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 $w$ 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 = \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 $\psi_{-1}$.

In terms of mathematics, this statement means that we need to set $|\psi_1(-\infty)|^2 = 1$ and $|\psi_{-1}(-\infty)|^2 = 0$, and then evaluate $w = |\psi_{-1}(\infty)|^2 = 1 - |\psi_{1}(\infty)|^2$. But how does one do this? I need to pick $c_1$ and $c_2$ to satisfy these boundary conditions at $-\infty$, but so far I haven't managed to figure out how to do 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}]
  • $\begingroup$ Is there any reason to choose the basis of solutions of the ODE for parabolic cylinder functions in the form you are using? (it does not look like the best choice to me) $\endgroup$ Mar 26 '19 at 21:52
  • $\begingroup$ No, the only reason I made that choice is because it was what came out of Mathematica. I didn't think about it very much I have to say. $\endgroup$ Mar 26 '19 at 21:54
  • 1
    $\begingroup$ The (2-dimensional) basis of solutions can be chosen in infinitely many ways but you need only a one-dimensional subspace of solutions (for $\psi_{-1}$) determined by the condition of vanishing at $x=-\infty$. It is generated by $D_{\frac{-B^2-A}{A}}(-x/\sqrt A)$. $\endgroup$ Mar 26 '19 at 22:14
  • $\begingroup$ I see! I will give it a try with that. In that case the problem probably comes down to choosing the coefficient such that $|\psi_1(\infty)|^2$ evaluates to 1. $\endgroup$ Mar 27 '19 at 9:00

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