Holographic Fisher information metric in Schrödinger spacetime

Abstract

In this paper, we study the Fisher information metric on the space of the coupling constants on both sides of the duality between non-relativistic dipole field theories and string theory in Schrödinger spacetime. We consider the following setup. In the gauge theory side one can deform a given conformal field theory by a proper scalar operator and compute the quantum information metric via the two-point correlation function between two such operators. On the string side the deformation corresponds to a scalar field probing the background. In the large N limit of the theory the probing can be done without backreaction on the original spacetime, thus one can construct a perturbative scheme for the calculation of the dual holographic Fisher information metric as shown by [1]. Considering the asymptotic behaviour of the holographic Fisher information metric close to the boundary of the Schrödinger spacetime we show that its divergence structure exactly matches its dual quantum counterpart up to the leading order, thus extending the holographic setup up to the non-relativistic case. One should note that the existence of other terms is not seen from the boundary theory to this level of approximation. Their behaviour near the boundary, however, is pointing what kind of information from the boundary theory is missing to be able to reconstruct the bulk. Obviously, more work is needed to refine and elucidate their meaning and interrelations in holographic setup.

Appendices

A short note on how to generate Schrödinger spacetimes and their dual CFT theories

The symmetry of the free Schrödinger equation

$$\begin{aligned} \frac{\partial ^2}{\partial \mathbf {r}^{\,2}}\phi -2im \frac{\partial }{\partial t}\phi =0, \end{aligned}$$

(A.1)

is the so-called Schrödinger group. In n-dimensional spacetime the group consists of spatial translations indicated by $\mathbf {A}$, rotations given by the matrix $\Omega$, and Galilean boosts with velocity $\mathbf {v}$,

$$\begin{aligned} t\,\rightarrow \, t'=\frac{at+b}{ct+d}, \qquad \mathbf {r}\,\rightarrow \, \mathbf {r}\,'=\frac{\Omega \mathbf {r} + \mathbf {v}t + \mathbf {A}}{ct+d},\quad ad-bc=1. \end{aligned}$$

(A.2)

In addition, one has dilatation, where time and space scale differently

$$\begin{aligned} t\rightarrow \lambda ^2 t,\quad \mathbf {r}\rightarrow \lambda \mathbf {r}, \end{aligned}$$

(A.3)

and one additional special conformal transformation

$$\begin{aligned} t\rightarrow \frac{t}{1+\lambda t},\quad \mathbf {r} \rightarrow \frac{\mathbf {r}}{1+\lambda t}. \end{aligned}$$

(A.4)

From group theory point of view, the Schrödinger group can be thought of as a non-relativistic analogue of the conformal group. In fact, the Schrödinger group can be embedded into the relativistic conformal group $SO(2,n+2)$ in $n+1$ dimensions [3, 4, 66], as well as a particular contraction of the conformal group.

For purposes of the holographic correspondence it is important to consider spaces with the Schrödinger group being the maximal group of isometries.¹² Such spaces are called Schrödinger spaces. There is a specific way to obtain the spacetime geometry equipped with this symmetry via TsT (T-duality-shift-T-duality) transformations.

Our starting point is the AdS metric, which is invariant under the whole conformal group and deform it to reduce the symmetry down to the Schrödinger group.

Being a particular case of the so-called Drinfel’d-Reshetikhin twist, the TsT procedure has been used for generating many backgrounds keeping partial or full integrability of the system. Specific point in generating Schrödinger backgrounds via TsT transformations is to include one of the light-cone variables. This particular deformation is also known as the null-Melvin twist. This procedure can be implemented by the following key steps:

• Represent the theory in light-cone coordinates and identify a Killing direction, say $\psi$,

• Perform a T-duality along the chosen Killing direction $\psi$,

• Boost the geometry in the Killing direction by $\hat{\mu }$, i.e. $x^- \rightarrow x^- - \hat{\mu } \tilde{\psi }$, where $\tilde{\psi }$ is the T-dualized coordinate $\psi$,

• Finally, perform a T-duality back to IIA/IIB along ${\tilde{\psi }}$.

In order to accomplish the desired result consider general background in the form $AdS_n\times X^m$,

$$\begin{aligned} ds^2=\ell ^2\,\frac{2dx^+dx^-+dx^idx_i +dz^2}{z^2}+ds^2_{X^m}, \end{aligned}$$

(A.5)

Now we perform a null Melvin twist along a Killing vector ${\mathcal {K}}$ on $X^m$. The result is

$$\begin{aligned} ds^2=ds^2_{Schr_n}+ds^2_{X^m}, \end{aligned}$$

(A.6)

where the Schrödinger metric yields

$$\begin{aligned} ds^2_{Schr_n}=-\frac{\Omega }{z^4}+\frac{1}{z^2}(2dx^{+}dx^{-}+d\mathbf {x}^2+dz^2), \quad \Omega = \left\| {\mathcal {K}}\right\| ^2=*({\mathcal {K}} \wedge *{\mathcal {K}}). \end{aligned}$$

(A.7)

Here the Hodge star operator $*$ is taken with respect to the metric on $X^m$. It is clear that $\Omega$ is nonnegative being a square length of a Killing vector.¹³ For particular examples see [20, 21, 67,68,69].¹⁴

An important remark is that in order to make holographic sense of these solutions one has to impose some conditions. In particular for these to be holographic duals to non-relativistic field theories the light-cone coordinate $x^-$ should be periodic, $x^-\sim {x^-} +2\pi r_{x^-}$ [3, 4, 19]. The momentum along this compact direction is quantized in units of the inverse radius $r_{x^-}^{-1}$.

The dual theory is conjectured to be a specific field theory, namely dipole field theory. The TsT transformation, which produces the bulk theory corresponds to a particular deformation (Drinfel’d-Reshetikhin twist) on field theory side. It translates to the dual theory as a star product. When the directions involved in TsT are transverse to the stack of branes from which geometry descends, it produces a twist in the field theory having ordinary product. If however, one of the directions is along the branes, then the star product is non-trivial and the theory becomes dipole one

$$\begin{aligned} (\Phi _1\star \Phi _2)(x)= \Phi _1(x+L_1)\Phi _2(x-L_2), \end{aligned}$$

(A.8)

where $L=L_1+L_2$ is the dipole length associated with R-charges. The big advantage of holographic model with Schrödinger symmetry is that they could be integrable. Proving that in the case of $Schr_5\times S^5$ for example, the authors of [70] were able to map the composite operators of monomial form to a spin chain. To study field theory side one must just replace the ordinary product with the star one

$$\begin{aligned} {\mathcal {O}}= {\text {tr}}(\Phi _1\star \Phi _2\star \cdots ). \end{aligned}$$

(A.9)

Here, one can choose to work either using Seiberg–Witten map or working directly with the star product. A nice analysis has been presented in [70].

Derivation of the CFT quantum geometric tensor

Let $|\varphi \rangle$ be a generic state, inserted at $\tau =0$. The overlap between the original ground state $|\psi _0\rangle$ at $\tau \rightarrow -\infty$ and this new state at $\tau \rightarrow 0$ is given by

$$\begin{aligned} \langle {\tilde{\varphi }}|\psi _0 \rangle =\frac{1}{\sqrt{Z_0}}\int \limits ^{\tilde{\varphi }}{\mathcal {D}}\varphi \,e^{-\int \limits _{-\infty }^0 d\tau \int d^d x {\mathcal {L}}_0}, \end{aligned}$$

(B.1)

where ${\tilde{\varphi }}=\varphi (\tau =0)$ and the partition function of the initial undeformed theory is defined by

$$\begin{aligned} Z_0=\int {\mathcal {D}}\varphi \, e^{-\int \limits _{-\infty }^\infty d\tau \int d^d x {\mathcal {L}}_0}. \end{aligned}$$

(B.2)

In a similar fashion, one can consider the evolution from $\tau = 0$, where $|{\tilde{\varphi }}\rangle$ is inserted, to $\tau \rightarrow \infty$, where we are placing the perturbed state $|\psi _1\rangle$,

$$\begin{aligned} \langle \psi _1|{\tilde{\varphi }} \rangle =\frac{1}{\sqrt{Z_1}}\int \limits _{\tilde{\varphi }}{\mathcal {D}}\varphi \,e^{-\int \limits _{0}^\infty d\tau \int d^d x {\mathcal {L}}_1}=\frac{1}{\sqrt{Z_1}}\int \limits _{\tilde{\varphi }}{\mathcal {D}}\varphi \,e^{-\int \limits _{0}^\infty d\tau \int d^d x( {\mathcal {L}}_0+\delta \lambda ^a{\mathcal {O}}_a)}. \end{aligned}$$

(B.3)

As usually,

$$\begin{aligned} Z_1=\int {\mathcal {D}}\varphi \,e^{-\int \limits _{-\infty }^\infty d\tau \int d^d x( {\mathcal {L}}_0+\delta \lambda ^a{\mathcal {O}}_a)} \end{aligned}$$

(B.4)

is the partition function of the deformed theory. Now, the overlap between both states in the above notations can be formally written as

$$\begin{aligned} \langle \psi _1|\psi _0 \rangle =\int \limits _{\varphi (\tau =0)={\tilde{\varphi }}}{\mathcal {D}}\varphi \langle \psi _1| \varphi \rangle \langle \varphi |\psi _0 \rangle =\frac{1}{\sqrt{Z_0 Z_1}} \int {\mathcal {D}}\varphi \,e^{-\int d^d x \left[ \int \limits _{-\infty }^0 d\tau {\mathcal {L}}_0+\int \limits _{0}^\infty d\tau ( {\mathcal {L}}_0+\delta \lambda ^a{\mathcal {O}}_a)\right] }.\nonumber \\ \end{aligned}$$

(B.5)

Let us write this overlap in a more convenient way [71]

$$\begin{aligned} \langle \psi _1|\psi _0 \rangle =\frac{\left\langle \exp \left( {-\int \limits _{0}^\infty d\tau \int \limits _{V_{{\mathbb {R}}^d}} d^d x \delta \lambda ^a {\mathcal {O}}_a}\right) \right\rangle }{\sqrt{\frac{Z_1}{ Z_0}}}=\frac{\left\langle \exp \left( {-\int \limits _{0}^\infty d\tau \int \limits _{V_{{\mathbb {R}}^d}} d^d x \delta \lambda ^a {\mathcal {O}}_a}\right) \right\rangle }{\sqrt{\left\langle \exp \left( -\int \limits _{-\infty }^\infty d\tau \int \limits _{V_{{\mathbb {R}}^d}}d^d x\delta \lambda ^a\mathcal O_a\right) \right\rangle }},\nonumber \\ \end{aligned}$$

(B.6)

therefore, one has

$$\begin{aligned} |\langle \psi _1|\psi _0\rangle |^2 = \frac{{\left\langle {\exp \left( { - \int \limits _0^\infty d \tau \int \limits _{V_{{\mathbb {R}}^d}} {{d^d}} x\delta {\lambda ^a}{\mathcal{O}_a}(\tau )} \right) } \right\rangle \left\langle {\exp \left( { - \int \limits _{ - \infty }^0 d \tau \int \limits _{V_{{\mathbb {R}}^d}} {{d^d}} x\delta {\lambda ^a}{\mathcal{O}_a}(\tau )} \right) } \right\rangle }}{{\left\langle {\exp \left( { - \int \limits _{ - \infty }^\infty {d\tau \int \limits _{V_{{\mathbb {R}}^d}} {{d^d}} } x\delta {\lambda ^a}{\mathcal{O}_a(\tau )}} \right) } \right\rangle }}.\nonumber \\ \end{aligned}$$

(B.7)

We can now easily expand this expression in power series

$$\begin{aligned} |\langle \psi _1|\psi _0\rangle |^2&= 1 + \frac{1}{2}\int \limits _{V_{{\mathbb {R}}^d}} {{d^d}x_1} \int \limits _{V_{{\mathbb {R}}^d}} {{d^d}x_2} \left[ {\int \limits _0^\infty {d{\tau _1}} \int \limits _0^\infty {d{\tau _2}} \left\langle {{\mathcal{O}_a}({\tau _1}){\mathcal{O}_b}({\tau _2})} \right\rangle } \right. \nonumber \\&\quad + \int \limits _{ - \infty }^0 {d{\tau _1}} \int \limits _{ - \infty }^0 {d{\tau _2}} \left\langle {{\mathcal{O}_a}({\tau _1}){\mathcal{O}_b}({\tau _2})} \right\rangle - \int \limits _{ - \infty }^\infty {d{\tau _1}} \int \limits _{ - \infty }^\infty {d{\tau _2}} \left\langle {{\mathcal{O}_a}({\tau _1}){\mathcal{O}_b}({\tau _2})} \right\rangle \nonumber \\&\quad + \left. 2{\int \limits _0^\infty {d{\tau _1}} \int \limits _{ - \infty }^0 {d{\tau _2}} \left\langle {{\mathcal{O}_a}({\tau _1})} \right\rangle \left\langle {{\mathcal{O}_b}({\tau _2})} \right\rangle } \right] \delta {\lambda ^a}\delta {\lambda ^b} + \mathcal{O}(\delta {\lambda ^3}). \end{aligned}$$

(B.8)

Since we have time reversal symmetry of the correlator,

$$\begin{aligned} \langle {\mathcal {O}}_a(-\tau _1){\mathcal {O}}_b(-\tau _2) \rangle =\langle {\mathcal {O}}_a(\tau _1){\mathcal {O}}_b(\tau _2) \rangle , \end{aligned}$$

(B.9)

and the fact that

$$\begin{aligned} \int \limits _{ - \infty }^\infty {d{\tau _1}} \int \limits _{ - \infty }^\infty {d{\tau _2}} = \int \limits _{ - \infty }^0 {d{\tau _1}} \int \limits _{ - \infty }^0 {d{\tau _2}} + \int \limits _{ - \infty }^0 {d{\tau _1}} \int \limits _0^\infty {d{\tau _2}} + \int \limits _0^\infty {d{\tau _1}} \int \limits _{ - \infty }^0 {d{\tau _2}} + \int \limits _0^\infty {d{\tau _1}} \int \limits _0^\infty {d{\tau _2}} ,\nonumber \\ \end{aligned}$$

(B.10)

one finds

$$\begin{aligned}&|\langle \psi _1|\psi _0\rangle |^2= 1 - \frac{1}{2}\int \limits _{V_{{\mathbb {R}}^d}} {{d^d}x_1} \int \limits _{V_{{\mathbb {R}}^d}} {{d^d}x_2} \int \limits _{ - \infty }^0 {d{\tau _1}} \nonumber \\&\quad \int \limits _0^\infty {d{\tau _2}} \left( {\left\langle {{\mathcal{O}_a}({\tau _1}){\mathcal{O}_b}({\tau _2})} \right\rangle - \left\langle {{\mathcal{O}_a}({\tau _1})} \right\rangle \left\langle {{\mathcal{O}_b}({\tau _2})} \right\rangle } \right) \delta {\lambda ^a}\delta {\lambda ^b}. \end{aligned}$$

(B.11)

Using the definition of the quantum fidelity, we extract the expression for QIM from Eq. (2.2).

Computation of the dual CFT quantum Fisher metric

One has to compute the following integral expression

$$\begin{aligned} G_{\lambda \lambda }^{(CFT)} = \frac{i\Delta M^{\Delta -1} 2^{-\Delta } e^{- \frac{i\pi \Delta }{2}} }{\pi ^{d/2}\, \Gamma \left( \Delta - \frac{d}{2} - 1\right) } \int \!\!d^d x_1 \int \!\!d^d x_2 \int \limits _{ - \infty }^{ - \epsilon }\!\!d\tau _1 \int \limits _\epsilon ^\infty \!\! d\tau _2 \, \frac{1}{(\tau _2 - \tau _1)^{\Delta }} \, e^{i\frac{M(1+i \varepsilon )}{2} \frac{(\mathbf {x}_2 - \mathbf {x}_1)^2}{\tau _2 - \tau _1}} ,\nonumber \\ \end{aligned}$$

(C.1)

which is a Gaussian integral over $\mathbf {x}$ space, i.e.

$$\begin{aligned} {I_x} = \int \!\! {{d^{d }}{x_1}\int \!\! {{d^{d}}{x_2} \, {e^{i\frac{M(1+i \varepsilon )}{2}\frac{{{{({{\mathbf {x}}_2} - {{\mathbf {x}}_1})}^2}}}{{{\tau _2} - {\tau _1}}}}}} } = {e^{\frac{{i\pi }}{2}(1 - d/2)}}{(2\pi )^{\frac{{d}}{2}}}{M^{\frac{{- d}}{2}}(1+i \varepsilon )^{-\frac{d}{2}}} (\tau _2-\tau _1)^\frac{d}{2}{V_{{{\mathbb {R}}^{d}}}},\nonumber \\ \end{aligned}$$

(C.2)

where $V_{{\mathbb {R}}^{d}}$ is the volume of ${\mathbb {R}}^{d}$ space and we have resorted to the standard Gaussian integral [72]:

$$\begin{aligned} \int \! {{d^\delta }x} \, {e^{i\beta {{\mathbf {x}}^2} - 2i\mathbf {q} \cdot \mathbf {x}}} = {e^{\frac{{i\pi }}{2}\left( {1 - \frac{\delta }{2}} \right) }}{\pi ^{\delta /2}}{\beta ^{ - \delta /2}}{e^{ - i\frac{{{{\mathbf {q}}^2}}}{\beta }}}. \end{aligned}$$

(C.3)

We can check this by setting $\xi = \frac{M(1+i \varepsilon )}{{2({\tau _2} - {\tau _1})}}$, thus

$$\begin{aligned} {I_x}&= \int \! {{d^d}{x_1} \int \! {{d^d}{x_2} \, {e^{i\xi {{({{\mathbf {x}}_2} - {{\mathbf {x}}_1})}^2}}}} } = \int \! {{d^d}{x_1} \int \! {{d^d}{x_2} \,{e^{i\xi (\mathbf {x}_1^2 - 2\mathbf {x}_1.{{\mathbf {x}}_2} + \mathbf {x}_2^2)}}} } \\&= \int \! {{d^d}{x_1} \, {e^{i\xi \mathbf {x}_1^2}} \underbrace{\int \! {{d^d}{x_2} \, {e^{i\xi \mathbf {x}_2^2 - 2i\xi \mathbf {x}_1.{{\mathbf {x}}_2}}}} }_{{e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) }}{\pi ^{d/2}}{\xi ^{ - d/2}}{e^{ - i\frac{{{{(\xi \mathbf {x}_1^{})}^2}}}{\xi }}}}} = {e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) }}{\pi ^{d/2}}{\xi ^{ - d/2}}\underbrace{\int \! {{d^d}{x_1} \, {e^{i\xi \mathbf {x}_1^2}} \, {e^{ - i\xi \mathbf {x}_1^2}}} }_{{V_{{{\mathbb {R}}^d}}}}\\&= {e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) }}{2^{d/2}}{\pi ^{d/2}}{M^{ - d/2}}(1+i \varepsilon )^{-\frac{d}{2}}{({\tau _2} - {\tau _1})^{d/2}}{V_{{{\mathbb {R}}^d}}}. \end{aligned}$$

The integrals over $\tau _1$ and $\tau _2$ now take the form

$$\begin{aligned} {I_t} = \int \limits _{ - \infty }^{ - \epsilon }\! {d{\tau _1} \int \limits _\epsilon ^\infty \! {d{\tau _2} \, {{({\tau _2} - {\tau _1})}^{\frac{d}{2} - \Delta }}} } = \frac{{{2^{\frac{d}{2} - \Delta + 4}}}}{{(d - 2\Delta + 2)(d - 2\Delta + 4)}}{\epsilon ^{\frac{d}{2} - \Delta + 2}}, \end{aligned}$$

(C.4)

where the integrals are convergent only if

$$\begin{aligned} 2+\frac{d}{2}-\Delta <0. \end{aligned}$$

(C.5)

Therefore, QIM in the dual CFT to Schrödinger spacetime yields

$$\begin{aligned} {G_{\lambda \lambda }^{(CFT)}} = \frac{{\Delta {2^{d - 2\Delta + 3}}{e^{- \frac{{i\pi }}{4}(d +2\Delta )}}{M^{ \Delta - \frac{d}{2} - 1}}{V_{{{\mathbb {R}}^d}}}}}{{(d - 2\Delta + 4) \,\Gamma \left( {\Delta - \frac{d}{2}} \right) {(1+i \varepsilon )^{\frac{d}{2}}}}}{\,\epsilon ^{2 + \frac{d}{2} - \Delta }}=C {\epsilon ^{2 + \frac{d}{2} - \Delta }}, \end{aligned}$$

(C.6)

where at the final expression we have removed the regulator $\varepsilon$ from the correlation function.

Computation of the bulk holographic Fisher metric

1.1 Computation of the fields $\phi _{1,2}$

We will be integrating the bulk-to-boundary propagator $K(r,\mathbf {x},\tau ;{{\mathbf {x}}_1},{\tau _1})$ over $x_1$ and $\tau _1$, when $\tau _1<\tau$. Therefore, the field $\phi _1$ is given by¹⁵

$$\begin{aligned} {\phi _1}(r,\mathbf {x},\tau )&= \delta \lambda \int \! {{d^d}{x_1}\int \limits _{ -\infty }^\tau \! {d{\tau _1} \, K(r,\mathbf {x},\tau ;{{\mathbf {x}}_1},{\tau _1})} \, {{\hat{\phi }} _0(\mathbf {x}_1,\tau _1)}} \\&= \delta \lambda \, {c_\Delta }{r^\Delta }\int \limits _{ -\infty }^\tau \! {d{\tau _1} \, \frac{1}{{{{(\tau - {\tau _1})}^\Delta }}}\, {e^{\frac{i\, r^2 M}{2(\tau - \tau _1) }}}} \, {e^{\frac{i\, M\mathbf {x}^{2} }{2(\tau - {\tau _1})}}} \int \! {{d^d}{x_1}} \, {e^{\frac{i M }{2(\tau - \tau _1)}(\mathbf {x}_1^2 - 2\mathbf {x}_1.{{\mathbf {x}}})}} \\&=\delta \lambda {c_\Delta }{r^\Delta }\int \limits _{-\infty }^\tau \! d{\tau _1} \, \frac{1}{{{{(\tau - {\tau _1})}^\Delta }}}\, e^{\frac{i\, r^2 M}{2(\tau - \tau _1)}} \, e^{\frac{i\, M \mathbf {x}^2}{2(\tau - \tau _1)}} \, e^{\frac{ -i\,M\mathbf {x}^2}{2(\tau - \tau _1)}} \, e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) } {\pi ^{d/2}} \\&\quad \times \left( \frac{M}{2(\tau - \tau _1)} \right) ^{ - d/2}\\&=\delta \lambda \, c_\Delta \, r^{\Delta } \,2^{d/2} M^{-d/2}\, e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) }{\pi ^{d/2}}\int \limits _{ -\infty }^\tau \! {d\tau _1\, {} e^{\frac{i\, \mu }{\tau - \tau _1}}} {(\tau - {\tau _1})^{\frac{d}{2} - \Delta }}, \end{aligned}$$

where we have introduced the notation

$$\begin{aligned} \mu = \frac{M r^2}{2}. \end{aligned}$$

(D.1)

Let us calculate the last integral. For this purpose, we change the variables to $y=\tau -\tau _1$ with boundaries

$$\begin{aligned} y = \tau - {\tau _1} = \left\{ \begin{array}{l} 0,\quad {\tau _1} \rightarrow \tau ,\\ \infty ,\quad {\tau _1} \rightarrow -\infty , \end{array} \right. \end{aligned}$$

(D.2)

hence,

$$\begin{aligned} \int \limits _0^\infty \! dy\, e^{\frac{i\mu }{y}} \, y^{\frac{d}{2} - \Delta } = \Gamma \left( \Delta - \frac{d}{2} - 1 \right) (-i\mu )^{\frac{d}{2} - \Delta + 1}, \end{aligned}$$

(D.3)

where one has the convergence condition

$$\begin{aligned} \Delta>\frac{d}{2}+1,\quad M>0. \end{aligned}$$

(D.4)

The field $\phi _1$ now becomes

$$\begin{aligned} {\phi _1}(r,\mathbf {x},\tau ) \equiv {\phi _1}(r) = i\delta \lambda \, e^{ - \frac{{i\pi d}}{2}} {r^{d - \Delta + 2}}. \end{aligned}$$

(D.5)

We can do a similar computation for the field $\phi _2$, namely

$$\begin{aligned} {\phi _2}(r,\mathbf {x},t)&=\delta \lambda \, c_\Delta \, r^{\Delta } \,2^{d/2} M^{-d/2}\, {e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) }}{\pi ^{d/2}} \int \limits _{0 }^\tau \! d\tau _1\, e^{-\frac{{{{\tilde{\mu }}}}}{{\tau - {\tau _1}}}} {(\tau - {\tau _1})^{\frac{d}{2} - \Delta }}\nonumber \\&=\delta \lambda \, c_\Delta \, r^{\Delta } \,2^{d/2} M^{-d/2}\, {e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) }}{\pi ^{d/2}} \int \limits _0^\tau \! d{y}\, e^{-\frac{{{{\tilde{\mu }}}}}{{y}}} {y^{\frac{d}{2} - \Delta }} \nonumber \\&=\delta \lambda \, c_\Delta \, r^{\Delta } \,2^{d/2} M^{-d/2}\, e^{\frac{{i\pi }}{2}\left( {1 - \frac{d}{2}} \right) } \pi ^{d/2}\, {\tilde{\mu }}^{\frac{d}{2}-\Delta +1}\!\int \limits _{ \tilde{\mu }/\tau }^\infty \! d{x}\, e^{-x} {x^{a-1 }} \nonumber \\&=\frac{i \delta \lambda \, e^{-\frac{i \pi d}{2}} r^{d-\Delta +2}}{\Gamma (a)} \Gamma \left( a,\frac{\mu }{t}\right) \!, \end{aligned}$$

(D.6)

where at the end we have returned to real time $\tau \rightarrow -i t$, and we have defined the parameter

$$\begin{aligned} {\tilde{\mu }} = -i\frac{M r^2}{2}. \end{aligned}$$

(D.7)

1.2 Computation of the HFIM integrals

In order to compute Eqs. (4.7) and (4.8) we have to consider the metric $\gamma _{ab}$, $a,b=0,\dots , d$, on the ($d+1$)-dimensional boundary of the Schrödinger background, namely

$$\begin{aligned} ds_{{{(\partial Schr)}_{d + 1}}}^2 = - {} {L^2}\frac{{d{t^2}}}{{{r^4}}} + {L^2}\frac{{d{{\mathbf {x}}^{{} 2}}}}{{{r^2}}}, \end{aligned}$$

(D.8)

where $\mathbf {x}$ is a d-dimensional vector. Therefore, one finds

$$\begin{aligned} \sqrt{|\gamma |}=\frac{L^{d+1}}{r^{d+2}}. \end{aligned}$$

(D.9)

On the other hand, only the $n_r$ component of the normal vector to the boundary contributes, which can be explicitly calculated from

$$\begin{aligned} 1=g_{\mu \nu }n^\mu n^\nu =g_{rr} n^r n^r=\frac{L^2}{r^2}(n^r)^2. \end{aligned}$$

(D.10)

Hence $n^r=r/L$ and consequently $n_r=g_{rr} n^r=L/r$. Finally, one has

$$\begin{aligned} \sqrt{|\gamma |}\, n_\mu g^{\mu \nu }=\sqrt{|\gamma |}\, n_r g^{rr}={L^d}\, r^{-1-d}. \end{aligned}$$

(D.11)

Now, we can proceed with the computation of the bulk holographic Fisher information metric, which can be written by

$$\begin{aligned} {G_{\lambda \lambda }^{Bulk}} = -\, \frac{1}{{\delta {\lambda ^2}}}\left( \frac{\delta I_1}{2} - \delta I_2 \right) = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2\kappa }}\mathop {\lim }\limits _{r \rightarrow {\tilde{\varepsilon }} } \left( {{c_0}{J_0}\, r^{d-2 \Delta +2} + {c_1}{J_1} + {c_2}{J_2}\, r^{d-2 \Delta +2}} \right) , \end{aligned}$$

(D.12)

where the coefficients $c_i$ are given by

$$\begin{aligned} c_0 = \frac{e^{-i \pi d}}{2} (d-\Delta +2), \quad c_1 = \frac{M^a \, e^{-id\pi } }{2^{a-1}\, \Gamma ^2(a)} , \quad c_2 = -\, \frac{2 c_0}{\Gamma ^2(a)}. \end{aligned}$$

(D.13)

Consequently, one has to compute the following integrals

$$\begin{aligned}&{J_0} = \int \limits _ {-T}^{-\epsilon }\! dt\, +\, \int \limits _{\epsilon }^{T}\! dt = 2(T - \epsilon ), \end{aligned}$$

(D.14)

$$\begin{aligned}&{J_1} = \int \limits _{\epsilon }^{T}\! dt \, t^{-a} \, e^{-\frac{\mu }{t}}\, \Gamma \left( {a,\frac{\mu }{{t}}} \right) =\int \limits _{1/T}^{1/\epsilon }\! dx\, x^{a-2}\, e^{-\mu x}\, \Gamma (a,\mu x), \end{aligned}$$

(D.15)

$$\begin{aligned}&{J_2} = \int \limits _{\epsilon }^{T}\! dt \,\, \Gamma ^2\! \left( a,\frac{\mu }{t} \right) . \end{aligned}$$

(D.16)

In order to solve $J_1$ we transform the incomplete gamma function in the following way

$$\begin{aligned} \Gamma (a,\mu x) = (a-1)(a-2)\, \Gamma (a-2,\mu x) \,+\, (a-1)\, \mu ^{a-2}\, x^{a-2}\, e^{-\mu x} \,+\, \mu ^{a-1}\, x^{a-1}\, e^{-\mu x}, \end{aligned}$$

(D.17)

which follows directly from

$$\begin{aligned} \Gamma (a,z) = \frac{{\Gamma (a)}}{{\Gamma (a - n)}}\, \Gamma (a-n,z) \,+\, z^{a-1}\, e^{-z} \sum \limits _{k = 0}^{n - 1} {\frac{{\Gamma (a)}}{{\Gamma (a - k)}}}\, z^{ - k} \end{aligned}$$

(D.18)

for $n=2$ and we have also used $\Gamma (a)=(a-1) \Gamma (a-1)$. Hence, the integral $J_1$ now becomes

$$\begin{aligned} {J_1}&= (a-1)(a-2) \int \limits _{1/T}^{1/\epsilon }\! e^{ - \mu x}\, x^{a - 2}\, \Gamma (a-2,\mu x)\, dx \nonumber \\&\quad + (a-1)\, \mu ^{a-2}\! \int \limits _{1/T}^{1/\epsilon }\! x^{2a - 4}\, e^{ - 2\mu x}\, dx \,+\, \mu ^{a-1}\! \int \limits _{1/T}^{1/\epsilon }\! x^{2a - 3}\, e^{ - 2\mu x}\, dx \nonumber \\&= \frac{1}{\mu }(a-1)(a-2)\, T^{2-a}\, e^{-\frac{\mu }{T}}\, \Gamma \! \left( a-2,\frac{\mu }{T} \right) \,+\, \frac{1}{2\mu ^{a-1}} (a-1)(a-2)^2 \, \Gamma ^2 {\left( {a - 2,\frac{\mu }{T}} \right) }\nonumber \\&\quad - \frac{1}{\mu }(a-1)(a-2)\, \epsilon ^{2-a}\, e^{-\frac{\mu }{\epsilon }}\, \Gamma \! \left( a-2,\frac{\mu }{\epsilon } \right) \,-\, \frac{1}{2\mu ^{a-1}} (a-1)(a-2)^2\, \Gamma ^2 {\left( {a - 2,\frac{\mu }{\epsilon }} \right) }\nonumber \\&\quad + \frac{4^{2-a}}{\mu ^{a-1}} (a-1)(a-2)\, \Gamma \! \left( 2a-4,\frac{2\mu }{\epsilon } \right) \,-\, \frac{4^{2 - a}}{\mu ^{a-1}} (a-1)(a-2)\, \Gamma \! \left( 2a-4, \frac{2\mu }{T} \right) \nonumber \\&\quad + \frac{2^{3-2a}}{\mu ^{a-1}} (a - 1)\, \Gamma \! \left( {2a - 3,\frac{{2\mu }}{T}} \right) \,-\, \frac{2^{3-2a}}{\mu ^{a-1}} (a - 1)\, \Gamma \! \left( {2a - 3,\frac{{2\mu }}{\epsilon }} \right) \nonumber \\&\quad + \frac{4^{1-a}}{\mu ^{a-1}}\, \Gamma \! \left( 2a-2, \frac{2\mu }{T} \right) \,-\, \frac{4^{1-a}}{\mu ^{a-1}}\, \Gamma \! \left( 2a-2, \frac{2\mu }{\epsilon } \right) . \end{aligned}$$

(D.19)

The integral $J_2$ is more complicated, but also analytically solvable if we use the following power reduction formula [73]:

$$\begin{aligned} \Gamma ^2 (a,z) = \Gamma ^2 (a) \,-\, 2\int \limits _0^{1/2}\! d\omega {} \, \,\gamma \left( 2a,\frac{z}{\omega } \right) \omega ^{a-1} (1 - \omega )^{a-1} . \end{aligned}$$

(D.20)

Let $\epsilon =T_1$ and $T=T_2$, hence

$$\begin{aligned} J_2&= \int \limits _{T_1}^{T_2}\! dt\,\, \Gamma ^2\! \left( {a,\frac{\mu }{t}} \right) = \int \limits _{T_1}^{T_2}\! dt \left[ \Gamma ^2 (a) \,-\, 2\int \limits _0^{1/2}\! d\omega \, \gamma \left( 2a,\frac{\mu }{\omega t} \right) \omega ^{a-1} \left( 1-\omega \right) ^{a - 1} \right] \\&= \left( T_2 - T_1 \right) \Gamma ^2 (a) \,-\, 2\int \limits _0^{1/2}\! d\omega \, \omega ^{a-1} (1 - \omega )^{a - 1} \, \int \limits _{T_1}^{T_2}\! dt\, \gamma \left( 2a,\frac{\mu }{\omega t} \right) \\&= \left( T_2-T_1 \right) \Gamma ^2 (a) \,-\, 2\int \limits _0^{1/2}\! d\omega \, \omega ^{a-1} (1- \omega )^{a-1} \left( \,\int \limits _{T_1}^{T_2}\! dt\, \Gamma (2a) - \int \limits _{T_1}^{T_2}\! dy\, \Gamma \! \left( 2a,\frac{\mu }{\omega t} \right) \right) \\&= \left( T_2-T_1 \right) \left[ \Gamma ^2 (a) - 2\Gamma (2a)\, {\mathcal {B}}_{1/2} (a,a) \right] \,+\, 2\int \limits _0^{1/2}\! d\omega \, \omega ^{a-1} (1- \omega )^{a - 1} {\mathcal {J}}(\omega ) . \end{aligned}$$

The inner integral is

$$\begin{aligned} {\mathcal {J}} (\omega )&= \int \limits _{T_1}^{T_2}\! dt\, \Gamma \! \left( 2a,\frac{\mu }{\omega t} \right) \nonumber \\&= T_2\, \Gamma \! \left( 2a,\frac{\mu }{\omega T_2} \right) - T_1\, \Gamma \! \left( 2a,\frac{\mu }{\omega T_1} \right) + \frac{\mu }{\omega } \left[ \Gamma \! \left( 2a - 1,\frac{\mu }{\omega T_1} \right) - \Gamma \! \left( 2a - 1,\frac{\mu }{\omega T_2} \right) \right] , \end{aligned}$$

(D.21)

thus

$$\begin{aligned} J_2 = \left( T_2 - T_1 \right) \left[ \Gamma ^2 (a) - 2\Gamma (2a)\, {\mathcal {B}}_{1/2} (a,a) \right] + {I_1} + {I_2} + {I_3} + {I_4}, \end{aligned}$$

(D.22)

where

$$\begin{aligned}&I_1 = -\, 2T_1 \int \limits _0^{1/2}\! d\omega \, \omega ^{a-1} (1-\omega )^{a-1}\, \Gamma \! \left( 2a,\frac{\mu }{\omega T_1} \right) , \end{aligned}$$

(D.23)

$$\begin{aligned}&I_2 = 2T_2 \int \limits _0^{1/2}\! d\omega \, \omega ^{a-1} (1 - \omega )^{a - 1}\, \Gamma \! \left( 2a,\frac{\mu }{\omega T_2} \right) , \end{aligned}$$

(D.24)

$$\begin{aligned}&I_3 = 2\mu \int \limits _0^{1/2}\! d\omega \, \omega ^{a - 2} (1 - \omega )^{a - 1}\, \Gamma \! \left( 2a - 1,\frac{\mu }{\omega T_1 } \right) , \end{aligned}$$

(D.25)

$$\begin{aligned}&I_4 = -\, 2\mu \int \limits _0^{1/2}\! d\omega \, \omega ^{a - 2} (1 - \omega )^{a - 1}\, \Gamma \! \left( 2a - 1,\frac{\mu }{\omega T_2 } \right) . \end{aligned}$$

(D.26)

The solution is the following. Let ${\mathfrak {b}}_{1,2}=\mu /T_{1,2}$, then

$$\begin{aligned} I_{1,2}&= {\mp } 2T_{1,2} \int \limits _0^{1/2}\! d\omega \, \omega ^{a - 1} (1 - \omega )^{a - 1}\, \Gamma \! \left( {2a,\frac{{\mathfrak {b}}_{1,2}}{\omega }} \right) \\&= {\mp } 2T_{1,2} \int \limits _0^{1/2}\! d\omega \, \omega ^{a - 1} (1-\omega )^{a - 1} {\left( \frac{ {\mathfrak {b}}_{1,2}}{\omega } \right) }^{2a} e^{ -\frac{{\mathfrak {b}}_{1,2}}{\omega } } \int \limits _0^\infty \frac{e^{ - \frac{{\mathfrak {b}}_{1,2}}{\omega }z} }{(1 + z)^{1 - 2a} } \,dz\\&= {\mp } 2T_{1,2} \int \limits _0^\infty \! dz\, (1 + z)^{2a-1}\, {\mathfrak {b}}_{1,2}^{2a} \int \limits _0^{1/2}\! d\omega \, \frac{1}{\omega ^2} \left( {\frac{1}{\omega } - 1} \right) ^{a - 1} e^{-\frac{{\mathfrak {b}}_{1,2}}{\omega }(1 + z)}\, \\&= {\mp } 2T_{1,2} \int \limits _0^\infty \! dz\, (1 + z)^{2a-1}\, {\mathfrak {b}}_{1,2}^{2a} \int \limits _0^{1/2}\! d\left( \frac{ -1}{\omega } \right) \! \left( \frac{1}{\omega } - 1 \right) ^{a-1} e^{- \frac{{\mathfrak {b}}_{1,2}}{\omega }(1 + z)} \\&= \left\{ {x = \frac{1}{\omega } = \left\{ \begin{array}{l} 2,\quad \,\,\, \omega = 1/2,\\ \infty ,\quad \omega = 0 \end{array} \right. \quad } \right\} \\&= {\mp } 2T_{1,2} \int \limits _0^\infty \! dz\, (1 + z)^{2a - 1}\, {\mathfrak {b}}_{1,2}^{2a} \int \limits _2^\infty \! dx\, \left( x - 1\right) ^{a - 1} e^{ - {\mathfrak {b}}_{1,2} (1 + z)x} \\&= {\mp } 2T_{1,2} \int \limits _0^\infty \! dz\, (1 + z)^{a - 1}\, {\mathfrak {b}}_{1,2}^a\, e^{ - {\mathfrak {b}}_{1,2}(1 + z)}\, \Gamma \! \left( a,{\mathfrak {b}}_{1,2}(1 + z) \right) \\&= \pm T_{1,2} \int \limits _0^\infty \! dz\, \frac{d}{dz} \Gamma ^2\! \left( a,{\mathfrak {b}}_{1,2}(1 + z) \right) \\&= \pm T_{1,2}\, \Gamma ^2 \left( a,{\mathfrak {b}}_{1,2}(1 + z) \right) \left| {\begin{array}{*{20}{l}} {^{z \rightarrow \infty }}\\ {_{z \rightarrow 0}} \end{array}} \right. = {\mp } T_{1,2}\, \Gamma ^2\left( {a,{\mathfrak {b}}_{1,2}} \right) . \end{aligned}$$

Therefore, one finds

$$\begin{aligned} {I_{1,2}} = {\mp } T_{1,2}\, \Gamma ^2\! \left( a,\frac{\mu }{T_{1,2}} \right) . \end{aligned}$$

(D.27)

In the previous computations we have used the following integral representation of the incomplete gamma function

$$\begin{aligned} \Gamma (a,t) = t^a\, e^{-t} \int \limits _0^\infty \frac{ e^{ -zt}}{(1 + z)^{1-a} }dz. \end{aligned}$$

(D.28)

With similar calculations for $I_{3,4}$, one finds

$$\begin{aligned} I_{3,4}&= \pm 2\mu \int \limits _0^{1/2}\! d\omega \, \omega ^{a-2}\, (1-\omega )^{a-1} \,\Gamma \! \left( 2a-1,\frac{{\mathfrak {b}}_{1,2}}{\omega } \right) \\&= \pm 2\mu \int \limits _0^{1/2}\! d\omega \, \omega ^{a-2}\, (1-\omega )^{a-1} \left( \frac{{\mathfrak {b}}_{1,2}}{\omega } \right) ^{2a-1} e^{ -\frac{{\mathfrak {b}}_{1,2}}{\omega }} \int \limits _0^\infty \frac{e^{ - \frac{{\mathfrak {b}}_{1,2}}{\omega }z}}{(1 + z)^{2 - 2a} } \,dz\\&= \pm 2\mu \int \limits _0^\infty \! dz\, (1+z)^{2a-2}\, {\mathfrak {b}}_{1,2}^{2a-1} \int \limits _0^{1/2}\! d\omega \, \frac{1}{\omega ^2} \left( \frac{1}{\omega } - 1 \right) ^{a - 1} e^{ -\frac{{\mathfrak {b}}_{1,2}}{\omega }(1 + z)} \, \\&= \pm 2\mu \int \limits _0^\infty \! dz\, (1+z)^{2a-2}\, {\mathfrak {b}}_{1,2}^{2a-1} \int \limits _0^{1/2}\! d\left( \frac{-1}{\omega } \right) \! \left( \frac{1}{\omega } -1 \right) ^{a-1} e^{ -\frac{{\mathfrak {b}}_{1,2}}{\omega }(1+z)} \\&= \left\{ {x = \frac{1}{\omega } = \left\{ \begin{array}{l} 2,\quad \,\,\, \omega = 1/2,\\ \infty ,\quad \omega = 0 \end{array} \right. } \right\} \\&= \pm 2\mu \int \limits _0^\infty \! dz\, (1+z)^{2a-2}\, {\mathfrak {b}}_{1,2}^{2a-1} \int \limits _2^\infty \! dx\, \left( x-1 \right) ^{a-1} e^{ -{\mathfrak {b}}_{1,2}(1+z)x} \\&= \pm 2\mu \int \limits _0^\infty \! dz\, (1+z)^{a-2}\, {\mathfrak {b}}_{1,2}^{a-1}\, e^{ - {\mathfrak {b}}_{1,2}(1 + z)}\, \Gamma \! \left( a,{\mathfrak {b}}_{1,2} (1+z) \right) . \end{aligned}$$

Now we use

$$\begin{aligned}&\Gamma (a,{\mathfrak {b}}(z + 1)) = {e^{ - {\mathfrak {b}}(z + 1)}}{{\mathfrak {b}}^{a - 1}}{(z + 1)^{a - 1}} \nonumber \\&\quad - \frac{1}{2}(a - 1){(z + 1)^{2 - a}}{{\mathfrak {b}}^{1 - a}}{e^{{\mathfrak {b}}(z + 1)}}\frac{d}{{dz}}\Gamma ^2 (a - 1,{\mathfrak {b}}(z + 1)), \end{aligned}$$

(D.29)

thus

$$\begin{aligned} I_{3,4}&= {\mp } (a-1)\mu \int \limits _0^\infty \! dz\, \frac{d}{dz} \Gamma ^2 (a-1,{\mathfrak {b}}_{1,2} (z + 1)) \,\pm \, 2\mu {\mathfrak {b}}_{1,2}^{2a-2} \int \limits _0^\infty \! dz\, (z + 1)^{2a-3}\, e^{ -2{\mathfrak {b}}_{1,2} (z + 1)}\\&= {\mp } (a - 1)\mu \, \Gamma ^2\! \left( a-1, {\mathfrak {b}}_{1,2}(z+1)\right) \left| \begin{array}{l} ^{z \rightarrow \infty }\\ _{z \rightarrow 0} \end{array} \right. \pm \, 2^{3-2a}\mu \, \Gamma (2a-2,2{\mathfrak {b}}_{1,2})\\&= \pm (a - 1)\mu \, \Gamma ^2 (a - 1,{\mathfrak {b}}_{1,2}) \,\pm \, 2^{3 - 2a}\mu \, \Gamma (2a - 2,2{\mathfrak {b}}_{1,2}). \end{aligned}$$

Therefore, one has

$$\begin{aligned} I_{3,4} = \pm \mu \left[ (a-1)\, \Gamma ^2\! \left( a-1,\frac{\mu }{T_{1,2}} \right) + 2^{3-2a}\, \Gamma \! \left( 2a-2,\frac{2\mu }{T_{1,2}} \right) \right] . \end{aligned}$$

(D.30)

Putting everything together we get the complete expression for $J_2$, namely

$$\begin{aligned} J_2&= (T-\epsilon )\left( \Gamma ^2 (a) - 2\Gamma (2a)\, {\mathcal {B}}_{1/2}(a,a) \right) \,+\, T\,\Gamma ^2\! \left( a,\frac{\mu }{T} \right) - \epsilon \, \Gamma ^2\! \left( a,\frac{\mu }{\epsilon } \right) \nonumber \\&\quad + \mu \left( (a-1)\, \Gamma ^2\! \left( a-1,\frac{\mu }{\epsilon } \right) + 2^{3-2a}\, \Gamma \! \left( 2a-2,\frac{2\mu }{\epsilon } \right) \right) \nonumber \\&\quad - \mu \left( (a - 1)\, \Gamma ^2\! \left( a-1,\frac{\mu }{T} \right) + 2^{3-2a}\, \Gamma \! \left( 2a-2, \frac{2\mu }{T} \right) \right) . \end{aligned}$$

(D.31)

The final expression for the holographic Fisher metric yields

$$\begin{aligned} {G_{\lambda \lambda }^{Bulk}}&= \frac{a_0}{\mu ^{a}} (T - \epsilon ) \,+\, a_1 \frac{T^{2 - a} e^{ -\frac{\mu }{T}}}{\mu } \, \Gamma \! \left( a-2,\frac{\mu }{T} \right) \,+\, \frac{a_2}{\mu ^{a-1}}\, \Gamma ^2\! \left( a-2,\frac{\mu }{T} \right) \nonumber \\&\quad + \frac{a_3}{\mu ^{a-1}}\, \Gamma \! \left( 2a - 4,\frac{2\mu }{T} \right) \,+\, \frac{a_4}{\mu ^{a-1}}\, \Gamma \! \left( 2a-2,\frac{2\mu }{T} \right) \,+\, \frac{a_5}{\mu ^{a-1}}\, \Gamma \! \left( 2a-3,\frac{2\mu }{T} \right) \nonumber \\&\quad + a_6 \frac{T}{\mu ^a}\, \Gamma ^2\! \left( a,\frac{\mu }{T} \right) \,+\, \frac{a_7}{\mu ^{a-1}}\, \Gamma ^2\! \left( a-1,\frac{\mu }{T} \right) \nonumber \\&\quad + b_1 \frac{\epsilon ^{2-a}}{\mu }\, e^{ -\frac{\mu }{\epsilon }}\, \Gamma \! \left( a-2,\frac{\mu }{\epsilon } \right) \,+\, \frac{b_2}{\mu ^{a-1}}\, \Gamma ^2\! \left( a-2,\frac{\mu }{\epsilon } \right) \nonumber \\&\quad + \frac{b_3}{\mu ^{a-1}}\, \Gamma \! \left( 2a-4,\frac{2\mu }{\epsilon } \right) \,+\, \frac{b_4}{\mu ^{a-1}}\, \Gamma \! \left( 2a-2,\frac{2\mu }{\epsilon } \right) \,+\, \frac{b_5}{\mu ^{a-1}}\, \Gamma \! \left( 2a-3,\frac{2\mu }{\epsilon } \right) \nonumber \\&\quad + b_6 \frac{\epsilon }{\mu ^{a}}\, \Gamma ^2\! \left( a,\frac{\mu }{\epsilon } \right) \,+\, \frac{b_7}{\mu ^{a-1}}\, \Gamma ^2\! \left( a-1,\frac{\mu }{\epsilon } \right) , \end{aligned}$$

(D.32)

with coefficients $a_i=-b_i$, $i=1,\dots , 7$, where

$$\begin{aligned} \begin{aligned}&{a_0} = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{2^{a+1}\kappa } M^a \left[ {{2c_0} + {c_2}\left( {{\Gamma ^2}\left( a \right) - 2\Gamma (2a)\, {\mathcal {B}}_{1/2} (a,a)} \right) } \right] , \\&{a_1} = - {b_1} = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2\kappa }}{c_1}(a - 1)(a - 2), \\&{a_2} = - {b_2} = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{4\kappa }}{c_1}{(a - 1)}(a - 2)^2, \\&{a_3} = - {b_3} = - \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2^{2a-3}\kappa }} c_1(a - 1)(a - 2), \\&{a_4} = - {b_4} = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2^{3a-1}\kappa }} \left( 2^a {c_1}-2 {c_2} M^a\right) , \\&{a_5} = - {b_5} = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2^{2a-2}\kappa }} c_1 (a - 1), \\&{a_6} = - {b_6} = \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2^{a+1}\kappa }} M^{a} c_2,\\&{a_7} = - {b_7} = - \frac{{{L^{d }}{V_{{{\mathbb {R}}^d}}}}}{{2^{a+1}\kappa }} M^{a} c_2(a - 1), \end{aligned} \end{aligned}$$

(D.33)

where ${\mathcal {B}}$ is the incomplete beta function.