An algorithm for exact analytical solutions for tilted anisotropic Dirac materials

Abstract

In this article, we obtain the exact solutions for bound states of tilted anisotropic Dirac materials under the action of external electric and magnetic fields with translational symmetry. In order to solve the eigenvalue equation that arises from the effective Hamiltonian of these materials, we describe an algorithm that allows us to decouple the differential equations that are obtained for the spinor components.

Appendices

Appendix A The matrix $\mathbb {S}$

If the matrix $\mathbb {K}$ described in section 3 becomes constant, it can be expressed as

$$\begin{aligned} \mathbb {K}= \begin{pmatrix} -a_2&{}-ia_1\\ &{}\\ -ia_1&{}a_2 \end{pmatrix}. \end{aligned}$$

(A.1)

Then, the characteristic polynomial of $\mathbb {K}$ is given by

$$\begin{aligned} p_{\mathbb {K}}(t)=t^2-(a_2^2-a_1^2). \end{aligned}$$

(A.2)

Its corresponding eigenvalues are $\lambda _1=\lambda$ and $\lambda _2=-\lambda$ with $\lambda =\sqrt{a_2^2-a_1^2}$. Consequently, the corresponding eigenvectors $\vec {v_j}$ for $j=1,2$ turn out to be

$$\begin{aligned} \vec {v_1}=w_1 \begin{pmatrix} \frac{a_1}{a_2+\lambda }\\ &{}\\ i \end{pmatrix},\quad \vec {v_2}=w_2 \begin{pmatrix} -i\\ &{}\\ \frac{a_1}{a_2+\lambda } \end{pmatrix}, \end{aligned}$$

(A.3)

where $w_1$, $w_2$ are two non-zero complex constants that we can choose at will. In this way, the matrix $\mathbb {S}=(\vec {v_1},\vec {v_2})$ reads as

$$\begin{aligned} \mathbb {S}=\frac{a_1(w_1+w_2)}{2(a_2+\lambda )}\sigma _0+i\frac{w_1-w_2}{2}\sigma _x+\frac{w_1+w_2}{2}\sigma _y+\frac{a_1(w_1-w_2)}{2(a_2+\lambda )}\sigma _z, \end{aligned}$$

(A.4)

while its inverse matrix $\mathbb {S}^{-1}$ and its conjugate transpose matrix $\mathbb {S}^{\dagger }$ are represented as follows:

$$\begin{aligned} \mathbb {S}^{-1}&=-\frac{a_2+\lambda }{2\lambda w_1 w_2} \left( \frac{a_1(w_1+w_2)}{2(a_2+\lambda )}\sigma _0-i\frac{w_1-w_2}{2}\sigma _x-\frac{w_1+w_2}{2}\sigma _y-\frac{a_1(w_1-w_2)}{2(a_2+\lambda )}\sigma _z\right) ,\nonumber \\ \mathbb {S}^{\dagger }&=\frac{a_1(\bar{w}_1+\bar{w}_2)}{2(a_2+\lambda )}\sigma _0-i\frac{\bar{w}_1-\bar{w}_2}{2}\sigma _x+\frac{\bar{w}_1+\bar{w}_2}{2}\sigma _y+\frac{a_1(\bar{w}_1-\bar{w}_2)}{2(a_2+\lambda )}\sigma _z. \end{aligned}$$

(A.5)

It is straightforward to prove that $\mathbb {K}$ is diagonalizable by the similarity transformation $\mathbb {S}^{-1}\mathbb {K}\mathbb {S}$, i.e.,

$$\begin{aligned} \mathbb {M}=\mathbb {S}^{-1}\mathbb {K}\mathbb {S}= \begin{pmatrix} \lambda &{}0\\ &{}\\ 0&{}-\lambda \end{pmatrix}. \end{aligned}$$

(A.6)

We have to highlight that the form of $\mathbb {S}$ depends not only on $w_1,w_2$ but also on the order of $\vec {v_1},\vec {v_2}$, i.e., $\mathbb {S}=(\vec {v_2},\vec {v_1})$ is a suitable similarity transformation that diagonalizes $\mathbb {K}$. However, in that case the result has to be $\mathbb {M}=\text{ diag }(-\lambda ,\lambda )$.

As we have mentioned, $\mathbb {S}$ has no specific form. Nevertheless, in order to simplify the calculations, in this work we will consider the special case when $w_1=w_2=1$ in Eq. (A.4). This case can be understood as the one leading to

$$\begin{aligned} \mathbb {S}&=\frac{a_1}{(a_2+\lambda )}\sigma _0+\sigma _y,\nonumber \\ \mathbb {S}^{-1}&=-\frac{a_1}{2\lambda }\sigma _0+\frac{a_2+\lambda }{2\lambda }\sigma _y,\nonumber \\ \mathbb {S}^{\dagger }&=\frac{a_1}{(a_2+\lambda )}\sigma _0+\sigma _y. \end{aligned}$$

(A.7)

Finally, it can be observed that if $a_1=0$, $\mathbb {S}^{-1}=\mathbb {S}^{\dagger }$, which is because for such a value $\mathbb {K}$ becomes Hermitian and normal.

Appendix B Probability and current densities

The eigenstates of the Hamiltonian obtained from (16) are stationary. Hence

$$\begin{aligned} \rho _n(x,y,t)=\rho _n(x)=|\mathcal {N}_n|^2\bar{\Phi }^{\dagger }_n(z)\mathbb {S}^{\dagger }\mathbb {S}\bar{\Phi }_n(z), \end{aligned}$$

(B.1)

where the functions $\bar{\Phi }^{\dagger }_n(z),\bar{\Phi }_n(z)$ depend on x since z is a change of variable and $\mathcal {N}_n$ represents a normalization constant. By considering the matrix $\mathbb {S}$ given in Eq. (A.7) we get

$$\begin{aligned} \rho _n(x)=\frac{2|\mathcal {N}_n|^2}{a_2+\lambda }\bar{\Phi }^{\dagger }_n(z)\left( a_2\sigma _0+a_1\sigma _y\right) \bar{\Phi }_n(z). \end{aligned}$$

(B.2)

Thus, the normalization constant can be chosen as $\mathcal {N}_n=\sqrt{\frac{a_2+\lambda }{2(a_2+a_1I_n)}}$ where

$$\begin{aligned} I_n=2^{\delta _{0n}}\sqrt{\frac{v_x}{v_y}}(1-\delta _{0n})\int _{z^-}^{z^+}\frac{\phi _n^+(z)\phi _{n-1}^-(z)}{g(z)}\textrm{d}z. \end{aligned}$$

(B.3)

and then

$$\begin{aligned} \rho _n(x)=\frac{1}{(a_2+a_1I_n)} \left( a_2|\bar{\Phi }_n(z)|^2+2^{\delta _{0n}}(1-\delta _{0n})a_1\phi _n^+(z)\phi _{n-1}^-(z)\right) . \end{aligned}$$

(B.4)

In a similar way, we can compute the components of the current density $\vec {\mathcal {J}_n}(x,y,t)$, which are given by

$$\begin{aligned} \mathcal {J}_{x,n}(x,y,t)&=\mathcal {J}_{x,n}(x) =\nu v_x|\mathcal {N}_n|^2\bar{\Phi }^{\dagger }_n(z)\left( \mathbb {S}^{\dagger }\sigma _x\mathbb {S}\right) \bar{\Phi }_n(z),\nonumber \\ \mathcal {J}_{y,n}(x,y,t)&=\mathcal {J}_{y,n}(x)=\nu |\mathcal {N}_n|^2\bar{\Phi }^{\dagger }_n(z) \left[ \mathbb {S}^{\dagger }\left( v_x\sigma _x+v_t\sigma _0\right) \mathbb {S}\right] \bar{\Phi }_n(z). \end{aligned}$$

(B.5)

After some calculations, they turn out to be

$$\begin{aligned} \mathcal {J}_{x,n}(x)&=-\frac{\nu v_x\lambda }{a_2+a_1I_n} \bar{\Phi }^{\dagger }_n(z) \sigma _x\bar{\Phi }(z)_n,\nonumber \\ \mathcal {J}_{y,n}(x)&=\frac{\nu }{a_2+a_1I_n} \left[ \left( a_1 v_y +a_2 v_t\right) |\bar{\Phi }_n(z)|^2 +2^{\delta _{0n}}(1-\delta _{0n})\left( a_1 v_t + a_2 v_y\right) \phi _n^+(z)\phi _{n-1}^-(z)\right] . \end{aligned}$$

(B.6)

We have to mark that the above expressions for the probability and current densities are only valid for the eigenstates of the Hamiltonian (2). On the other hand, the $x-$component of the current $\mathcal {J}_n$ will be zero if the components $\phi _{n}^{\pm }$ are real functions.