# Electron and holes in tight binding Hamiltonian on two sublattices

Background

Say that I have a tight binding Hamiltonian (with spinless fermions) of the form $$H = - 2t \sum_{ij} (c_i^{\dagger} c_j + h.c) - 2\mu \sum_i c^{\dagger}_i c_i$$ where we only sum over nearest neighbours and $$\mu$$ is the chemical potential. If there is only one sublattice I can diagonalize this Hamiltonian using a Fourier transformation $$c_i = 1/N \sum_k e^{ikx}c_k$$. Furthermore, to make the particle-hole symmetry explicit I symmetrize the operators by writing $$\begin{equation} c^{\dagger}_k c_k = \frac{1}{2}\left( c^{\dagger}_k c_k + c^{\dagger}_k c_k\right)= \frac{1}{2}\left( c^{\dagger}_{k} c_k - c_{-k} c_{-k}^{\dagger}\right) + constant \end{equation}$$ and ignore the constant. I then obtain the Hamiltonian $$\begin{equation} H = \sum_k \left[c^{\dagger}_k\quad c_{-k}\right]\begin{bmatrix} \gamma_k - \mu & 0 \\ 0 & -\gamma_k + \mu \end{bmatrix} \begin{bmatrix} c_k \\ c^{\dagger}_{-k} \end{bmatrix}. \end{equation}$$ Here $$\gamma_k = -t \sum_{\delta} e^{i \mathbf{k}\cdot {\delta}}$$ where $$\delta$$ is the vector connecting nearest neighbours.

Thus the eigenvalues of the electrons is $$\epsilon_e = \gamma_k - \mu$$ and the eigenvalues of the holes is $$\epsilon_h = -\gamma_k + \mu$$ and the chemical potential takes the role of the Fermi energy $$\mu = E_f$$.

The problem

Ok, now let's say that there are two sublattices $$A$$ and $$B$$ with creation/annhilation operators $$A_i$$ and $$B_i$$. By following the procedure outlined above: first Fourier transformation, then a symmetrization of all operators; I obtain the Hamiltonian

$$\begin{equation} H = \sum_k \left[A^{\dagger}_k\quad B^{\dagger}_k \quad A_{-k} \quad B_{-k}\right]\begin{bmatrix} -\mu & \gamma_k & 0 & 0 \\ \gamma_k & -\mu & 0 & 0 \\ 0 & 0 & \mu & -\gamma_k \\ 0 & 0 & -\gamma_k & \mu \end{bmatrix} \begin{bmatrix} A_k \\ B_{k} \\ A^{\dagger}_{-k} \\ B^{\dagger}_{-k} \end{bmatrix}. \end{equation}$$ In this case I find the 4 eigenvalues $$\begin{equation} \begin{split} &\epsilon_1 = \gamma_k + \mu\\ &\epsilon_2 = \gamma_k - \mu\\ &\epsilon_3 = -\gamma_k + \mu\\ &\epsilon_4 = -\gamma_k - \mu. \end{split} \end{equation}$$

My question is: what kind of particles do the eigenvalues $$\epsilon_{1,2,3,4}$$ describe physically? To me it seems that $$\epsilon_{2,3}$$ describe electrons and particles as above. But what on earth is the physical interpretation of $$\epsilon_{1,4}$$? I think they are strange because if I set $$\mu = E_f$$ then the eigenvalues describe particles that never cross the Fermi energy $$E_f$$.

I do of course understand that I get two additional eigenvalues because there are degrees of freedom associated with the two sublattices. However, how do I reconcile the two ways of looking at the tight-binding Hamiltonian presented here? Shouldn't I get the same answer if I set the two sublattices equal by letting $$B \rightarrow A$$?

The context of the question is that I want to look at Andreev reflection in a normal-metal - superconductor hybrid junction with two sublattices. Which one of these excitations are holes and contribute to the Andreev reflection?

I suggest the following: let's not double the degrees of freedom at the beginning, as it is not necessary until we introduce superconductivity. We can always do it after we understand the normal spectrum (I wrote a little bit about the doubling of the degrees of freedoms in this answer). Alternatively - you can think of it as diagonalizing each of the $$2\times 2$$ separate blocks in your $$4\times 4$$ matrix separately.

So we have the band Hamiltonian $$H = \sum_k h_k$$ with $$h_k = \left(\begin{array}{cc} A^{\dagger}_k & B^{\dagger}_k \end{array}\right) \left( \begin{array}{cc} -\mu & \gamma_k \\ \gamma_k & -\mu \end{array} \right) \left( \begin{array}{c} A_k \\ B_k \end{array} \right)$$ which we can readily diagonalize, getting the eigenoperators $$c^{\dagger}_{\pm, k} =\frac{A^{\dagger}_k \pm B^{\dagger}_k}{\sqrt{2}}$$ with the eigenenergies $$\epsilon_{\pm, k} = \pm \gamma_k -\mu$$.

The content of this excitations is clear from the point of view of the lattice - they create a particle with wave number $$k$$ that lives on both lattices $$A$$ and $$B$$ with different phases between the sublattices.

Now we can turn to the particle/hole discussion: we assume that we fill up all the states with negative energies. Then the ground state is $$|{\rm gs}\rangle = \prod_{\epsilon_{\nu,k}<0} c^{\dagger}_{\nu,k}|{\rm vac}\rangle$$. By adjusting $$\mu$$ we can control the filling of the bands. Note that for tight-binding $$\gamma_k=-2t\cos(k)$$ and $$\mu=0$$ (half-filling) both types of $$\pm$$ particles have both positive and negative energies, and the $$+$$ particles will be occupied for $$-\pi/2 < k < \pi/2$$ while the $$-$$ particles will be occupied for $$|k| > \pi/2$$ (I take the range of $$k$$ to be $$-\pi < k \leq \pi$$).

Now we can double the degrees of freedom artificially and get explicitly that $$c_{+,k}$$, for example, creates a hole for $$-\pi/2 < k < \pi/2$$ and it will have negative energy in this scenario.

Alternatively, we can deplete the system by setting $$\mu = -2t$$ in which case all excitations have positive energies, and we have no hole-like excitations.

• thx this is a very good answer. But there is one loose end. To me it seems that the fermi energy $E_f = 0$ even when $\mu = t/2$. I was under the impression that at zero temperature $\mu = E_f$. So what is the relationship between $\mu$ and $E_f$? Now it seems that the chemical potential is a free parameter used to control the degree of filling, while the fermi energy is fixed at $E = 0$ Nov 14, 2019 at 18:10
• Maybe this answer could help? physics.stackexchange.com/a/512706/245141
– user245141
Nov 19, 2019 at 9:28