Effective steepest descent approximation

From the path integral for the Steigl/Hinterleitner operator orderings, we are effectively quantizing an effective Lagrangian on a curved background, with the effective Lagrangian being

(1)
\begin{align} L_{\rm eff} = \int dt \left[ \frac{1}{2} a \dot a ^{2} - \left( V \left( a \right) + \frac{\hbar^2}{2} p \left( i + k \right) a^{-3} \right) \right] \end{align}

where

(2)
\begin{align} V \left( a \right) = \frac{1}{2} \left( -a + \frac{\Lambda}{3} a^3 \right), \quad p \left( i+k \right) = - \frac{7}{16} + \left( i+k \right) - \frac{ \left( i+k \right)^2}{4} \end{align}

To get an idea of how the quantum potential might modify the ground state wavefunction at small values of $a$, we can try a steepest descent approximation on this effective Lagrangian. The Hamilton-Jacobi equation is obtained using

(3)
\begin{align} H &= \frac{a \dot a ^2}{2} + \left( V \left( a \right) + \frac{\hbar^2}{2} p \left( i+k \right) a^{-3} \right) \\ &= \frac{p^2}{2a} + \left( V \left( a \right) + \frac{\hbar^2}{2} p \left( i+k \right) a^{-3} \right) \end{align}

so we have

(4)
\begin{align} \frac{1}{2a} \left( \frac{dS}{da} \right)^2 = \frac{1}{2} \left( a - \frac{\Lambda}{3} a^3 \right) - \frac{\hbar^2}{2} p \left( i+k \right) a^{-3} \\ \implies S = \pm \int \left( a^2 - \frac{\Lambda}{3} a^4 - \hbar^2 p \left( i+k \right) a^{-2} \right) ^{1/2} da \end{align}

Letting $u=a^2$,

(5)
\begin{align} S &= \pm \int \frac{\left( u - \frac{\Lambda}{3}u^2 - \hbar^2 p \left( i+k \right) u^{-1} \right) ^{1/2}}{2u^{1/2}} du \\ &= \pm \frac{1}{2} \int \frac{u^2 - \frac{\Lambda}{3} u^3 - \hbar^2 p \left( i+k \right) }{u \sqrt{u^2 - \frac{\Lambda}{3} u^3 - \hbar^2 p \left( i+k \right) }} du \\ &= \pm \frac{1}{2} \int \frac{ - \frac{\Lambda}{3} u^3 + u^2 - \hbar^2 p \left( i+k \right) }{u \sqrt{ - \frac{\Lambda}{3} \left( u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) \right) }} \end{align}

In order to convert these integrals to a sum of elementary functions and elliptic integrals in the three canonical forms, we need to find the roots of the cubic polynomial

(6)
\begin{align} u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) \end{align}

For a cubic

(7)
$$ax^3 + bx^2 + cx + d$$

the real root is given by

(8)
\begin{align} r_* = \left( q + \left( q^2 + \left( r - p^2 \right)^3 \right)^{1/2} \right)^{1/3} + \left( q - \left( q^2 + \left( r - p^2 \right)^3 \right)^{1/2} \right)^{1/3} + p \end{align}

where

(9)
\begin{align} p = - \frac{b}{3a}, \quad q = p^3 + \frac{bc - 3ad}{6a^2}, \quad r = \frac{c}{3a}. \end{align}

In our case $a = 1, \quad b = - \frac{3}{\Lambda}, \quad c = 0, \quad d = \frac{3 \hbar^2}{\Lambda} p \left( i+k \right)$, so that

(10)
\begin{align} p = \Lambda^{-1}, \quad q = \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right) }{2 \Lambda}, \quad r = 0. \end{align}
(11)
\begin{align} \implies r_*& = \left( \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right) }{2 \Lambda} + \left[ \left( \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right) }{2 \Lambda} \right)^2 - \Lambda^{-6} \right] ^{1/2} \right)^{1/3} + \\ & \quad \left( \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right) }{2 \Lambda} - \left[ \left( \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right) }{2 \Lambda} \right)^2 - \Lambda^{-6} \right] ^{1/2} \right)^{1/3} + \Lambda^{-1} \\ & = \left( \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right)}{2 \Lambda} + \left[ - \frac{3 \hbar^2 p \left( i+k \right)}{ \Lambda^4} + \frac{9 \hbar^4 p^2 \left( i+k \right)}{4 \Lambda^2} \right]^{1/2} \right)^{1/3} + \\ & \quad \left( \Lambda^{-3} - \frac{3 \hbar^2 p \left( i+k \right)}{2 \Lambda} - \left[ - \frac{3 \hbar^2 p \left( i+k \right)}{ \Lambda^4} + \frac{9 \hbar^4 p^2 \left( i+k \right)}{4 \Lambda^2} \right]^{1/2} \right)^{1/3} + \Lambda^{-1} \end{align}

(as expected this is $3 \Lambda^{-1}$ when $p \left( i+k \right) = 0$).

For the remaining roots, divide the cubic $u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right)$ by $\left( u-r_* \right)$, where $r_*$ is the root found above, to obtain the quadratic $u^2 + \left( - \frac{3}{\Lambda} + r_* \right) u + \left( r_* ^2 - \frac{3}{\Lambda} r_* \right)$, with roots

(12)
\begin{align} r_{\pm} = \frac{\left( \frac{3}{\Lambda} - r_* \right) \pm \sqrt{ \frac{9}{\Lambda^2} + \frac{6}{\Lambda} r_* - 3 r_*^2 }}{2} \\ \end{align}

Note that we can determine the nature of the roots (real or complex) of the cubic by examining the discriminant, which for a cubic $ax^3 +bx^2 +cx + d$ is given by

(13)
\begin{align} \Delta = 18abcd - 4b^3 d + b^2 c^2 - 4ac^3 - 27a^2 d^2. \end{align}

Since in our case $a= - \frac{\Lambda}{3}$, $b=1$, $c=0$, and $d=- \hbar^2 p \left( i+k \right)$, we have

(14)
\begin{align} \Delta = \hbar^2 p \left( i+k \right) \left( 4 - 3 \Lambda^2 \hbar^2 p \left( i+k \right) \right). \end{align}

Since $\Delta \ge 0$ implies all three roots are real and $\Delta < 0$ implies one real and two complex conjugate roots, for all except very extreme values of $p \left( i+k \right)$ the sign of $p \left( i+k \right)$ determines whether the roots are real or complex.

Following the technique of Whitaker/Watson, we can now reduce (5) to elementary functions plus sums of the three standard elliptic integrals (see e.g. Wolfram MathWorld):

(15)
\begin{align} F \left( \phi, k \right) &= \int_0 ^{\phi} \frac{d \theta}{\sqrt{1 - k^2 \sin^2 \theta}} d \theta = \int_0 ^{\sin \phi} \frac{dt}{\sqrt{ \left( 1 - k^2 t^2 \right) \left( 1 - t^2 \right)}} \\ &= \int_0 ^{\tan \phi} \frac{dv}{\sqrt{ \left( 1 + v^2 \right) \left( 1 + k' ^2 v^2 \right) }} \ \text{where} \ k' ^2 = 1 - k^2 \\ E \left( \phi, k \right) &= \int_0 ^{\phi} \sqrt{1 - k^2 \sin^2 \theta} d \theta = \int_0 ^{\sin \theta} \sqrt{\frac{1 - k^2 t^2}{1 - t^2}} dt \\ \Pi \left( n; \phi, k \right) &= \int_0 ^{\phi} \frac{d \theta}{ \left( 1 - n \sin^2 \theta \right) \sqrt{1 - k^2 \sin^2 \theta}} = \int_0 ^{\sin \phi} \frac{dt}{\left( 1 - n t^2 \right) \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) }} \end{align}

We have

(16)
\begin{align} &\pm \frac{1}{2} \int \frac{ - \frac{\Lambda}{3} u^3 + u^2 - \hbar^2 p \left( i+k \right) }{u \sqrt{ - \frac{\Lambda}{3} \left( u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) \right) }} = \\ &\pm \frac{i}{2} \sqrt{ \frac{3}{\Lambda} } \int \frac{ \left( -\frac{\Lambda}{3} u^3 +u^2 - \hbar^2 p \left( i+k \right) \right) du }{u \sqrt{ \left( A_1 \left( u- \alpha \right)^2 + B_1 \left( u - \beta \right)^2 \right) \left( A_2 \left( u-\alpha \right)^2 + B_2 \left( u - \beta \right)^2 \right)}} \end{align}

where $A_1, B_1, A_2, B_2, \alpha, \beta$ are chosen so that

(17)
\begin{align} u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) = \left( A_1 \left( u- \alpha \right)^2 + B_1 \left( u - \beta \right)^2 \right) \left( A_2 \left( u-\alpha \right)^2 + B_2 \left( u - \beta \right)^2 \right). \end{align}

This can be done by first factorizing

(18)
\begin{align} u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) &= S_1 S_2, \\ &S_1 = u^2 + \left( - \frac{3}{\Lambda} + r_* \right) u + \left( r_*^2 - \frac{3}{\Lambda} r_* \right) \\ &S_2 = u - r_* \end{align}

The technique:

Write in general form $S_1 = a_1u^2 + 2b_1u+c_1, S_2 = 2b_2u+c_2$. Then for some constant $\lambda$, $S_1 - \lambda S_2$ will be a perfect square in$u$ if

(19)
\begin{align} a_1 \left( c_1 - \lambda c_2 \right) - \left( b_1 - \lambda b_2 \right)^2 = 0. \end{align}

Let $\lambda_1, \lambda_2$ be the roots of (19). [Note: We need to add a condition for $\lambda_1, \lambda_2$ to be real. This will be discussed momentarily.] Then there exist $\alpha, \beta$ such that

(20)
\begin{align} S_1 - \lambda_1 S_2 &= a_1 \left( u - \alpha \right)^2 \\ S_1 - \lambda_2 S_2 &= a_1 \left( u - \beta \right)^2 \\ \\ \implies &S_1 = A_1 \left( u - \alpha \right)^2 + B_1 \left( u - \beta \right)^2 \\ &S_2 = A_2 \left( u - \alpha \right)^2 + B_2 \left( u - \beta \right)^2 \end{align}

The implication is by solving for $S_1, S_2$ above to get

(21)
\begin{align} &A_1 = \frac{\lambda_2 a_1}{\lambda_2 - \lambda_1}, &&B_1 = - \frac{\lambda_1 a_1}{\lambda_2 - \lambda_1} \\ &A_2 = \frac{a_1}{\lambda_2 - \lambda_1}, &&B_2 = - \frac{ a_1}{\lambda_2 - \lambda_1} \end{align}

Condition for $\lambda_1, \lambda_2$ to be real: $\lambda_1, \lambda_2$ are the roots of the polynomial (19), which has discriminant $\left( 2 b_1 b_2 - a_1 c_2 \right)^2 + 4 b_2^2 \left( a_1 c_1 - b_1^2 \right)$. Note that the second term is the negative of the discriminant of $S_1$, so that if $S_1$ has complex roots, $\lambda_1, \lambda_2$ are guaranteed real. If on the other hand the roots $r_{+}, r_{-}$ of $S_1$ are real, then the discriminant of (19) can be written in terms of $r_*, r_{+}, r_{-}$ as $\left( 2 a_1 b_2 \right)^2 \left( r_* - r_{+} \right) \left( r_* - r_{-} \right)$, so the condition for $\lambda_1, \lambda_2$ to be real is that the roots of $S_1$ do not interleave with that of $S_2$.

In our case of interest, we can check that the roots should not interleave for any “reasonable” value of $p \left( i+k \right)$. As above,

(22)
\begin{align} r_*& = \Lambda^{-1} \left( 1 - \frac{3 \hbar^2 \Lambda^2 p \left( i+k \right)}{2} + \left[ -3 \hbar^2 \Lambda^2 p \left( i+k \right) + \frac{9 \hbar^4 \Lambda^4 p^2 \left( i+k \right)}{4} \right]^{1/2} \right)^{1/3} \\ &+ \Lambda^{-1} \left( 1 - \frac{3 \hbar^2 \Lambda^2 p \left( i+k \right)}{2} - \left[ - 3 \hbar^2 \Lambda^2 p \left( i+k \right) + \frac{9 \hbar^4 \Lambda^4 p^2 \left( i+k \right)}{4} \right]^{1/2} \right)^{1/3} \\ &+ \Lambda^{-1} \\ & \sim \Lambda^{-1} \left( 1 + \mathcal{O} \left( \hbar \Lambda \right) \right)^{1/3} + \Lambda^{-1} \left( 1 + \mathcal{O} \left( \hbar \Lambda \right) \right)^{1/3} + \Lambda^{-1} \end{align}

and from the binomial series

(23)
\begin{align} \left( 1 + \mathcal{O} \left( \hbar \Lambda \right) \right)^{1/3} = 1 + \frac{1}{3} \mathcal{O} \left( \hbar \Lambda \right) + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{-2}{3} \mathcal{O} \left( \hbar \Lambda \right)^2 + \cdots \end{align}

we can write $r_* \sim \frac{3}{\Lambda} + \mathcal{O} \left( \hbar \right)$. Now we can estimate $r_{\pm}$:

(24)
\begin{align} r_{\pm} & \sim \frac{ \frac{3}{\Lambda} - \left( \frac{3}{\Lambda} + \mathcal{O} \left( \hbar \right) \right) \pm \sqrt{\frac{9}{\Lambda^2} + \frac{6}{\Lambda} \left( \frac{3}{\Lambda} + \mathcal{O} \left( \hbar \right) \right) - 3 \left( \frac{3}{\Lambda} + \mathcal{O} \left( \hbar \right) \right)^2 } }{2} \\ & \sim \frac{- \mathcal{O} \left( \hbar \right) \pm \sqrt{ - \frac{12}{\Lambda} \mathcal{O} \left( \hbar \right) - 3 \mathcal{O} \left( \hbar \right)^2 } }{2} \end{align}

So $r_*$ is a deviation of order $\hbar$ from $\frac{3}{\Lambda}$ while $r_{\pm}$ are deviations of order $\hbar$ from $0$, and thus for any value of $p \left( i+k \right)$ that is not of order $\hbar^{-2}$ (and barring extremely large values of $\Lambda \sim \hbar^{-1}$), the roots will not interleave and we can conclude that $\lambda_1, \lambda_2$ are real.

In our case we have factored the cubic

(25)
\begin{align} u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) \end{align}

into

(26)
\begin{align} S_1 = a_1 u^2 + 2b_1 u + c_1 \\ S_2 = 2b_2 u + c_2 \end{align}

where

(27)
\begin{align} S_1 &= u^2 + \left( - \frac{3}{\Lambda} + r_* \right) u + \left( r_*^2 - \frac{3}{\Lambda} r_* \right) \\ S_2 &= u - r_* \\ \implies a_1 = 1, \quad & b_1 = - \frac{3}{2 \Lambda} + \frac{r_*}{2}, \quad c_1 = r_*^2 - \frac{3}{\Lambda} r_* \\ & b_2 = \frac{1}{2}, \quad c_2 = - r_* \end{align}

We can thus write (19) as

(28)
\begin{align} - \frac{1}{4} \lambda^2 + \frac{3}{2} \left( r_* - \Lambda^{-1} \right) \lambda + \left( \frac{3 r_*^2}{4} - \frac{3}{2 \Lambda} r_* - \frac{9}{4 \Lambda^2} \right), \end{align}

which has roots

(29)
\begin{align} \lambda_1 = \left( 3 r_* - \frac{3}{\Lambda} \right) - \sqrt{12 r_* \left( r_* - \frac{2}{\Lambda} \right) } \\ \lambda_2 = \left( 3 r_* - \frac{3}{\Lambda} \right) + \sqrt{12 r_* \left( r_* - \frac{2}{\Lambda} \right) } \end{align}

Since $S_1 - \lambda_1 S_2 = a_1 \left( u - \alpha \right)^2$ and $S_1 - \lambda_2 S_2 = a_1 \left( u - \beta \right)^2$, we can use

(30)
\begin{align} \left( 2 b_1 - \lambda_1 2 b_2 \right) = - 2 a_1 \alpha \\ \left( 2 b_1 - \lambda_2 2 b_2 \right) = - 2 a_1 \beta \end{align}

to get

(31)
\begin{align} \alpha = r_* - \sqrt{3 r_* \left( r_* - \frac{2}{\Lambda} \right) } \\ \beta = r_* + \sqrt{3 r_* \left( r_* - \frac{2}{\Lambda} \right) } \end{align}

Notice that we can now rewrite $\lambda_1$ and $\lambda_2$ in terms of $\alpha$ and $\beta$ as

(32)
\begin{align} \lambda_1 = r_* - \frac{3}{\Lambda} + 2 \alpha \\ \lambda_2 = r_* - \frac{3}{\Lambda} + 2 \beta \end{align}

We can now also obtain $A_1, B_1, A_2$ and $B_2$ using the expressions for $\lambda_{1,2}$:

(33)
\begin{align} A_1 &= \frac{ \lambda_2 a_1}{\lambda_2 - \lambda_1} &&= && \frac{\left( 3 r_* - \frac{3}{\Lambda} \right) + \sqrt{12 r_* \left( r_* - \frac{2}{\Lambda} \right) } }{2 \sqrt{12 r_* \left( r_* - \frac{2}{\Lambda} \right) } } \\ B_1 &= \frac{- \lambda_1 a_1}{ \lambda_2 - \lambda_1} &&= && \frac{ - \left( 3 r_* - \frac{3}{\Lambda} \right) + \sqrt{ 12 r_* \left( r_* - \frac{2}{\Lambda} \right) } }{2 \sqrt{12r_* \left( r_* - \frac{2}{\Lambda} \right) } } \\ A_2 &= \frac{ - a_1}{\lambda_1 - \lambda_2 } &&= && \frac{1}{2 \sqrt{ 12 r_* \left( r_* - \frac{2}{\Lambda} \right) } } \\ B_2 &= \frac{a_1}{\lambda_1 - \lambda_2} &&= && \frac{-1}{2 \sqrt{12r_* \left( r_* - \frac{2}{\Lambda} \right) } } \end{align}

Thus we have

(34)
\begin{align} S &= \pm \frac{1}{2} \int \frac{ - \frac{\Lambda}{3} u^3 + u^2 - \hbar^2 p \left( i+k \right) }{u \sqrt{ - \frac{\Lambda}{3} \left( u^3 - \frac{3}{\Lambda} u^2 + \frac{3 \hbar^2}{\Lambda} p \left( i+k \right) \right) }} \\ &= \pm \frac{i}{2} \sqrt{ \frac{3}{\Lambda}} \int \frac{- \frac{\Lambda}{3} u^3 + u^2 - \hbar^2 p \left( i+k \right)}{ u \sqrt{ S_1 \cdot S_2} } \\ &= \pm \frac{i}{2} \sqrt{\frac{3}{\Lambda}} \int \frac{- \frac{\Lambda}{3} u^3 + u^2 - \hbar^2 p \left( i+k \right)}{ u \sqrt{ \left( A_1 \left( u - \alpha \right)^2 + B_1 \left( u - \beta \right)^2 \right) \left( A_2 \left( u - \alpha \right)^2 + B_2 \left( u - \beta \right)^2 \right) } } \end{align}

Note that using the substitution $t = \frac{u - \alpha}{u - \beta}$, we get

(35)
\begin{align} \frac{\left( \alpha - \beta \right)^{-1} dt}{\left\{ \left( A_1 t^2 + B_1 \right) \left( A_2 t^2 + B_2 \right) \right\}^{1/2}} = \frac{du}{ \left\{ \left( A_1 \left( u - \alpha \right)^2 + B_1 \left( u - \beta \right)^2 \right) \left( A_2 \left( u - \alpha \right)^2 + B_2 \left( u - \beta \right)^2 \right) \right\} ^{1/2} } \end{align}

Using $u = \frac{\beta t - \alpha}{t - 1}$, we get

(36)
\begin{align} \frac{- \frac{\Lambda}{3} u^3 + u^2 - \hbar^2 p \left( i+k \right) }{u} = -\frac{\Lambda}{3} \left( \frac{\beta t - \alpha}{t - 1} \right)^2 + \left( \frac{\beta t - \alpha}{t - 1} \right) - \hbar^2 p \left( i+k \right) \left( \frac{t - 1}{\beta t - \alpha} \right) \end{align}

Write each term as the sum of a rational function of $t^2$ and $t$ times a rational function of $t^2$:

(37)
\begin{align} - \frac{\Lambda}{3} \left( \frac{\beta t - \alpha}{t - 1} \right)^2 =& - \frac{\Lambda}{3} \left[ \left( \beta^2 + \frac{\left( \alpha^2 - 4 \alpha \beta + 3 \beta^2 \right)}{\left( t^2 - 1 \right)} + \frac{2 \left( \alpha - \beta \right)^2}{\left( t^2 - 1 \right)^2} \right) \right. \\ & \left. + \frac{\left( 2 \beta^2 - 2 \alpha \beta \right) t^3 + \left( -2 \alpha \beta + 2 \alpha^2 \right) t }{\left( t^2 - 1 \right)^2} \right] \\ \left( \frac{\beta t - \alpha }{t - 1} \right) =& \left( \beta + \frac{\beta - \alpha}{\left( t^2 - 1 \right)} \right) + \frac{\left( \beta - \alpha \right) t}{\left( t^2 - 1 \right)} \\ - \hbar^2 p \left( i+k \right) \left( \frac{t - 1}{\beta t - \alpha} \right) =& - \hbar^2 p \left( i+k \right) \left[ \left( \beta^{-1} + \frac{\alpha^2 \beta^{-1} - \alpha}{\beta^2 t^2 - \alpha^2} \right) + \frac{\left( \alpha - \beta \right) t}{\beta^2 t^2 - \alpha^2} \right] \end{align}

So

(38)
\begin{align} S = \pm \frac{i}{2} \sqrt{ \frac{3}{\Lambda} } & \left\{ C_1 \int \frac{dt}{ \sqrt{w \left( t^2 \right) } } + C_2 \int \frac{dt}{ \left( t^2 - 1 \right) \sqrt{ w \left( t^2 \right) } } + C_3 \int \frac{dt}{ \left( t^2 - 1 \right)^2 \sqrt{ w \left( t^2 \right) } } + \right. \\ & C_4 \int \frac{dt}{\left( t^2 - \frac{\alpha^2}{\beta^2} \right) \sqrt{ w \left( t^2 \right) } } + \\ & K_1 \int \frac{t^3 dt}{ \left( t^2 -1 \right)^2 \sqrt{ w \left( t^2 \right) }} + K_2 \int \frac{t dt}{ \left( t^2 -1 \right)^2 \sqrt{ w \left( t^2 \right) } } - \int \frac{t dt}{ \left( t^2 -1 \right) \sqrt{ w \left( t^2 \right) } } + \\ & \left. K_4 \int \frac{t dt}{ \left( t^2 - \frac{\alpha^2}{\beta^2} \right) \sqrt{ w \left( t^2 \right) } } \right\} \end{align}

where $w \left( t^2 \right) = \left( A_1 t^2 + B_1 \right) \left( A_2 t^2 + B_2 \right)$ and

(39)
\begin{align} C_1 &= \frac{ - \frac{\Lambda}{3} \beta^2 + \beta - \hbar^2 p \left( i+k \right) \beta^{-1} }{\alpha - \beta} \\ C_2 &= - 1 - \frac{\Lambda}{3} \left( \alpha - 3 \beta \right) \\ C_3 &= - \frac{ 2 \Lambda}{3} \left( \alpha - \beta \right) \\ C_4 &= - \hbar^2 p \left( i+k \right) \frac{\alpha}{\beta^3} \\ K_1 &= \frac{ 2 \Lambda}{3} \beta \\ K_2 &= - \frac{2 \Lambda}{3} \alpha \\ K_4 &= - \frac{\hbar^2 p \left( i+k \right)}{ \beta^2 } \end{align}

The integrals having coefficients $C_i$ are elliptic integrals; the others are elementary integrals. Since we only want some solution to the Hamilton-Jacobi equation, the lower limit does not matter since it will only result in an additive constant, so we consider the lower limit on all integrals to be 0 and the upper limit we denote as $y$. Note that the original scale factor $a \in \left[ 0, \infty \right)$, so $u = a^2 \in \left[ 0, \infty \right)$, and $t = \frac{u - \alpha}{u - \beta}$ where

(40)
\begin{align} \alpha &= r_* - \sqrt{3 r_* \left( r_* - \frac{2}{\Lambda} \right) } \\ \beta &= r_* + \sqrt{3 r_* \left( r_* - \frac{2}{\Lambda} \right) } \end{align}
(41)
\begin{align} \implies t = \frac{u - \alpha}{u - \beta} = \frac{u - r_* + \sqrt{3 r_* \left( r_* - \frac{2}{\Lambda} \right) } }{u - r_* - \sqrt{3 r_* \left( r_* - \frac{2}{\Lambda} \right) } } \end{align}

The graph shows $t$ plotted as a function of $u$. The vertical asymptote is at $u=r_* + \sqrt{3r_* \left( r_* - \frac{2}{\Lambda} \right) }$ and the u-intercept is at $u=r_* - \sqrt{3r_* \left( r_* - \frac{2}{\Lambda} \right) }$. Where $u=0$ falls depends on the value of $p \left( i+k \right)$. Note that when $p \left( i+k \right) = 0$, we have $r_* = \frac{3}{\Lambda}$ and thus $r_* - \sqrt{3r_* \left( r_* - \frac{2}{\Lambda} \right) } = 0$, so that $u=0$ is exactly at the intercept. Note that we can exclude consideration of values $u \ge r_*$ (classically forbidden because the radius is too large). Since $u=r_*$ corresponds to $t=-1$, this means we only need consider the upper limit $y$ in the integration as ranging in $(-1, 1)$.

The elliptic integrals all have the piece $\sqrt{w \left( t^2 \right)}$ in the denominator, which can be rewritten as

(42)
\begin{align} \sqrt{ w \left( t^2 \right) } &= \sqrt{ \left( A_1 t^2 + B_1 \right) \left( A_2 t^2 + B_2 \right) } \\ &= \sqrt{A_1 A_2} \sqrt{ \left( t^2 + \frac{B_1}{A_1} \right) \left( t^2 + \frac{B_2}{A_2} \right) } \\ &= \sqrt{A_1 A_2} \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) } \end{align}

where $\gamma = - \frac{\lambda_1}{\lambda_2}$. Notice that $\gamma >0$ should correspond to $p \left( i+ k \right) <0$, since this then yields no new zeroes in the denominator of the integrands, whereas $\gamma <0$ makes for a new zero which corresponds to the fact that for $p \left( i + k \right) > 0$, we get a new classically forbidden region. Note that

(43)
\begin{align} A_1 A_2 &= \left( \frac{\lambda_2 a_1}{\lambda_2 - \lambda_1} \right) \left( \frac{ - a_1}{\lambda_1 - \lambda_2} \right) \\ &= \frac{\lambda_2 a_1^2}{\left( \lambda_2 - \lambda_1 \right)^2 } = \frac{\lambda_2}{\left( \lambda_2 - \lambda_1 \right)^2} \end{align}

Since

(44)
\begin{align} \lambda_2 \sim \left( \frac{6}{\Lambda} + \mathcal{O} \left( \hbar \right) \right) + \sqrt{ 12 \left( \frac{3}{\Lambda} + \mathcal{O} \left( \hbar \right) \right) \left( \frac{1}{\Lambda} + \mathcal{O} \left( \hbar \right) \right) } > 0, \end{align}

$\sqrt{ A_1 A_2 }$ is real.

Thus we can write

(45)
\begin{align} S = \pm \frac{1}{2 \sqrt{ A_1 A_2 } } \sqrt{ \frac{3}{ \Lambda } } \left( S_E + S_{Elem} \right) \end{align}

where

(46)
\begin{align} S_E = & C_1 \int_0^y \frac{dt}{ \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + C_2 \int_0^y \frac{dt}{ \left( t^2 - 1 \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ & + C_3 \int_0^y \frac{dt}{ \left( t^2 - 1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left(1 - t^2 \right) } } + C_4 \int_0^y \frac{dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ = & C_1 E_1 + C_2 E_2 + C_3 E_3 + C_4 E_4 \end{align}

and

(47)
\begin{align} S_{Elem} = & K_1 \int_0^y \frac{t^3 dt}{ \left( t^2 -1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + K_2 \int_0^y \frac{t dt}{ \left( t^2 -1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ & - \int_0^y \frac{t dt}{ \left( t^2 -1 \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + K_4 \int_0^y \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ = & K_1 I_1 + K_2 I_2 - I_3 + K_4 I_4 \end{align}

where $q = \frac{\alpha^2}{\beta^2}$.

Case 1: $\gamma > 0$

Let $x = \sqrt{ \frac{ y^2 \left( 1 + \gamma \right) }{ y^2 + \gamma } }$, let $k = \frac{ 1 }{ \sqrt{ 1 + \gamma } }$, let $k' = \sqrt{ 1 - k^2 }$, let $j^2 = \frac{ q + \gamma }{ q \left( 1 + \gamma \right) }$. Rewrite $E_4$ as

(48)
\begin{align} E_4 &= \int_0 ^y \frac{dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ &= \int_0 ^{-1} \frac{dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + \int_{-1} ^y \frac{dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \end{align}

Also rewrite $I_4$ as

(49)
\begin{align} I_4 &= \int_0 ^y \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} \\ &= \int_0 ^{-1} \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} + \int_{-1} ^y \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} \end{align}

In both cases the first term must be evaluated as a Cauchy principal value.

To pick up the classically allowed region, we must let the integration limit $y$ range from $y = \frac{\alpha}{\beta}$ ($u = 0$) to $y = -1$ ($u = r_*$). Note that for $\gamma > 0$, we have $\lambda_1 <0$ (since $\gamma = - \frac{\lambda_1}{\lambda_2}$ and $\lambda_2 >0$). So

(50)
\begin{align} r_* - \frac{3}{\Lambda} + 2 \alpha < 0. \end{align}

Observe that when $p \left( i + k \right) < 0$, for large $a$ the quantum potential has the effect of slightly decreasing the original potential, making the root $r_*$ occur at a slightly larger value than the original $\frac{3}{\Lambda}$. Thus

(51)
\begin{align} 0 > \left( r_* - \frac{3}{\Lambda} \right) + 2 \alpha \implies \alpha < 0 \implies \frac{\alpha}{\beta} <0 \text{ since } \beta > 0 \end{align}

Using Mathematica to integrate the elementary integrals (see the notebook attached to this wiki as the file elem_assumptions_gamma_pos.nb), we get

(52)
\begin{align} I_1 &= \int_0^y \frac{t^3 dt}{ \left( t^2 -1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ &= - \frac{-y^2 \sqrt{ \gamma + y^2 } + \gamma \left( 2 - 3 y^2 \right) \sqrt{ \gamma + y^2 } - 2 \left( \gamma - \gamma y^2 \right) ^{3/2} }{ 3 \left( 1 + \gamma \right)^2 \left( 1 - y^2 \right)^{3/2} } \\ \\ I_2 &= \int_0^y \frac{t dt}{ \left( t^2 -1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ &= - \frac{ 3 \sqrt{ \gamma } \left( 1 - y^2 \right)^{3/2} - \gamma \sqrt{ \gamma + y^2 } + \sqrt{ \gamma + y^2 } \left( -3 + 2 y^2 \right) + \left( \gamma - \gamma y^2 \right)^{3/2} }{ 3 \left( 1 + \gamma \right)^2 \left( 1 - y^2 \right)^{3/2} } \\ \\ I_3 &= \int_0^y \frac{t dt}{ \left( t^2 -1 \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } = \frac{ y^2 }{ \sqrt{ \gamma } \left( y^2 - 1 \right) - \sqrt{ \left( \gamma + y^2 \right) \left( 1 - y^2 \right) } } \\ \\ I_4 &= \int_0 ^{-1} \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} + \int_{-1} ^y \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} \\ &= \int_0 ^{-1} \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} - \sqrt{ \frac{ 1 }{ \left( 1 - q \right) \left( \gamma + q \right) } } \tanh^{-1} \sqrt{ \frac{ \left( \gamma + q \right) \left( 1 - y^2 \right) }{ \left( 1 - q \right) \left( \gamma + y^2 \right) } } \end{align}

Using integration formulas from Byrd and Friedman, we can compute

(53)
\begin{align} E_1 &= k F \left( \sin^{-1} x, k \right) \\ E_2 &= k \left( F \left( \sin^{-1} x , k \right) - E \left( \sin^{-1} x , k \right) + \sqrt{ 1 - k^2 x^2 } + \frac{ x }{ \sqrt{ 1 - x^2 } } + \frac{ \pi }{2} -1 \right) \\ E_3 &= \frac{-k}{3} \left[ \left( 2 + k^2 \right) F \left( \sin^{-1} x, k \right) - 2 \left( 1 + k^2 \right) E \left( \sin^{-1} x , k \right) + \left( 2 + 2 k^2 + \left( k' \right)^2 \left( \frac{ x }{ 1 - x^2 } \right) \sqrt{ \frac{ 1 - k^2 x^2 }{ 1 - x^2 } } \right) - \left( 1 + k^2 \right) \pi \right] \\ E_4 &= \frac{1}{ \left( q - 1 \right) \sqrt{ \gamma + 1} } \left[ \Pi \left( \frac{\pi}{2} , \frac{1}{1 - q}, \sqrt{ \frac{1}{1 + \gamma} } \right) - \Pi \left( \cos^{-1} \left( - y \right) , \frac{1}{1 - q}, \sqrt{ \frac{1}{1 + \gamma} } \right) \right] \end{align}

Case 2. $\gamma < 0$

In this case we adjust the lower limit of integration to be $- \sqrt{ - \gamma }$, and take the range of the upper limit $y$ from $- \sqrt{ - \gamma }$ to $-1$, so that we will pick up just the classically allowed regions. Then we have

(54)
\begin{align} S_E = & C_1 \int_{- \sqrt{ - \gamma }}^y \frac{dt}{ \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + C_2 \int_{- \sqrt{ - \gamma }}^y \frac{dt}{ \left( t^2 - 1 \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ & + C_3 \int_{- \sqrt{ - \gamma }}^y \frac{dt}{ \left( t^2 - 1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + C_4 \int_{- \sqrt{ - \gamma }}^y \frac{dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ = & C_1 E_1 + C_2 E_2 + C_3 E_3 + C_4 E_4 \end{align}

and

(55)
\begin{align} S_{Elem} = & K_1 \int_{- \sqrt{ - \gamma }}^y \frac{t^3 dt}{ \left( t^2 -1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + K_2 \int_{- \sqrt{ - \gamma }}^y \frac{t dt}{ \left( t^2 -1 \right)^2 \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ & - \int_{- \sqrt{ - \gamma }}^y \frac{t dt}{ \left( t^2 -1 \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } + K_4 \int_{- \sqrt{ - \gamma }}^y \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( 1 - t^2 \right) } } \\ = & K_1 I_1 + K_2 I_2 - I_3 + K_4 I_4 \end{align}

where with the same definitions of $x$, $k$ and $j^2$ as in Case 1,

(56)
\begin{align} E_1 = - F \left( \sin^{-1} \left( \frac{1}{x} \right) , \frac{1}{k} \right), \quad -1 < y < - \sqrt{ - \gamma } \end{align}
(57)
\begin{align} E_2 = k^2 \left[ F \left( \sin^{-1} \left( \frac{1}{x} \right) , \frac{1}{k} \right) - E \left( \sin^{-1} \left( \frac{1}{x} \right) , \frac{1}{k} \right) + \sqrt{ \frac{ 1 - \frac{ 1 }{ k^2 x^2 } }{ x^2 - 1 } } + \frac{\pi}{2} \right], \quad -1 < y < - \sqrt{ - \gamma } \end{align}
(58)
\begin{align} E_3 = - \frac{1}{3 \left( 1 + \gamma \right)^2} \left[ \left( 2 + \frac{1}{k^2} \right) \sin^{-1} \left( \frac{1}{x} \right) - 2 \left( 1 + \frac{1}{k^2} \right) E \left( \sin^{-1 } \left( \frac{1}{x} \right) , \frac{1}{k} \right) + \left( 2 + \frac{2}{k^2} + \left( 1 - \frac{1}{k^2} \right) \frac{x^2}{x^2 - 1} \right) \sqrt{ \frac{1 - \frac{1}{k^2 x^2}}{ x^2 - 1 } } - \left( 1 + k^2 \right) \pi \right] \end{align}
(59)
\begin{align} E_4 = \frac{j^2}{ \left( q + \gamma \right) } \left[ \frac{1}{k^2} F \left( \sin^{-1} \left( \frac{1}{x} \right), \frac{1}{k} \right) + \left( \frac{1}{j^2} - \frac{1}{k^2} \right) \Pi \left( \frac{1}{j^2}; \sin^{-1} \left( \frac{1}{x} \right) , \frac{1}{k} \right) \right] \end{align}

## 2. Analysis of behavior of approximate wavefunction for small values of the scale factor

Case 1. $\gamma > 0$

Our approximate wavefunction is given by $e^{-S/ \hbar}$ where

(60)
\begin{align} S &= \pm \frac{1}{2 \sqrt{ A_1 A_2 } } \sqrt{ \frac{3}{\Lambda} } \left( S_E + S_{Elem} \right), \\ S_E &= C_1 E_1 + C_2 E_2 + C_3 E_3 + C_4 E_4 \\ S_{Elem} &= K_1 I_1 + K_2 I_2 - I_3 + K_4 I_4 \end{align}

From (52) it can be seen that $I_1$, $I_2$, and $I_3$ have no singularity at $a=0$ ($y= - \sqrt{q} = \alpha / \beta$). On the other hand

(61)
\begin{align} I_4 = \int_0 ^{-1} \frac{t dt}{ \left( t^2 - q \right) \sqrt{ \left( t^2 + \gamma \right) \left( t^2 - 1 \right) }} - \frac{ 1 }{ \sqrt{ \left( 1 - q \right) \left( \gamma + q \right) } } \tanh^{-1} \sqrt{ \frac{ \left( \gamma + q \right) \left( 1 - y^2 \right) }{ \left( 1 - q \right) \left( \gamma + y^2 \right) } }. \end{align}

The first term will only contribute a multiplicative constant in front of the whole wavefunction, so we neglect it and study the behavior of the second term. Using the fact that

(62)
\begin{align} \tanh^{-1} \left( \frac{a}{b} \right) = \frac{1}{2} \ln \left( \frac{b+a}{b-a} \right), \end{align}

the second term becomes

(63)
\begin{align} \frac{1}{2} \frac{ 1 }{ \sqrt{ \left( 1 - q \right) \left( \gamma + q \right) } } \ln \left( \frac{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } - \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } }{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } + \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } } \right). \end{align}

Multiplying in

(64)
\begin{align} K_4 = - \frac{ \hbar^2 p \left( i + k \right) }{ \beta^2 } \end{align}

and the overall multiplicative constant

(65)
\begin{align} \pm \frac{1}{2 \sqrt{ A_1 A_2 } } = \pm \frac{1}{2} \frac{ \left( \lambda_2 - \lambda_1 \right) }{ \sqrt{ \lambda_2 } } \sqrt{ \frac{3}{ \Lambda } }, \end{align}

the divergent piece of $S$ coming from $I_4$ is

(66)
\begin{align} -\frac{1}{4} \frac{ \left( \lambda_2 - \lambda_1 \right) }{ \sqrt{ \lambda_2 } } \sqrt{ \frac{ 3 }{ \Lambda } } \frac{ \hbar^2 p \left( i+k \right) }{ \beta^2 } \frac{ 1 }{ \sqrt{ \left( 1 - q \right) \left( \gamma + q \right) } } \ln \left( \frac{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } - \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } }{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } + \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } } \right) \\ = -\frac{1}{4} \sqrt{ \frac{3}{\Lambda} } \frac{ \hbar^2 p \left( i+k \right) }{ r_* \sqrt{ r_* - \frac{3}{\Lambda} } } \ln \left( \frac{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } - \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } }{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } + \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } } \right) \\ = \frac{ \hbar \sqrt{ - p } }{4} \ln \left( \frac{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } - \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } }{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } + \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } } \right) \end{align}

where in the final line we have used the fact that $r_*$ is a root of $u^3 - \frac{3}{\Lambda} u^2 + \frac{ 3 \hbar^2 p \left( i+k \right) }{ \Lambda }$. Negating and exponentiating, the divergent piece of the wavefunction $e^{-S / \hbar}$ coming from $I_4$ is

(67)
\begin{align} \left( \frac{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } - \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } }{ \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } + \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } } \right)^{ - \frac{ \hbar \sqrt{ - p } }{4} }. \end{align}

Recalling that $y = \frac{ a^2 - \alpha }{ a^2 - \beta }$ and denoting the function of $y$ (and hence of $a$) in parentheses by $f \left( a \right)$, we can see that as $a \to 0$, $f \left( a \right)$ is asymptotically equivalent to

(68)
\begin{align} \frac{ \left( \alpha - \beta \right) \alpha }{ 2 \beta^3 } \frac{ 1 + \gamma }{ \left( 1 - q \right) \left( \gamma + q \right) } a^2 \end{align}

because

(69)
\begin{align} \lim_{a \to 0} \frac{ f \left( a \right) }{ \frac{ \left( \alpha - \beta \right) \alpha }{ 2 \beta^3 } \frac{ 1 + \gamma }{ \left( 1 - q \right) \left( \gamma + q \right) } a^2 } = \lim_{a \to 0} \frac{f' \left( a \right)}{ \frac{ \left( \alpha - \beta \right) \alpha }{ \beta^3 } \frac{ 1 + \gamma }{ \left( 1 - q \right) \left( \gamma + q \right) } a }, \end{align}

and

(70)
\begin{align} f' \left( a \right) = 4 \sqrt{ \left( \gamma + q \right) \left( 1 - q \right) } \left( \alpha - \beta \right) \frac{ a }{ \left( a^2 - \beta \right)^2 } \frac{ y \left( \sqrt{ \frac{ 1 - y^2 }{ \gamma + y^2 } } + \sqrt{ \frac{ \gamma + y^2 }{ 1 - y^2 } } \right) }{ \left( \sqrt{ \left( 1 - q \right) \left( \gamma + y^2 \right) } + \sqrt{ \left( \gamma + q \right) \left( 1 - y^2 \right) } \right)^2 }, \end{align}

and the limit of the above expression leaving out the factor of $a$ in the numerator is

(71)
\begin{align} \frac{ \left( \alpha - \beta \right) \alpha }{ \beta^3 } \frac{ 1 + \gamma }{ \left( 1 - q \right) \left( \gamma + q \right) }. \end{align}

Thus the asymptotic behavior of the divergent piece of the wavefunction coming from $I_4$ goes as $a^{- \frac{ \hbar \sqrt{-p} }{2}}$.

Similarly as for (52), we can see that for (53), $E_1$, $E_2$, and $E_3$ have no singularity at $a=0$ ($y= - \sqrt{q} = \alpha / \beta$), while $E_4$ has a singularity which can be analyzed as follows:

(72)
\begin{align} E_4 = \frac{1}{ \left( q - 1 \right) \sqrt{ \gamma + 1} } \left[ \Pi \left( \frac{\pi}{2} , \frac{1}{1 - q}, \sqrt{ \frac{1}{1 + \gamma} } \right) - \Pi \left( \cos^{-1} \left( - y \right) , \frac{1}{1 - q}, \sqrt{ \frac{1}{1 + \gamma} } \right) \right] \end{align}

As in the case of $I_4$, we neglect the first term, since that merely contributes a multiplicative constant to the overall wavefunction. Thus we are interested in the divergent behavior of

(73)
\begin{align} \frac{-1}{ \left( q - 1 \right) \sqrt{ \gamma + 1 }} \Pi \left( \cos^{-1} \left( -y \right) , \frac{1}{1 - q} , \sqrt{ \frac{1}{1 + \gamma} } \right). \end{align}

We put the elliptic integral into the Jacobian form given by

(74)
\begin{align} \Pi \left( \phi, \ell^2, k \right) = \int_0^x \frac{dt}{\left( 1 - \ell^2 t^2 \right) \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) }} \end{align}

where $x = \sin \phi$, so that designating $\ell = 1/ \sqrt{ 1 - q}$ and $k = 1 / \sqrt{ 1 + \gamma }$

(75)
\begin{align} \Pi \left( \cos^{-1} \left( -y \right) , \ell^2 , k \right) = \int_0^{ \sqrt{1 - y^2} } \frac{ dt }{ \left( 1 - \ell^2 t^2 \right) \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) } } \end{align}

and we are interested in the behavior of

(76)
\begin{align} \frac{-1}{ \left( q - 1 \right) \sqrt{ \gamma + 1 }} \int_0^{ \sqrt{1 - y^2} } \frac{ dt }{ \left( 1 - \ell^2 t^2 \right) \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) } } \end{align}

as $y \nearrow - \sqrt{q}$ or equivalently $t \nearrow \sqrt{1 - q} = 1 / \ell$. To do so, break the integrand of the elliptic integral into a piece which diverges as $t \nearrow 1 / \ell$ and a piece which does not:

(77)
\begin{align} \frac{1}{\left( 1 - \ell^2 t^2 \right) \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) }} &= \frac{A}{\left( 1 - \ell^2 t^2 \right) } + \frac{B \left(t \right)}{ \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) }} \\ \implies B \left( t \right) &= \frac{1 - A \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) } }{\left( 1 - \ell^2 t^2 \right)} \end{align}

In order for $B \left( t \right)$ to have a finite limit at $t = 1/ \ell$, we must have

(78)
\begin{align} 1 - A \sqrt{ \left( 1 - t^2 \right) \left( 1 - k^2 t^2 \right) } = 0 \end{align}

so that $B \left( t \right)$ is an indeterminate form of type $0/0$. Thus using the definitions of $\ell$, $q$, and $\gamma$ we obtain

(79)
\begin{align} A = - \frac{\beta^2}{\alpha} \frac{1}{\sqrt{ r_* \left( r_* - \frac{3}{\Lambda} \right) }}, \end{align}

and using L'Hopital's rule it is easily verified that $\lim_{t \to 1 / \ell} B \left( t \right)$ is finite. We must thus study

(80)
\begin{align} - \frac{1}{\left( q -1 \right) \sqrt{ \gamma + 1 }} \int_0^{\sqrt{1-y^2}} \frac{A dt}{\left( 1 - \ell^2 t^2 \right) } = - \frac{A}{2 \sqrt{ \left( 1 - q \right) \left( \gamma + 1 \right) }} \ln \left( \frac{ \sqrt{ 1 - q } - \sqrt{ 1 - y^2 }}{ \sqrt{ 1 - q } + \sqrt{ 1 - y^2 } } \right) \end{align}

so that multiplying in the overall constant $-1/ \left( 2 \sqrt{ A_1 A_2 } \right)$ and $C_4$ and exponentiating, the wavefunction receives a factor of

(81)
\begin{align} \exp & \left\{ \frac{1}{2 \sqrt{A_1 A_2}} \sqrt{\frac{3}{\Lambda}} \left( - \hbar^2 p \frac{\alpha}{\beta^3} \right) \frac{A}{2 \sqrt{ \left( 1 - q \right) \left( \gamma + 1 \right) }} \ln \left( \frac{ \sqrt{ 1 - q } - \sqrt{ 1 - y^2 }}{ \sqrt{ 1 - q } + \sqrt{ 1 - y^2 } } \right) \right\} \\ &= \exp \left\{ \frac{1}{4} \sqrt{ \frac{3}{\Lambda} } \frac{ \hbar^2 p }{ r_* \sqrt{ r_* - \frac{3}{\Lambda} } } \ln \left( \frac{ \sqrt{ 1 - q } - \sqrt{ 1 - y^2 }}{ \sqrt{ 1 - q } + \sqrt{ 1 - y^2 } } \right) \right\} \end{align}

By a similar process to that used in the case of $I_4$, it can be seen that the argument of the logarithm above is asymptotically equivalent to

(82)
\begin{align} \frac{ \alpha }{ 2 \beta \left( \alpha + \beta \right) } a^2, \end{align}

and thus the divergent behavior of the wavefunction coming from $E_4$ is

(83)
\begin{align} a^{- \frac{ \sqrt{ -p } }{2}}, \end{align}

so combining with the blowup contributed by $I_4$ we see that as $a \to 0$, the wavefunction blows up as

(84)
\begin{align} a^{- \sqrt{-p}}. \end{align}

## Description of attached files

 File name Description Dependencies effpot.nb Graphs effective potential ($V \left( a \right) + QP$) Takes values of $\Lambda, \hbar^2, p \left( i+k \right)$ from e.g. elliptic_gamma_pos_unnorm.nb elem_assumptions_gamma_pos.nb Performs elementary integrals to get $I_1, I_2, I_3, I_4$ None elliptic_gamma_pos.nb Calculates $E_{i} \left( a \right), I_{i} \left( a \right), S_{Elem} \left( a \right), S_{E} \left( a \right), S \left( a \right)$ for given $\Lambda, \hbar^2, p \left( i + k \right)$. WARNING: This file is not finished - it has a technical issue with calculating Cauchy principal values of complete integrals. None elliptic_gamma_pos_unnorm.nb Same as elliptic_gamma_pos.nb, but it leaves out the complete integrals in elliptic_gamma_pos.nb since these will only contribute an overall multiplicative constant to the wavefunction. None t-plot.png Image of the graph of the variable $t$ as a function of $u$. None wavefnplots.nb Plots the various components of the wavefunction calculated e.g. in elliptic_gamma_pos_unnorm.nb Takes everything from e.g. elliptic_gamma_pos_unnorm.nb
page revision: 452, last edited: 11 Mar 2015 18:11