# Background

Consider a normal metal (N) - superconductor (S) junction. Assume N is in the region $$x<0$$ and S is in the region $$x>0$$. In addition, let us assume that the system is invariant in the $$y$$ and $$z$$ directions such that the momentum parallel to the interface is conserved.

The system can be described by the BdG Hamiltonian of the form $$\begin{equation} \begin{bmatrix} H_e & \Delta\\ \Delta & -(H_e)^* \end{bmatrix} \begin{bmatrix} u(x,y,z)\\ v(x,y,z) \end{bmatrix} = E\begin{bmatrix} u(x,y,z)\\ v(x,y,z) \end{bmatrix}, \end{equation}$$ where $$\Delta = \Delta_0\theta(x)$$, $$H_e = -\frac{\hbar^2}{2m}\nabla^2 - \mu(x)$$, and $$\mu(x) = \mu>0$$. Here $$\Delta_0$$, $$m$$, and $$\mu$$ are positive constants.

## Question

Let us assume that an electron is incident from N. By solving this equation for $$x<0$$ and $$x>0$$ we can obtain the reflection coefficient for normal reflection and Andreev reflection. I have read many places that the normal reflection is specular, while the Andreev reflection is retroreflective. By normal reflection I mean that the angle of incidence is equal to the angle of reflection. By retroreflection I mean the process sketched in the image below: The filled red circle is an electron with spin up, the hollow dashed circle is a hole with spin down.

My question is why and how can one show from the BdG equation that the hole is always retroreflected? Alternatively Why is specular Andreev reflection forbidden in this simple case?

I'm asking this question because I would like to understand why one can have specular andreev reflection in graphene:

Andreev reflection and Klein tunneling in graphene

but not in the simple model above.

The solution is actually quite simple. Let us for simplicity consider two dimensions $$(x,y)$$. The electron and hole dispersions are $$\begin{equation} \mathbf{k}^2 = \frac{2m}{\hbar}\left(\mu + E\right)\, \mathrm{and}\, \mathbf{q}^2 = \frac{2m}{\hbar}\left(\mu - E\right) \end{equation}$$ respectively. Therefore, the group velocities are $$\begin{equation} \mathbf{v}^e = \frac{\hbar^2}{m} \mathbf{k}\, \mathrm{and}\,\mathbf{v}^h = -\frac{\hbar^2}{m} \mathbf{q}. \end{equation}$$ Note that the (electrons) holes velocity is in the (same) opposite direction of its wavenumber. This is a consequence of the opposite curvature of the electron and hole bands. In addition, there is translational invariance in the $$y$$ direction so the corresponding wavenumber has to be conserved. This means that if the electron enters with a wavenumber $$\mathbf{k}=(k_x,k_y)$$ the y-component of velocity of the reflected electron will be in the $$+\hat{y}$$-direction, while the y-component of velocity for the reflected hole will be in the $$-\hat{y}$$-direction.
Furthermore, if we denote the electron angle of incidence by $$\alpha$$, the angle of electron reflection by $$\beta$$, and the angle of hole reflection by $$\gamma$$ in k-space we can write $$\begin{equation} |\mathbf{k}|\sin \alpha = |\mathbf{k}|\sin \beta = |\mathbf{q}|\sin \gamma. \end{equation}$$ This means that $$\beta = \alpha$$, and the electron will always be perfectly specularly reflected.
For the hole we have $$\begin{equation} \gamma = \arcsin\left(\frac{|\mathbf{k}|}{|\mathbf{q}|} \sin\alpha\right). \end{equation}$$ This means that the hole is only perfectly retro reflected if $$|\mathbf{k}| = |\mathbf{q}|$$. If we assume that $$\mu\gg E$$ (known as the Andreev approximation) we obtain exactly this condition $$\begin{equation} \begin{split} &|\mathbf{k}^2| = \frac{2m \mu}{\hbar}\left(1 + E/\mu\right) \approx \frac{2m \mu}{\hbar},\\ &|\mathbf{q}^2| = \frac{2m \mu}{\hbar}\left(1 - E/\mu\right) \approx \frac{2m \mu}{\hbar}. \end{split} \end{equation}$$
In undoped $$(\mu=0)$$ graphene the dispersion of electrons and holes are identical, so the relationship between velocity and wavenumber is identical. Hence, in graphene the Andreev reflection can be specular.