In this paper the author studies the BdG equation for a metal-superconductor-metal system. The left metal is electron doped and is in the region $x<0$, the supercondcutor is in the region $0<x<d$, and the right metal is hole doped and located in the region $x>d$. The BdG equation is

\begin{equation} \begin{bmatrix} v_F \vec{p}\cdot\vec{\sigma} + U(r) &\Delta(r) \sigma_0\\ \Delta^*(r) \sigma_0& -v_F \vec{p}\cdot\vec{\sigma} - U(r) \end{bmatrix}\Psi(r) = \epsilon \Psi(r), \end{equation} Here we defined the fermi velocity $v_f$, the momentum operator $\vec{p} = - i \hbar (\partial_x,\partial_y)$, the Pauli matrices $\vec{\sigma} = (\sigma_x,\sigma_y)$, the superconducting gap $\Delta(r) = \Delta_0 e^{i \phi}$, and a gate voltage $U$ used to control the Fermi energy.

Due to the translational invariance we can write that the momentum in the y-direction is conserved: \begin{equation} k_y = k\sin \alpha = k' \sin\alpha'. \end{equation} Here $\hbar v_F k = \mu + E $, and $\hbar v_F k' = |\mu - E|$, and $\alpha$ is the angle of incidence.

If we define the probability density as $\rho = \Psi^{\dagger} \Psi$ we can obtain the continuity equation \begin{equation} \frac{\partial\rho}{\partial t} + \partial_x J_x + \partial_y J_y = 0 \end{equation} where the probability density in the x direction is given by \begin{equation} J_x = v_F \Psi^{\dagger}\left( \sigma_z \otimes \sigma_x \right)\Psi. \end{equation}

Scattering problem

To evaluate the conductance it is necessary to formulate a scattering problem. In the region $x>d$ the author writes that the solution is \begin{equation} \Psi(x) = t_{ee}\left(1,e^{-i \sigma \alpha '},0,0\right)e^{-i \sigma k' x \cos \alpha'} \end{equation} if the angle of incidence $\alpha$ satisfies \begin{equation} \alpha < \alpha_c \equiv\arcsin\left(\frac{|\mu-\epsilon|}{\mu + \epsilon}\right). \end{equation} If I use this state to evaluate the transmission probability I obtain \begin{equation} T_{ee} = \frac{J_x^{\mathrm{transmitted}}}{J_x^{\mathrm{in}}} = |t_{ee}|^2 \frac{\cos \alpha'}{\cos \alpha}. \end{equation}

In the paper (Eq. (3)) the author writes that the corresponding contribution to the conductance is \begin{equation} \frac{\partial I}{\partial V} = g_0 \int d\epsilon \left(-\frac{\partial f}{\partial \epsilon}\right) \int d\alpha \left( \frac{k'}{k} |t_{ee}|^2 \cos \alpha' + \dots\right). \end{equation} My question is where does the factor $k'/k$ come from? Or alternatively, how can we derive this?

Attempt at solution

I thought the current could be defined as something like \begin{equation} I = 2e \int dk_y L_y \int dk_x v(k_x) T_{ee}F \end{equation} where $L_y$ is the length in the $y$-direction, $v(k_x) = \frac{d E}{dk_x}$, and $F$ is some combination of Fermi distributions giving rise to the factor $\left(-\frac{\partial f}{\partial \epsilon}\right)$. However, if I do this I end up with \begin{equation} I \sim \int d\epsilon \int d \alpha k |t_{ee}|^2 \cos \alpha' \end{equation} where I miss the factor $k'/k$. I guess therefore that my expression for the current is not correct. Is there anything I have forgotten in my starting equation for $I$?


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