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In this answer to a question, it is mentioned that in the AdS/CFT correspondence, on-shell amplitudes on the AdS side are related to off-shell correlators on the CFT side.

Can somebody explain this to me in some more (technical) details, maybe by an explanatory example?

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3 Answers 3

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First, a reference article, by Witten, http://arxiv.org/abs/hep-th/9802150v2.pdf

I'll try to expose the basic idea, with a flat space-time. Suppose you have a relativistic scalar field theory, on a flat space-time domain, with boundary. The equation of the field is :

$$\square \Phi(x) = 0$$ (fields on-shell)

Now, define the partition function

$$Z = e^{−S(\Phi)}$$, where $$S(\Phi) = \int d^nx \,\partial_i \Phi(x)\,\partial^i \Phi(x)$$ is the action for the field $\Phi$

After this, you make a integration by parts (using the above fied equation) , and Stokes theorem, and you get:

$$S(\Phi) = \int d^nx \, \partial_i \Phi(x)\,\partial^i \Phi(x) = \int d^nx \,\partial_i(\Phi(x)\,\partial^i \Phi(x))$$ $$= \int_{Boundary} d \sigma_i \,(\Phi(x)\,\partial^i \Phi(x)) $$

Now, suppose that the field $\Phi(x)$ has the value $\Phi_0(x)$ on the boundary. Then, you can see that $S$ and $Z$ could be considered as functionals of $\Phi_0$, so we could write $Z(\Phi_0)$:

$$ Z(\Phi_0) = e^{ \,( -\int_{Boundary} d \sigma_i \,(\Phi(x)\,\partial^i \Phi(x)))}$$

Now, the true calculus is not with flat space-time, but with Ads or euclidean Ads,so in your calculus, you must involve the correct metrics, but the idea is the same.

The last step is to say that there is a relation between, the Generating function of correlation functions of CFT operators $O(x)$ living on the boundary, and the partition function $Z$

$$<e^{\int_{Boundary} \Phi_0(x) O(x)}>_{CFT} = Z(\Phi_0)$$

The RHS and LHS terms of this equation should be seen as functionals of $\Phi_0$ You can make a development of these terms in powers of $\Phi_0$, and so you got all the correlations functions for the CFT operators $O(x)$

$$<O(x_1)O(x_2)...O(x_n)> \sim \frac{\partial^n Z}{\partial \Phi_0(X_1)\partial \Phi_0(X_2)...\partial \Phi_0(X_n)}$$

So, ADS side, we are using on-shell partitions functions (because field equations for $\Phi$ are satisfied)

Now, CFT/QFT side, the correlations functions $<O(x_1)O(x_2)...>$ are, by definition, off-shell correlation functions (by Fourier transforms, there is no constraint about momentum). To get scattering amplitudes, we simply need to put the external legs on-shell.

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    $\begingroup$ Comment to the answer (v1): In the future, please link to abstract page rather than pdf file if possible, e.g. arxiv.org/abs/hep-th/9802150 $\endgroup$
    – Qmechanic
    Commented May 9, 2013 at 10:50
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The statement can be understood in terms of the GKPW-formula (named after Gubser, Klebanov, Polyakov and Witten), which does exactly that: it relates correlation functions on the CFT side (boundary) to string amplitudes on the AdS side (bulk). Assume that $\phi(\vec{x},z)$ is some field in the bulk, where $z$ is the so-called "holographic coordinate", which reaches its CFT-value $\phi_0$ at the boundary at $z=0$. For some Operator $\mathcal{O}(\vec{x})$, the GKPW-formula is given by $$\langle e^{\int d^4x\,\phi_0(\vec{x})\mathcal{O}(\vec{x})}\rangle_{CFT}=\mathcal{Z}(\phi(\vec{x},z)|_{z=0}=\phi_0(\vec{x})).$$

On the left hand side, you have the correlation function for the CFT and on the right hand side, you find the partition function of the corresponding string theory, evaluated for the boundary value of the field.

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  • $\begingroup$ Brunner Is there a typo in your 2nd RHS? Shouldn't that $\phi_0 (\vec{x})$ be in the subscript giving the boundary condition for the bulk Z? $\endgroup$
    – user6818
    Commented May 6, 2013 at 18:15
  • $\begingroup$ It is intended as it is, but one could also write it your way. $\endgroup$ Commented May 6, 2013 at 20:10
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TL;DR: The (asymptotically$^1$) AdS/CT correspondence is$^2$ $$\begin{align} Z_{CT}[J] \quad\leftrightarrow&\quad Z_{AdS}[J],\cr W^c_{CT}[J] \quad\leftrightarrow&\quad W^c_{AdS}[J],\cr J(x) \quad\leftrightarrow&\quad \lim_{\epsilon_{UV}\to 0}\frac{\Phi(x,\epsilon_{UV})}{\left(\frac{\epsilon_{UV}}{R}\right)^{\Delta_-}}, \cr \langle {\cal O}(x)\rangle_J ~=~ -2\nu O(x)\quad\leftrightarrow&\quad -2\nu\lim_{z\to 0}\frac{\langle\Phi(x,z) \rangle_J}{\left(\frac{z}{R}\right)^{\Delta_+}} ~=~ \frac{\delta W_{AdS}^c[J]}{\delta J(x)}, \cr \langle {\cal O}(x_1)\ldots {\cal O}(x_n)\rangle_J^c \quad\leftrightarrow&\quad (-2\nu)^n\lim_{z_1\to 0}\ldots\lim_{z_n\to 0}\frac{\langle\Phi(x_1,z_1) \ldots \Phi(x_n,z_n)\rangle_J^c}{\left(\frac{z_1}{R}\right)^{\Delta_+}\ldots \left(\frac{z_n}{R}\right)^{\Delta_+}}\cr \|\qquad\qquad\qquad &\qquad\qquad\qquad \| \cr -\hbar^{n-1} \frac{\delta^n W_{CT}^c[J]}{\delta J(x_1)\ldots\delta J(x_n)} \quad\leftrightarrow&\quad -\hbar^{n-1} \frac{\delta^n W_{AdS}^c[J]}{\delta J(x_1)\ldots\delta J(x_n)} ,\end{align}\tag{1}$$ cf. the Gubser-Klebanov-Polyakov-Witten (GKPW) dictionary.

  • The IR$^3$ volume divergence in AdS as $z\to 0$ corresponds in principle to a UV singularity in the conformal theory (CT), so both sides are renormalized.

  • When $g_s\to0$ and $\alpha^{\prime}\to 0$ the (rescaled effective) Plank constant $\hbar\to 0$ goes to zero, so that the semi-classical WKB expansion with an on-shell action applies to the string/gravity/$AdS$ side.

Now let us give more details.

I) The conformal theory (CT) has a $N\times N$ matrix-valued field $M(x)$. Let ${\cal O}(x)={\rm tr}f(M)$ denote a local single-trace operator with conformal dimension $\Delta_+$. The (Euclidean) CT partition function is $$\begin{align} \exp\left[-\frac{1}{\hbar}W^c_{CT}[J]\right]~=~&Z_{CT}[J]\cr ~=~\int \!{\cal D}\frac{M}{\sqrt{\hbar}}~&\exp\left[-\frac{1}{\hbar}S_{CT}[M] +\frac{1}{\hbar}\int\!d^dx~J(x){\cal O}(x) \right].\end{align}\tag{2}$$

II) On the (Euclidean) $AdS_{d+1}$ side, we will use Poincare coordinates $(x^{\mu},z)$ where the metric tensor components $$\begin{align} ds^2 ~=~& R^2\frac{dz^2 + dx_{\mu}dx^{\mu}}{z^2},\cr g_{MN}~=~&\frac{R^2}{z^2}\delta_{MN}, \end{align}\tag{3}$$ do not depend explicitly on the $x^{\mu}$ coordinates. The conformal boundary corresponds to $z=0$ and $z=\infty$, cf. e.g. my Phys.SE answer here. Let us for simplicity only consider a single scalar field $(x,z)\mapsto\Phi(x,z)$, and that the (Euclidean) action does not depend explicitly on the $x^{\mu}$ coordinates. The (Euclidean) path integral is according to the Witten prescription $$\begin{align}\exp\left[-\frac{1}{\hbar}W^c_{AdS}[J]\right]~=~& Z_{AdS}[J]~=~\langle {\rm finite} ,\infty \mid \left(\frac{\epsilon_{UV}}{R}\right)^{\Delta_-}J ,\epsilon_{UV} \rangle^{\rm ren}\cr~ \stackrel{\rm WKB}{\sim}~&\exp\left[-\frac{1}{\hbar}S_{\text{on-shell}}^{\rm ren}[{\rm finite} ,\infty ; \left(\frac{\epsilon_{UV}}{R}\right)^{\Delta_-}J ,\epsilon_{UV}]\right]\cr &\quad{\rm for}\quad\hbar~\to~0,\end{align} \tag{4}$$ where the (renormalized) bra-ket is$^4$ $$\begin{align} \langle \phi_{IR} ,\epsilon_{IR} \mid \phi_{UV} ,\epsilon_{UV} \rangle^{\rm ren} ~:=~& \int_{\Phi(\cdot,\epsilon_{UV})=\phi_{UV}}^{\Phi(\cdot,\epsilon_{IR})=\phi_{IR}} \!{\cal D}\frac{\Phi}{\sqrt{\hbar}}~\exp\left[-\frac{1}{\hbar}S^{\rm ren}[\Phi;\epsilon_{UV},\epsilon_{IR}]\right]\cr ~\stackrel{\rm WKB}{\sim}~&\exp\left[-\frac{1}{\hbar}S_{\text{on-shell}}^{\rm ren}[\phi_{IR} ,\epsilon_{IR} ; \phi_{UV} ,\epsilon_{UV}]\right]\cr &\quad{\rm for}\quad\hbar~\to~0,\cr S^{\rm ren}[\Phi;\epsilon_{UV},\epsilon_{IR}]~=~&S[\Phi;\epsilon_{UV},\epsilon_{IR}]+S^{\rm ct}[\Phi(\cdot,\epsilon_{UV})],\cr S[\Phi;\epsilon_{UV},\epsilon_{IR}]~=~&\int_{\epsilon_{UV}}^{\epsilon_{IR}}\!dz~ L[\Phi(\cdot,z)],\cr 0~<~&\epsilon_{UV} ~<~\epsilon_{IR}, \end{align}\tag{5}$$ and where the counterterm (ct) should be an $x$-local functional.

  1. Let $(x,z)\mapsto\Phi_0(x,z)$ denote the solution to the (possibly non-linear) Euler-Lagrange (EL) equation with Dirichlet boundary conditions (DBC) $$\begin{align} \Phi(x,\epsilon_{UV})~=~\phi_{UV}(x)\quad\text{and}&\quad\Phi(x,\epsilon_{IR})~=~\phi_{IR}(x)\cr \quad\text{where}&\quad 0~<~\epsilon_{UV} ~<~\epsilon_{IR}.\end{align}\tag{6}$$

  2. Let $(x,z)\mapsto\Phi_{00}(x,z)$ denote the solution to the linearized EL equation, i.e. without interaction terms/coupling constants $g_i=0$ are zero.

  3. By a possible field redefinition, we may assume that the linearized EL equation is homogeneous $$\left( \frac{1}{\sqrt{|g|}}\partial_M \sqrt{|g|}g^{MN}\partial_N -m^2 \right)\Phi(x,z)~=~0.\tag{7}$$

  4. We assume that the full solution $$\Phi_0(x,z)~=~\Phi_{00}(x,z)+g_i\underbrace{\Phi_{0i}(x,z)}_{={\cal O}(\Phi_{00})}+\ldots\tag{8}$$ is perturbative in the coupling constants $g_i$ and a solution $\Phi_{00}(x,z)$ to the linearized problem (7).

  5. If a boundary profile $\phi$ is infinitesimal$^2$, then interaction terms with higher-order powers of $\Phi$ are suppressed, so that we expect that $$ \Phi_0(x,z)~\approx~\Phi_{00}(x,z) \quad\text{for}\quad z~\approx~\epsilon,\tag{9}$$ i.e. that the full solution $\Phi_0(x,z)$ is given by the linearized solution $\Phi_{00}(x,z)$ near the boundary.

  6. Since there are $x^{\mu}$-derivatives but no explicit $x^{\mu}$ coordinate dependence, it is convenient to partially Fourier transform $x^{\mu}\leftrightarrow k_{\mu}$. Then the linearized EL equation becomes a 2nd-order ODE in the $z$-coordinate, $$\left( z^{1+d}\partial_z z^{1-d}\partial_z -z^2k^2 -m^2R^2\right)\widetilde{\Phi}(k,z)~=~0,\tag{10}$$ with 2 linearly independent solutions $$\begin{align} \widetilde{\Phi}_{00}(k,z)~=~&\widetilde{J}(k)\widetilde{\Phi}_-(k,z)+\widetilde{O}(k)\widetilde{\Phi}_+(k,z),\cr \widetilde{\Phi}_{00}(k,z)~=~&\widetilde{\phi}_{UV}(k)\frac{\widetilde{\Phi}_-(k,z)\widetilde{\Phi}_+(k,\epsilon_{IR})-\widetilde{\Phi}_+(k,z)\widetilde{\Phi}_-(k,\epsilon_{IR})}{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,\epsilon_{IR})-\widetilde{\Phi}_+(k,\epsilon_{UV})\widetilde{\Phi}_-(k,\epsilon_{IR})}\cr ~+~&\widetilde{\phi}_{IR}(k)\frac{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,z)-\widetilde{\Phi}_+(k,\epsilon_{UV})\widetilde{\Phi}_-(k,z)}{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,\epsilon_{IR})-\widetilde{\Phi}_+(k,\epsilon_{UV})\widetilde{\Phi}_-(k,\epsilon_{IR})},\cr \Phi_-(x,z)~\sim~&\left(\frac{z}{R}\right)^{\Delta_-}\quad\text{for}\quad z~\ll ~R, \cr \Phi_+(x,z)~\sim~&\left(\frac{z}{R}\right)^{\Delta_+}\quad\text{for}\quad z~\ll ~R.\cr \Delta_{\pm} ~=~& \frac{d}{2}\pm\nu, \cr \nu~:=~& \sqrt{\frac{d^2}{4}+(Rm)^2}~\geq~0,\cr \Delta_-~<~&\frac{d}{2}~<~\Delta_+. \end{align} \tag{11}$$ Here $$ \begin{matrix} \text{irrelevant}\cr \text{marginal}\cr \text{relevant}\end{matrix}\quad \Delta_+\begin{Bmatrix} >\cr =\cr <\end{Bmatrix}d \quad\Leftrightarrow\quad m^2\begin{Bmatrix} >\cr =\cr <\end{Bmatrix}0. \tag{12}$$

  7. Note that $$ |\Phi_-(x,z)| ~\gg~|\Phi_+(x,z)| \quad\text{for}\quad z~\ll ~R. \tag{13}$$

  8. In other words, there is a bijective map $J\leftrightarrow \Phi_0$ between infinitesimal sources $J(x)$ and infinitesimal solutions $\Phi_0(x,z)$ near the boundary $z\ll R$.

III) For $\epsilon_{UV}\ll\epsilon_{IR}\ll R$, and an infinitesimal source, then interaction terms with higher-order powers of $\Phi$ are suppressed, so that we can replace the action $S$ with the free quadratic action $$\begin{align} S_2[\Phi;\epsilon_{UV},\epsilon_{IR}]~=~&\int_{\epsilon_{UV}}^{\epsilon_{IR}}\!dz~ L_2[\Phi(\cdot,z)],\cr 2L_2[\Phi(\cdot,z)] ~=~&\int d^dx\sqrt{|g|}\left(\partial_M\Phi(x,z)g^{MN}\partial_N\Phi(x,z) +m^2\Phi(x,z)^2 \right) \cr ~=~&\left(\frac{R}{z}\right)^{d-1}\int\! d^dx \cr &\{ (\partial_z\Phi(x,z))^2 +\partial_{\mu}\Phi(x,z)\partial^{\mu}\Phi(x,z)\cr &+\left(\frac{Rm}{z}\right)^2\Phi(x,z)^2 \} \cr ~=~&\left(\frac{R}{z}\right)^{d-1}\int \frac{d^dk}{(2\pi)^d}\cr &\{\partial_z\widetilde{\Phi}(k,z)\partial_z\widetilde{\Phi}(-k,z) \cr &+\left(k^2+\frac{R^2m^2}{z^2}\right)\widetilde{\Phi}(k,z)\widetilde{\Phi}(-k,z)\},\cr 2S_{2,\text{on-shell}}&[\phi_{IR} ,\epsilon_{IR} ; \phi_{UV} ,\epsilon_{UV}]\cr ~=~&2S_2[\Phi_{00};\epsilon_{UV},\epsilon_{IR}]\cr ~\stackrel{\text{IPB}}{=}~&\int\!d^dx\int_{\epsilon_{UV}}^{\epsilon_{IR}}\!dz~ \partial_z\left[\Phi_{00}(x,z)\left(\frac{R}{z}\right)^{d-1}\partial_z\Phi_{00}(x,z)\right],\cr ~=~&\int\frac{d^dk}{(2\pi)^d}\int_{\epsilon_{UV}}^{\epsilon_{IR}}\!dz~ \partial_z\left[\widetilde{\Phi}_{00}(-k,z)\left(\frac{R}{z}\right)^{d-1}\partial_z\widetilde{\Phi}_{00}(k,z)\right],\cr ~\approx~&\int\frac{d^dk}{(2\pi)^d}\widetilde{\phi}_{UV}(-k)\left(\frac{R}{z}\right)^{d-1}\cr &\left(\widetilde{\phi}_{UV}(k)\frac{\partial_z\widetilde{\Phi}_-(k,z)\widetilde{\Phi}_+(k,\epsilon_{IR})-\partial_z\widetilde{\Phi}_+(k,z)\widetilde{\Phi}_-(k,\epsilon_{IR})}{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,\epsilon_{IR})}\right. \cr ~+~&\left.\widetilde{\phi}_{IR}(k)\frac{\widetilde{\Phi}_-(k,\epsilon_{UV})\partial_z\widetilde{\Phi}_+(k,z)-\widetilde{\Phi}_+(k,\epsilon_{UV})\partial_z\widetilde{\Phi}_-(k,z)}{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,\epsilon_{IR})}\right)_{z=\epsilon_{UV}},\cr ~-~&\int\frac{d^dk}{(2\pi)^d}\widetilde{\phi}_{IR}(-k)\left(\frac{R}{z}\right)^{d-1}\cr &\left(\widetilde{\phi}_{UV}(k)\frac{\partial_z\widetilde{\Phi}_-(k,z)\widetilde{\Phi}_+(k,\epsilon_{IR})-\partial_z\widetilde{\Phi}_+(k,z)\widetilde{\Phi}_-(k,\epsilon_{IR})}{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,\epsilon_{IR})}\right. \cr ~+~&\left.\widetilde{\phi}_{IR}(k)\frac{\widetilde{\Phi}_-(k,\epsilon_{UV})\partial_z\widetilde{\Phi}_+(k,z)-\widetilde{\Phi}_+(k,\epsilon_{UV})\partial_z\widetilde{\Phi}_-(k,z)}{\widetilde{\Phi}_-(k,\epsilon_{UV})\widetilde{\Phi}_+(k,\epsilon_{IR})}\right)_{z=\epsilon_{IR}}\cr ~\approx~&4\nu\int\frac{d^dk}{(2\pi)^d}\underbrace{\frac{\widetilde{\phi}_{UV}(k)}{\left(\frac{\epsilon_{UV}}{R}\right)^{\Delta_-}}}_{=\widetilde{J}(k)} \underbrace{\frac{\widetilde{\phi}_{IR}(-k)}{\left(\frac{\epsilon_{IR}}{R}\right)^{\Delta_+}}}_{=\widetilde{O}(-k)} +{\cal O}(\phi_{UV}^2) +{\cal O}(\phi_{IR}^2),\cr 2S^{\rm ct}[\phi_{UV}]~=~&\left.\left(\frac{R}{z}\right)^{d-1}\int \frac{d^dk}{(2\pi)^d} \frac{\partial_z\widetilde{\Phi}_-(k,z)}{\widetilde{\Phi}_-(k,z)} \widetilde{\phi}_{UV}(k)\widetilde{\phi}_{UV}(-k) \right|_{z=\epsilon_{UV}}, \end{align}\tag{14}$$

References:

  1. E. Witten, Anti De Sitter Space And Holography, arXiv:hep-th/9802150; p.23.

  2. J. Kaplan, 2013 Lectures on AdS/CFT from the Bottom Up; section 12.1. (NB: Some eqs. in the PDF file are corrupted by some left parentheses.)

  3. A. Zaffaroni, 2009 LACES lectures on AdS/CFT; sections 2.2 + 2.3.

  4. A.V. Ramallo, Intro to the AdS/CFT correspondence, arXiv:1310.4319; chap. 8 + 9.

  5. H. Liu, String Theory and Holographic Duality; MIT 2014 lectures 20 + 21.

  6. D. Harlow & D. Stanford, Operator Dictionaries and Wave Functions in AdS/CFT and dS/CFT, arXiv:1104.2621; chapter 2.

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$^1$ Let us for simplicity consider $AdS$ rather than asymptotically $AdS$. The conformal symmetry is broken away from fixed-points of the RG flow. Ref. 2 defines a conformal field theory (CFT) as a conformal theory (CT) with local SEM tensor.

$^2$ Sources $J$ are always assumed to be infinitesimal, i.e. much smaller than any characteristic scale of the problem.

$^3$ NB: Be aware that the words UV and IR will from now on align with the CT side (as opposed to the $AdS$ side), so that e.g. $z\to 0$ is from now on referred to as the UV.

$^4$A Wilsonian RG flow can be implemented via a (semi)group property $$\langle \phi_3 ,\epsilon_3 | \phi_1 ,\epsilon_1 \rangle^{\rm ren} ~=~\int \! {\cal D}\frac{\phi_2}{\sqrt{\hbar}} \langle \phi_3 ,\epsilon_3 | \phi_2 ,\epsilon_2 \rangle\langle \phi_2 ,\epsilon_2 | \phi_1 ,\epsilon_1 \rangle^{\rm ren}, \tag{15}$$ where the (unrenormalized) bra-ket is $$\begin{align} \langle \phi_2 ,\epsilon_2 \mid \phi_1 ,\epsilon_1 \rangle ~:=~&\int_{\Phi(\cdot,\epsilon_1)=\phi_1}^{\Phi(\cdot,\epsilon_2)=\phi_2} \!{\cal D}\frac{\Phi}{\sqrt{\hbar}}~\exp\left[-\frac{1}{\hbar}\int_{\epsilon_1}^{\epsilon_2}\!dz~ L[\Phi(\cdot,z)]\right] ,\cr 0~<~&\epsilon_1 ~<~\epsilon_2. \end{align}\tag{16}$$

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  • $\begingroup$ Notes for later: Bulk-boundary propagator: $\quad K_{\nu}(x,z)=\frac{\Gamma(\frac{d}{2}+\nu)}{\pi^{d/2}\Gamma(\nu)}\left(\frac{z}{x^2+z^2}\right)^{d/2+\nu}$, cf. physics.stackexchange.com/q/255257 Kernel satisfies modified Bessel DE. Fourier transform: $\endgroup$
    – Qmechanic
    Commented Mar 4 at 8:33
  • $\begingroup$ $\quad\widetilde{K}_{\nu}(k,z)=\int_{\mathbb{R}^d}\!d^dx~e^{-ik\cdot x}K_{\nu}(x,z)$ $=\frac{z^{d/2+\nu}}{\pi^{d/2}\Gamma(\nu)}\int_{\mathbb{R}^d}\!d^dx\int_{\mathbb{R}_+}\!d\alpha~\alpha^{d/2+\nu-1}e^{-ik\cdot x-\alpha(x^2+z^2)}$ $=\frac{z^{d/2+\nu}}{\Gamma(\nu)}\int_{\mathbb{R}_+}\!\frac{d\alpha}{\alpha}\alpha^{\nu}e^{-\frac{k^2}{4\alpha}-\alpha z^2}$ $=\frac{z^{d/2+\nu}}{\Gamma(\nu)}\int_{\mathbb{R}_+}\!\frac{d\alpha}{\alpha}\alpha^{-\nu}e^{-\frac{\alpha k^2}{4}-\frac{z^2}{\alpha}}$ $=\frac{2z^{d/2}}{\Gamma(\nu)}\left(\frac{k}{2}\right)^{\nu}K_{\nu}(kz)$ $\sim z^{d/2-\nu}$ for $z\to 0$. $\endgroup$
    – Qmechanic
    Commented Mar 4 at 11:37
  • $\begingroup$ Hypergeometric function: $\quad{}_2F_1(a,b;c;z):=\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$ $=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\int_{i\mathbb{R}}\!\frac{ds}{2\pi i}\frac{\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)}\Gamma(-s)(-z)^s$ where $-z\notin\mathbb{R}_+\cup\{0\}$. The Mellin-Barnes integral representation seems to ignore all other contributions than residues of $\Gamma(-s)$. $\quad (a)_n:=\frac{\Gamma(a+n)}{\Gamma(a)}$. $\quad{\rm Res}(\Gamma,-n)=\frac{(-1)^n}{n!}=-{\rm Res}(s\mapsto\Gamma(-s),n)$ for $n\in\mathbb{N}_0$. $\endgroup$
    – Qmechanic
    Commented Mar 5 at 11:54
  • $\begingroup$ Mellin transform:$\quad({\cal M}f)(s)=\int_{\mathbb{R}_+}\!dx~x^{s-1}f(x)$. $\quad({\cal M}^{-1}f)(x)=\int_{i\mathbb{R}+c}\frac{ds}{2\pi i}x^{-s}f(s)$. Related to two-sided/bilateral Laplace transform. $\quad({\cal M}e^{-(\cdot)})(s)=\Gamma(s)$. $\quad e^{-z}=\oint_{\mathbb{R}_-}\!\frac{ds}{2\pi i}\Gamma(s)z^{-s}$. dlmf.nist.gov/15.8 Bulk-bulk propagator: $\quad G(x,z;x^{\prime},z^{\prime})=\frac{\Gamma(\Delta)}{2\pi^{d/2}\Gamma(\nu+1)}(2\xi)^{-\Delta}{}_2F_1(a,b;c;\xi^{-2})$, $\endgroup$
    – Qmechanic
    Commented Mar 6 at 11:08
  • $\begingroup$ where $\xi:=\frac{(x-x^{\prime})^2+z^2+z^{\prime 2}}{2zz^{\prime}}=\xi_x+\xi_z=2\zeta+1$, $a:=\frac{\Delta}{2}$, $b:=\frac{\Delta+1}{2}$, $c:=\nu\!+\!1$. Fourier transform: $\quad\widetilde{G}(k,z;z^{\prime}) =\int_{\mathbb{R}^d}\!d^d(x\!-\!x^{\prime})~e^{-ik\cdot(x-x^{\prime})}G(x,z;x^{\prime},z^{\prime})$ $=\frac{\Gamma(\Delta)}{2^{\Delta+1}\Gamma(\nu+1)}\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\frac{(2zz^{\prime})^{\Delta+2n}}{n!\Gamma(\Delta+2n)}\int_{\mathbb{R}_+}\!d\alpha~\alpha^{\nu+2n-1}e^{-\frac{k^2}{4\alpha}-\alpha(z^2+z^{\prime 2})}$. $\endgroup$
    – Qmechanic
    Commented Mar 7 at 8:00

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