1
$\begingroup$

In the famous article Physical Review B 92, 064520 (2015), a theoretical model was proposed to realize the chiral Majorana zero mode.

$$H_{\mathrm{BdG}}=\left(\begin{array}{cc}{H_{0}(\mathbf{k})-\mu} & {\Delta_{\mathbf{k}}} \\ {\Delta_{\mathbf{k}}^{\dagger}} & {-H_{0}^{*}(-\mathbf{k})+\mu}\end{array}\right)$$

Here, $$ \mathcal{H}_{\mathrm{BdG}}=\sum_{\mathbf{k}} \Psi_{\mathbf{k}}^{\dagger} H_{\mathrm{BdG}} \Psi_{\mathbf{k}} / 2 ,$$ where $$\Psi_{\mathbf{k}}=\left[\left(c_{\mathbf{k} \uparrow}^{t}, c_{\mathbf{k} \downarrow}^{t}, c_{\mathbf{k} \uparrow}^{b}, c_{\mathbf{k} \downarrow}^{b}\right),\left(c_{-\mathbf{k} \uparrow}^{t \dagger}, c_{-\mathbf{k} \downarrow}^{t \dagger}, c_{-\mathbf{k} \uparrow}^{b \dagger}, c_{-\mathbf{k} \downarrow}^{b \dagger}\right)\right]^{T}.$$

Obviously, the Hamiltonian contains both $k_x$ and $k_y$.

My question is how to discretize the Hamiltonian and apply it to a square lattice ribbon (of finite size in the $y$ direction but periodic in the $x$ direction)?

$\endgroup$
0
$\begingroup$

Introduce an ultraviolet cutoff. The inverse of this cutoff gives you your lattice spacing. Momentum will be discretized in the periodic direction; in the nonperiodic direction, it will depend on your boundary conditions. Beyond that, I am not sure what you are asking.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Now the given Hamiltonian contains kx and ky. I want to know how to get the Hamlitonian for nanoribbons(square lattice, periodic in x direction with open boundary condition in y direction). Since ky will not be a good quantum number, the Hamiltonian should contain c(i) which annihilates an electron on site i. $\endgroup$ – OliverShen Oct 18 '19 at 2:22
0
$\begingroup$

A tight-binding Hamiltonian $$ H = -t\sum_{j} c^{\dagger}_j c_{j+1} +{\rm h.c.}$$ in the momentum representation is $H=-2t\sum_k\cos(k)c^{\dagger}_k c_k$ and when we expand for small $k$ we get the quadratic dispersion of a free particle. In a similar manner, if we will hop with $it$ instead of $t$ we will get a $\sin(k)$ dispersion that for small $k$ will give the linear behavior you are looking for. So to answer your question - you discretize by writing a tight-binding Hamiltonian where quadratic dispresion are real tight-binding hopping and linear terms in $k$ are tight-binding with hopping $i$. Note that you must adjust the chemical potential $\mu$ accordingly. For example, in 1d the tight-binding Hamiltonian $$ H = -t\sum_{j} c^{\dagger}_j c_{j+1} +{\rm h.c.} - \mu\sum_j c^{\dagger}_j c_j = -\sum_k[\mu+2t\cos(k)]c^{\dagger}_kc_k$$ and one chooses $\mu=-2t$ to get the bottom of the band to be near zero energy, which is $\mu=0$ for the continuous model.

HOWEVER, and this is a big however, one should proceed with care as to the phase diagram, since now you have also $k=\pi$ as low energy states, which in the continuous model will be high-energy states! In general it should work at least for a limited part of the phase diagram, but you should check for this specific model what you get.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.