We have Bogoliubov-De Gennes (BdG) equation, $$\left(\begin{array}{cc} \mathbf{p}\cdot\boldsymbol{\sigma} - V & \Delta_{0}e^{i\phi} \\ \Delta_{0}e^{-i\phi} & V - \mathbf{p}\cdot\boldsymbol{\sigma} \end{array} \right) \left( \begin{array}{c}u\\v\end{array}\right) = \mathcal{E}\left( \begin{array}{c}u\\v\end{array}\right)$$ with definition, $$\mathbf{p}\cdot\boldsymbol{\sigma} = -i\hbar v_{F}(\partial_{x}\sigma_x + \partial_y\sigma_y)$$ We solve the equation through ansatz, $$\Phi = {\rm Const}\cdot \exp(iqy + ik_{0}x + \kappa x)$$ for $k_{0} > 0$ and $\kappa > 0$. The question is: How to use the assumption that $V >> \Delta_{0}, \mathcal{E}$ in the calculations to get nice looking formulas like in Beenakker (A13)?
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$\begingroup$ Did you at least tried to find $k_{0}$, $q$ and $\kappa$ ? The assumption come from their expressions ... $\endgroup$– FraSchelleCommented Sep 2, 2014 at 9:14
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$\begingroup$ Yes I did, but it involves double square root (nested). You can find Yourself expressions for $k_{0}$ and $\kappa$ in Mathematica which depend on $q$. $\endgroup$– WoofDoggyCommented Sep 2, 2014 at 20:04
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2 Answers
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When you write down the formulas A15-A18, apply this assumption to these definitions and put it into wave functions, which are first calculated by solving BdG equation. You can reach the Beenakker results.
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This is actually exactly the same question I was asking myself a while ago, and it took me quite some time to figure it out. What I ended up doing was:
- diagonalise the BdG Hamiltonian in Mathematica
- solve the expressions for the eigenvalues for $\kappa$
- neglect terms proportional to $\kappa^2$, $\kappa \Delta$ in the expressions for the eigenvectors, expand them to first order in $\kappa$, insert the expression for $\kappa$ you got from the eigenvalues
- simplify everything (using the expression for $\epsilon$ and approximate $\hbar v k \approx E_F+U$) and arrive at Beenakker's simple-looking expressions (tedious)
I hope that helps some, in case you haven't figured it out yourself in the meantime. If you need more info/details don't hesitate to ask, I can elaborate some more if needed!
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$\begingroup$ If you're still willing, please could you elaborate? This is still causing me trouble! $\endgroup$– xzd209Commented Mar 30, 2022 at 14:20