Thermality of horizon through near horizon instability: a path integral approach

Abstract

Recent investigations revealed that the near horizon Hamiltonian of a massless, chargeless outgoing particle, for its particular motion in static as well as stationary black holes, is effectively $\sim xp$ kind. This is unstable by nature and has the potential to explain a few interesting physical phenomena. From the path integral kernel, we first calculate the density of states. Also, following the idea of Singh and Padmanabhan (Phys Rev D 85:025011, 2012. https://doi.org/10.1103/PhysRevD.85.025011. arXiv:1112.6279 [hep-th]) here, in the vicinity of the horizon, we calculate the effective path corresponding to its Schrodinger version of Hamiltonian through the path integral approach. The latter result appears to be complex in nature and carries the information of escaping the probability of the particle through the horizon. In both ways, we identify the correct expression of Hawking temperature. Moreover, here we successfully extend the complex path approach to a more general black hole like Kerr spacetime. We feel that such a complex path is an outcome of the nature of near horizon instability provided by the horizon and, therefore, once again bolstered the fact that the thermalization mechanism of the horizon may be explained through the aforesaid local instability.

Appendices

Appendices

Appendix A: Alternative approach to DOS

In order to evaluate (9) we first transform Hamiltonian (7) to that of an inverted harmonic oscillator (IHO)($H=(\kappa /2)(P^2-X^2)$) by going into a new canonical variable

$$\begin{aligned} x=\frac{1}{\sqrt{2}}(P-X); \,\,\,\ p = \frac{1}{\sqrt{2}}(P+X). \end{aligned}$$

(A1)

The propagator of the IHO can be obtained from that of harmonic oscillator by replacing frequency $\omega \rightarrow i\omega$. For (A1) frequency is identified as $\omega =\kappa$ and hence the propagator in this case comes out to be

$$\begin{aligned} K(x_2=X,t_2=t; x_1=X,t_1=0) = \Big (\frac{i}{-2\pi \sinh (\kappa t)}\Big )^{1/2}\exp \Big [ibX^2\Big ], \end{aligned}$$

(A2)

where $b=\frac{1}{\sinh (\kappa t)}\{\cosh (\kappa t) - 1\}$ (to find (A2) substitute $\omega = \kappa \rightarrow i\omega = i\kappa$ and $m=1/\kappa$ as well as $x_f=x_2=x_i=x_1=X, t_i=t_1=0, t_f=T=t_2=t$ in Eq. 3.81 combined with Eq. 3.66 of [26]). Then we have

$$\begin{aligned} \int _{-\infty }^\infty dX~G(X,E)= & {} \int _0^\infty dt e^{-\frac{i E t}{\hbar }} \int _0^\infty dx~K(x_2=X,t_2=t; x_1=X,t_1=0)\nonumber \\= & {} \int _0^\infty dt~ \frac{e^{-\frac{i E t}{\hbar }}}{2\sinh (\frac{\kappa t}{2})}. \end{aligned}$$

(A3)

In the above the X integration has been done by replacing $b\rightarrow b+i\epsilon$ for $\epsilon >0$ and at the end $\epsilon \rightarrow 0$ has been taken.

The final t integration can be done by replacing $t\rightarrow t-i\delta$ with $\delta >0$ and then choosing the contour in the fourth quadrant in complex t plane. This will yield (13). The explicit steps are as follows. First change the integration variable as $t\rightarrow 2t/\kappa$. This yields

$$\begin{aligned} \dfrac{1}{2}\int _{0}^{\infty }~dt~\dfrac{e^{-\frac{iEt}{\hbar }}}{\sinh (\frac{\kappa t}{2})}=\dfrac{1}{\kappa }\int _{0}^{\infty }~dt~\dfrac{e^{-i\frac{2 E}{\hbar \kappa }t}}{\sinh (t)}. \end{aligned}$$

(A4)

Fig. 1

Contour plot for integration. The integration is done considering t as a complex variable. The integration in the range $[0,\infty )$ is infinitesimally shifted from the real axis for the integral to be convergent

The integration is done using a suitable contour, as shown in Fig. 1. The motivation for choosing the contour can be found in [28]. There are no poles inside the contour and hence, the total integral over the closed contour vanishes. However, the integration over the small semi-circles is non-zero. Hence, our required integral is negative of the sum of these integrals over the semi-circle:

$$\begin{aligned} \int _{{\mathcal {C}}}dt~f(t)+\int _{{\mathcal {C}}'}dt~ f(t)+\sum _{n=1}\int _{{\mathcal {C}}_{n}}dt ~f(t)+\int _{{\mathcal {D}}}dt~f(t)=0;~~f(t)=\dfrac{e^{-i\frac{2Et}{\kappa \hbar }}}{\sinh (t)}.\nonumber \\ \end{aligned}$$

(A5)

In the above ${\mathcal {D}}$ is the path on the imaginary axis between to consecutive poles. If we look at the integration over a portion of the contour along the imaginary axis, it looks as follows

$$\begin{aligned} \int _{-i\pi +i\epsilon }^{0}\dfrac{e^{-i\frac{2Et}{\hbar \kappa }}}{\sinh (t)} dt=\int _{\pi -\epsilon }^{0}\dfrac{e^{-\frac{2Et}{\hbar \kappa }}}{\sin (t)} dt \end{aligned}$$

This is a Real quantity. Hence, the integration over ${\mathcal {D}}$ is real, and won’t contribute to the imaginary part. In this, the integral on the circular part, i.e, on the contour ${\mathcal {C}}'$ goes to zero as we take the limit of radius going to infinity as there is $\sinh (t)$ term in the denominator. We are interested in the imaginary part of the integration. This comes from the contribution made by integration over the infinitesimal semi-circles drawn over the poles. Using the Residue theorem, the integration on small semi-circular deformations will give us,

$$\begin{aligned} \int _{{\mathcal {C}}_{n}}dt~f(t)=-\pi i\text {Res}(f(t_{n}));~~~t_{n}=-in\pi ,~~n\in \mathbb {N}. \end{aligned}$$

(A6)

The negative sign is due to the direction in which the contour is traversed along. Note that $n=0$ is not included as for this we will have $t_0 = i\delta$ which is outside our chosen contour. Hence, we have

$$\begin{aligned} \Im \left( \int _{0}^{\infty }~dt~\dfrac{e^{-i\frac{2 E}{\hbar \kappa }t}}{\sinh (t)}\right) =\pi \sum _{n=1}^{\infty }\text {Res}(f(t_{n})). \end{aligned}$$

(A7)

The Residue at the poles are given as follows

$$\begin{aligned} \text {Res}(f(t_{n}))=(-1)^{n}e^{-\frac{2En\pi }{\hbar \kappa }};~~t_{n}=-in\pi . \end{aligned}$$

(A8)

Hence, we have,

$$\begin{aligned} \Im \left( \dfrac{1}{2}\int _{0}^{\infty }~dt~\dfrac{e^{-\frac{iEt}{\hbar }}}{\sinh (\frac{\omega t}{2})}\right) =\dfrac{\pi }{\omega }\sum _{n=1}^{\infty }\big (-e^{-\frac{2E\pi }{\hbar \kappa }}\big )^{n}=-\dfrac{\pi }{\kappa }\dfrac{e^{-\frac{2\pi E}{\hbar \kappa }}}{1+e^{-\frac{2E\pi }{\hbar \kappa }}}=-\dfrac{\pi }{\kappa }\dfrac{1}{e^{\frac{2E\pi }{\hbar \kappa }}+1}\nonumber \\ \end{aligned}$$

(A9)

and therefore DOS is given by (14).

Appendix B: Alternative way to effective potential

The solution of (25) provides the energy eigenfunction. This is given by

$$\begin{aligned} \psi _E(x)=C_0x^{\frac{iE}{\hbar \kappa }-\frac{1}{2}}, \end{aligned}$$

(B1)

where $C_0$ is the normalization constant. $C_0$ is determined through the nomalization condition $\int dx \psi _E'^*(x)\psi _E(x) = \delta (E-E')$ which yields $C_0=1/\sqrt{2\pi \hbar \kappa }$.

Now we ask the following question: What is the effective potential under which a massive particle with mass m will have energy eigen wavefunction, given by Eq. (B1), through the standard form of time-independent Schrodinger equation? In order to identify the effective potential, we have to substitute this wavefunction in the Schrodinger equation for a massive particle given by (27). Substitution of (B1) in (27) yields the form of potential as

$$\begin{aligned} V(x)={\mathcal {E}}+\dfrac{\hbar ^{2}}{8mx^{2}}\Big (3-\dfrac{8iE}{\hbar \kappa }-\dfrac{4E^{2}}{\hbar ^{2}\kappa ^{2}}\Big ). \end{aligned}$$

(B2)

This is the same as Eq. (28) for a zero energy $({\mathcal {E}})=0$ massive particle.

Just for the shake of completeness we provide a systematic steps to obtain the solution directly from the Schrodinger equation with the potential given by (28). Then the Hamiltonian of the system with a massive non-relativistic particle is given by,

$$\begin{aligned} H=\dfrac{p^{2}}{2m}-\dfrac{k}{x^{2}};~~~k=\dfrac{\hbar ^{2}}{8m}\left( \dfrac{8iE}{\hbar \kappa }-3+\dfrac{4E^{2}}{\hbar ^{2}\kappa ^{2}}\right) . \end{aligned}$$

(B3)

As the potential is time-independent, we can write $H{\left| {\psi }\right\rangle }={\mathcal {E}}{\left| {\psi }\right\rangle }$. Hence the Schrodinger equation is given by

$$\begin{aligned} -\dfrac{\hbar ^{2}}{2m}\dfrac{d^{2}\psi }{dx^{2}}-\dfrac{k}{x^{2}}={\mathcal {E}}\psi . \end{aligned}$$

(B4)

On rearranging the equation, we get,

$$\begin{aligned} x^{2}\psi ''+\psi \left[ \dfrac{2mk}{\hbar ^{2}}+\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}x^{2}\right] =0. \end{aligned}$$

(B5)

Let us simplify the equation by expressing the wavefunction as $\psi (x)=C x^{1/2}f(x)$, where C is a constant and f(x) is a function to be determined. Substituting this in the (B5) one finds

$$\begin{aligned} x^{2}f''+xf'+f\left[ -\dfrac{1}{4}+\dfrac{2mk}{\hbar ^{2}}+\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}x^{2}\right] =0. \end{aligned}$$

(B6)

Changing the variable $\sqrt{\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}}x=y$, and then substituting the value of k, we obtain

$$\begin{aligned} y^{2}\ddot{f}+y{\dot{f}}+f\left[ y^{2}-\left( 1-\dfrac{iE}{\hbar \kappa }\right) ^{2}\right] =0. \end{aligned}$$

(B7)

This is the standard Bessel equation and hence the general solution is given by

$$\begin{aligned} \psi (x)=C x^{1/2}\left( AJ_{\nu }\left( \sqrt{\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}}x\right) +BY_{\nu }\left( \sqrt{\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}}x\right) \right) . \end{aligned}$$

(B8)

In the above we denoted $\nu =1-\dfrac{iE}{\hbar \kappa }$ and A, B are integration constants. We now proceed to study this solution under two limiting cases.

1.1 1. Limiting Conditions

• (1) In the first limiting condition, we make the energy of the massive particle to be 0 i.e., ${\mathcal {E}}\rightarrow 0$. In this case, we know that the function must look like $x^{-1/2+\frac{iE}{\hbar \kappa }}$ (see Eq. (B1)). The Limiting nature of the Bessel functions when $\vert z\vert \rightarrow 0$ are [48],

  $$\begin{aligned}{} & {} J_{\nu }(z)\approx \dfrac{1}{\Gamma (\nu +1)}\Big (\dfrac{z}{2}\Big )^{\nu }~;\nonumber \\{} & {} Y_{\nu }(z)\approx -\dfrac{\Gamma (\nu )}{\pi }\Big (\dfrac{z}{2}\Big )^{-\nu }. \end{aligned}$$

  (B9)

  Then under the limit ${\mathcal {E}}\rightarrow 0$, (B8) yields

  $$\begin{aligned} \psi (x)\approx C x^{1/2}\left[ \dfrac{A}{\Gamma (\nu +1)}\left( \sqrt{\dfrac{m{\mathcal {E}}}{2\hbar ^{2}}}x\right) ^{1-\frac{iE}{\hbar \kappa }}-\dfrac{B\Gamma \big (1-\frac{iE}{\hbar \kappa }\big )}{\pi }\left( \sqrt{\dfrac{m{\mathcal {E}}}{2\hbar ^{2}}}x\right) ^{-1+\frac{iE}{\hbar \kappa }}\right] .\nonumber \\ \end{aligned}$$

  (B10)

  Note that the first term decays quickly as ${\mathcal {E}}\rightarrow 0$ and therefore will not contribute to this limiting case. Hence we have

  $$\begin{aligned} \psi (x)\approx - C \dfrac{B\Gamma \big (1-\frac{iE}{\hbar \kappa }\big )}{\pi }\left( \sqrt{\dfrac{m{\mathcal {E}}}{2\hbar ^{2}}}\right) ^{-1+\frac{iE}{\hbar \kappa }}x^{-\frac{1}{2}+\frac{iE}{\hbar \kappa }}. \end{aligned}$$

  (B11)

  Now in order to obtain the required form for ${\mathcal {E}}=0$ set $B=\Big (\sqrt{\dfrac{m{\mathcal {E}}}{2\hbar ^{2}}}\Big )^{1-\frac{iE}{\hbar \kappa }}$ and then one obtains

  $$\begin{aligned} \psi (x)\approx -C \dfrac{\Gamma \big (1-\frac{iE}{\hbar \kappa }\big )}{\pi }x^{-\frac{1}{2}+\frac{iE}{\hbar \kappa }}. \end{aligned}$$

  (B12)

  Finally use of normalization condition will yield (B1).

• (2) Even though the investigation is done near the horizon, where we find the effective model having the form given by Eq. (23), we can look at the inverse-squared potential as a separate system and study the asymptotic limits, i.e. $x\rightarrow \infty$, or in other words, far away from the potential source. In this case, the particle should behave similarly to a free particle. The asymptotic values of the Bessel functions are [48]:

  $$\begin{aligned} \begin{aligned} J_{\nu }(z)\approx \sqrt{\dfrac{2}{\pi z}}\cos \Big (z-\nu \dfrac{\pi }{2}-\dfrac{\pi }{4}\Big )~;\\ Y_{\nu }(z)\approx \sqrt{\dfrac{2}{\pi z}}\sin \Big (z-\nu \dfrac{\pi }{2}-\dfrac{\pi }{4}\Big ). \end{aligned} \end{aligned}$$

  (B13)

  Then (B8) reduces to

  $$\begin{aligned}{} & {} \psi (x)\approx C\left( \dfrac{2\hbar ^{2}}{m\pi ^{2}{\mathcal {E}}}\right) ^{1/4}\left[ A\cos \left( \sqrt{\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}}x-\dfrac{3\pi }{4}+\dfrac{i\pi E}{2\hbar \kappa }\right) \right. \nonumber \\{} & {} \left. \quad +B\sin \left( \sqrt{\dfrac{2m{\mathcal {E}}}{\hbar ^{2}}}x-\dfrac{3\pi }{4}+\dfrac{i\pi E}{2\hbar \kappa }\right) \right] . \end{aligned}$$

  (B14)

  This can be expanded as,

  $$\begin{aligned} \psi (x){} & {} \approx C\Big (\dfrac{2\hbar ^{2}}{m\pi ^{2}{\mathcal {E}}}\Big )^{1/4} \Big [A\dfrac{e^{-\frac{\pi E}{2\hbar \kappa }}e^{i(\sqrt{\frac{2m{\mathcal {E}}}{\hbar ^{2}}}x-\frac{3\pi }{4})}+e^{\frac{\pi E}{2\hbar \kappa }}e^{-i(\sqrt{\frac{2m{\mathcal {E}}}{\hbar ^{2}}}x-\frac{3\pi }{4})}}{2}\nonumber \\{} & {} \qquad +B\dfrac{e^{-\frac{\pi E}{2\hbar \kappa }}e^{i(\sqrt{\frac{2m{\mathcal {E}}}{\hbar ^{2}}}x-\frac{3\pi }{4})}-e^{\frac{\pi E}{2\hbar \kappa }}e^{-i(\sqrt{\frac{2m{\mathcal {E}}}{\hbar ^{2}}}x-\frac{3\pi }{4})}}{2i}\Big ]. \end{aligned}$$

  (B15)

  So both ingoing and outgoing free solutions are present.

Appendix C: Evaluating the ratios $(I_{1}+I_{2})/D,~I_{3}/D$

Before starting the calculations, we mention the integral identity used to simplify the equations [49].

$$\begin{aligned} \begin{aligned}&\int _{0}^{\infty } dx~x^{\lambda +1}e^{-\alpha x^{2}}J_{\mu }(\beta x)J_{\nu }(\gamma x)=\dfrac{\beta ^{\mu }\gamma ^{\nu }\alpha ^{-(\mu +nu+\lambda +2)/2}}{2^{\mu +nu+1}\Gamma (\nu +1)}\\&\quad \sum _{m=0}^{\infty }\Big [\dfrac{\Gamma (m+\frac{1}{2}(\nu +\mu +\lambda +2))}{m!\Gamma (m+\mu +1)} \Big (\dfrac{-\beta ^{2}}{4\alpha }\Big )^{m}\\&\quad \times F(-m,-\mu -m;\nu +1;\dfrac{\gamma ^{2}}{\beta ^{2}})\Big ]. \end{aligned} \end{aligned}$$

(C1)

The condition on $\mu ,\nu ,\lambda$ is $\text {Re}(\mu +\nu +\lambda )>-2$. In the present case we identify $\mu =\nu =ia=1-ia_{0},~a_{0}=E/\hbar \kappa$ and $\lambda =1$. Hence the above mentioned condition is satisfied. We now evaluate our integrals.

1.1 1. Calculation of $(I_{1}+I_{2})/D$

We start our calculations with evaluating the term $I_{1}$ (see Eq. (38)). The integration of this expression to be evaluated is as follows:

$$\begin{aligned}{} & {} \int _{-\infty }^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)=\int _{-\infty }^{0}dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)\nonumber \\{} & {} \qquad +\int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)\nonumber \\{} & {} \quad = \int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(-px)J_{ia}(-qx)+\int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)\nonumber \\{} & {} \quad = e^{2\pi a}\int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)+\int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)\nonumber \\{} & {} \quad = (1+e^{2\pi a})\underbrace{\int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)}_Y. \end{aligned}$$

(C2)

Using the identity Eq. (C1) we find

$$\begin{aligned} \begin{aligned} Y&= \int _{0}^{\infty }dx~x^{2}e^{i\lambda x^{2}} J_{ia}(px)J_{ia}(qx)\\&=\dfrac{p^{ia}q^{ia}(-i\lambda )^{-ia-3/2}}{2^{2ia+1}\Gamma (ia+1)} \sum _{m=0}^{\infty }\Bigg [\dfrac{\Gamma (m+ia+3/2)}{m!\Gamma (m+ia+1)}\Bigg (\dfrac{-p^{2}}{4i\lambda }\Bigg )^{m}\\&\quad \times F(-m,-ia-m;ia+1;\dfrac{q^{2}}{p^{2}})\Bigg ]. \end{aligned} \end{aligned}$$

(C3)

Next substituting the values of $p,q,\lambda$ from Eq. (32) in the above we obtain

$$\begin{aligned} \begin{aligned} Y&=\Bigg (\dfrac{2m\epsilon ^{2}}{i\hbar (t_{2}-t_{1})}\Bigg )^{ia}\dfrac{(-i\lambda )^{-3/2}}{2^{2ia+1} \Gamma (ia+1)}\sum _{m=0}^{\infty }\Bigg [\dfrac{\Gamma (m+ia+3/2)}{m!\Gamma (m+ia+1)}\Bigg (\dfrac{-m\epsilon ^{2}(t_{2}-t)}{2\hbar i(t_{2}-t_{1})(t-t_{1})}\Bigg )^{m}\\&\quad \times F\Bigg (-m,-ia-m;ia+1;\Bigg (\dfrac{t-t_{1}}{t_{2}-t}\Bigg )^{2}\Bigg )\Bigg ]. \end{aligned} \end{aligned}$$

(C4)

In the limit $\epsilon \rightarrow 0$, only $m=0$ term survives and then one obtains

$$\begin{aligned} Y= & {} \Bigg (\dfrac{2m\epsilon ^{2}}{i\hbar (t_{2}-t_{1})}\Bigg )^{ia}\dfrac{(-i\lambda )^{-3/2}}{2^{2ia+1}\Gamma (ia+1)}\dfrac{\Gamma (ia+3/2)}{\Gamma (ia+1)}\nonumber \\{} & {} \quad \times F\Bigg (0,-ia;ia+1;\Bigg (\dfrac{t-t_{1}}{t_{2}-t}\Bigg )^{2}\Bigg ). \end{aligned}$$

(C5)

Later, we take the limit $t_{2}\rightarrow \infty$ in which case, the Hypergeometric function becomes $F(0,-ia;ia+1;0)=1$. Hence we find

$$\begin{aligned} I_{1}=\dfrac{e^{-2\pi a}(1+e^{2\pi a})}{\sinh ^{2}(\pi a)}\Bigg (\dfrac{2m\epsilon ^{2}}{i\hbar (t_{2}-t_{1})}\Bigg )^{ia}\dfrac{(-i\lambda )^{-3/2}}{2^{2ia+1}\Gamma (ia+1)}\dfrac{\Gamma (ia+3/2)}{\Gamma (ia+1)}. \end{aligned}$$

(C6)

Finally using the property $\Gamma (1-z)\Gamma (z)=\dfrac{\pi }{\sin \pi z}$ one finds

$$\begin{aligned} I_{1}=e^{\pi a/2}(1+e^{-2\pi a})\Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )^{ia}(-i\lambda )^{-3/2}\dfrac{[\Gamma (-ia)]^{2}\Gamma (ia+3/2)}{2\pi ^{2}}. \end{aligned}$$

(C7)

In the integral $I_{2}$, there is $J_{-ia}$ instead of $J_{ia}$ (see Eq. (39)). Hence, we can replace $a\rightarrow -a$ and get $I_{2}$

$$\begin{aligned} I_{2}= & {} e^{-2\pi a}\times I_{1}(a\rightarrow -a)=e^{-\pi a/2}(1+e^{-2\pi a})\nonumber \\{} & {} \quad \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )^{-ia}(-i\lambda )^{-3/2} \dfrac{[\Gamma (ia)]^{2}\Gamma (-ia+3/2)}{2\pi ^{2}}. \end{aligned}$$

(C8)

Till this point, the calculations are in agreement with the calculations done in [27]. As mentioned earlier, we have,

$$\begin{aligned} a=-a_{0}-i;~~a_{0}=\dfrac{E}{\hbar \kappa }. \end{aligned}$$

(C9)

Then for our case

$$\begin{aligned} \begin{aligned} I_{1}&= ie^{-\pi a_{0}/2}(1+e^{2\pi a_{0}})\dfrac{(-i\lambda )^{-3/2}}{2\pi ^{2}}\exp \Bigg [\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\\&\quad -ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{1}+\theta _{2})\Bigg ]\\&\quad \times \vert \Gamma (-1+ia_{0})\vert ^{2}\vert \Gamma (5/2-ia_{0})\vert \end{aligned} \end{aligned}$$

(C10)

where,

$$\begin{aligned} \theta _{1}=\arg (\Gamma (-1+ia_{0}));~~\theta _{2}=\arg (\Gamma (5/2-ia_{0})) \end{aligned}$$

(C11)

and,

$$\begin{aligned}{} & {} \vert \Gamma (5/2-ia_{0})\vert =\Bigg \vert \dfrac{3}{2}-ia_{0}\Bigg \vert \Bigg \vert \dfrac{1}{2}-ia_{0}\nonumber \\{} & {} \Bigg \vert \vert \Gamma (1/2-ia_{0})\vert =\Bigg \vert \dfrac{3}{2}-ia_{0}\Bigg \vert \Bigg \vert \dfrac{1}{2}-ia_{0}\Bigg \vert \sqrt{\dfrac{\pi }{\cosh (\pi a_{0})}}~;\nonumber \\{} & {} \vert \Gamma (-1+ia_{0})\vert ^{2}=\dfrac{\pi }{a_{0}\sinh (\pi a_{0})}\dfrac{1}{1+a_{0}^{2}}. \end{aligned}$$

(C12)

Hence, using the above identities, we express $I_{1}$ as follows,

$$\begin{aligned} \begin{aligned} I_{1}&= ie^{-\pi a_{0}/2}(1+e^{2\pi a_{0}})\dfrac{(-i\lambda )^{-3/2}}{2\pi ^{2}}\exp \Bigg [\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\\&\quad -ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{1}+\theta _{2})\Bigg ]\\&\quad \times \Bigg \vert \dfrac{3}{2}-ia_{0}\Bigg \vert \Bigg \vert \dfrac{1}{2}-ia_{0}\Bigg \vert \sqrt{\dfrac{\pi }{\cosh (\pi a_{0})}}\times \dfrac{\pi }{a_{0}\sinh (\pi a_{0})}\dfrac{1}{1+a_{0}^{2}}. \end{aligned} \end{aligned}$$

(C13)

Similarly, we can now express $I_{2}$ as

$$\begin{aligned} I_{2}{} & {} =-ie^{\pi a_{0}/2}(1+e^{2\pi a_{0}})\dfrac{(-i\lambda )^{-3/2}}{2\pi ^{2}}\exp \Bigg [-\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\nonumber \\{} & {} \quad +ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{3}+\theta _{4})\Bigg ]\nonumber \\{} & {} \quad \times \vert \Gamma (1-ia_{0})\vert ^{2}\vert \Gamma (1/2+ia_{0})\vert , \end{aligned}$$

(C14)

where

$$\begin{aligned} \theta _{3}=\arg (\Gamma (1-ia_{0}));~~\theta _{4}=\arg (\Gamma (1/2+ia_{0}) \end{aligned}$$

(C15)

and,

$$\begin{aligned} \vert \Gamma (1/2+ia_{0})\vert =\sqrt{\dfrac{\pi }{\cosh (\pi a_{0})}}~; \,\,\,\ \vert \Gamma (1-ia_{0})\vert ^{2}=\dfrac{a_{0}\pi }{\sinh (\pi a_{0})}. \end{aligned}$$

(C16)

Hence we have

$$\begin{aligned} I_{2}{} & {} =-ie^{\pi a_{0}/2}(1+e^{2\pi a_{0}})\dfrac{(-i\lambda )^{-3/2}}{2\pi ^{2}}\exp \Bigg [-\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\nonumber \\{} & {} \quad +ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{3}+\theta _{4})\Bigg ]\nonumber \\{} & {} \quad \times \sqrt{\dfrac{\pi }{\cosh (\pi a_{0})}}\dfrac{a_{0}\pi }{\sinh (\pi a_{0})}. \end{aligned}$$

(C17)

Now that we have derived the expressions for $I_{1},I_{2}$, we add them:

$$\begin{aligned} I_{1}+I_{2}{} & {} =-ie^{\pi a_{0}/2}(1+e^{2\pi a_{0}})\dfrac{(-i\lambda )^{-3/2}}{2\pi ^{2}}\exp \Bigg [-\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\nonumber \\{} & {} \quad +ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{3}+\theta _{4})\Bigg ]\nonumber \\{} & {} \quad \times \sqrt{\dfrac{\pi }{\cosh (\pi a_{0})}}\dfrac{a_{0}\pi }{\sinh (\pi a_{0})}\Bigg \{-1+e^{-\pi a_{0}} \exp \Bigg [2\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\nonumber \\{} & {} \quad -2ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{1}+\theta _{2}\nonumber \\{} & {} \quad -2\theta _{3}-\theta _{4})\Bigg ]\times \Bigg \vert \dfrac{3}{2}-ia_{0}\Bigg \vert \Bigg \vert \dfrac{1}{2}-ia_{0}\Bigg \vert \dfrac{1}{a^{2}_{0}(1+a_{0}^{2})}\Bigg \}. \end{aligned}$$

(C18)

In the limit $\epsilon \rightarrow 0$ there will be divergent terms in $I_{2}$. These will be cancelled when we take the ratio with D.

The denominator (given by (41)), in the limiting case, is expressed as,

$$\begin{aligned} D= & {} -\dfrac{ie^{\pi a_{0}}}{\pi }\Bigg (-e^{-\pi a_{0}}\Gamma (-1+ia_{0})\Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )^{ia}+\Gamma (1-ia_{0}) \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )^{-ia}\Bigg )\nonumber \\= & {} -\dfrac{ie^{\pi a_{0}}}{\pi }\Bigg (-e^{-\pi a_{0}}\sqrt{\dfrac{\pi }{a_{0}\sinh (\pi a_{0})}\dfrac{1}{1+a_{0}^{2}}}\nonumber \\{} & {} \quad \exp \Bigg [\Bigg (\ln \dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg ) -ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i\theta _{1}\Bigg ]\nonumber \\{} & {} \quad +\sqrt{\dfrac{a_{0}\pi }{\sinh (\pi a_{0})}}\exp \Bigg [-\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg ) +ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i\theta _{3}\Bigg ]\Bigg )\nonumber \\= & {} -\dfrac{ie^{\pi a_{0}}}{\pi }\sqrt{\dfrac{a_{0}\pi }{\sinh (\pi a_{0})}}\exp \Bigg [-\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg ) +ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i\theta _{3}\Bigg ]\nonumber \\{} & {} \quad \times \Bigg (-e^{-\pi a_{0}}\dfrac{1}{a_{0}}\sqrt{\dfrac{1}{1+a_{0}^{2}}}\exp \Bigg [2\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\end{aligned}$$

(C19)

$$\begin{aligned}{} & {} \quad -2ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(\theta _{1}-\theta _{3})\Bigg ]+1\Bigg ). \end{aligned}$$

(C20)

Now, upon taking the ratio of Eq. (C20) and Eq. (C18), we get,

$$\begin{aligned} \dfrac{I_{1}+I_{2}}{D}=e^{-\pi a_{0}/2}(1+e^{2\pi a_{0}})\dfrac{(-i\lambda )^{-3/2}}{2}\sqrt{\dfrac{a_{0}}{\cosh (\pi a_{0})\sinh (\pi a_{0})}}e^{i(\theta _{3}+\theta _{4})}\times \zeta \nonumber \\ \end{aligned}$$

(C21)

where,

$$\begin{aligned} \zeta =\dfrac{-1+e^{-\pi a_{0}} \exp \Bigg [2\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )-2ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(2\theta _{1}+\theta _{2} -2\theta _{3}-\theta _{4})\Bigg ]\dfrac{\vert \frac{3}{2}-ia_{0}\vert \vert \frac{1}{2}-ia_{0}\vert }{a^{2}_{0}(1+a_{0}^{2})}}{-e^{-\pi a_{0}}\dfrac{1}{a_{0}}\sqrt{\dfrac{1}{1+a_{0}^{2}}}\exp \Bigg [2\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )-2ia_{0}\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )+i(\theta _{1}-\theta _{3})\Bigg ]+1}.\nonumber \\ \end{aligned}$$

(C22)

In the limit $\epsilon \rightarrow 0$ we have $\exp \Bigg [\ln \Bigg (\dfrac{m\epsilon ^{2}}{2\hbar (t_{2}-t_{1})}\Bigg )\Bigg ]\rightarrow 0$ and hence, $\lim _{\epsilon \rightarrow 0}\zeta =-1$. Then (C21) reduces to the simple form Eq. (43).

1.2 2. Evaluation of $I_{3}/D$

$I_3$ is given by (40). Use of the identity Eq. (C1) yields

$$\begin{aligned} I_{3}= & {} -\dfrac{2e^{-\pi a}}{\sinh ^{2}(\pi a)}\Bigg [\dfrac{(p/q)^{ia}(-i\lambda )^{-3/2}}{2\Gamma (1-ia)}\sum _{m=0}^{\infty } \dfrac{\Gamma (m+\frac{3}{2})}{m!\Gamma (m+ia+1)}\Bigg (\dfrac{p^{2}}{4i\lambda }\Bigg )^{m}\nonumber \\{} & {} \quad F \Bigg (-m,-ia-m;-ia+1;\dfrac{q^{2}}{p^{2}}\Bigg )+(a\rightarrow -a)\Bigg ]. \end{aligned}$$

(C23)

In the above expression, $(a\rightarrow -a)$ means, replace “a" in the terms before positive sign which are inside the third bracket by “$- a$". Under the limits the above reduces to

$$\begin{aligned} I_{3}= & {} -\dfrac{2e^{-\pi a}}{\sinh ^{2}(\pi a)}\Bigg [\dfrac{(-i\lambda )^{-3/2}\Gamma (3/2)}{2\Gamma (1-ia)\Gamma (ia+1)} \Bigg (-\dfrac{t_{2}-t}{t-t_{1}}\Bigg )^{ia}\nonumber \\{} & {} \quad +\dfrac{(-i\lambda )^{-3/2}\Gamma (3/2)}{2\Gamma (1+ia) \Gamma (1-ia)}\Bigg (-\dfrac{t_{2}-t}{t-t_{1}}\Bigg )^{-ia}\Bigg ] \end{aligned}$$

(C24)

Note that $I_{3}$ independent of $\epsilon$. Then it is clearly visible that, due to the divergent part for the limit $\epsilon \rightarrow 0$ in D (see Eq. (C20)), $\lim _{\epsilon \rightarrow 0} (1/D)=0$. Hence, in the limit $\epsilon \rightarrow 0$, $I_{3}/D$ vanishes and so we have (44).