# Background

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 $$0, and the right metal is hole doped and located in the region $$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 $$I$$?