1 Introduction

Recent progress in the holographic superconductivity [1,2,3], based on the gauge gravity duality [4,5,6], made an essential contribution in understanding the symmetry broken phase of AdS/CFT by constructing a dynamical symmetry breaking mechanism. While the symmetry breaking in the Abelian Higgs model in flat space is adhoc by assuming the presence of the potential having a Maxican hat shape, the symmetry breaking of the abelian Higgs model in AdS can be done by the gravitational instability of near horizon geometry to create a haired black hole, thereby the model is equipped with a fully dynamical mechanism of the symmetry breaking. The observables’ dependence on \(\Delta \) is interesting because \(\Delta \) depends on the strength of the interaction.

After initial stage of the model building [1, 2] where probe limit of the gravity background was used, full back reacted version [3] worked out. It turns out that although there are significant differences in the zero temperature limit between the probe limit and the full back reacted version, the former captures the physics [7] correctly near the critical temperature \(T_{c}\), which is expected because the back reaction cannot be large when the condensation just begin to appear.

The analytic expressions of observables within the probe approximation were also obtained in [8, 9]. One problem is that [8,9,10] the critical temperature is divergent at the \(\Delta =1/2\), which does not seems to make physical sense and it has not been understood as far as we know. This was also noticed as a problem [7] but the reason for it has not been cleared yet.

In this paper, we consider the problem by recomputing \(T_{c}\) and physical observables analytically near the \(T_{c}\), where the probe approximation is a good one. We apply Pincherle’s theorem [11] to handle the Heun’s equation which appears in the computation of the critical temperature in the blackhole background. We find that the region of \(1/2\le \Delta <1\) for AdS\(_{4}\) does not have a well defined eigenvalue and therefore does not have well defined critical temperature either. See Fig. 1. We will also see that, in this same regime, the AC conductivity gap \(\omega _{g}\) does not exist either, giving us another confidence in concluding the absence of the superconductivity in this regime. The situation remind us the physics of the pseudo gap regime where one can find Cooper pairs but not superconductivity due to the absence of the phase alignment of the pairs. Similar phenomena exists for \(1<\Delta <3/2\) for AdS\(_{5}\), which is described in the Appendix B.

2 Set up

We start with the action [1],

$$\begin{aligned} S=\int d^{d+1}x \sqrt{-|g|}\left( -\frac{1}{4} F_{\mu \nu }^{2}- |D_{\mu }\Phi |^{2}-m^{2}|\Phi |^{2} \right) ,\nonumber \\ \end{aligned}$$
(2.1)

where \(|g|=\det g_{ij}\), \(D_{\mu }\Phi =\partial _{\mu }-igA_{\mu }\) and \(F=dA\). Following the Ref. [1], we use the fixed metric of AdS\(_{d+1}\) blackhole,

$$\begin{aligned} ds^2= -f(r)dt^2+\frac{dr^2}{f(r)}+r^2d\mathbf {x}^2,\quad f(r)=r^2\left( 1-\frac{r_h^d}{r^d}\right) .\nonumber \\ \end{aligned}$$
(2.2)

The AdS radius is set to be 1 and \(r_h\) is the radius of the horizon. The Hawking temperature is \(T=\frac{d}{4\pi } r_h. \) In the coordinate \(z= r_h/r\), the field equations are

$$\begin{aligned}&\frac{d^2 \Psi }{d z^2} -\frac{d-1+z^d}{z(1-z^d)}\frac{d \Psi }{d z}+\left( \frac{g^2 \Phi ^2}{r_h^2(1-z^d)^2}-\frac{m^2}{z^2(1-z^d)}\right) \Psi =0, \nonumber \\&\frac{d^2 \Phi }{d z^2}- \frac{d-3}{z}\frac{d \Phi }{d z } -\frac{2g^2\Psi ^2}{z^2(1-z^d)}\Phi =0. \end{aligned}$$
(2.3)

Here, \(\Psi (z)\) is the scalar field and \(\Phi (z)\) is an electrostatic scalar potential \(A_{t}\).

Near the boundary \(z=0\), we have

$$\begin{aligned} \Psi (z)= & {} z^{\Delta _{-}}\Psi ^{(-)}(z)+z^{\Delta _{+}}\Psi ^{(+)}(z), \nonumber \\ \Phi (z)= & {} \mu - ({\rho }/{r_h^{d-2}})z^{d-2} +\cdots \end{aligned}$$
(2.4)

where \(\Delta _{\pm } = {d/2\pm \sqrt{(d/2)^2+m^2}} \), \(\mu \) is the chemical potential and \(\rho \) is the charge density. By \(\Delta \) we mean \(\Delta _{+}\).

We examine the range \(\frac{d-1}{2} \le \Delta < d\) only, because the regime \(0< \Delta <\frac{d-1}{2} \) is not physical. Notice also that \(\Delta =d/2\) is the value for which \(\Delta _{+}=\Delta _{-}\) and \(\Delta =d\) is the value where \(m^{2}=0\). We request the boundary conditions at the horizon \(z=1\): \(\Phi (1)=0\) and the finiteness of \(\Psi (1)\). Then the condensate of the Cooper pair operator \({\mathcal {O}}_{\Delta }\) dual to the field \(\Psi \) is given by \( \langle {\mathcal {O}}_{\Delta }\rangle =\lim _{{r\rightarrow \infty }}\sqrt{2}r^{\Delta }\Psi (r) \) under the assumption that the source is zero.

3 Critical temperature \(T_{c}\) in AdS4

At \(T= T_c\), \(\Psi =0\), and Eq. (2.3) is integrated [8] to give

$$\begin{aligned} \Phi (z)= \lambda _{d} r_c (1-z^{d-2}) \quad \text{ with } \ \lambda _{d}= {\rho }/{r_c^{d-1}}, \end{aligned}$$
(3.1)

where \(r_c\) is the horizon radius at \( T_c\). As \(T\rightarrow T_c\), the field equation of \(\Psi \) becomes

$$\begin{aligned}&-\frac{d^2 \Psi }{d z^2} +\frac{d-1+z^d}{z(1-z^d)}\frac{d \Psi }{d z}+ \frac{m^2}{z^2(1-z^d)} \Psi \nonumber \\&\quad =\frac{\lambda _{g,d}^2(1-z^{d-2})^2}{(1-z^d)^2}\Psi \end{aligned}$$
(3.2)

where \(\lambda _{g,d}= g\lambda _d\). Our result for the critical temperature is given by

$$\begin{aligned} T_c =\frac{d}{4\pi } \left( \frac{g \rho }{\lambda _{g,d}}\right) ^{\frac{1}{d-1}}, \end{aligned}$$
(3.3)

which is a part of the first line of Table 1. Details of deriving this result is in Sects. 3.1 and 3.2.

For \(\Delta =1\) and 2 in AdS\(_{4}\), we have \(T_c/(g^{1/2}\sqrt{\rho } )=0.2256 \) and 0.1184 respectively. If we set our coupling \(g=1\), these are in good agreement with the numerical data of [1] confirming the validity of our method.

To find the \(\Delta \)-dependence of the \(T_{c}\), we first calculate \(\lambda _{g,d}\). The procedures are rather involved both analytically and numerically. Here, we display the analytic structure of the calculated data of \(\lambda _{g,d}\) leaving the details to the Sect. 3.1 and Appendix B.1.1:

$$\begin{aligned} \lambda _{g,3}&= 1.96 \Delta ^{4/3} - 0.87 \quad \text{ at }\ 1\le \Delta \le 3, \nonumber \\ \lambda _{g,4}&= 1.18 \Delta ^{4/3} - 0.97\quad \text{ at }\ 3/2\le \Delta \le 4. \end{aligned}$$
(3.4)

Here, we used the Pincherle’s Theorem with matrix-eigenvalue algorithm [11]. Notice that the variational method used in [8] is not applicable near the singularity \(\Delta =(d-2)/2\).

3.1 Matrix algorithm and Pincherle’s Theorem

At the critical temperature \(T_c\), \(\Psi =0\), so Eq. (2.3) tells us \(\Phi ^{\prime \prime }=0\). Then, we can set

$$\begin{aligned} \Phi (z)= \lambda _3 r_c (1-z)\quad \text{ where }\ \lambda _3 =\frac{\rho }{r_c^2} \end{aligned}$$
(3.5)

here, \(r_c\) is the radius of the horizon at \(T=T_c\). As \(T\rightarrow T_c\), the field equation \(\Psi \) approaches to

$$\begin{aligned}&-\frac{d^2 \Psi }{d z^2} +\frac{2+z^3}{z(1-z^3)}\frac{d \Psi }{d z}+ \frac{m^2}{z^2(1-z^3)} \Psi \nonumber \\&\quad =\frac{\lambda _{g,3}^2}{(z^2+z+1)^2}\Psi \end{aligned}$$
(3.6)

where \(\lambda _{g,3}= g\lambda _3\). Factoring out the behavior near the boundary \(z=0\) and the horizon, we define

$$\begin{aligned}&\Psi (z)= \frac{\langle {\mathcal {O}}_{\Delta }\rangle }{\sqrt{2}r_h^{\Delta }}z^{\Delta }F(z) \nonumber \\&\quad \text{ where }\ F(z)= (z^2+z+1)^{-\lambda _{g,3}/\sqrt{3}}y(z) \end{aligned}$$
(3.7)

Then, F is normalized as \(F(0)=1\) and we obtain

$$\begin{aligned}&\frac{d^2 y }{d z^2} +\frac{(1-\frac{4}{\sqrt{3}} \lambda _{g,3} +2\Delta )z^3+\frac{2\lambda _{g,3}}{\sqrt{3}}z^2 +\frac{2\lambda _{g,3}}{\sqrt{3}}z+2(1-\Delta )}{z( z^3-1)}\frac{d y}{d z} \nonumber \\&\quad +\frac{(3\Delta ^2-4\sqrt{3}\Delta \lambda _{g,3}+4\lambda _{g,3}^2)z^2-(4\lambda _{g,3}^2-2\sqrt{3}\Delta \lambda _{g,3}+\sqrt{3}\lambda _{g,3})z-2\sqrt{3}(1-\Delta )\lambda _{g,3}}{3z( z^3-1)}y=0. \end{aligned}$$
(3.8)

Notice that this is the generalized Heun’s equation [12] that has five regular singular points at \(z=0,1,\frac{-1\pm \sqrt{3}i}{2},\infty \). Substituting \(y(z)= \sum _{n=0}^{\infty } d_n z^{n}\) into (3.8), we obtain the following four term recurrence relation:

$$\begin{aligned}&\alpha _n\; d_{n+1}+ \beta _n \;d_n + \gamma _n \;d_{n-1}+\delta _n\;d_{n-2}=0 \nonumber \\&\quad \hbox { for } n \ge 2, \end{aligned}$$
(3.9)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha _n=-3(n+1)(n+2\Delta -2) \\ \beta _n=2\sqrt{3}\lambda _{g,3}(n+\Delta -1) \\ \gamma _n=\sqrt{3}(2n+2\Delta -3)\lambda _{g,3}-4\lambda _{g,3}^2 \\ \delta _n=3(n-\frac{2}{\sqrt{3}}\lambda _{g,3}+\Delta -2)^2. \end{array}\right. } \end{aligned}$$
(3.10)

The first four \(d_{n}\)’s are given by \(\alpha _0 d_1+ \beta _0 d_0=0\), \(\alpha _1 d_2+ \beta _1 d_1+ \gamma _1 d_0=0 \), \(d_{-1}=0\) and \(d_{-2}=0\). Equations (3.7), (3.9) and (3.10) give us the following boundary condition

$$\begin{aligned} F^{\prime }(0)=0. \end{aligned}$$
(3.11)

Since the 4 term relation can be reduced to the 3 term relation, we first review for a minimal solution of the three term recurrence relation

$$\begin{aligned} \alpha _n\; d_{n+1}+ \beta _n \;d_n + \gamma _n \;d_{n-1} =0 \quad \hbox {for } n \ge 1, \end{aligned}$$
(3.12)

with \(\alpha _0 d_1+ \beta _0 d_0=0\) and \(d_{-1}=0\). Equation (3.12) has two linearly independent solutions X(n), Y(n). We recall that \(\{X(n)\}\) is a minimal solution of Eq. (3.12) if not all \(X(n)=0\) and if there exists another solution Y(n) such that \(\lim _{n\rightarrow \infty }X(n)/Y(n)=0\). Now \((d_n)_{n\in {\mathbb {N}}}\) is the minimal solution if \(\alpha _0 \ne 0\) and

$$\begin{aligned} \beta _0 +\frac{-\alpha _0 \gamma _1}{\beta _1 -\frac{\alpha _1 \gamma _2}{\beta _2-\frac{\alpha _2 \gamma _3}{\beta _3 -\cdots }}}=0. \end{aligned}$$
(3.13)

One should remember that \(\alpha _{n},\beta _{n},\gamma _{n}\)’s are functions of \(\lambda \) so that above equation should be read as equation for \(\lambda \).

As we mentioned above, we can transform the four term recurrence relations into three-term recurrence relations by the Gaussian elimination steps. More explicitly, the transformed recurrence relation is

$$\begin{aligned} \alpha _n^{\prime }\; d_{n+1}+ \beta _n^{\prime } \;d_n + \gamma _n^{\prime } \;d_{n-1} =0 \quad \hbox {for}\ n \ge 1, \end{aligned}$$
(3.14)

where

$$\begin{aligned} \alpha _n^{\prime }=\alpha _n,\quad \beta _n^{\prime }=\beta _n,\quad \gamma _n^{\prime }= \gamma _n \quad \hbox {for}\ n =0,1 \end{aligned}$$

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \delta _n^{\prime } =0\\ \alpha _n^{\prime }=\alpha _n \\ \beta _n^{\prime }= \beta _n-\frac{\alpha _{n-1}^{\prime }\delta _n}{\gamma _{n-1}^{\prime }} \\ \gamma _n^{\prime }= \gamma _n-\frac{\beta _{n-1}^{\prime }\delta _n}{\gamma _{n-1}^{\prime }}\quad \hbox { for } n \ge 2 \end{array}\right. } \end{aligned}$$
(3.15)

and \(\alpha _0^{\prime } d_1+ \beta _0^{\prime } d_0=0\) and \(d_{-1}=0\). Now the minimal solution is determined by

$$\begin{aligned} \beta _0^{\prime } +\frac{-\alpha _0^{\prime } \gamma _1^{\prime }}{\beta _1^{\prime } -\frac{\alpha _1^{\prime } \gamma _2^{\prime }}{\beta _2^{\prime }-\frac{\alpha _2^{\prime } \gamma _3^{\prime }}{\beta _3^{\prime } -\cdots }}} =0 \end{aligned}$$
(3.16)

which, in terms of the unprimed parameters, is equivalent to

$$\begin{aligned} det\left( M_{N\times N}\right)= & {} \begin{vmatrix} \beta _0&\alpha _0&&&\\ \gamma _1&\beta _1&\alpha _1&&&\\ \delta _2&\gamma _2&\beta _2&\alpha _2&&\\&\delta _3&\gamma _3&\beta _3&\alpha _3&&\\&\delta _4&\gamma _4&\beta _4&\alpha _4&\\&&\ddots&\ddots&\ddots&\ddots&\\&&\delta _{N-1}&\gamma _{N-1}&\beta _{N-1}&\alpha _{N-1} \\&&&\delta _{N}&\gamma _{N}&\beta _{N} \end{vmatrix} \nonumber \\= & {} 0, \end{aligned}$$
(3.17)

or

$$\begin{aligned} d_N =0 \end{aligned}$$
(3.18)

in the limit \(N\rightarrow \infty \).

We now show why y(z) is convergent at \(z=1\) if \(d_n\) in Eq. (3.9) is a minimal solution. We rewrite Eq. (3.9) as

$$\begin{aligned} d_{n+1}+ A_n\;d_n + B_n\;d_{n-1}+ C_n\;d_{n-2}=0, \end{aligned}$$
(3.19)

where \(A_n\), \(B_n\) and \(C_n\) have asymptotic expansions of the form

$$\begin{aligned} {\left\{ \begin{array}{ll} A_n= \frac{\beta _n}{\alpha _n} \sim \sum _{j=0}^{\infty }\frac{a_j}{n^j} \\ B_n= \frac{\gamma _n}{\alpha _n} \sim \sum _{j=0}^{\infty }\frac{b_j}{n^j} \\ C_n= \frac{\delta _n}{\alpha _n} \sim \sum _{j=0}^{\infty }\frac{c_j}{n^j} \end{array}\right. } \end{aligned}$$
(3.20)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} a_0= 0, \quad a_1= -\frac{2\lambda _{g,3}}{\sqrt{3}}, &{}\quad a_2= \frac{2\Delta \lambda _{g,3} }{\sqrt{3} } \\ b_0= 0, \quad b_1= -\frac{2\lambda _{g,3}}{\sqrt{3}}, &{}\quad b_2= \frac{(2\Delta +1)\lambda _{g,3} }{\sqrt{3} } \\ c_0= -1, \quad c_1= 3+\frac{4\lambda _{g,3}}{\sqrt{3}}, &{}\quad c_2= -3-\frac{4\lambda _{g,3}}{\sqrt{3}}-(\frac{2\lambda _{g,3}}{\sqrt{3}}+\Delta )^2 . \end{array}\right. }\nonumber \\ \end{aligned}$$
(3.21)

The radius of convergence, \(\rho \), satisfies characteristic equation associated with Eq. (3.19) [13,14,15]:

$$\begin{aligned} \rho ^3+ a_0\rho ^2+b_0\rho ^1+c_0\rho =\rho ^3 -1= 0, \end{aligned}$$
(3.22)

whose roots are given by

$$\begin{aligned} \rho _1= & {} 1,\quad \rho _2 = \frac{-1+\sqrt{3}i}{2},\nonumber \\ \rho _3= & {} \frac{-1-\sqrt{3}i}{2}. \end{aligned}$$
(3.23)

So for a four-term recurrence relation in Eq. (3.9), the radius of convergence is 1 for all three cases. Since the solutions should converge at the horizon, y(z) should be convergent at \(|z|\le 1\). According to Pincherle’s Theorem [16], we have a convergent solution of y(z) at \(|z|=1\) if only if the four term recurrence relation Eq. (3.9) has a minimal solution. Since we have three different roots \(\rho _i\)’s, so Eq. (3.19) has three linearly independent solutions \(d_1(n)\), \(d_2(n)\), \(d_3(n)\). One can show that [16] for the large n,

$$\begin{aligned} d_i(n)\sim \rho _i^n n^{\alpha _i}\sum _{r=0}^{\infty } \frac{\tau _i(r)}{n^r},\quad i=1,2,3 \end{aligned}$$
(3.24)

with

$$\begin{aligned} \alpha _i=\frac{a_1\rho _i^2+b_1\rho _i+c_1}{a_0\rho _i^2+2b_0\rho _i+ 3 c_0 },\quad i=1,2,3 \end{aligned}$$
(3.25)

and \(\tau _i(0)=1\). In particular, we obtain

$$\begin{aligned}&\tau _i(1)= \frac{(-a_0 \rho _i^2+3c_0)\alpha _i^2+(a_0\rho _i^2-2b_1\rho _i-3c_0-4c_1)\alpha _i+2(a_2\rho _i^2+b_2\rho _i+c_2)}{2(a_0\rho _i^2+2b_0\rho _i+3c_0)\alpha _i-2((a_0+a_1)\rho _i^2+(b_1+2b_0)\rho _i+c_1+3c_0)}, \quad i=1,2,3 \end{aligned}$$
(3.26)

Substituting Eqs. (3.23) and (3.21) into Eqs. (3.24)–(3.26), we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} d_1(n)\sim n^{-1}\left( 1+\frac{-3\Delta ^2+(3\sqrt{3}-4\lambda _{g,3})\lambda _{g,3}}{9n} \right) \\ d_2(n)\sim \left( \frac{-1+\sqrt{3}i}{2} \right) ^n n^{-1-\frac{2\lambda _{g,3}}{\sqrt{3}}}\left( 1+\frac{\varpi -\chi i}{n} \right) \\ d_3(n)\sim \left( \frac{-1-\sqrt{3}i}{2} \right) ^n n^{-1-\frac{2\lambda _{g,3}}{\sqrt{3}}}\left( 1+\frac{\varpi +\chi i}{n} \right) \end{array}\right. } \end{aligned}$$
(3.27)

with

$$\begin{aligned} {\left\{ \begin{array}{ll} \varpi =\frac{-6\Delta ^2-12\sqrt{3}\Delta \lambda _{g,3}+(15\sqrt{3}+16\lambda _{g,3})\lambda _{g,3}}{18} \\ \chi = \frac{(3+4\sqrt{3})\lambda _{g,3}}{18}.\end{array}\right. } \end{aligned}$$
(3.28)

Since \(\lambda _{g,3}>0\),

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{ d_2(n)}{ d_1(n)}=0, \quad \lim _{n\rightarrow \infty }\frac{ d_3(n)}{ d_1(n)}=0. \end{aligned}$$
(3.29)

Therefore \(d_2(n)\) and \(d_3(n)\) are minimal solutions. Also,

$$\begin{aligned} {\left\{ \begin{array}{ll} \sum |d_1(n)|\sim \sum \frac{1}{n}\rightarrow \infty \\ \sum |d_2(n)|\sim \sum n^{-1-\frac{2\lambda _{g,3}}{\sqrt{3}}}<\infty \\ \sum |d_3(n)|\sim \sum n^{-1-\frac{2\lambda _{g,3}}{\sqrt{3}}} <\infty . \end{array}\right. } \end{aligned}$$
(3.30)

Therefore, \(y(z)=\sum _{n=0}^{\infty }d_n z^n\) is convergent at \(z=1\) if only if we take \(d_2\) and \(d_{3}\) which are minimal solutions.

Equation (3.17) becomes a polynomial of degree N with respect to \(\lambda _{g,3}\). The algorithm to find \(\lambda _{g,3}\) for a given \(\Delta \) is as follows:

  1. 1.

    Choose an N.

  2. 2.

    Define a function returning the determinant of system Eq. (3.17).

  3. 3.

    Find the roots of interest of this function.

  4. 4.

    Increase N until those roots become constant to within the desired precision [11].

3.2 Presence of unphysical regime: \(\frac{1}{2}<\Delta <1\)

We numerically compute the determinant to locate its roots. We are only interested in smallest positive real roots of \(\lambda _{g,3}\). Taking \(N=32\), we first compute the roots and then find an approximate fitting function, which turns out to be given by

$$\begin{aligned} \lambda _{g,3} \approx 1.96 \Delta ^{4/3} - 0.87\quad \text{ for }\ 1\le \Delta \le 3. \end{aligned}$$
(3.31)

However, for \(1/2< \Delta <1\), we will see that there is no convergent solution, because there are three branches so that it is impossible to get an unique value \(\lambda _{g,3}\) no matter how large N is. See the Fig. 1b. Notice, however, that these three branches merge to the single value \(\lambda _{g,3} \approx 1\) as N increases as Fig. 1b shows.

We now want to understand analytically why three branches occur near \(\Delta =1/2\) regardless of the size of N. Equation (3.17) can be simplified using the formula for the determinant of a block matrix,

$$\begin{aligned} \det \begin{pmatrix} A &{} B \\ C &{} D \end{pmatrix}= \det (A)\det (D-C A^{-1} B), \quad \text {with} \end{aligned}$$
(3.32)
$$\begin{aligned}&A= \begin{pmatrix} \beta _0 &{} \alpha _0 \\ \gamma _1 &{} \beta _1 \end{pmatrix} , \quad B=\begin{pmatrix} 0 &{} 0 &{} \cdots &{} 0 \\ \alpha _1 &{} 0 &{} \cdots &{} 0 \end{pmatrix} , \quad C=\begin{pmatrix} \delta _2 &{} \gamma _2 \\ 0 &{} \delta _3 \\ 0 &{} 0 \\ 0 &{} 0 \\ \vdots &{} \vdots \\ 0 &{} 0 \end{pmatrix} ,\\&\quad D=\begin{pmatrix} \beta _2 &{} \alpha _2 &{} &{} &{} &{} \\ \gamma _3 &{} \beta _3 &{} \alpha _3 &{} &{} &{} \\ \delta _4 &{} \gamma _4 &{} \beta _4 &{} \alpha _4 &{} &{} \\ &{} \ddots &{} \ddots &{} \ddots &{} \ddots &{} \\ &{} &{} \delta _{N-1} &{} \gamma _{N-1} &{} \beta _{N-1} &{} \alpha _{N-1} \\ &{} &{} &{} \delta _{N} &{} \gamma _{N} &{} \beta _{N} \end{pmatrix}. \end{aligned}$$

By explicit computation, we can see the factor \(\det (A)=9 \lambda ^2\) at \(\Delta =1/2\) so that the minimal real root is \(\lambda _{g,3}=0\).

Near \(\Delta =1/2\), we can expand the determinant as a series in \( \varepsilon =\Delta - 1/2 \ll 1\) and \(0< \lambda _{g,3}\ll 1\). After some calculations, we found that \(d_N =0\) gives following results:

  1. 1.

    For \(N=3m\) with positive integer m,

    $$\begin{aligned}&\lambda _{g,3}^3 \sum _{n=0}^{N-2} \alpha _{1,n} \lambda _{g,3}^n +\varepsilon \lambda _{g,3} \sum _{n=0}^{N} \beta _{1,n} \lambda _{g,3}^n \nonumber \\&\quad + {\mathcal {O}}(\varepsilon ^2) =0. \end{aligned}$$
    (3.33)

    This leads us \(\lambda _{g,3} \sim \varepsilon ^{1/2} \sim (\Delta -1/2)^{1/2}\) as far as \(\alpha _{1,0}\beta _{1,0}\ne 0\), which can be confirmed by explicit computation. This result does not depends on the size of N. Similarly,

  2. 2.

    For \(N=3m+1\),

    $$\begin{aligned}&\lambda _{g,3}^2 \sum _{n=0}^{N-1} \alpha _{2,n} \lambda _{g,3}^n +\varepsilon \lambda _{g,3} \sum _{n=0}^{N} \beta _{2,n} \lambda _{g,3}^n \nonumber \\&\quad + {\mathcal {O}}(\varepsilon ^2) =0, \end{aligned}$$
    (3.34)

    giving us \(\lambda _{g,3} \sim (\Delta -1/2)\).

  3. 3.

    For \(N=3m+2\),

    $$\begin{aligned} \lambda _{g,3}^3 \sum _{n=0}^{N-2} \alpha _{3,n} \lambda _{g,3}^n +\varepsilon \sum _{n=0}^{N+1} \beta _{3,n} \lambda _{g,3}^n + {\mathcal {O}}(\varepsilon ^2) =0,\nonumber \\ \end{aligned}$$
    (3.35)

    leading to \(\lambda _{g,3} \sim (\Delta -1/2)^{1/3}\).

These results prove the presence of three branches near \(\Delta =1/2\) (Fig. 2).

We numerically calculated 101 different values of \(\lambda _{g,3}\)’s at various \(\Delta \) and the result is the red colored curve in Fig. 3. These data fits well by above formula.

Fig. 1
figure 1

a \(\lambda _{g,3}\) vs \(\Delta \): green, blue and red colored curves for \(\lambda _{g,3}\) are obtained by letting \(d_N =0\), with \(N=30, 31, 32 \) respectively. There are three branches in \(1/2<\Delta <1\). And such branches merge for \(\Delta \ge 1\). b Zoom in of figure a for the regime \(1/2<\Delta <1\)

Fig. 2
figure 2

a Shaded area should be eliminated due to the non-uniqueness of eigenvalues \(\lambda _{g,3}\) shown in a, b at Fig. 1. This naturally eliminates \(\Delta =1/2\) where \(T_{c}\) diverges. We have set \(g=\rho =1\). b Similarly, shaded area at \(1 \le \Delta < 3/2\) should be eliminated in \(AdS_5\) due to the non-uniqueness of eigenvalues \(\lambda _{g,4}\). This naturally eliminates \(\Delta =1\) where \(T_{c}\) diverges. There are two branches in \(AdS_5\) unlike \(AdS_4\)

Fig. 3
figure 3

\(\lambda _{g,d}\) vs \(\Delta \): red line for \(\lambda _{g,3}\) and blue line for \(\lambda _{g,4}\). The results of variational method in Ref. [8] is denoted by red and blue dotted lines for AdS\(_{4}\) and AdS\(_{5}\) respectively

The authors of Ref. [8] got \(\lambda _{g,3}\)’s by using variational method using the fact that the eigenvalue \(\lambda _{g,3}\) minimizes the expression

$$\begin{aligned} \lambda _{g,3}^2 = \frac{\int _{0}^{1}dz\; z^{2\Delta -2}\left( (1-z^3)\left[ F^{\prime }(z)\right] ^2+\Delta ^2 z \left[ F(z)\right] ^2\right) }{\int _{0}^{1}dz\; z^{2\Delta -2}\frac{1-z}{1+z+z^2}\left[ F(z)\right] ^2} .\nonumber \\ \end{aligned}$$
(3.36)

for \(\Delta >1/2\). The integral does not converge at \(\Delta =1/2\) because of \(\ln (z)\). The trial function used is \(F(z)=1-\alpha z^2\) where \(\alpha \) is the variational parameter. Their result is given by the red dotted line in Fig. 3. While the variational method tells us that there are numerical values of \(\lambda _{g,3}\) for \(1/2<\Delta <1\), our method tells us that this region does not allow well defined value of \(\lambda _{g,3}\), hence \(T_{c}\) is not defined there.

The critical temperature is given by \( T_c =\frac{3}{4\pi }\sqrt{\frac{\rho }{\lambda _3}} \), so that it can be calculated by once \(\lambda \) is given. Notice that \(T_c\) is a monotonically decreasing function of \(\Delta \). See Fig. 2a. Similary see Fig. 2b for AdS5 case.

Similar statements are true for AdS5: Depending on even-ness or odd-ness of n, there are two branches if \(1<\Delta <1.5\). Two branches merge in \(\Delta \ge 1.5\) for AdS5. For more detail, see Appendix B.1.2.

4 The condensation near critical temperature

Substituting Eq. (3.7) into Eq. (2.3), the field equation \(\Phi \) becomes

$$\begin{aligned} \frac{d^2 \Phi }{d z^2} = \frac{g^2\langle {\mathcal {O}}_{\Delta }\rangle ^2}{ r_h^{2\Delta }}\frac{z^{2(\Delta -1)} F^{2}(z)}{1-z^3 } \Phi , \end{aligned}$$
(4.1)

where \( {g\langle {\mathcal {O}}_{\Delta }\rangle ^2}/{ r_h^{2\Delta }}\) is small because \(T\approx T_c\). The above equation have the expansion around Eq. (3.5) with small correction [8]:

$$\begin{aligned} \frac{ \Phi }{r_h} = \lambda _3(1-z)+ \frac{g^2\langle {\mathcal {O}}_{\Delta }\rangle ^2}{ r_h^{2\Delta }}\chi _1(z). \end{aligned}$$
(4.2)

We have \(\chi _1(1)=\chi _1^{\prime }(1)=0\) due to the boundary condition \(\Phi (1)=0 \). Taking the derivative of Eq. (4.2) twice with respect to z and using the result in Eq. (4.1),

$$\begin{aligned} \chi _1^{\prime \prime }&=\frac{z^{2(\Delta -1)}F^2(z)}{1-z^3}\left\{ \lambda _3(1-z)+ \frac{g^2\langle {\mathcal {O}}_{\Delta }\rangle ^2}{ r_h^{2\Delta }}\chi _1 \right\} \nonumber \\&\approx \frac{\lambda _3 z^{2(\Delta -1)}F^2(z)}{z^2+z+1}. \end{aligned}$$
(4.3)

Integrating Eq. (4.3) gives us

$$\begin{aligned} \chi _1^{\prime }(0) = -\lambda _3 {\mathcal {C}}_3 \quad \hbox {for}\ {\mathcal {C}}_3 = \int _{0}^{1}dz\;\frac{ z^{2(\Delta -1)}F^2(z)}{z^2+z+1}. \end{aligned}$$
(4.4)

Equation (3.7) with Eq. (3.10) shows

$$\begin{aligned} F(z)&= (z^2+z+1)^{-\lambda _{g,3}/\sqrt{3}}y(z)\nonumber \\&\approx \left( z^2+z+1\right) ^{-\frac{\lambda _{g,3}}{\sqrt{3}}} \sum _{n=0}^{15}d_n z^n . \end{aligned}$$
(4.5)

Here, we ignore \(d_n z^n\) terms if \(n\ge 16\) because \(0<|d_n|\ll 1\) numerically and y(z) converges for \(0\le z\le 1\).

We can calculate the numerical value of \(\sqrt{1/{\mathcal {C}}_3}\) by putting Eqs. (3.31) and (3.10) into Eqs. (4.4) and (4.5). We calculated 102 different values of \(\sqrt{1/{\mathcal {C}}_3}\)’s at various \(\Delta \), which is drawn as dots in Fig. 4. Then we tried to find an approximate fitting function. The result is given as follows,

$$\begin{aligned} \sqrt{ \frac{1}{{\mathcal {C}}_3} } \approx \frac{\Delta ^6 +120 \Delta ^{1.5}}{84}. \end{aligned}$$
(4.6)
Fig. 4
figure 4

a \(\sqrt{1/{\mathcal {C}}_3}\) data by Eqs. (4.4) and (4.5) with Eq. (3.31), as functions of \(\Delta \). Red colored curve is the plot of Eq. (4.6). b \(\sqrt{1/{\mathcal {C}}_4}\) data by Eqs. (B.25) and (B.26) with Eq. (B.19), as functions of \(\Delta \). Blue colored curve is the plot of Eq. (B.27)

Figure 4 shows how the data fits by above formula.

From the Eqs. (4.2) and (2.4), we have

$$\begin{aligned} \frac{\rho }{r_h^2}=\lambda _3\left( 1+\frac{{\mathcal {C}}_3 g^2 \langle {\mathcal {O}}_{\Delta }\rangle ^2}{r_h^{2\Delta }}\right) . \end{aligned}$$
(4.7)

Putting \(T=\frac{3}{4\pi } r_h \) with \(\lambda _3=\frac{\rho }{r_c^2}\) into Eq. (4.7), we obtain the condensate near \(T_c\):

$$\begin{aligned} g\frac{\langle {\mathcal {O}}_{\Delta }\rangle }{T_c^{\Delta }}\approx {\mathcal {M}}_3 \; \sqrt{1-\frac{T}{T_c}} \quad \hbox {for } {\mathcal {M}}_3 = \left( \frac{4\pi }{3}\right) ^{\Delta }\sqrt{\frac{2}{{\mathcal {C}}_3}}. \end{aligned}$$
(4.8)

In Ref. [8] it was argued that \(\lim _{\Delta \rightarrow d}{\mathcal {C}}_d=0\), which would lead to the divergence of the condensation in Eq. (4.4). However, our result shows that \(\lim _{\Delta \rightarrow d}{\mathcal {C}}_d =finite \) so that Eq. (4.4) is finite, which can be confirmed in the Fig. 5. The condensate is an increasing function of the \(\Delta \) but it decreases with increasing T.

Fig. 5
figure 5

\( {g^{\frac{1}{ \Delta }}\langle O_{\Delta } \rangle ^{\frac{1}{\Delta }}}/{T_c}\) vs \(\Delta \) for a few T’s near \(T_{c}\). Sold curves are for AdS\(_{4}\) and dotted ones are for AdS\(_{5}\)

As we substitute Eq. (3.3) into Eq. (4.8), we obtain

$$\begin{aligned} g\frac{\langle {\mathcal {O}}_{\Delta }\rangle }{(g \rho )^{\frac{\Delta }{2}} }\approx \lambda _{g,3}^{-\frac{\Delta }{2}} \sqrt{\frac{2}{{\mathcal {C}}_3}} \; \sqrt{1-\frac{T}{T_c}}. \end{aligned}$$
(4.9)

The square root temperature dependence is typical of a mean field theory [1, 8, 17]. Our main interest here is the \(\Delta \) dependence of the \(\mathcal{M}_{3}\), especially the singular dependence through \(\mathcal{C}_{3}\) whose values for some particular value of \(\Delta \) was obtained before: for \(\Delta =1\), we have \({\mathcal {M}}_3 =8.53\) which is in good agreement with the \({{\mathcal {M}}}_3 =9.3\) [1]. For \(\Delta =2\), we have \({{\mathcal {M}}}_3 =119.17\) which roughly agrees with the results \({{\mathcal {M}}}_3 =119\) of Ref. [18] and \({{\mathcal {M}}}_3 =144\) of Ref. [1]. We obtained the approximate results for general \({\mathcal {C}}_d\). See Eqs. (4.6) and (B.27) in the Appendix. For large \(\Delta \), \({\mathcal {C}}_d \sim \Delta ^{-(d+9)}\). We conclude that we do not have a singular dependence of the condensation anywhere for the s-wave holographic superconductivity, which is different from the result of Ref. [8]. See the Fig. 5.

5 The AC conductivity for \(\Delta =1,2\) in \(2+1\)

The Maxwell equation for the planar wave solution with zero spatial momentum and frequency \( \omega \) is

$$\begin{aligned}&r_{+}^2 (1-z^3)^2 \frac{d^2 A_x }{d z^2} -3 r_{+}^2 z^2 (1-z^3)\frac{d A_x }{d z}\nonumber \\&\quad +\left( \omega ^2-V(z) \right) A_x=0, \end{aligned}$$
(5.1)

where \(A_x\) is the perturbing electromagnetic potential and

$$\begin{aligned} V(z)= \frac{g^2 \langle {\mathcal {O}}_{\Delta }\rangle ^2}{r_{+}^{2\Delta -2}}(1-z^3) z^{2\Delta -2} F(z)^2, \end{aligned}$$

with F defined as before. To request the ingoing boundary conditions at the horizon, \(z=1\), we introduce G(z) by \(A_x(z)=(1-z)^{-\frac{i}{3} {\hat{\omega }}} G(z)\) where \({\hat{\omega }}=\omega /r_{+}\). Then the wave equation (5.1) reads

$$\begin{aligned}&(1-z^3) \frac{d^2 G }{d z^2} + \left( -3 z^2+\frac{2i {\hat{\omega }}}{3} (1+z+z^2) \right) \frac{d G }{d z}\\&\quad +\left( \frac{(2+z)(4+z+z^2)}{9(1+z+z^2)}{\hat{\omega }}^2 + \frac{i{\hat{\omega }}}{3} (1+2z) \right. \nonumber \\&\quad \left. - \frac{g^2\langle {\mathcal {O}}_{\Delta }\rangle ^2}{r_{+}^{2\Delta }} z^{2\Delta -2} F^2(z) \right) G=0\nonumber \end{aligned}$$
(5.2)

If the asymptotic behaviour of the Maxwell field at large r is given by

$$\begin{aligned} A_x = A_x^{(0)}+\frac{ A_x^{(1)}}{r}+\cdots , \end{aligned}$$
(5.3)

then the conductivity is given by

$$\begin{aligned} \sigma (\omega ) =\frac{1}{i \omega } \frac{A_x^{(1)}}{A_x^{(0)}} = \frac{1}{i {\hat{\omega }}} \frac{\frac{d G(0) }{d z}+\frac{i{\hat{\omega }}}{3}G(0)}{G(0)} . \end{aligned}$$
(5.4)

Near the \(T=0\), the Eq. (5.2) is simplified to

$$\begin{aligned}&\frac{d^2 G }{d z^2} +\frac{2i {\hat{\omega }} }{3}\frac{d G }{d z} \nonumber \\&\quad +\left( \frac{8}{9} {\hat{\omega }}^2 +\frac{i {\hat{\omega }}}{3} - \frac{g^2\langle {\mathcal {O}}_{\Delta }\rangle ^2}{r_{+}^{2\Delta }} z^{2\Delta -2} F(z)^2 \right) G=0. \end{aligned}$$
(5.5)

For \(\Delta =1\), \(F(z)\approx 1\) so that the solution of Eq. (5.5) is

$$\begin{aligned}&G(z)= \exp \left( i z\left( -\frac{\omega }{3}+\sqrt{ \omega ^2+\frac{i \omega }{3}-\frac{g^2\langle {\mathcal {O}}_{1}\rangle ^2}{r_{+}^{2}}}\right) \right) \\&\quad +R\; \exp \left( i z\left( -\frac{\omega }{3}-\sqrt{ \omega ^2+\frac{i \omega }{3}-\frac{g^2\langle {\mathcal {O}}_{1}\rangle ^2}{r_{+}^{2}}}\right) \right) . \end{aligned}$$

Here, R is a constant called reflection coefficient. Taking the zero temperature limit \(T \rightarrow 0\) is equivalent to sending the horizon to infinity. Then the in-falling boundary condition corresponds to \(R=0\). Then it gives the conductivities,

$$\begin{aligned} \sigma (\omega ) =\frac{g\langle {\mathcal {O}}_{1}\rangle }{\omega } \sqrt{\left( 1+\frac{i r_{+}}{3 \omega } \right) \left( \frac{\omega }{g\langle {\mathcal {O}}_{1}\rangle } \right) ^2-1} . \end{aligned}$$
(5.6)

Compare Fig. 6a with Fig. 6c. Similarly, for \(\Delta =2\), we can obtain the conductivity given as follow,

$$\begin{aligned} \sigma (\omega ) =\frac{3i\sqrt{g\langle {\mathcal {O}}_{2}\rangle }}{ \sqrt{2}\omega } \frac{\Gamma \left( 0.24-\frac{4i}{9}\sqrt{P(\omega )} \right) }{\Gamma \left( -0.26-\frac{4i}{9}\sqrt{P(\omega )} \right) } \frac{\Gamma \left( 1.26-\frac{4i}{9}\sqrt{P(\omega )} \right) }{\Gamma \left( 0.76-\frac{4i}{9}\sqrt{P(\omega )} \right) } \nonumber \\ \end{aligned}$$
(5.7)

where

$$\begin{aligned} P(\omega ) = \frac{9}{8}\left( 1+\frac{i r_{+}}{3 \omega } \right) \left( \frac{\omega }{\sqrt{g\langle {\mathcal {O}}_{2}\rangle }}\right) ^2 -1. \end{aligned}$$

This result fits the numerical data almost exactly as one can see in Fig. 6b. And it is consistent with the result of Ref. [7]; compare Fig. 6b with Fig. 6d. For derivation of these results, the Appendix A.2.1.

Fig. 6
figure 6

Analytic results (real lines) vs numerical results (dotted lines). a, b Plots of \(Re (\sigma (\omega ))\) in Eqs. (5.6) and (5.7). Dot points are numerical results in Fig. 3 of Ref. [1]. c, d Plots of \(Im (\sigma (\omega ))\). Dot points are from Fig. 4 of Ref. [1]. In all cases \(T/T_c=0.1\)

To request the ingoing boundary conditions at the horizon, \(z=1\), we introduce H(z) by \(A_x(z)=(1-z^3)^{-\frac{i}{3} {\hat{\omega }}} H(z)\) where \({\hat{\omega }}=\omega /r_{+}\). Then the wave equation (5.1) reads

$$\begin{aligned}&(1-z^3) \frac{d^2 H }{d z^2} -3 \left( 1-\frac{2i {\hat{\omega }}}{3} \right) z^2 \frac{d H }{d z}\nonumber \\&\quad +\left( \frac{{\hat{\omega }}^2(1+z)(1+z^2)}{1+z+z^2} +2i{\hat{\omega }}z - \frac{g^2\langle {\mathcal {O}}_{\Delta }\rangle ^2}{r_{+}^{2\Delta }} z^{2\Delta -2} F^2(z) \right) H=0.\nonumber \\ \end{aligned}$$
(5.8)

The boundary conditions at the horizon are [19]

$$\begin{aligned} H(1)=1, \quad \lim _{z\rightarrow 1}(1 -z^3)^{-\frac{i}{3}{\hat{\omega }}}H^{\prime }(z)=0. \end{aligned}$$

To evaluate the conductivities at low frequency, it is enough to obtain H(z) up to first order in \(\omega \),

$$\begin{aligned} H(z)= H_0(z) + \omega H_1(z) + {\mathcal {O}}(\omega ^2) . \end{aligned}$$
(5.9)

Inserting this into Eq. (5.8), \(H_0(z)\) and \(H_1(z)\) satisfy

$$\begin{aligned}&(1-z^3) H_0^{\prime \prime }-3 z^2 H_0^{\prime }- \Delta ^2 b^{2\Delta } z^{2\Delta -2} F(z)^2 H_0 =0, \end{aligned}$$
(5.10)
$$\begin{aligned}&(1-z^3) H_1^{\prime \prime }-3 z^2 H_1^{\prime }- \Delta ^2 b^{2\Delta } z^{2\Delta -2} F(z)^2 H_1\nonumber \\&\quad = -2iz(z H_0^{\prime }+ H_0 ) . \end{aligned}$$
(5.11)

where \(b^{\Delta } = \frac{g\langle {\mathcal {O}}_{\Delta }\rangle }{\Delta r_{+}^{ \Delta }}\). Near the \(T=0\) we can simplify two coupled equations (5.10) and (5.11) as

$$\begin{aligned} H_0^{\prime \prime } - \Delta ^2 b^{2\Delta } z^{2\Delta -2} H_0= & {} 0, \end{aligned}$$
(5.12)
$$\begin{aligned} H_1^{\prime \prime } - \Delta ^2 b^{2\Delta } z^{2\Delta -2} H_1= & {} -2i z H_0 . \end{aligned}$$
(5.13)

The conductivity is given by

$$\begin{aligned} \sigma (\omega ) =\frac{1}{i \omega } \frac{A_x^{(1)}}{A_x^{(0)}} = \frac{1}{i {\hat{\omega }}} \frac{\frac{d H(0) }{d z} }{H(0)} . \end{aligned}$$
(5.14)

The solution of Eq. (5.14) is given in Eq. (A.1). Here, \(n_s\) is the coefficient of the pole in the imaginary part \(\Im {\sigma (\omega )}\sim n_s/\omega \) as \(\omega \rightarrow 0\). For derivation of these results, see the Appendix A.2.1. For the \(\Delta \) values other than 1 or 2, there is no analytic result available at this moment.

6 Discussion

One problem is that [8,9,10] the critical temperature is divergent at the \(\Delta =1/2\), which does not seems to make physical sense and it has not been understood as far as we know. This was also noticed as a problem [7] but the reason for it has not been cleared yet.

In this paper, we consider the problem of divergence of the critical temperature at \(\Delta =1/2\) by recalculating \(T_{c}\) using Pincherle’s theorem [11] to handle the Heun’s equation. We find that the region of \(1/2\le \Delta <1\) for AdS\(_{4}\) does not have well defined critical temperature. Similar phenomena also occur in AdS5. We also computed the AC conductivity gap \(\omega _{g}\) and in this same regime, it does not exist either. The situation is similar to the physics of the pseudo gap where Cooper pairs are formed but the phase alignment of the pairs are absent. In the future work, we will work out the same phenomena in other background and also for non s-wave situation, to confirmed the universality of the phenomena.