K}^{b}]=\epsilon^{abc}\hat{\cal K}_{c}\;,$ (3.11)

and the action of $\hat{\cal K}^{a}$ leaves the defining relations (3.1) and
(3.2) invariant.

In defining the Laplacian and Dirac operators on quantized $EAdS_{2}$ it is
convenient to introduce left and right acting operators $\hat{X}^{La}$ and
$\hat{X}^{Ra}$, respectively. Their action on functions $\hat{\cal F}$ on
quantized $EAdS_{2}$ is defined to be

$\hat{X}^{La}\hat{\cal F}=\hat{X}^{a}\hat{\cal
F}\;,\quad\quad\hat{X}^{Ra}\hat{\cal F}=\hat{\cal F}\hat{X}^{a}\;.$ (3.12)

We note that the right action is an anti-involuton, i.e., for two right acting
operators $\hat{A}^{R}$ and $\hat{B}^{R}$,
$(\hat{A}\hat{B})^{R}=\hat{B}^{R}\hat{A}^{R}.$ As a result there will be a
sign difference between commutators of right acting operators and the
corresponding commutators of left acting operators. In addition, right acting
operators commute with a left acting operators. In other words, $\hat{A}^{R}$
and $\hat{B}^{R}$ are elements of the opposite algebra, ${\cal A}^{op}$. Thus
$\hat{X}^{La}$ and $\hat{X}^{Ra}$ satisfy

$\displaystyle[\hat{X}^{La},\hat{X}^{Lb}]=i\alpha\epsilon^{abc}\hat{X}^{L}_{c}\;,\quad\quad[\hat{X}^{Ra},\hat{X}^{Rb}]=-i\alpha\epsilon^{abc}\hat{X}^{R}_{c}\;,\quad\quad[\hat{X}^{La},\hat{X}^{Rb}]=0\;,$
(3.13) (3.14)
$\displaystyle\hat{X}^{La}\hat{X}_{a}^{L}=\hat{X}^{Ra}\hat{X}_{a}^{R}=-{\mathbb{1}}\;.$
(3.15)

The quantum version of the Killing vectors can be expressed as

$\hat{\cal K}^{a}=-\frac{i}{\alpha}(\hat{X}^{La}-\hat{X}^{Ra})\;.$ (3.16)

The obvious choice for the Laplacian operator on quantized $EAdS_{2}$ is
$\hat{\Delta}=\hat{\cal K}^{a}\hat{\cal K}_{a}$. It can be rewritten as

$\hat{\Delta}=\,-\frac{1}{\alpha^{2}}(\hat{X}^{La}-\hat{X}^{Ra})(\hat{X}^{L}_{a}-\hat{X}^{R}_{a})\,=\,\frac{2}{\alpha^{2}}(\hat{X}^{Ra}\hat{X}^{L}_{a}+{\mathbb{1}})\;.$
(3.17)

So the equation of motion for a massless scalar field $\hat{\Phi}$ on
quantized $EAdS_{2}$ can be written in a purely algebraic way

$\frac{\alpha^{2}}{2}\hat{\Delta}\hat{\Phi}\;=\;\hat{X}_{a}\hat{\Phi}\hat{X}^{a}\,+\,\hat{\Phi}\;=\;0\;.$
(3.18)

This equation reduces to its commutative analogue (2.54) after requiring that
the commutator of two functions $\hat{\cal F}$ and $\hat{\cal G}$ on quantized
$EAdS_{2}$ goes to $i\alpha$ times the associated Poisson bracket in the limit
$\alpha\rightarrow 0$

$[\hat{\cal F},\hat{\cal G}]\rightarrow i\alpha\\{\hat{\cal F},\hat{\cal
G}\\}\;,\quad{\rm as}\;\;\alpha\rightarrow 0\;.$ (3.19)

Before writing down the Dirac operator on quantized $EAdS_{2}$, we examine the
chirality operator $\hat{\gamma}$. It is required to i) commute with the
algebra ${\cal A}$, ii) square to unity, and iii) it should reduce to $\gamma$
in the commutative limit $\alpha\rightarrow 0$. For i), we note that although
$\hat{X}^{La}$ does not commute with ${\cal A}$, $\hat{X}^{Ra}$ does. Using
associativity,

$(\hat{X}^{Ra}\hat{\cal F})\hat{\cal G}=\hat{X}^{Ra}(\hat{\cal F}\hat{\cal
G})=\hat{\cal F}\hat{\cal G}\hat{X}^{a}\;,$

for any two function $\hat{\cal F},\hat{\cal G}\in{\cal A}$, while we get the
same result from

$(\hat{\cal F}\hat{X}^{Ra})\hat{\cal G}=\hat{\cal F}(\hat{X}^{Ra}\hat{\cal
G})=\hat{\cal F}\hat{\cal G}\hat{X}^{a}\;.$

Therefore $\hat{\gamma}$ should be constructed from $\hat{X}^{Ra}$, and since
we want to recover (2.23) in the commutative limit, a natural candidate would
be $\hat{X}^{Ra}\tau_{a}$. However, this doesn’t square to unity, and quantum
corrections must be introduced to satisfy ii). A chirality operator satisfying
i-iii) was obtained by Fakhri and Imaanpur[10]:

$\displaystyle\hat{\gamma}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\,(\hat{X}^{Ra}\tau_{a}-\frac{\alpha}{2}{\mathbb{1}})\;.$
(3.20)

Upon taking its square

$\hat{\gamma}^{2}=\frac{1}{{1+\frac{\alpha^{2}}{4}}}\,\Bigl{(}\hat{X}^{Ra}\hat{X}^{Rb}\,\tau_{a}\tau_{b}-\alpha\hat{X}^{Ra}\tau_{a}+\frac{\alpha^{2}}{4}\Bigr{)}\;=\;{\mathbb{1}}\;.$
(3.21)

In writing down the Dirac operator $\hat{D}$ on quantized $AdS_{2}$, we should
require $a)$ that it anti-commutes with the chirality operator $\hat{\gamma}$
and $b)$ that it reduces to the Dirac operator $D$ on $EAdS_{2}$ (2.33) in the
commutative limit. A natural candidate for $\hat{D}$ that is consistent with
the commutative limit is $-i\tau_{a}\hat{\cal K}^{a}-{\mathbb{1}}$, however
this does not satisfy $a)$. On the other hand, $a)$ is satisfied for
$\hat{X}^{Ra}(\epsilon_{abc}\hat{\cal K}^{b}\tau^{c}+\tau_{a})$

$[\,\hat{X}^{Ra}(\epsilon_{abc}\hat{\cal
K}^{b}\tau^{c}+\tau_{a})\,,\,\hat{X}^{Rd}\tau_{d}-\frac{\alpha}{2}{\mathbb{1}}\,]_{+}=0\,,$
(3.22)

In order to satisfy $b)$, one can multiply by $-\hat{\gamma}$. The Dirac
operator is then[10],

$\displaystyle\hat{D}$ $\displaystyle=$
$\displaystyle-\hat{\gamma}\hat{X}^{Ra}(\epsilon_{abc}\hat{\cal
K}^{b}\tau^{c}+\tau_{a})=\frac{i}{\alpha}\,\epsilon_{abc}\hat{\gamma}\hat{X}^{Ra}\hat{X}^{Lb}\tau^{c}\;.$
(3.23)

Upon expanding out the chirality operator one gets

$\hat{D}=\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\,\Bigl{(}-i\tau_{a}\hat{\cal
K}^{a}-{\mathbb{1}}+i\alpha\hat{X}^{R}_{a}\hat{\cal
K}^{a}-\frac{\alpha}{2}(\epsilon_{abc}\hat{X}^{Ra}\hat{\cal
K}^{b}+\hat{X}^{R}_{c})\tau^{c}-i\hat{X}^{R}_{a}\tau^{a}\,\hat{X}^{R}_{b}\hat{\cal
K}^{b}\,\Bigr{)}\;.$ (3.24)

We note that $\hat{X}^{R}_{a}\hat{\cal K}^{a}$ acting on ${\cal A}$ is of
order $\alpha$, and so to leading order in $\alpha$,
$\hat{D}\sim-i\tau_{a}\hat{\cal K}^{a}-{\mathbb{1}}$, which is consistent with
the requirement $b)$.

(3.23) is the quantization of the Dirac operator $D$ (2.33) written in the
covariant basis. In Appendix B we obtain the quantization of the Dirac
operator $D$ (2.17) written in the noncovariant basis. This is done by finding
the quantum analogue of the $2\times 2$ matrix $U$ in (2.26), and applying it
to write down the inverse of the map (2.28).

Using (3.23), the Dirac equation for a massless particle on quantized
$EAdS_{2}$ is

$\hat{D}\hat{\Psi}=\frac{i}{\alpha}\,\epsilon_{abc}\hat{\gamma}\hat{X}^{Ra}\hat{X}^{Lb}\tau^{c}\hat{\Psi}=0\;,$
(3.25)

or simply

$\epsilon_{abc}\tau^{a}\hat{X}^{b}\hat{\Psi}\hat{X}^{c}=0\;.$ (3.26)

This equation reduces to its commutative analogue (2.55), after again
requiring (3.19). Like the equation for the massless scalar field (3.18), the
field equation for a massless spinor on quantized $EAdS_{2}$ has an algebraic
form. Both of these equations have exact solutions which we show in the next
subsection.

###  3.2 Exact solutions

Here we find the exact solutions to the quantized massless Klein-Gordon (3.18)
and Dirac (3.26) equations. We will start with the scalar case and then we
will show how to map the scalar solutions to the spinor ones. Because the
exposition is rather technical, we will structure it as the sequence of
several propositions.

As the first step we want to establish the quantum analogue of the Fefferman-
Graham coordinates and their relation to the quantum embedding coordinates.
Classically, this relation is given in (2.22). Inverting this we have for the
commutative Fefferman-Graham coordinates

$z=\frac{1}{X^{2}-X^{0}}\ ,\qquad t=-zX^{1}\equiv-\frac{X^{1}}{X^{2}-X^{0}}\
.$ (3.27)

As a quantization of these relations we will take the symmetric ones

$\displaystyle\hat{z}$ $\displaystyle=$
$\displaystyle(\hat{X}^{2}-\hat{X}^{0})^{-1}\;,$ (3.28) $\displaystyle\
\hat{t}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}(\hat{z}\hat{X}^{1}+\hat{X}^{1}\hat{z})\equiv-\frac{1}{2}\Bigl{(}(\hat{X}^{2}-\hat{X}^{0})^{-1}\hat{X}^{1}+\hat{X}^{1}(\hat{X}^{2}-\hat{X}^{0})^{-1}\Bigr{)}\quad$
(3.30)

and show below that this is a consistent choice. Because
$[\hat{z},\hat{X}^{a}]=-\hat{z}[\hat{z}^{-1},\hat{X}^{a}]\hat{z}$, it is easy
to find all the commutators $[\hat{z},\hat{X}^{a}]$ using
$\hat{z}^{-1}=\hat{X}^{2}-\hat{X}^{0}$ and (3.1)

$\displaystyle[\hat{z},\hat{X}^{1}]=i\alpha\hat{z}\ ,$
$\displaystyle[\hat{z},\hat{X}^{0}]=-i\alpha\hat{z}\hat{X}^{1}\hat{z}\equiv
i\alpha\hat{z}\hat{t}-\frac{\alpha^{2}}{2}\hat{z}^{2}\equiv
i\alpha\hat{t}\hat{z}+\frac{\alpha^{2}}{2}\hat{z}^{2}\equiv\frac{i\alpha}{2}(\hat{z}\hat{t}+\hat{t}\hat{z})\
,$ $\displaystyle[\hat{z},\hat{X}^{2}]=[\hat{z},\hat{X}^{0}]\ .$ (3.31)

From these commutators and the definition of $\hat{t}$ in (3.28) we
immediately get

$[\hat{t},\hat{z}]=i\alpha\hat{z}^{2}\ ,\qquad[\hat{t},\hat{z}^{-1}]=-i\alpha\
,$ (3.32)

i.e. the correct quantization of (2.47) and (2.48) without any quantum
corrections.

To find the commutators for $\hat{t}$ and $\hat{X}^{a}$, we first want to find
the quantum analogue of (2.22). It turns out that the quantization is non-
trivial, compared to (3.28). While for $\hat{X}^{1}$ one trivially has from
(3.28)

$\hat{X}^{1}=-\frac{1}{2}(\hat{z}^{-1}\hat{t}+\hat{t}\hat{z}^{-1})\equiv-\hat{t}\hat{z}^{-1}-\frac{i\alpha}{2}\equiv-\hat{z}^{-1}\hat{t}+\frac{i\alpha}{2}\
,$ (3.33)

the similar relations for $\hat{X}^{0,2}$ are less trivial. Using (3.2) we
have

$\hat{z}^{-1}(\hat{X}^{2}+\hat{X}^{0})=(\hat{X}^{2}-\hat{X}^{0})(\hat{X}^{2}+\hat{X}^{0})=-1-(\hat{X}^{1})^{2}+i\alpha\hat{X}^{1}\
.$ (3.34)

Then combining this with (3.33) we get

$\hat{X}^{2}+\hat{X}^{0}=-\kappa(\alpha)^{2}\hat{z}-\hat{t}\hat{z}^{-1}\hat{t}\
,$ (3.35)

where we define

$\kappa(\alpha):=\sqrt{1+\frac{\alpha^{2}}{4}}$ (3.36)

Notice that the same factor appears in the denominator of the quantum
chirality operator (3.20) and Dirac operator (3.23). Furthermore, from (3.9),
this factor can be related to the label $k$ of the discrete series
representation $D^{+}(k)$ in a simple way:
$\frac{\kappa(\alpha)}{\alpha}=-k-\frac{1}{2}\;.$ (Recall, $k<-1$.)

Combining (3.35) with the definition of $\hat{z}$ we finally get the full set
of relations

$\displaystyle\hat{X}^{1}=-\frac{1}{2}(\hat{z}^{-1}\hat{t}+\hat{t}\hat{z}^{-1})\equiv-\hat{t}\hat{z}^{-1}-\frac{i\alpha}{2}\equiv-\hat{z}^{-1}\hat{t}+\frac{i\alpha}{2}\
,$
$\displaystyle\hat{X}^{0}=-\frac{1}{2}(\kappa(\alpha)^{2}\hat{z}+\hat{t}\hat{z}^{-1}\hat{t}+\hat{z}^{-1})\
,$
$\displaystyle\hat{X}^{2}=-\frac{1}{2}(\kappa(\alpha)^{2}\hat{z}+\hat{t}\hat{z}^{-1}\hat{t}-\hat{z}^{-1})\
.$ (3.37)

Note the appearance of a non-trivial deformation factor, $\kappa(\alpha)^{2}$,
which would be impossible to guess assuming some minimal quantization, as it
was done in (3.28). It is easy to verify explicitly that this factor is
essential to guarantee the correct algebraic relations (3.1) starting with the
fundamental commutation relations (3.32).

Using the result (3.2), it is an easy exercise to find the commutators
$[\hat{t},\hat{X}^{a}]$

$\displaystyle[\hat{t},\hat{X}^{1}]=i\alpha\hat{t}\ ,$
$\displaystyle[\hat{t},\hat{X}^{0}]=\frac{i\alpha}{2}(\hat{t}^{2}-\kappa(\alpha)^{2}\hat{z}^{2}+1)\
,$
$\displaystyle[\hat{t},\hat{X}^{2}]=\frac{i\alpha}{2}(\hat{t}^{2}-\kappa(\alpha)^{2}\hat{z}^{2}-1)\
.$ (3.38)

Now we focus on solutions to the quantum scalar field equation of motion
(3.18), and formulate our first proposition:

######  Proposition 1.

$\hat{z}$ and $\hat{t}$, as defined in (3.28), are exact solutions of the
quantum scalar field equation of motion (3.18).

This is the analogue of the obvious fact that $z$ and $t$ solve the classical
field equation (2.6). The proof of the proposition is as follows:

###### Proof.

Using (3.2) it easy to show that (3.18) can be equivalently rewritten as

$\hat{X}_{a}\hat{\Phi}\hat{X}^{a}+\hat{\Phi}=0\ \Leftrightarrow\
\hat{X}_{a}[\hat{\Phi},\hat{X}^{a}]=[\hat{X}_{a},\hat{\Phi}]\hat{X}^{a}=0\ .$
(3.39)

Then $\hat{X}_{a}[\hat{z},\hat{X}^{a}]=0$ is an immediate consequence of
(3.2), which shows that $\hat{z}$ is an exact solution.

For $\hat{t}$ we can proceed the same way, using (3.2), but let us prove it
slightly differently by establishing some very useful criteria for a solution.
Namely, if we know that some $\hat{A}$ is a solution, then $\hat{A}\hat{B}$
will be a solution for some $\hat{B}$ if and only if
$\hat{X}_{a}\hat{A}[\hat{B},\hat{X}^{a}]=0$. This is trivially verified. Then
one has that $\hat{t}=-\hat{z}\hat{X}^{1}+\frac{i\alpha}{2}\hat{z}$ will be a
solution if $\hat{X}_{a}\hat{z}[\hat{X}^{1},\hat{X}^{a}]=0$, which is
trivially true after some simple algebra. ∎

The classical solution (2.19) is given in terms of arbitrary functions of
either ${\zeta}$ or $\bar{{\zeta}}$. Motivated by this and by the possibility
that the quantum $\hat{z}$ “coordinate” could be re-scaled, as in (3.2), we
will look for the general solutions of (3.18) in the form of an arbitrary
polynomial in the quantum variable $\hat{\Xi}=C\hat{z}+i\hat{t}$, where $C$ is
a constant to be fixed later. First of all, in view of Proposition 1,
$\hat{\Xi}$ is an exact solution for arbitrary $C$. The next proposition fixes
this constant.

######  Proposition 2.

$\hat{\Xi}^{2}$ is an exact solution of the quantum scalar equation of motion
(3.18) iff $C=\pm\sqrt{1+\frac{\alpha^{2}}{4}}\equiv\pm\kappa(\alpha)$.

###### Proof.

First of all, from the observation made in the proof of Proposition 1 and from
the fact that $\hat{\Xi}$ is a solution, we know that $\hat{\Xi}^{2}$ will be
a solution if and only if
$\hat{X}_{a}\hat{\Xi}[\hat{\Xi},\hat{X}^{a}]=[\hat{X}^{a},\hat{\Xi}]\hat{\Xi}\hat{X}_{a}=0$.
Using $[\hat{X}^{a},\hat{\Xi}]\hat{X}_{a}=0$ (because $\hat{\Xi}$ is a
solution), this is equivalent to

$[\hat{X}_{a},\hat{\Xi}][\hat{X}^{a},\hat{\Xi}]=0\ .$ (3.40)

Each commutator in (3.40) we can easily calculate from (3.2) and (3.2) giving

$\displaystyle[\hat{X}^{1},\hat{\Xi}]=-i\alpha\hat{\Xi}\ ,$
$\displaystyle[\hat{X}^{0},\hat{\Xi}]=-\frac{i\alpha}{2}C(\hat{z}\hat{t}+\hat{t}\hat{z})+\frac{\alpha}{2}(\hat{t}^{2}-\kappa(\alpha)^{2}\hat{z}^{2}+1)\
,$
$\displaystyle[\hat{X}^{2},\hat{\Xi}]=-\frac{i\alpha}{2}C(\hat{z}\hat{t}+\hat{t}\hat{z})+\frac{\alpha}{2}(\hat{t}^{2}-\kappa(\alpha)^{2}\hat{z}^{2}-1)\
.$ (3.41)

Plugging this in (3.40) we get

$0=[\hat{X}_{a},\hat{\Xi}][\hat{X}^{a},\hat{\Xi}]=\alpha^{2}(\kappa(\alpha)^{2}-C)\hat{z}^{2}$
(3.42)

from where the result follows. ∎

So, we see that $\hat{\Xi}_{\pm}:=\pm\kappa(\alpha)\hat{z}+i\hat{t}$ are the
quantum analogues of the classical ${\zeta}$ and $-\bar{{\zeta}}$. Using the
found value for $C$, the relations (3.2) could be written in a more compact
and closed form

$\displaystyle[\hat{X}^{1},\hat{\Xi}]=-i\alpha\hat{\Xi}\ ,$
$\displaystyle[\hat{X}^{0},\hat{\Xi}]=-\frac{\alpha}{2}(\hat{\Xi}^{2}-1)\ ,$
$\displaystyle[\hat{X}^{2},\hat{\Xi}]=-\frac{\alpha}{2}(\hat{\Xi}^{2}+1)\ .$
(3.43)

Now this will allow us to prove the claimed result.

######  Proposition 3.

An arbitrary polynomial in $\hat{\Xi}$ (either of $\hat{\Xi}_{\pm}$) is an
exact solution of the quantum scalar equation of motion (3.18).

###### Proof.

The proof is by induction. We already know that $\hat{\Xi}$ and
$\hat{\Xi}^{2}$ are solutions (we need to start with $\hat{\Xi}^{2}$, because
only on this level do we get a non-trivial condition on $C$). Assume that
$\hat{\Xi}^{n-1}$ is also a solution. Then we know that
$\hat{\Xi}^{n}\equiv\hat{\Xi}^{n-1}\hat{\Xi}$ will be a solution iff
$\hat{X}_{a}\hat{\Xi}^{n-1}[\hat{\Xi},\hat{X}^{a}]=0$. But from (3.2) we know
that $\hat{\Xi}^{n-1}$ commutes with all $[\hat{\Xi},\hat{X}^{a}]$, so we can
write

$\hat{X}_{a}\hat{\Xi}^{n-1}[\hat{\Xi},\hat{X}^{a}]=\hat{X}_{a}[\hat{\Xi},\hat{X}^{a}]\hat{\Xi}^{n-1}=0\
,$ (3.44)

where we used $\hat{X}_{a}[\hat{\Xi},\hat{X}^{a}]=0$ because $\hat{\Xi}$ is a
solution. ∎

So, we see that the general expression (2.19) for the classical solution has a
non-trivial exact quantum counterpart. The general solution of the quantum
scalar equation of motion (3.18) can be written in terms of $\hat{\Xi}_{+}$
and $\hat{\Xi}_{-}$, or equivalently, $\hat{\Xi}=\hat{\Xi}_{+}$ and
$\hat{\Xi}^{\dagger}$. It is given by

$\hat{\Phi}=\hat{F}(\hat{\Xi})+\hat{G}(\hat{\Xi}^{\dagger})\ ,$ (3.45)

where $\hat{F}$ and $\hat{G}$ are arbitrary elements of the polynomial
algebras (or of their closures) generated by $\hat{\Xi}$ and
$\hat{\Xi}^{\dagger}$, respectively.

Now we want to ask whether the analogous exact solutions can be established
for the quantum Dirac equation (3.26). The main motivation is the classical
result (2.35), which could be interpreted as arbitrary spinorial solution
written in terms of arbitrary scalar solutions. The natural quantization of
(2.35) is

$\hat{\Psi}=\frac{1}{2}\pmatrix{1-\hat{\Xi}&-1-\hat{\Xi}^{\dagger}\cr
1+\hat{\Xi}&-1+\hat{\Xi}^{\dagger}}\pmatrix{\hat{F}(\hat{\Xi})\cr\hat{G}(\hat{\Xi}^{\dagger})}\
.$ (3.46)

We want to see if it is a solution to (3.26). Due to the linearity of (3.26),
it is enough to consider the case $\hat{F}(\hat{\Xi})\neq 0$ and
$\hat{G}(\hat{\Xi}^{\dagger})=0$. Then the general case will trivially follow.
Also by the linearity, it suffices to work only with the scalar solutions
given by monomials $\hat{F}_{n}:=\hat{\Xi}^{n}$. So, we want to show that

$\hat{\Psi}_{n}:=\frac{1}{2}\pmatrix{(1-\hat{\Xi})\hat{F}_{n}\cr(1+\hat{\Xi})\hat{F}_{n}}\equiv\frac{1}{2}\pmatrix{\hat{F}_{n}-\hat{F}_{n+1}\cr\hat{F}_{n}+\hat{F}_{n+1}}$
(3.47)

is an exact solution of (3.26). The following proposition will prove to be
crucial in the demonstration

######  Proposition 4.

Let $\hat{\Psi}^{\prime}_{n}:=\pmatrix{\hat{F}_{n}\cr\hat{F}_{n}}$, then
(3.47) is given by

$2\alpha n\hat{\Psi}_{n}=\tau_{a}[\hat{X}^{a},\hat{{\Psi}}^{\prime}_{n}]\ ,$
(3.48)

where $\tau_{a}$ are defined after (2.23).

###### Proof.

From the definition of $\tau_{a}$ we have

$\tau_{a}[\hat{X}^{a},\hat{{\Psi}}^{\prime}_{n}]=\pmatrix{[\hat{X}^{0}+i\hat{X}^{1}+\hat{X}^{2},\hat{F}_{n}]\cr[-\hat{X}^{0}+i\hat{X}^{1}-\hat{X}^{2},\hat{F}_{n}]}\
.$ (3.49)

Now, the relations (3.2) can be easily generalized (e.g. by recursion) to

$\displaystyle[\hat{X}^{1},\hat{\Xi}^{n}]=-in\alpha\hat{\Xi}^{n}\ ,$
$\displaystyle[\hat{X}^{0},\hat{\Xi}^{n}]=-n\frac{\alpha}{2}(\hat{\Xi}^{n+1}-\hat{\Xi}^{n-1})\
,$
$\displaystyle[\hat{X}^{2},\hat{\Xi}^{n}]=-n\frac{\alpha}{2}(\hat{\Xi}^{n+1}+\hat{\Xi}^{n-1})\
.$ (3.50)

From here we have

$[i\hat{X}^{1}\pm(\hat{X}^{0}+\hat{X}^{2}),\hat{\Xi}^{n}]=n\alpha(\hat{\Xi}^{n}\mp\hat{\Xi}^{n+1})\
,$ (3.51)

which immediately proves the claim. ∎

Now we finally can prove the main result for the spinorial case.

######  Proposition 5.

The spinor (3.47) (and, as a trivial consequence, the general one given in
(3.46)) is an exact solution of the quantum Dirac equation (3.26).

###### Proof.

Using the result of Proposition 4, we need to prove that

$\epsilon_{abc}\tau^{c}\hat{X}^{a}\tau_{k}[\hat{X}^{k},\hat{{\Psi}}^{\prime}_{n}]\hat{X}^{b}=0\
.$ (3.52)

Using (2.24) and

$\epsilon_{abc}\epsilon^{c}_{\ kl}=-(\eta_{ak}\eta_{bl}-\eta_{al}\eta_{bk})$
(3.53)

we have

$\epsilon_{abc}\tau^{c}\hat{X}^{a}\tau_{k}[\hat{X}^{k},\hat{{\Psi}}^{\prime}_{n}]\hat{X}^{b}=i\tau_{b}\hat{X}_{a}[\hat{X}^{a},\hat{{\Psi}}^{\prime}_{n}]\hat{X}^{b}-i\tau_{a}\hat{X}^{a}[\hat{X}^{b},\hat{{\Psi}}^{\prime}_{n}]\hat{X}_{b}-\epsilon_{abk}\hat{X}^{a}[\hat{X}^{k},\hat{{\Psi}}^{\prime}_{n}]\hat{X}^{b}\
.$ (3.54)

Because each component of $\hat{{\Psi}}^{\prime}_{n}$ is an exact solution for
the scalar case, cf. (3.39), the first two terms are equal to zero. The
remaining term we can manipulate as follows

$\displaystyle\epsilon_{abk}\hat{X}^{a}[\hat{X}^{k},\hat{{\Psi}}^{\prime}_{n}]\hat{X}^{b}$
$\displaystyle=$
$\displaystyle\epsilon_{abk}\hat{X}^{a}\hat{X}^{k}\hat{{\Psi}}^{\prime}_{n}\hat{X}^{b}-\epsilon_{abk}\hat{X}^{a}\hat{{\Psi}}^{\prime}_{n}\hat{X}^{k}\hat{X}^{b}$
(3.55) $\displaystyle=$ $\displaystyle
i\alpha{X}_{b}\hat{{\Psi}}^{\prime}_{n}\hat{X}^{b}-i\alpha{X}^{a}\hat{{\Psi}}^{\prime}_{n}\hat{X}_{a}=0\
,$

where we used $\epsilon_{abc}\hat{X}^{b}\hat{X}^{c}=-i\alpha\hat{X}_{a}$. This
proves (3.52). ∎

Consequently (3.46) solves (3.26).∥∥∥An alternative proof is presented in
Appendix B. There we construct the quantum version $\hat{U}$ of the matrix $U$
in (2.26), and use it to map the Dirac operator to the analogue of the
noncovariant basis. The solutions for massless spinors in that basis are
trivially obtained from the classical solutions (2.19) by replacing $\zeta$ by
$\Xi$, with a particular choice of ordering for the $\sqrt{\hat{z}}$ factor.
Upon using $\hat{U}$ to map the solution back to the covariant basis, we
recover (3.46). Therefore, the main conclusion of this section is that the
exact solutions to the quantized massless Klein-Gordon (3.18) and Dirac (3.26)
equations are given by essentially the same expressions as in the classical
case with the non-trivial quantization of the variable ${\zeta}$ given by
$\hat{\Xi}$.

##  4 Exact boundary correlation functions

Boundary correlation functions are generated from the on-shell bulk action.
Before obtaining the boundary two-point correlation functions for operators
sourced by massless scalars and spinors on $EAdS_{2}$ on quantized $EAdS_{2}$,
we briefly recall how the correspondence works on the classical $EAdS_{2}$
bulk manifold in the first subsection. We give the quantized version for
scalars and spinors in the following two subsections. As a result of the exact
solutions of the previous section, our results are exact, and in fact,
remarkably simple.

###  4.1 Two-point correlators on the boundary of $EAdS_{2}$

We begin by reviewing the calculation of the two-point correlation function
for operators sourced by the massless scalar field on $EAdS_{2}$, and then for
operators sourced by massless spinors. For the latter, we use the
unconventional choice of the covariant description of the Dirac operator.

####  4.1.1 Massless scalar field

For the case of the massless scalar field, one starts with the action

$\displaystyle S[\Phi]$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{{\mathbb{R}}^{2}_{+}}dtdz\,\,\Bigl{\\{}(\partial_{z}\Phi)^{2}\,+\,(\partial_{t}\Phi)^{2}\Bigr{\\}}\;.$
(4.1)

Variations $\delta\Phi$ of $\Phi$ in (4.1) give

$\displaystyle\delta S[\Phi]$ $\displaystyle=$
$\displaystyle-\int_{{\mathbb{R}}^{2}_{+}}dtdz\,\delta\Phi\,(\partial_{z}^{2}+\partial_{t}^{2})\Phi-\int_{{\mathbb{R}}}dt\,(\partial_{z}\Phi\,\delta\Phi)\Big{|}_{z=0}\;.$
(4.2)

Extremizing the action with Dirichlet boundary conditions yields the field
equation (2.6). Since the equation is second order we should impose two
boundary conditions to fix the solution (2.7). Solutions which are everywhere
(and in particular at $z\rightarrow\infty$) regular can be expressed in terms
of the boundary value of the field $\phi_{0}(t)$, using the boundary-to-bulk
propagator.[3, 4] Actually, in the free case, the full boundary-to-bulk
propagator is not needed for the calculation of the boundary two-point
function. The solution (2.7) can be expanded in powers of $z$, as was done in
(2.8), and we only need to write the ${\cal O}(z)$ term, $\phi_{1}(t)$, in
terms of $\phi_{0}(t)$. For this we can use******This comes from the Cauchy
formula. For this define a (clockwise) closed contour $C$ in the complex plane
(parametrized by $\zeta^{\prime}=z^{\prime}+it^{\prime}$) which goes from
$t^{\prime}=-R$ to $t^{\prime}=R$ along the $t^{\prime}$ axis and then returns
to $t^{\prime}=-R$ along the semi-circle $\zeta^{\prime}=Re^{i\phi}$, where
$\phi$ runs from $\frac{\pi}{2}$ to $-\frac{\pi}{2}$. Then
$f(\zeta)=-\frac{1}{2\pi
i}\oint_{C}\frac{d\zeta^{\prime}f(\zeta^{\prime})}{\zeta^{\prime}-\zeta}\;,\quad\quad
g(\bar{\zeta})=\frac{1}{2\pi
i}\oint_{C}\frac{d\bar{\zeta}^{\prime}g(\bar{\zeta}^{\prime})}{\bar{\zeta}^{\prime}-\bar{\zeta}}\;.$
Upon assuming that $f(Re^{i\phi})$ and $g(Re^{-i\phi})$ vanish in the limit
$R\rightarrow\infty$, we get that
$f(\zeta)=\frac{1}{2\pi}\int_{-\infty}^{\infty}dt^{\prime}\frac{f(it^{\prime})}{\zeta-
it^{\prime}}\;,\quad\quad
g(\bar{\zeta})=\frac{1}{2\pi}\int_{-\infty}^{\infty}dt^{\prime}\frac{g(-it^{\prime})}{\bar{\zeta}+it^{\prime}}\;.$
(4.3) follows from setting $z=0$.

$\displaystyle f(it)=\frac{1}{2\pi i}\int
dt^{\prime}\,\frac{f(it^{\prime})}{t-t^{\prime}}\,,$ $\displaystyle
g(-it)=-\frac{1}{2\pi i}\int
dt^{\prime}\,\frac{g(-it^{\prime})}{t-t^{\prime}}\,,$ (4.3)

and consequently

$\displaystyle f^{\prime}(it)=\frac{1}{2\pi}\int
dt^{\prime}\,\frac{f(it^{\prime})}{(t-t^{\prime})^{2}}\;,$ $\displaystyle
g^{\prime}(-it)=\frac{1}{2\pi}\int
dt^{\prime}\,\frac{g(-it^{\prime})}{(t-t^{\prime})^{2}}\;.$ (4.4)

From (2.8) we then get the limiting value of $\partial_{z}\Phi$

$\partial_{z}\Phi\rightarrow\phi_{1}(t)=\frac{1}{2\pi}\int
dt^{\prime}\,\frac{\phi_{0}(t^{\prime})}{(t-t^{\prime})^{2}}\;,\qquad{\rm
as}\;\;z\rightarrow 0\;.$ (4.5)

We next substitute this back in the action (4.1), which can be re-written as

$\displaystyle S[\Phi]$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\int_{{\mathbb{R}}^{2}_{+}}dtdz\,\Phi(\partial_{z}^{2}+\partial_{t}^{2})\Phi\;-\frac{1}{2}\int_{{\mathbb{R}}}dt\,(\Phi\partial_{z}\Phi)\Big{|}_{z=0}\;.$
(4.6)

Only the boundary term survives in the on-shell action. Using (4.5), the
result is

$S[\Phi]|_{\rm on-
shell}=-\frac{1}{2\pi}\int_{{\mathbb{R}}}dt\int_{{\mathbb{R}}}dt^{\prime}\,\frac{\phi_{0}(t)\phi_{0}(t^{\prime})}{(t-t^{\prime})^{2}}\;\;.$
(4.7)

In the $AdS/CFT$ correspondence one identifies $S[\Phi|_{\rm on-shell}$ with
the generating functional of the $n-$point connected correlation functions for
the operator ${\cal O}$ associated with $\phi_{0}$. Here, both ${\cal O}$ and
$\phi_{0}$ are functions of only $t$,

$<{\cal O}(t_{1})\cdots{\cal O}(t_{n})>=\frac{\delta^{n}S[\Phi]|_{\rm on-
shell}}{\delta\phi_{0}(t_{1})\cdots\delta\phi_{0}(t_{n})}\bigg{|}_{\phi_{0}=0}\
.$ (4.8)

So the two-point function in this example is

$<{\cal O}(t){\cal
O}(t^{\prime})>\;=\;-\frac{1}{\pi}\,\frac{1}{(t-t^{\prime})^{2}}\ .$ (4.9)

####  4.1.2 Massless spinor field

Since we presented two different formulations for spinors on the classical
$EAdS_{2}$ bulk manifold, we can proceed with two different derivations of the
corresponding two-point correlator on the boundary. These two approaches
cannot be easily mapped from one to another because the map (2.28) between the
Dirac operators is singular at the $z\rightarrow 0$ boundary. Nevertheless, it
can be checked that both approaches yield the same result for the boundary
correlation function. Below we work with the Dirac operator $D$ (2.33) of the
covariant description.

The bulk action for massless spinors in the covariant description is given by

${S}_{B}[\overline{\Psi},\Psi]\;=\;-\int_{{\mathbb{R}}^{2}_{+}}\frac{dzdt}{z^{2}}\,\overline{\Psi}D\Psi\;=\;\int_{{\mathbb{R}}^{2}_{+}}\frac{dzdt}{z^{2}}\,\overline{\Psi}(i\tau_{a}{\cal
K}^{a}+{\mathbb{1}})\Psi\;,$ (4.10)

where the conjugate spinor is defined by
$\overline{\Psi}={\Psi}^{\dagger}\sigma_{3}$. With this definition one gets
that $\overline{\Psi}\Psi$ is a scalar under $SU(1,1)$ transformations,
$\Psi\rightarrow M\Psi$, where the $2\times 2$ transformation matrix satisfies
$M^{\dagger}\sigma_{3}M=\sigma_{3}$.††††††Another standard choice is
$M^{\dagger}\sigma_{2}M=\sigma_{2}$. In this case we should define
$\overline{\Psi}={\Psi}^{\dagger}\sigma_{2}$. This choice would lead to the
appearance of $\sigma_{2}$ instead of $\sigma_{3}$ in the formula for the two-
point correlation function below, (4.23).  The action obviously leads to the
field equation $D\Psi=0$, and its hermitean conjugate
$\overline{\Psi}\overleftarrow{D}=0$.

As is well known,[19, 20] the action for spinor fields should be supplemented
with a boundary term. From [20], because the field equation is first order in
derivatives, we cannot fix all components of the spinor at spatial infinity
(corresponding to $z\rightarrow+0$), as well as demand that the fields are
everywhere regular in the bulk. Therefore some spinor degrees of freedom at
spatial infinity remain dynamical. The variations of the action with regards
to these boundary degrees of freedom, as well as the bulk degrees of freedom,
should be consistently made to vanish. This is not possible if the action
consists only of (4.10). It is for this reason that boundary term should be
included in the total action. Note, that this situation differs for the case
of a scalar field, where $\delta\Phi$ could be consistently subjected to the
Dirichlet boundary conditions in (4.2).

To obtain the boundary term consider an arbitrary variation in $\Psi$ in the
action (4.10)

$\displaystyle\delta{S}_{B}[\overline{\Psi},\Psi]$ $\displaystyle=$
$\displaystyle-\int_{{\mathbb{R}}^{2}_{+}}\frac{dzdt}{z^{2}}\,\overline{\Psi}D\delta\Psi$
(4.11) $\displaystyle=$ $\displaystyle\Sigma_{B}-\lim_{z\rightarrow
0}\int_{{\mathbb{R}}}{dt}\,\overline{\Psi}\Upsilon\delta\Psi\;,$ (4.13)

where $\Sigma_{B}$ are the terms that vanish on-shell, and $\Upsilon$ is
defined by

$\Upsilon=-\frac{i}{z}\pmatrix{t&t+i\cr-t+i&-t}\;.$ (4.14)

The boundary contribution in (4.13) resulted from $\partial_{z}$ terms in
(2.29). $\Upsilon$ can also be expressed in terms of the imbedding coordinates
$X^{a}$ and the chirality operator $\gamma$

$\Upsilon=\epsilon_{abc}X^{a}\partial_{t}X^{b}\,\gamma\tau^{c}\;.$ (4.15)

The asymptotic form of the solution for $\Psi$ was given in (2.44), while for
$\bar{\Psi}$ we have

$\displaystyle\overline{\Psi}(\zeta,\bar{\zeta})$ $\displaystyle=$
$\displaystyle\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})+iz\partial_{t}\Bigl{(}\,\overline{\psi_{1}}(t)(\sigma_{3}+it{\mathbb{1}})\Bigr{)}+{\cal
O}(z^{2})\;,$ (4.16)

where $\psi_{0}$ and $\psi_{1}$ were defined in (2.45). We shall regard
$\psi_{0}(t)$ (and $\overline{\psi_{0}}(t)$) as non-dynamical, i.e., source
fields, and so variations in (4.13) are to be carried out with respect to the
remaining boundary fields $\psi_{1}(t)$. The boundary term in (4.13) is
removed upon adding the following term to the bulk action (4.10)

$\displaystyle{S}_{\partial B}[\overline{\Psi},\Psi]$ $\displaystyle=$
$\displaystyle\lim_{z\rightarrow 0}\int_{\partial
B}{dt}\,\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\Upsilon\Psi\;.$
(4.17)

The total action is then $S_{\rm total}=S_{B}+S_{\partial B}$.

Only the boundary term (4.17) survives in the on-shell action. To evaluate it
we can rewrite the asymptotic expression for the solution (2.44) using (4.3)
and (4.4):

$\Psi(\zeta,\bar{\zeta})\;=\;\frac{1}{2\pi}\int
dt^{\prime}\,\Biggl{(}-\frac{i}{t-t^{\prime}}(\sigma_{3}-it{\mathbb{1}})\psi_{1}(t^{\prime})\,+\,\frac{z}{(t-t^{\prime})^{2}}(\sigma_{3}-it^{\prime}{\mathbb{1}})\psi_{0}(t^{\prime})\,+\,{\cal
O}(z^{2})\Biggr{)}\;.$ (4.18)

So evaluating $S_{\rm total}$ on-shell gives

$\displaystyle S_{\rm total}[\bar{\Psi},\Psi]|_{\tt on-shell}$
$\displaystyle=$
$\displaystyle\int{dtdt^{\prime}}\,\frac{z}{2\pi(t-t^{\prime})^{2}}\,\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\Upsilon(\sigma_{3}-it^{\prime}{\mathbb{1}})\psi_{0}(t^{\prime})$
(4.19)

The $z$ independent term in the integrand of (4.18) does not contribute to the
on-shell action due to the result that
$\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\Upsilon(\sigma_{3}-it{\mathbb{1}})\psi_{1}(t^{\prime})$
is identically zero. Notice also that the factor $z$ in (LABEL:inthsznt)
cancels with the $1/z$ appearing in $\Upsilon$ (4.15). Finally, use the
identity

$\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\Upsilon(\sigma_{3}-it^{\prime}{\mathbb{1}})\psi_{0}(t^{\prime})=-\frac{i}{z}(t-t^{\prime})\,\overline{\psi_{0}}(t)\sigma_{3}\psi_{0}(t^{\prime})\;,$
(4.21)

to simplify the result to

$\displaystyle S_{\rm total}[\bar{\Psi},\Psi]|_{\tt on-shell}$
$\displaystyle=$
$\displaystyle-\frac{i}{2\pi}\int_{{\mathbb{R}}}dt\int_{{\mathbb{R}}}dt^{\prime}\,\overline{\psi_{0}}(t)\frac{\sigma_{3}}{t-t^{\prime}}\psi_{0}(t^{\prime})\;.$
(4.22)

The resulting two point function correlation function on the boundary is

$<{\cal O}_{{}_{\overline{\psi_{0}}}}\,(t){\cal
O}_{\psi_{0}}(t^{\prime})>\;=\;\frac{\delta^{2}S_{\rm
total}[\bar{\Psi},\Psi]|_{\tt on-
shell}}{\delta\overline{\psi_{0}}(t)\delta\psi_{0}(t^{\prime})}\bigg{|}_{\overline{\psi_{0}}=\psi_{0}=0}\;=\;\frac{1}{2\pi
i}\,\frac{\sigma_{3}}{t-t^{\prime}}\;.$ (4.23)

###  4.2 Scalar two-point correlator on the boundary of quantized $EAdS_{2}$

The non-commutative generalization of the action (4.1) is

$\hat{S}[\hat{\Phi}]=-\frac{1}{2}{\rm
Tr}\,\Bigl{\\{}[\hat{X}^{\mu},\hat{\Phi}][\hat{X}_{\mu},\hat{\Phi}]\Bigr{\\}}\
,$ (4.24)

where Tr denotes a trace. The field equation (3.18) follows from extremizing
the action with respect to variations in $\hat{\Phi}$.

Here it is convenient to introduce the star product realization of the
operator product. The Moyal-Weyl star product can be employed for this
purpose, provided one transforms to an appropriate (canonical) pair of
coordinates $(x,y)$, as was discussed in [6, 7]. From (2.48), that canonical
pair can be taken to be $z^{-1}$ and $t$. As $(z^{-1},t)$ span
${\mathbb{R}}^{2}_{+}$, appropriate boundary conditions should be imposed at
$z^{-1}=0$ (which recall is not the asymptotic boundary of $EAdS_{2}$), i.e.,
all functions should vanish sufficiently rapidly as $z\rightarrow\infty$.
Alternatively, one can employ a pair of canonical coordinates, denoted by
$(x,y)$, that span the entire plane, as was done in [6, 7]. Then the map back
to Fefferman Graham coordinates is

$\begin{array}[]{l}t=ye^{-x}\\\ z=e^{-x}\end{array}\ .$ (4.25)

In the quantum theory $x$ and $y$ get promoted to operators $\hat{x}$ and
$\hat{y}$, satisfying the standard canonical commutation relation

$[\hat{x},\hat{y}]=i\alpha{\mathbb{1}}\;$ (4.26)

The quantization of the map (4.25) is obtained using symmetric ordering:

$\begin{array}[]{l}\hat{t}=\frac{1}{2}(\hat{y}e^{-\hat{x}}+e^{-\hat{x}}\hat{y})\\\
\hat{z}=e^{-\hat{x}}\end{array}\;.$ (4.27)

Then from (4.26) and (4.25) we recover the fundamental commutation relations
for $\hat{z}$ and $\hat{t}$ (3.32).

The product of operators $\hat{\cal F}$ and $\hat{\cal G}$ can be expressed in
terms of the Moyal-Weyl star product of their symbols (which we also denote by
$\hat{\cal F}$ and $\hat{\cal G}$, respectively), which when expressed in
terms of canonical variables $(x,y)$ [or $(z^{-1},t)$], is defined in the
standard way

$[\hat{\cal F}\star\hat{\cal G}](x,y)=\hat{\cal
F}(x,y)\,\exp{\Bigl{\\{}\,\frac{i\alpha}{2}\,(\overleftarrow{{\partial_{x}}}\,\overrightarrow{{\partial_{y}}}\,-\,\overleftarrow{{\partial_{y}}}\,\overrightarrow{{\partial_{x}}})\,\Bigr{\\}}}\;\hat{\cal
G}(x,y)\ .$ (4.28)

Furthermore, the trace on the algebra, Tr, becomes
$\frac{1}{\alpha^{2}}\int_{{\mathbb{R}}^{2}}dxdy$. Then the operator equations
(4.27) are replaced by the corresponding equations for the for the symbols
$\hat{t}$ and $\hat{z}$

$\begin{array}[]{l}\hat{t}=\frac{1}{2}(\hat{y}\star
e^{-\hat{x}}+e^{-\hat{x}}\star\hat{y})\\\ \hat{z}=e^{-\hat{x}}\end{array}\ ,$
(4.29)

and we recover (3.32), where $[\hat{\cal F},\hat{\cal G}]$ now denotes a star
commutator $\hat{\cal F}\star\hat{\cal G}-\hat{\cal G}\star\hat{\cal F}.$ The
star commutator goes to $i\alpha$ times the corresponding Poisson bracket in
the commutative limit $\alpha\rightarrow 0$, (3.19). From (3.32) it follows
that given any two functions $\hat{\cal F}$ and $\hat{\cal G}$ that are well-
behaved when $z\rightarrow 0$, their star commutator must vanish in the
boundary limit, with the leading order term in $z$ agreeing with the
commutative limit,

$[\hat{\cal F},\hat{\cal G}]\rightarrow i\alpha
z^{2}\Bigl{(}\partial_{t}\hat{\cal F}\partial_{z}\hat{\cal
G}-\partial_{t}\hat{\cal G}\partial_{z}\hat{\cal
F}\Bigr{)}=i\alpha\\{\hat{\cal F},\hat{\cal G}\\}\qquad{\rm as}\;z\rightarrow
0\;.$ (4.30)

Moreover, the star product of any two functions with a well-behaved
$z\rightarrow 0$ limit reduces to the point-wise product in the boundary
limit.

The expressions for the noncommutative analogues $\hat{X}^{a}$ of the
embedding coordinates in term of $\hat{t}$ and $\hat{z}$, given in (3.2), can
be realized in terms of the star product. From them we recover (3.1) and
(3.2), the latter now written as
$\hat{X}^{a}\star\hat{X}_{a}=-{\mathbb{1}}\;$. Upon expanding (3.2) in
$\alpha$, it can be shown that embedding coordinates $X^{0}$ and $X^{2}$ only
pick up a second order correction in $\alpha$ upon quantization (in fact, the
corrections to $X^{0}$ and $X^{2}$ are identical), while $X^{1}$ is undeformed

$\hat{X}^{0}=X^{0}-\frac{\alpha^{2}}{8}z\;,\quad\quad\hat{X}^{1}=X^{1}\;,\quad\quad\hat{X}^{2}=X^{2}-\frac{\alpha^{2}}{8}z\;.$
(4.31)

We see that the quantum corrections vanish in the boundary limit $z\rightarrow
0$. Recall that the quantum analogues $\hat{\cal K}^{a}$ of Killing vectors
${\cal K}^{a}$ were constructed from commutators (or star commutators) with
$\hat{X}_{a}$ (3.10). Since $\hat{X}^{a}$ reduces to embedding coordinates
$X^{a}$ in the boundary limit, and the star commutator goes to $i\alpha$ times
the corresponding Poisson bracket in the limit (4.30), it follows that
$\hat{\cal K}^{a}$ must reduce to the classical Killing vectors ${\cal K}^{a}$
(2.50) on the boundary. This led to the conclusion in [6, 7] that quantized
$EAdS_{2}$ is an asymptotically anti-de Sitter space.

From the above we can write the (4.24) according to

$\hat{S}[\hat{\Phi}]=-\frac{1}{2\alpha^{2}}\int_{{\mathbb{R}}^{2}_{+}}\frac{dtdz}{z^{2}}\,\Bigl{\\{}[\hat{X}^{\mu},\hat{\Phi}]\star[\hat{X}_{\mu},\hat{\Phi}]\Bigr{\\}}\
,$ (4.32)

where again $[\,,\,]$ now denotes a star commutator. The action can be
expressed as a sum of two terms, where one term vanishes on-shell, and the
other term, which we denote by $\hat{S}_{\partial D}[\hat{\Phi}]$, is only
defined on the asymptotic boundary. As was argued in [6, 7], the expression
for the boundary action is identical in form to that appearing in the
commutative theory (4.6),

$\hat{S}_{\partial D}[\hat{\Phi}]=-\frac{1}{2}\lim_{z\rightarrow 0}\int
dt\,\hat{\Phi}\partial_{z}\hat{\Phi}\;.$ (4.33)

The exact solution to the quantized scalar field equation is given by (3.45).
Near the boundary $z=0$ the solution behaves as

$\hat{\Phi}(\hat{\Xi},\hat{\Xi}^{\dagger})=\phi_{0}(t)+\kappa(\alpha)z\,\phi_{1}(t)+{\cal
O}(z^{2})\;,$
$\phi_{0}(t)={\hat{F}(it)+\hat{G}(-it)}\;\qquad\;\phi_{1}(t)=\Bigl{(}{\hat{F}^{\prime}(\hat{\Xi})+\hat{G}^{\prime}(\hat{\Xi}^{\dagger})}\Bigr{)}|_{z=0}\;,$
(4.34)

the prime denoting a derivative. Here we used
$z\,\phi_{1}(t)=z\star\phi_{1}(t)+{\cal O}(z^{2})=\phi_{1}(t)\star z+{\cal
O}(z^{2})$. So, like with the classical solution, $\hat{\Phi}$ tends to
$\phi_{0}(t)$ as $z\rightarrow 0$, while $\partial_{z}\hat{\Phi}(z,t)$ picks
up a factor of $\kappa(\alpha)$:

$\displaystyle\partial_{z}\hat{\Phi}(z,t)$ $\displaystyle\rightarrow$
$\displaystyle\kappa(\alpha)\,\partial_{z}\Phi(z,t)|_{z\rightarrow 0}$ (4.35)
$\displaystyle\rightarrow$ $\displaystyle\,\frac{\kappa(\alpha)}{\pi}\int
dt^{\prime}\frac{\phi_{0}(t^{\prime})}{(t-t^{\prime})^{2}}\;,$ (4.37)

where we used (4.5). Upon substituting into (4.33), the result for the on-
shell action is

$S[\Phi]|_{\rm on-
shell}=-\frac{{\kappa(\alpha)}}{2\pi}\int_{{\mathbb{R}}}dt\int_{{\mathbb{R}}}dt^{\prime}\,\frac{\phi_{0}(t)\phi_{0}(t^{\prime})}{(t-t^{\prime})^{2}}\;\;.$
(4.38)

So the boundary two-point function for massless scalar fields on quantized
$EAdS_{2}$ picks up an overall factor of $\kappa(\alpha)$

$<{\cal O}(t){\cal
O}(t^{\prime})>\;=\;-\frac{{\kappa(\alpha)}}{\pi}\,\frac{1}{(t-t^{\prime})^{2}}\
.$ (4.39)

This agrees up to ${\cal O}(\alpha^{4})$ with the leading order perturbative
result found in [6, 7].

###  4.3 Spinor two-point correlator on the boundary of quantized $EAdS_{2}$

As in the previous subsection we can utilize the star product realization of
the operator product to write down the equations of motion and the action.
Concerning the latter, the bulk action for a massless spinor on quantized
$EAdS_{2}$ can be taken to be

$\displaystyle\hat{S}_{B}[\overline{\hat{\Psi}},\hat{\Psi}]=-\int_{B}\frac{dzdt}{z^{2}}\,\overline{\hat{\Psi}}\star\hat{D}\hat{\Psi}$
(4.40) (4.41)
$\displaystyle\;=-\frac{i}{\alpha}\int_{B}\frac{dzdt}{z^{2}}\,\epsilon_{abc}\overline{\hat{\Psi}}\star\hat{\gamma}\Bigl{(}\tau^{c}\hat{X}^{b}\star\hat{\Psi}\star\hat{X}^{a}\Bigr{)}$
(4.42) (4.43)
$\displaystyle\;=-\frac{i}{\alpha\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{B}\frac{dzdt}{z^{2}}\epsilon_{abc}\overline{\hat{\Psi}}\star\Bigl{(}\tau_{d}\tau^{c}\hat{X}^{b}\star\hat{\Psi}\star\hat{X}^{a}\star\hat{X}^{d}-\frac{\alpha}{2}\tau^{c}\hat{X}^{b}\star\hat{\Psi}\star\hat{X}^{a}\Bigr{)}\;.$
(4.44) (4.45)

Like in the commutative case, the conjugate spinor is defined by
$\overline{\hat{\Psi}}={\hat{\Psi}}^{\dagger}\sigma_{3}$. Variations of
$\overline{\hat{\Psi}}$ in the action lead to the equations of motion in the
bulk (3.26). Using the cyclic property of the Moyal-Weyl star product, $\int
dxdy\,{\cal F}\star{\cal G}\star{\cal H}=\int dxdy\,\hat{\cal H}\star\hat{\cal
F}\star\hat{\cal G}\,+\,$boundary terms, the action (3.26) can also be written
as

$\hat{S}_{B}[\overline{\hat{\Psi}},\hat{\Psi}]\;=\;-\,\frac{i}{\alpha\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{B}\frac{dzdt}{z^{2}}\epsilon_{abc}\,\hat{X}^{a}\star\Bigl{(}\hat{X}^{d}\star\overline{\hat{\Psi}}\tau_{d}-\frac{\alpha}{2}\overline{\hat{\Psi}}\Bigr{)}\star\hat{X}^{b}\tau^{c}\star\hat{\Psi}\,+\,B.T.$
$\;=\frac{i}{\alpha\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{B}\frac{dzdt}{z^{2}}{\epsilon_{abc}}\biggl{(}\hat{X}^{d}\star\hat{X}^{a}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau^{c}\tau_{d}-\frac{\alpha}{2}\hat{X}^{a}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau^{c}\biggr{)}\star\hat{\Psi}+B.T$
$=\frac{i}{\alpha}\int_{B}\frac{dzdt}{z^{2}}\,{\epsilon_{abc}}\hat{\gamma}\Bigl{(}\hat{X}^{a}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau^{c}\Bigr{)}\star\hat{\Psi}\,+\,B.T.\;,\qquad\qquad\qquad\qquad\qquad\qquad$
(4.46)

where $B.T.$ denotes boundary terms, and we used the identity

$\epsilon_{abc}\hat{X}^{a}\star\hat{X}^{d}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau_{d}\tau^{c}=-\epsilon_{abc}\hat{X}^{d}\star\hat{X}^{a}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau^{c}\tau_{d}+\alpha\epsilon_{abc}\hat{X}^{a}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau^{c}\;.$
(4.47)

Then variations of $\hat{\Psi}$ in the action lead to the equations of motion

${\epsilon_{abc}}\hat{X}^{a}\star\overline{\hat{\Psi}}\star\hat{X}^{b}\tau^{c}=0\;.$
(4.48)

This is the hermitean conjugate of the Dirac equation (3.26). To show this we
can use the identity $\tau_{a}^{\dagger}=\sigma_{3}\tau_{a}\sigma_{3}$.

The exact solution to the Dirac equation (3.26) is given by (3.46). As before,
we only need the leading and next to leading behavior as $z\rightarrow 0$ to
obtain the two-point correlator on the boundary. The expansion of (3.46) near
$z=0$ is

$\displaystyle\hat{\Psi}(\hat{\Xi},\hat{\Xi}^{\dagger})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}1-it\\\ 1+it\\\
\end{array}\right)(\hat{F}(it)-\hat{G}(-it))$ (4.51) $\displaystyle+$
$\displaystyle\frac{\kappa(\alpha)z}{2}\left(\begin{array}[]{c}-\hat{F}(it)-\hat{G}(-it)+(1-it)\left(\hat{F}^{\prime}(it)-\hat{G}^{\prime}(-it)\right)\\\
\;\;\,\hat{F}(it)+\hat{G}(-it)+(1+it)\left(\hat{F}^{\prime}(it)-\hat{G}^{\prime}(-it)\right)\\\
\end{array}\right)+O\left(z^{2}\right)\,.$ (4.55)

It, along with its hermitean conjugate, can be re-expressed as

$\displaystyle\hat{\Psi}(\hat{\Xi},\hat{\Xi}^{\dagger})$ $\displaystyle=$
$\displaystyle(\sigma_{3}-it{\mathbb{1}})\,\psi_{0}(t)-iz\kappa(\alpha)\,\partial_{t}\Bigl{(}(\sigma_{3}-it{\mathbb{1}})\,\psi_{1}(t)\Bigr{)}+{\cal
O}(z^{2})\;,$ (4.57)
$\displaystyle\overline{\hat{\Psi}}(\hat{\Xi},\hat{\Xi}^{\dagger})$
$\displaystyle=$
$\displaystyle\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})+iz\kappa(\alpha)\,\partial_{t}\Bigl{(}\,\overline{\psi_{1}}(t)(\sigma_{3}+it{\mathbb{1}})\Bigr{)}+{\cal
O}(z^{2})\;,$ (4.59)

where $\psi_{0}$ and $\psi_{1}$ are the quantum analogues of (2.45).

As was true in the commutative case, the action should be supplemented with a
boundary term so that variations of the total action can be consistently made
to vanish. A general variation in $\hat{\Psi}$ induces the following change in
the bulk action (4.45):

$\displaystyle\delta\hat{S}_{B}$ $\displaystyle=$
$\displaystyle-\frac{i}{\alpha}\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{B}\frac{dzdt}{z^{2}}\epsilon_{abc}\overline{\hat{\Psi}}\star\Bigl{(}\tau_{d}\tau^{c}\hat{X}^{b}\star\delta\hat{\Psi}\star\hat{X}^{a}\star\hat{X}^{d}-\frac{\alpha}{2}\tau^{c}\hat{X}^{b}\star\delta\hat{\Psi}\star\hat{X}^{a}\Bigr{)}$
(4.60) $\displaystyle=$
$\displaystyle-\frac{i}{\alpha}\,\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{B}\frac{dzdt}{z^{2}}\epsilon_{abc}\hat{X}^{a}\star\Bigl{(}\hat{X}^{d}\star\overline{\hat{\Psi}}\tau_{d}-\frac{\alpha}{2}\overline{\hat{\Psi}}\Bigr{)}\star\hat{X}^{b}\tau^{c}\star\delta\hat{\Psi}\;+\;b.\,v.\;,$
(4.62)

where the boundary variation is

$b.\,v.=-\frac{i}{\alpha}\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{B}\frac{dzdt}{z^{2}}\epsilon_{abc}\Biggl{(}\Bigl{[}\overline{\hat{\Psi}}\star\tau_{d}\tau^{c}\hat{X}^{b}\star\delta\hat{\Psi}\,,\,\hat{X}^{a}\star\hat{X}^{d}\Bigr{]}-\frac{\alpha}{2}\Bigl{[}\overline{\hat{\Psi}}\star\tau^{c}\hat{X}^{b}\star\delta\hat{\Psi}\,,\,\hat{X}^{a}\Bigr{]}\Biggr{)}\;.$
(4.64)

The boundary term results from the integral of a star commutator, and as shown
in appendix A, such an integral is given explicitly by

$\int_{D}\frac{dzdt}{z^{2}}\,[\hat{\cal F},\hat{\cal
G}](z,t)=-i\alpha\int{dt}\,(\partial_{t}\hat{\cal F}\,\hat{\cal G})|_{z=0}\;,$
(4.65)

for any two functions $\hat{\cal F}$ and $\hat{\cal G}$ on quantized
$EAdS_{2}$. Applying this to (4.64) gives

$b.\,v.=-\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int
dt\,\epsilon_{abc}\Biggl{(}(\hat{X}^{a}\star\hat{X}^{d})\partial_{t}\Bigl{(}\overline{\hat{\Psi}}\star\hat{X}^{b}\star\tau_{d}\tau^{c}\delta\hat{\Psi}\Bigr{)}-\frac{\alpha}{2}\hat{X}^{a}\,\partial_{t}\Bigl{(}\overline{\hat{\Psi}}\star\hat{X}^{b}\star\tau^{c}\delta\hat{\Psi}\Bigr{)}\Biggr{)}\Bigg{|}_{z=0}\;.$
(4.66)

Using the result that the star product reduces to the pointwise product in the
boundary limit, and that $\hat{X}^{a}$ go to the embedding coordinates $X^{a}$
in the boundary limit, this simplifies to

$\displaystyle b.\,v.$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int
dt\,\epsilon_{abc}X^{a}\partial_{t}X^{b}\;\overline{\hat{\Psi}}\Bigl{(}X^{d}\tau_{d}-\frac{\alpha}{2}\Bigr{)}\tau^{c}\delta\hat{\Psi}\Big{|}_{z=0}$
(4.67) $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int
dt\,\overline{\hat{\Psi}}\Bigl{(}{\mathbb{1}}-\frac{\alpha}{2}\gamma\Bigr{)}\Upsilon\delta\hat{\Psi}\Big{|}_{z=0}\;,$
(4.69)

where $\gamma$ is the chirality operator on the classical manifold (2.23), and
$\Upsilon$ was defined in (4.15). We see that the boundary term reduces to its
commutative counterpart (4.13) in the limit $\alpha\rightarrow 0$.

As in section 4.1.2, we need to add a term to the action that will cancel the
boundary variation (4.69). For this we shall again assume that ${\psi_{0}}$ is
the source field. Then (4.69) is eliminated upon adding

$\displaystyle\hat{S}_{\partial B}[\overline{\hat{\Psi}},\hat{\Psi}]$
$\displaystyle=$ $\displaystyle\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int
dt\,\,\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\Bigl{(}{\mathbb{1}}-\frac{\alpha}{2}\gamma\Bigr{)}\Upsilon\hat{\Psi}\Big{|}_{z=0}$
(4.70)

to the bulk action (4.45). Upon using the identity,

$\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\gamma\Upsilon=-\overline{\psi_{0}}(t)\;,$
(4.71)

(4.70) simplifies to

$\displaystyle\hat{S}_{\partial B}[\overline{\hat{\Psi}},\hat{\Psi}]$
$\displaystyle=$ $\displaystyle\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int
dt\,\overline{\psi_{0}}(t)\biggl{(}(\sigma_{3}+it{\mathbb{1}})\Upsilon+\frac{\alpha}{2}\biggr{)}\hat{\Psi}\Big{|}_{z=0}\;.$
(4.72)

Once again only the boundary term survives in the total action $\hat{S}_{\rm
total}=\hat{S}_{B}+\hat{S}_{\partial B}$ when evaluating it on-shell. To
evaluate it we can rewrite the asymptotic expression for the solution for
$\hat{\Psi}$ (4.59) using (4.3) and (4.4):

$\hat{\Psi}(\hat{\Xi},\hat{\Xi}^{\dagger})\;=\;\frac{1}{2\pi}\int
dt^{\prime}\,\Biggl{(}-\frac{i}{t-t^{\prime}}(\sigma_{3}-it{\mathbb{1}})\psi_{1}(t^{\prime})\,+\,\frac{\kappa(\alpha)z}{(t-t^{\prime})^{2}}(\sigma_{3}-it^{\prime}{\mathbb{1}})\psi_{0}(t^{\prime})\,+\,{\cal
O}(z^{2})\Biggr{)}\;.$ (4.73)

Substituting this result into $\hat{S}_{\partial B}$, and regarding $\psi_{0}$
as the source field, we get

$\displaystyle\hat{S}_{\rm total}[\overline{\hat{\Psi}},\hat{\Psi}]|_{\tt on-
shell}$ $\displaystyle=$
$\displaystyle\frac{\kappa(\alpha)}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int{dtdt^{\prime}}\,\frac{z}{2\pi(t-t^{\prime})^{2}}\,\overline{\psi_{0}}(t)(\sigma_{3}+it{\mathbb{1}})\Upsilon(\sigma_{3}-it^{\prime}{\mathbb{1}})\psi_{0}(t^{\prime})\,.$
(4.74)

Like in the commutative case, this term survives after taking the
$z\rightarrow 0$ limit because $\Upsilon$ (4.15) contains a factor of $1/z$,
while the term that is linear in $\alpha$ in the integrand of (4.72) does not
survive. Using the identity (4.21), we then get

$\displaystyle\hat{S}_{\rm total}[\overline{\hat{\Psi}},\hat{\Psi}]|_{\tt on-
shell}$ $\displaystyle=$
$\displaystyle-\frac{i}{2\pi}\frac{\kappa(\alpha)}{\sqrt{1+\frac{\alpha^{2}}{4}}}\int_{{\mathbb{R}}}dt\int_{{\mathbb{R}}}dt^{\prime}\,\overline{\psi_{0}}(t)\frac{\sigma_{3}}{t-t^{\prime}}\psi_{0}(t^{\prime})\;.$
(4.76)

But we saw in (3.36) that ${\kappa(\alpha)}={\sqrt{1+\frac{\alpha^{2}}{4}}}$,
so the on-shell action and resulting two-point function on the boundary are
identical to the commutative case to all orders in $\alpha$!

##  5 Concluding remarks

In this paper we have continued our examination of the non-commutative
$AdS/CFT$ correspondence. Though here we restricted our attention to the case
of massless fields, we have extended previous results in two directions. First
of all, by considering spinors, we have gone beyond the case of scalar fields
examined in [6, 7, 8]. An essential tool for the study of spinor fields on the
quantized $EAdS_{2}$ was the map (2.28) to the covariant formulation of the
Dirac operator, which we were able to extend to the quantized space [cf.
appendix B]. The second direction of this paper makes what we regard as a
notable advance in the study of fields on quantized spaces-times. In our
previous works we obtained results only to leading non-trivial order in the
non-commutative parameter $\alpha$, while here we were able to solve the free
scalar and spinor equations exactly, i.e., non-perturbatively in $\alpha$. The
results suggest more generally that there may be a simple prescription for
finding exact solutions on this quantum space. It is to replace functions of
the complex variables $\zeta=z+it$ and $\bar{\zeta}=z-it$ with functions of
the quantum operators $\hat{\Xi}=\kappa(\alpha)\hat{z}+i\hat{t}$ and
$\hat{\Xi}^{\dagger}=\kappa(\alpha)\hat{z}-i\hat{t}$, respectively, where
$\hat{z}$ and $\hat{t}$ are the noncommutative analogues of the Fefferman-
Graham coordinates $z$ and $t$. Thus it appears that the $z$ coordinate (which
is the inverse of the radial coordinate) gets re-scaled by $\kappa(\alpha)$
upon quantization. The $z$ coordinate is said to be associated with the energy
scale of the corresponding $CFT$. The implication then is that the $CFT$
energy scale gets re-scaled by the quantization of the bulk.

We conjectured in our previous works that boundary conformal symmetry survives
the quantization of the $AdS$ bulk space-time. By having exact results we have
now been able to prove the conjecture for free massless fields. Furthermore,
we have shown that the boundary two-point function is undeformed for the case
of a spinor source, while it gets re-scaled by $\kappa(\alpha)$ for the case
of a scalar source.

There are a number of extensions of this work that can be considered, some of
them being crucial (in particular, item 2. below) to the study of the non-
commutative correspondence. Among them are:

  1. 1.

One obvious extension of our work is to obtain exact solutions for the case of
massive scalars and spinors on quantized $EAdS_{2}$. Exact results are
possible and will likely involve special functions, with the domain spanned by
operators $\hat{\Xi}$ and $\hat{\Xi}^{\dagger}$. Exact results for the
corresponding boundary two-point functions can then be deduced from the
asymptotic properties of these solutions, as was done here.

  2. 2.

Another obvious extension is to include interactions in scalar and spinor
field theory on quantized $EAdS_{2}$, such as $\lambda\Phi^{3}$ or a Yukawa
coupling. Partial results for the $\lambda\Phi^{3}$ interaction were obtained
in [7], but because we didn’t have exact solutions they required a rather
involved expansion in two parameters, $\lambda$ and the non-commutativity
parameter $\alpha$. A simpler one parameter expansion can now be pursued by
exploiting the exact solution we have found here. It should admit leading
order expressions of $n(>2)$ point correlators on the boundary. A question
raised previously was whether or not the re-scaling of the boundary two-point
function (in the case of scalars) could be due to a trivial field
renormalization. The study of interactions using the exact results found here
should resolve this issue.

  3. 3.

Natural generalizations of quantized $EAdS_{2}$ were obtained in [8]. They
correspond to the quantization of ${\mathbb{CP}}^{p,q}$. As with $EAdS_{2}$,
the quantization is uniquely given upon demanding that it preserves the
isotropies. A construction of the Laplacian and Dirac operators on these
spaces should be possible, from which one can search for exact solutions to
the equations for massless scalar and spinor fields analogous to the ones
found in this article.

We plan to address these and related topics in forthcoming articles.

Appendix A Boundary terms

The integral of the commutator of the Moyal-Weyl star product of any two
functions $\hat{\cal F}$ and $\hat{\cal G}$ on the plane vanishes provided
that the functions vanish sufficiently rapidly at infinity. The integral does
not vanish, however, if the domain $D$ (which say is two-dimensional) has a
boundary $\partial D$ and the functions are unconstrained on the boundary.
Below we show that if $D$ is the bulk region of $EAdS_{2}$, we get the result
(4.65).

Say that $D$ is some two-dimensional region with a boundary $\partial D$. From
the the definition of the Moyal-Weyl star product (4.28), one gets

$\int_{D}dxdy\,[\hat{\cal F},\hat{\cal
G}](x,y)=\int_{D}dxdy\,[\partial_{x}{\cal V}_{y}-\partial_{y}{\cal
V}_{x}](x,y)=\int_{\partial D}\,({\cal V}_{x}dx+{\cal V}_{y}dy)\;.$ (A.1)

Upon expanding ${\cal V}_{x}$ and ${\cal V}_{y}$ to leading and next to
leading orders in $\alpha$ one has

$\displaystyle{\cal V}_{x}$ $\displaystyle=$ $\displaystyle
i\alpha\biggl{(}-\partial_{x}\hat{\cal F}\,\hat{\cal
G}+\frac{\alpha^{2}}{24}\Bigl{(}\partial_{x}^{3}\hat{\cal
F\,}\partial^{2}_{y}\hat{\cal G}+\partial_{x}\partial^{2}_{y}\hat{\cal
F}\,\partial^{2}_{x}\hat{\cal G}-2\partial_{x}^{2}\partial_{y}\hat{\cal
F}\,\partial_{x}\partial_{y}\hat{\cal G}\Bigr{)}+{\cal
O}(\alpha^{4})\biggr{)}\ ,$ (A.2) $\displaystyle{\cal V}_{y}$ $\displaystyle=$
$\displaystyle i\alpha\biggl{(}-\partial_{y}\hat{\cal F}\,\hat{\cal
G}+\frac{\alpha^{2}}{24}\Bigl{(}\partial_{y}^{3}\hat{\cal
F}\,\partial^{2}_{x}\hat{\cal G}+\partial^{2}_{x}\partial_{y}\hat{\cal
F}\,\partial^{2}_{y}\hat{\cal G}-2\partial_{x}\partial^{2}_{y}\hat{\cal
F}\,\partial_{x}\partial_{y}\hat{\cal G}\Bigr{)}+{\cal
O}(\alpha^{4})\biggr{)}\ .$ (A.4)

Now let $D$ correspond to the bulk region of $EAdS_{2}$ with $\partial D$
corresponding to $z=0$, where the map between the canonical coordinates
$(x,y)$ to Fefferman-Graham coordinates can be taken to be (2.49). A constant
$z$ slice corresponds to $x=const$. So using $dy=\frac{dt}{z}$, we get the
following integral along the asymptotic boundary

$\int_{D}dxdy\,[\hat{\cal F},\hat{\cal G}](x,y)=\int dt\,\frac{{\cal
V}_{y}}{z}\Big{|}_{z=0}\;.$ (A.6)

${\cal V}_{y}$ is re-expressed in terms for Fefferman-Graham coordinates,
using

$\partial_{x}=-z\partial_{z}-t\partial_{t}\;,\quad\quad\partial_{y}=z\partial_{t}\;.$
(A.7)

Assuming that $\hat{\cal F}$ and $\hat{\cal G}$, along with their derivatives,
go to well defined functions of $t$ in the limit $z\rightarrow 0$, we get that
the order $\alpha^{3}$ correction terms in ${\cal V}_{y}$ go like $z^{3}$, and
so do not contribute in the limit. Thus only the ${\cal O}(\alpha)$ term
survives in the limit, giving the result (4.65).

Appendix B Quantum map to noncovariant basis

In section 2 we gave the map (2.28) between the covariant and noncovariant
formulations of the Dirac operator on $EAdS_{2}$. Here we give the quantum
analogue of the map. This will allow us to construct the Dirac operator on
quantized $EAdS_{2}$ in the noncovariant basis. We shall see that solutions to
the Dirac equation for massless spinors are trivial in this basis. As a check,
we can then map these solutions back to the covariant basis. The result agrees
with the exact solutions obtained in section 3, (3.46).

To obtain the quantum map we will need the quantum version of the $2\times 2$
matrix $U$ (2.26). For this we first introduce the left and right acting
operators $\hat{z}^{A}$ and $\hat{t}^{A}$, $A=L,R$, which act on functions
$\hat{\cal F}$ on quantized $EAdS_{2}$ according to

$\displaystyle\hat{z}^{L}\hat{\cal F}=\hat{z}\hat{\cal F}$
$\displaystyle\;,\quad\quad$ $\displaystyle\hat{t}^{L}\hat{\cal
F}=\hat{t}\hat{\cal F}\;,$ (B.1) $\displaystyle\hat{z}^{R}\hat{\cal
F}=\hat{\cal F}\hat{z}$ $\displaystyle\;,\quad\quad$
$\displaystyle\hat{t}^{R}\hat{\cal F}=\hat{\cal F}\hat{t}\ \;.$ (B.3)

From the commutation relations (3.32), the left acting operators $\hat{t}^{L}$
and $\hat{z}^{L}$ satisfy

$[\hat{t}^{L},\hat{z}^{L}]=i\alpha(\hat{z}^{L})^{2}\;,$ (B.4)

while the opposite sign appears in the commutator of $\hat{t}^{R}$ and
$\hat{z}^{R}$

$[\hat{t}^{R},\hat{z}^{R}]=-i\alpha(\hat{z}^{R})^{2}\;.$ (B.5)

All other commutators between $\hat{z}^{A}$ and $\hat{t}^{A}$ vanish. From
(B.5) it follows that

$[\hat{t}^{R},(\hat{z}^{R})^{-1}]=i\alpha{\mathbb{1}}\;.$ (B.6)

We can further introduce the right-acting version of the operator $\hat{\Xi}$
defined in section 3.2, along with its hermitean conjugate:

$\hat{\Xi}^{R}=\kappa(\alpha)\hat{z}^{R}+i\hat{t}^{R}\;,\quad\quad\hat{\Xi}^{\dagger
R}=\kappa(\alpha)\hat{z}^{R}-i\hat{t}^{R}\;.$ (B.7)

We now claim that the quantization of of the $2\times 2$ matrix (2.26) is

$\hat{U}^{R}=\pmatrix{-\hat{\Xi}^{R}+1&-\hat{\Xi}^{\dagger
R}-1\cr\hat{\Xi}^{R}+1&\hat{\Xi}^{\dagger
R}-1}\frac{1}{2\sqrt{\kappa(\alpha)\hat{z}^{R}}}\;.$ (B.8)

It satisfies the analogue of the identity (2.27), i.e.,

$\hat{\tilde{U}}^{\dagger R}\sigma_{3}\hat{U}^{R}=-\sigma_{3}\;,$ (B.9)

where ‡‡‡‡‡‡We can assume that hermitean conjugation of right acting operators
$\hat{A}^{R}$ is an anti-involution. For this define the action of
$(\hat{A}^{R})^{\dagger}$ on functions $\hat{\cal F}$ on quantized $AdS_{2}$
by  $\hat{\cal F}(\hat{A}^{R})^{\dagger}=(\hat{A}^{R}\hat{\cal
F}^{\dagger})^{\dagger}=(\hat{\cal
F}^{\dagger}\hat{A})^{\dagger}=\hat{A}^{\dagger}\hat{\cal F}\;.$ Then for two
right acting operators $\hat{A}^{R}$ and $\hat{B}^{R}$ $\hat{\cal
F}(\hat{A}^{R})^{\dagger}(\hat{B}^{R})^{\dagger}=(\hat{A}^{\dagger}\hat{\cal
F})(\hat{B}^{R})^{\dagger}=\hat{B}^{\dagger}\hat{A}^{\dagger}\hat{\cal
F}=(\hat{A}\hat{B})^{\dagger}\hat{\cal F}=\hat{\cal
F}((\hat{A}\hat{B})^{R})^{\dagger}=\hat{\cal
F}((\hat{B})^{R}(\hat{A})^{R})^{\dagger}\;.\qquad$

$\hat{U}^{\dagger
R}=\frac{1}{2\sqrt{\kappa(\alpha)\hat{z}^{R}}}\pmatrix{-\hat{\Xi}^{\dagger
R}+1&\hat{\Xi}^{\dagger R}+1\cr-\hat{\Xi}^{R}-1&\hat{\Xi}^{R}-1}\;.$ (B.10)

The inverse of $\hat{U}^{R}$ is

$(\hat{U}^{R})^{-1}=\frac{1}{2\sqrt{\kappa(\alpha)\hat{z}^{R}}}\pmatrix{\hat{\Xi}^{\dagger
R}-1&\hat{\Xi}^{\dagger R}+1\cr-\hat{\Xi}^{R}-1&-\hat{\Xi}^{R}+1}\;.$ (B.11)

It is easy to check that it is the left inverse of $\hat{U}^{R}$,
$(\hat{U}^{R})^{-1}\hat{U}^{R}={\mathbb{1}}$. To check that it is the right
inverse, $\hat{U}^{R}(\hat{U}^{R})^{-1}={\mathbb{1}}$, use

$[\hat{\Xi}^{R},(\hat{z}^{R})^{-1}]=-[\hat{\Xi}^{\dagger
R},(\hat{z}^{R})^{-1}]=-\alpha\;,$ (B.12)

which follows from (B.6).

From (2.27), we saw that $U$ maps the classical chirality operator
$\sigma_{3}$ of the noncovariant basis to the chirality operator $\gamma$ of
the covariant basis. The analogous property holds for $\hat{U}$ in the
quantized system. Upon computing $\hat{U}^{R}\sigma_{3}(\hat{U}^{R})^{-1}$, we
get

$\hat{U}^{R}\sigma_{3}(\hat{U}^{R})^{-1}=\frac{1}{\sqrt{1+\frac{\alpha^{2}}{4}}}\,(\hat{X}^{Ra}\tau_{a}-\frac{\alpha}{2}{\mathbb{1}})\;,$
(B.13)

which is the chirality operator $\hat{\gamma}$ on quantized $EAdS_{2}$. The
inverse map takes $\hat{\gamma}$ back to $\sigma_{3}$,

$(\hat{U}^{R})^{-1}\hat{\gamma}\hat{U}^{R}=\sigma_{3}\;.$ (B.14)

Now we apply the map to the Dirac operator $\hat{D}$, which in section 3 was
formulated in the covariant basis. We write this Dirac operator as

$\hat{D}=\frac{i}{\alpha}\,\hat{\gamma}\Delta_{c}\tau^{c}\;,\qquad\;\;\Delta_{c}=\,\epsilon_{abc}\hat{X}^{Ra}\hat{X}^{Lb}\;,$
(B.15)

where $\hat{X}^{Aa}$, $A=L,R$, give the left and right action of
$\hat{X}^{a}$, (3.12). Using (3.2), we can express them in terms of
$\hat{z}^{A}$ and $\hat{t}^{A}$ according to

$\displaystyle\hat{X}^{A0}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Bigl{(}\hat{t}^{A}(\hat{z}^{A})^{-1}\hat{t}^{A}+\kappa(\alpha)^{2}\hat{z}^{A}+(\hat{z}^{A})^{-1}\Bigr{)}\;,$
(B.16) $\displaystyle\hat{X}^{A1}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Bigl{(}\hat{t}^{A}(\hat{z}^{A})^{-1}+\hat{t}^{A}(\hat{z}^{A})^{-1}\Bigr{)}\;,$
(B.18) $\displaystyle\hat{X}^{A2}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Bigl{(}\hat{t}^{A}(\hat{z}^{A})^{-1}\hat{t}^{A}+\kappa(\alpha)^{2}\hat{z}^{A}-(\hat{z}^{A})^{-1}\Bigr{)}\;.$
(B.20)

To write down the Dirac operator in the noncovariant basis we apply a
similarity transformation to $\hat{D}$. We get

$\hat{D}^{(\hat{U}^{R})^{-1}}\;=\;(\hat{U}^{R})^{-1}\hat{D}\hat{U}^{R}\;=\;\frac{i}{\alpha}\sigma_{3}\,(\hat{U}^{R})^{-1}\Delta_{c}\tau^{c}\hat{U}^{R}\;,$
(B.21)

where we used (B.14). Substituting (B.8) and (B.11) into (B.21) gives

$\hat{D}^{(\hat{U}^{R})^{-1}}=\frac{i}{2\alpha\kappa(\alpha)}\sigma_{3}\,\frac{1}{\sqrt{\hat{z}^{R}}}\pmatrix{M_{11}&M_{12}\cr
M_{21}&M_{22}}\frac{1}{\sqrt{\hat{z}^{R}}}\;,$ (B.22)

where the matrix elements $M_{ab}$ are

$\displaystyle M_{11}$ $\displaystyle=$ $\displaystyle\hat{\Xi}^{\dagger
R}\Delta_{+}\hat{\Xi}^{R}+\Delta_{-}+i(\hat{\Xi}^{\dagger
R}\Delta_{1}-\Delta_{1}\hat{\Xi}^{R})\;,$ (B.23) $\displaystyle M_{12}$
$\displaystyle=$ $\displaystyle\hat{\Xi}^{\dagger
R}\Delta_{+}\hat{\Xi}^{\dagger R}-\Delta_{-}-i(\hat{\Xi}^{\dagger
R}\Delta_{1}+\Delta_{1}\hat{\Xi}^{\dagger R})\;,$ (B.25) $\displaystyle
M_{21}$ $\displaystyle=$
$\displaystyle-\hat{\Xi}^{R}\Delta_{+}\hat{\Xi}^{R}+\Delta_{-}-i(\hat{\Xi}^{R}\Delta_{1}+\Delta_{1}\hat{\Xi}^{R})\;,$
(B.27) $\displaystyle M_{22}$ $\displaystyle=$
$\displaystyle-\hat{\Xi}^{R}\Delta_{+}\hat{\Xi}^{\dagger
R}-\Delta_{-}+i(\hat{\Xi}^{R}\Delta_{1}-\Delta_{1}\hat{\Xi}^{\dagger R})\;,$
(B.29)

and $\Delta_{\pm}$ and $\Delta_{1}$ are defined according to

$\Delta_{\pm}=\Delta_{0}\pm\Delta_{2}=\epsilon_{AB}\hat{X}^{A\mp}\hat{X}^{B1}\;,\quad\quad\Delta_{1}=\frac{1}{2}\epsilon_{AB}\hat{X}^{A+}\hat{X}^{B-}\;.$
(B.30)

Here $\hat{X}^{A\pm}=\hat{X}^{A2}\pm\hat{X}^{A0}$, and $\epsilon_{AB}$ is
antisymmetric, with $\epsilon_{LR}=1$.

The rather involved expressions for the matrix elements can be simplified.
Starting with $M_{11}$, we can write

$\displaystyle M_{11}$ $\displaystyle=$
$\displaystyle\;\;\hat{X}^{L+}\Bigl{(}\hat{X}^{R1}+\frac{i}{2}(\hat{\Xi}^{\dagger
R}\hat{X}^{R-}-\hat{X}^{R-}\hat{\Xi}^{R})\,\Bigr{)}$ (B.35)
$\displaystyle+\hat{X}^{L-}\Bigl{(}\hat{\Xi}^{\dagger
R}\hat{X}^{R1}\hat{\Xi}^{R}-\frac{i}{2}(\hat{\Xi}^{\dagger
R}\hat{X}^{R+}-\hat{X}^{R+}\hat{\Xi}^{R})\,\Bigr{)}$
$\displaystyle-\hat{X}^{L1}\Bigl{(}\hat{\Xi}^{\dagger
R}\hat{X}^{R-}\hat{\Xi}^{R}+\hat{X}^{R+}\Bigr{)}\;.$

Some algebra shows that this is zero. [In fact, all three lines separately
vanish. For this we need to use (B.7) and (B.20).] The same result follows for
$M_{22}$, which is expanded as

$\displaystyle M_{22}$ $\displaystyle=$
$\displaystyle\;\;\hat{X}^{L+}\Bigl{(}-\hat{X}^{R1}+\frac{i}{2}(\hat{\Xi}^{R}\hat{X}^{R-}-\hat{X}^{R-}\hat{\Xi}^{\dagger
R})\,\Bigr{)}$ (B.40)
$\displaystyle-\hat{X}^{L-}\Bigl{(}\hat{\Xi}^{R}\hat{X}^{R1}\hat{\Xi}^{\dagger
R}+\frac{i}{2}(\hat{\Xi}^{R}\hat{X}^{R+}-\hat{X}^{R+}\hat{\Xi}^{\dagger
R})\,\Bigr{)}$
$\displaystyle+\hat{X}^{L1}\Bigl{(}\Xi^{R}\hat{X}^{R-}\hat{\Xi}^{\dagger
R}+\hat{X}^{R+}\Bigr{)}\;.$

Again, all three lines separately vanish. So as with the commutative Dirac
operator in the noncovariant basis (2.17), the diagonal matrix elements
vanish, $M_{11}=M_{22}=0$. Concerning the off-diagonal matrix elements, we
first expand $M_{21}$:

$\displaystyle M_{21}$ $\displaystyle=$
$\displaystyle\;\;\hat{X}^{L+}\Bigl{(}\hat{X}^{R1}-\frac{i}{2}(\hat{\Xi}^{R}\hat{X}^{R-}+\hat{X}^{R-}\hat{\Xi}^{R})\,\Bigr{)}$
(B.45)
$\displaystyle+\hat{X}^{L-}\Bigl{(}-\hat{\Xi}^{R}\hat{X}^{R1}\hat{\Xi}^{R}+\frac{i}{2}(\hat{\Xi}^{R}\hat{X}^{R+}+\hat{X}^{R+}\hat{\Xi}^{R})\,\Bigr{)}$
$\displaystyle+\hat{X}^{L1}\Bigl{(}\hat{\Xi}^{R}\hat{X}^{R-}\hat{\Xi}^{R}-\hat{X}^{R+}\Bigr{)}\;.$

With some work this can be simplified to

$\displaystyle M_{21}$ $\displaystyle=$
$\displaystyle\;\kappa(\alpha)\biggl{(}-i\hat{X}^{L+}-i\hat{X}^{L-}(\hat{\Xi}^{R})^{2}+2\hat{X}^{L1}\hat{\Xi}^{R}\Bigr{)}$
(B.46) $\displaystyle=$ $\displaystyle
i\kappa(\alpha)\Biggl{(}\kappa(\alpha)^{2}\hat{z}^{L}+\hat{t}^{L}(\hat{z}^{L})^{-1}\hat{t}^{L}-(\hat{z}^{L})^{-1}(\hat{\Xi}^{R})^{2}+i\Bigl{(}(\hat{z}^{L})^{-1}\hat{t}^{L}+\hat{t}^{L}(\hat{z}^{L})^{-1}\Bigr{)}\hat{\Xi}^{R}\Biggr{)}$
(B.48) $\displaystyle=$ $\displaystyle
i\kappa(\alpha)\,(\hat{z}^{L})^{-1}(\hat{\Xi}^{\dagger
L}+\hat{\Xi}^{R}-\alpha\hat{z}^{L})(\hat{\Xi}^{L}-\hat{\Xi}^{R})\;.$ (B.50)

Finally for $M_{12}$ we get

$\displaystyle M_{12}$ $\displaystyle=$
$\displaystyle\;-\hat{X}^{L+}\Bigl{(}\hat{X}^{R1}+\frac{i}{2}(\hat{\Xi}^{\dagger
R}\hat{X}^{R-}+\hat{X}^{R-}\hat{\Xi}^{\dagger R})\,\Bigr{)}$ (B.55)
$\displaystyle+\hat{X}^{L-}\Bigl{(}\hat{\Xi}^{\dagger
R}\hat{X}^{R1}\hat{\Xi}^{\dagger R}+\frac{i}{2}(\hat{\Xi}^{\dagger
R}\hat{X}^{R+}+\hat{X}^{R+}\hat{\Xi}^{\dagger R})\,\Bigr{)}$
$\displaystyle+\hat{X}^{L1}\Bigl{(}-\hat{\Xi}^{\dagger
R}\hat{X}^{R-}\hat{\Xi}^{\dagger R}+\hat{X}^{R+}\Bigr{)}\;,$

which simplifies to

$\displaystyle M_{12}$ $\displaystyle=$
$\displaystyle\kappa(\alpha)\Bigl{(}-i\hat{X}^{L+}-i\hat{X}^{L-}(\hat{\Xi}^{\dagger
R})^{2}-2\hat{X}^{L1}\hat{\Xi}^{\dagger R}\Bigr{)}$ (B.56) $\displaystyle=$
$\displaystyle
i\kappa(\alpha)\Biggl{(}\kappa(\alpha)^{2}\hat{z}^{L}+\hat{t}^{L}(\hat{z}^{L})^{-1}\hat{t}^{L}-(\hat{z}^{L})^{-1}(\hat{\Xi}^{\dagger
R})^{2}-i\Bigl{(}(\hat{z}^{L})^{-1}\hat{t}^{L}+\hat{t}^{L}(\hat{z}^{L})^{-1}\Bigr{)}\hat{\Xi}^{\dagger
R}\Biggr{)}$ (B.58) $\displaystyle=$ $\displaystyle
i\kappa(\alpha)\,(\hat{z}^{L})^{-1}(\hat{\Xi}^{\dagger
R}+\hat{\Xi}^{L}+\alpha\hat{z}^{L})(\hat{\Xi}^{\dagger L}-\hat{\Xi}^{\dagger
R})\;.$ (B.60)

So the transformed Dirac operator $\hat{D}^{(\hat{U}^{R})^{-1}}$ is

$\frac{1}{2\alpha\hat{z}^{L}\sqrt{\hat{z}^{R}}}\pmatrix{0&-(\hat{\Xi}^{\dagger
R}+\hat{\Xi}^{L}+\alpha\hat{z}^{L})(\hat{\Xi}^{\dagger L}-\hat{\Xi}^{\dagger
R})\cr(\Xi^{\dagger
L}+\hat{\Xi}^{R}-\alpha\hat{z}^{L})(\hat{\Xi}^{L}-\hat{\Xi}^{R})&0}\frac{1}{\sqrt{\hat{z}^{R}}}\;.$
(B.61)

It is the quantization of the standard Dirac operator on $EAdS_{2}$ (2.17).

When the Dirac operator (B.61) acts on a spinor
$\hat{\tilde{\Psi}}=\pmatrix{\hat{\tilde{\Psi}}_{1}\cr\hat{\tilde{\Psi}}_{2}}$
it gives

$\hat{D}^{(\hat{U}^{R})^{-1}}\hat{\tilde{\Psi}}=\frac{1}{2\alpha\hat{z}}\pmatrix{-[\hat{\Xi}^{\dagger},\hat{\tilde{\Psi}}_{2}\frac{1}{\sqrt{\hat{z}}}]\,\hat{\Xi}^{\dagger}-(\hat{\Xi}+\alpha\hat{z})\,[\hat{\Xi}^{\dagger},\hat{\tilde{\Psi}}_{2}\frac{1}{\sqrt{\hat{z}}}]\cr\;[\hat{\Xi},\hat{\tilde{\Psi}}_{1}\frac{1}{\sqrt{\hat{z}}}]\,\hat{\Xi}\;+\,(\hat{\Xi}^{\dagger}-\alpha\hat{z})\,[\hat{\Xi},\hat{\tilde{\Psi}}_{1}\frac{1}{\sqrt{\hat{z}}}]}\frac{1}{\sqrt{\hat{z}}}\;.$
(B.62)

For the case of massless spinors, the Dirac equation implies that
$[\hat{\Xi},\hat{\tilde{\Psi}}_{1}\frac{1}{\sqrt{\hat{z}}}]=[\hat{\Xi}^{\dagger},\hat{\tilde{\Psi}}_{2}\frac{1}{\sqrt{\hat{z}}}]=0$,
and therefore we get the simple solution

$\hat{\tilde{\Psi}}=\pmatrix{F(\hat{\Xi})\cr
G(\hat{\Xi}^{\dagger})}\sqrt{\hat{z}}\;,$ (B.63)

yielding the quantum version of (2.19). Once again, $\hat{F}$ and $\hat{G}$
are arbitrary elements of the polynomial algebras generated by $\hat{\Xi}$ and
$\hat{\Xi}^{\dagger}$, respectively. Finally, if we act on the solution (B.63)
with $\hat{U}$, we get

$\displaystyle\hat{\Psi}=\hat{U}^{R}\hat{\tilde{\Psi}}$ $\displaystyle=$
$\displaystyle\frac{1}{2\sqrt{\kappa(\alpha)}}\,\pmatrix{-\hat{\Xi}^{R}+1&-\hat{\Xi}^{\dagger
R}-1\cr\hat{\Xi}^{R}+1&\hat{\Xi}^{\dagger R}-1}\pmatrix{F(\hat{\Xi})\cr
G(\hat{\Xi}^{\dagger})}$ (B.64) $\displaystyle=$
$\displaystyle\frac{1}{2\sqrt{\kappa(\alpha)}}\,\pmatrix{(1-\hat{\Xi})F(\hat{\Xi})-(1+\hat{\Xi}^{\dagger})G(\hat{\Xi}^{\dagger})\cr(1+\hat{\Xi})F(\hat{\Xi})+(1-\hat{\Xi}^{\dagger})G(\hat{\Xi}^{\dagger})}\;,$
(B.66)

which agrees with the exact solution (3.46).

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