For context, I'm studying the paper Coulomb blockade in superconducting quantum point contacts by Averin from 1998. Specifically, I am trying to find how he obtains equation 11 from equation 10, which gives the Landau Zener probability of ending up in a specific branch of the Josephson potential of a superconducting QPC.

Equation 10, describing the Schrodinger equation of the problem in the specific limit under consideration, is given by a system of two coupled first order ODE's: \begin{equation} 2\sqrt{\frac{E_C}{\Delta}} \frac{\partial\psi_s}{\partial x} = -s x \psi_s/2 + \sqrt{R} \psi_{-s} \end{equation} where $s\pm1$.

If I (for convenience) now take $A = 2\sqrt{\frac{E_C}{\Delta}}$ and $B = \sqrt{R}$, then substituting the differential equations one finds a 2nd order ODE that is solved by parabolic cylinder functions. Specifically, \begin{equation} \psi_{-1} = c_1 D_{\frac{-B^2-A}{A}}\left(\frac{x}{\sqrt{A}}\right)+c_2 D_{\frac{B^2}{A}}\left(\frac{i x}{\sqrt{A}}\right) \end{equation}


\begin{equation} \psi_1 = -\frac{1}{B}\left(x c_2 D_{\frac{B^2}{A}}\left(\frac{i x}{\sqrt{A}}\right)+\sqrt{A} \left[c_1 D_{-\frac{B^2}{A}}\left(\frac{x}{\sqrt{A}}\right)+i c_2 D_{\frac{B^2+A}{A}}\left(\frac{i x}{\sqrt{A}}\right)\right]\right) \\ \end{equation}

Now, in his paper he then says that by evaluating the asymptotes of these functions, one can find the probability $w = |\psi_{-1}(\infty)|^2$ for the state $s=1$ starting at $x \rightarrow -\infty$ to end up in the state $s=-1$ at $x\rightarrow \infty$. This leads to equation 11, which he writes as \begin{equation} w = |\psi_{-1}(\infty)|^2 = \frac{1}{\Gamma(\lambda)}\sqrt{\frac{2\pi}{\lambda}}\left(\frac{\lambda}{e}\right)^\lambda \end{equation}

where $\lambda = \frac{R^2}{2\sqrt{E_C/\Delta}}$, and using our substitutions, we can identify $\lambda = B^2/A$, which we already saw occur in the solution for $\psi_{-1}$.

My question is how one obtains this. In terms of the mathematics, we need to choose our boundary conditions (setting $c_1$ and $c_2$) such that $|\psi_1(-\infty)|^2 = 1$, $|\psi_{-1}(-\infty)|^2 = 0$ and then evaluate $|\psi_{-1}(\infty)|^2$ to get $w$. But I can't seem to figure out how to properly find the values of $c_1$ and $c_2$. I imagined that taking the limit $x->-\infty$ would see one of the parabolic cylinder equations go to zero while the other does not, so as to satisfy $|\psi_{-1}(-\infty)|^2 = 0$, and then I need to normalize the remaining coefficient so that $|\psi_1(-\infty)|^2 = 1$ is satisfied. Then evaluating 1 - $|\psi_{1}(\infty)|^2$ should give the solution, but evaluating these asymptotes does not seem to be working out.

One hint is perhaps already found in the form of the answer itself, which clearly looks like some form of a Stirling approximation/Gamma function in $\lambda$, but I can't make sense of it. Moreover, I should note that this problem is essentially a Landau Zener tunneling problem for those who are familiar with that.

Some code for evaluating the ODE's:

ff = f[x] /. Solve[A*g'[x] == x*g[x]/2 + B*f[x], f[x]];
dff = D[ff, x];
gg = g[x] /. 
  DSolve[{A*f'[x] == -x*f[x]/2 + B*g[x] /. {f[x] -> ff, 
      f'[x] -> dff}}, g[x], x]
dgg = D[gg, x];
ff = FullSimplify[ff /. {g[x] -> gg, g'[x] -> dgg}]
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    $\begingroup$ Isn't the MMA component very limited here? Wouldn't this question be more appropriate for the math group at math.stackexchange.com? $\endgroup$ – MarcoB Mar 22 '19 at 18:19
  • $\begingroup$ You're probably right. I imagine that one could use Mathematica to evaluate these limits, that was why I posted it here. $\endgroup$ – user129412 Mar 22 '19 at 19:02
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    $\begingroup$ What's the relationship between $w$ and $ψ_s$? $\endgroup$ – xzczd Mar 23 '19 at 3:45
  • $\begingroup$ Ah, sorry. I wrote that in a rather convoluted way. $w = |\psi_{s_{-1}}(\infty)|^2$. I'll update the main post accordingly $\endgroup$ – user129412 Mar 23 '19 at 10:33

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