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Extended Blonder Tinkham Klapwijk in graphene

Physics Asked on February 18, 2021

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 $0<x<d$, 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}cdotvec{sigma} + U(r) &Delta(r) sigma_0
Delta^*(r) sigma_0& -v_F vec{p}cdotvec{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 = ksin alpha = k’ sinalpha’.
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{partialrho}{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 ‘},0,0right)e^{-i sigma k’ x cos alpha’}
end{equation}

if the angle of incidence $alpha$ satisfies
begin{equation}
alpha < alpha_c equivarcsinleft(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 depsilon left(-frac{partial f}{partial epsilon}right) int dalpha left( frac{k’}{k} |t_{ee}|^2 cos alpha’ + dotsright).
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 depsilon 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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