1 Introduction

Dirac materials [3] are lattice systems in which low-energy charge carriers behave like Dirac fermions. The development of these materials started with the isolation of graphene [7, 9], a two-dimensional monolayer of carbon atoms forming a hexagonal honeycomb lattice. Further examples for Dirac materials include d-wave superconductors [2, 12], superfluids [14, 21], and topological insulators [20, 23], just to name a few. Dirac materials exhibit many unusual properties, one of which is Klein tunneling. This phenomenon, referring to perfect transmission of Dirac fermions impinging perpendicularly to a potential barrier, was theoretically predicted some time ago [10] and experimentally observed in graphene [24]. The presence of Klein tunneling inhibits the existence of bound states within a Dirac material due to high mobility of the charge carriers. A variety of techniques have been proposed to achieve confinement of Dirac fermions, see [5, 6] for an overview. One of the techniques proposed in [6] is the introduction of a position-dependent mass function into the governing Dirac equation. It turned out that charge carriers associated with a spatially varying mass can undergo confinement, provided the mass function is chosen suitably. This was shown by an example featuring closed-form solutions in terms of Bessel functions, see [6] for details. In general, closed-form solutions of Dirac equations are rare and therefore hard to find. One of the most effective techniques to find such solutions is the Darboux transformation. The first version of this transformation [4] that applied to linear second-order equations, meanwhile has been generalized to be compatible with a variety of linear and nonlinear models [8, 13], including the Dirac equation. Particularly in the two-dimensional case governing Dirac materials, a method was devised [17] to adapt the first-order Darboux transformation, based on a result on Schrödinger models for quadratically energy-dependent potentials [11]. Besides the Dirac equation, these models can be linked to Korteweg-de Vries [22] and Klein–Gordon systems [18].

The purpose of the present work is to construct higher-order Darboux transformations for two-dimensional Dirac equations featuring a position-dependent mass in combination with a diagonal potential matrix. Our method uses a recent generalization [15, 16] of the results from [11], as well as a connection between two-dimensional Dirac equations and their Schrödinger counterparts for quadratically energy-dependent potentials. The remainder of this note is organized as follows: in Sect. 2, we summarize main results from [15]. Section 3 is devoted to the construction of the Darboux transformation, while we give an example in Sect. 4. Finally, “Appendix A” is devoted to stating a few excessively long expressions that result from our example.

2 Preliminaries

For the sake of completeness, we briefly review the principal results from [15]. Our goal is to establish a Darboux transformation between the following two Schrödinger equations with quadratically energy-dependent potentials:

$$\begin{aligned} \psi _0''(x)-[E^2+E~V_0(x)+U_0(x)]~\psi _0(x)= & {} 0 \end{aligned}$$
(1)
$$\begin{aligned} \psi _n''(x)-[E^2+E~V_n(x)+U_n(x)]~\psi _n(x)= & {} 0. \end{aligned}$$
(2)

The energy E is a real constant, the functions \(V_j\), \(U_j\), \(j=0,2\) are the potential terms that do not depend on E, and \(\psi _0\), \(\psi _n\) stand for the respective solutions, where the index n is a natural number. Assume that \(h_j\), \(j=0,\ldots ,n-1\), are auxiliary solutions to Eq. (1) at energies \(\lambda _j\), \(j=0,\ldots ,n-1\), respectively, such that the constants \(\lambda _0\), \(\lambda _1\),..., \(\lambda _{n-1},~E\) are pairwise different. Define functions \(v_j\), \(j=0,\ldots ,n-1\), by means of

$$\begin{aligned} v_j(x)= & {} \exp \left[ (E-\lambda _j)~x \right] ~ h_j(x),~~~j=0,\ldots ,n-1. \end{aligned}$$
(3)

Next, we introduce the n-th order Darboux transformation of the solution \(\phi _0\) to (1) as

$$\begin{aligned} D_{h_0,\ldots h_{n-1}}(\psi _0)(x)= & {} \frac{W_{v_0,\ldots ,v_{n-1},\psi _0}(x)}{\sqrt{{\hat{W}}_{n-1}(x)~W_{v_0,\ldots ,v_{n-1}}(x)}}, \end{aligned}$$
(4)

where \(W_{v_0,\ldots ,v_{n-1}}\) and \(W_{v_0,\ldots ,v_{n-1},\psi _0}\) denote the Wronskians of \(v_0,\ldots ,~v_{n-1}\) and of \(v_0,\ldots ,~v_{n-1},~\psi _0\), respectively. Furthermore, the quantities \({\hat{W}}_{j}\), \(j=0,\ldots ,n-1\), are defined recursively by the rules

$$\begin{aligned} {\hat{W}}_0(x)= & {} 2~v_0'(x)-v_0(x)~[V_0(x)+2E] \end{aligned}$$
(5)
$$\begin{aligned} {\hat{W}}_j(x)= & {} 2~\frac{{\hat{W}}_{j-1}'(x)~W_{v_0,\ldots ,v_j}(x)}{W_{j-1}(x)}-2~\frac{{\hat{W}}_{j-1}(x)~W_{v_0,\ldots ,v_j}'(x)}{W_{j-1}(x)} \nonumber \\&+ \frac{{\hat{W}}_{j-1}(x)~W_{v_0,\ldots ,v_j}(x)}{W_{j-1}(x)}~[V_0(x)+2E],~~~j=1,2,\ldots ,n-1. \end{aligned}$$
(6)

Under these conditions, the function \(\psi _n=D_{h_0,\ldots ,h_{n-1}}(\psi _0)\) solves Eq. (2), provided the potential terms comply with the following constraints

$$\begin{aligned} V_n(x)= & {} V_0(x) + \frac{\mathrm{d}}{\mathrm{d}x}~\log \left[ \frac{{\hat{W}}_{n-1}(x)}{W_{v_0,\ldots ,v_{n-1}}(x)} \right] \end{aligned}$$
(7)
$$\begin{aligned} U_n(x)= & {} U_0(x) -\frac{n}{2}~V_0'(x)+\frac{V_0(x)}{2} \left\{ \frac{\mathrm{d}}{\mathrm{d}x}~\log \left[ \frac{{\hat{W}}_{n-1}(x)}{W_{v_0,\ldots ,v_{n-1}}(x)} \right] \right\} + \frac{3~({\hat{W}}'_{n-1}(x))^2}{4~{\hat{W}}_{n-1}(x)^2} \nonumber \\&+\frac{3~(W_{v_0,\ldots ,v_{n-1}}'(x))^2}{4~W_{v_0,\ldots ,v_{n-1}}(x)^2}- \frac{{\hat{W}}_{n-1}'(x)~W_{v_0,\ldots ,v_{n-1}}'(x)}{2~{\hat{W}}_{n-1}(x)~W_{v_0,\ldots ,v_{n-1}}(x)} -\frac{{\hat{W}}_{n-1}''(x)}{2~{\hat{W}}_{n-1}(x)}\nonumber \\&-\frac{W_{v_0,\ldots ,v_{n-1}}''(x)}{2~W_{v_0,\ldots ,v_{n-1}}(x)}. \end{aligned}$$
(8)

The proof of these results and a separate consideration of the second-order case can be found in [15, 16], respectively.

3 Construction of the Darboux transformation

As mentioned above, we want to establish a Darboux transformation between two-dimensional Dirac equations with a position-dependent mass. To this end, we start out from our initial equation that we write in the form

$$\begin{aligned} -i~\sigma _1~\frac{\partial }{\partial x}~\varPsi (x,y)-i~\sigma _2~\frac{\partial }{\partial y}~\varPsi (x,y)+\left[ m(x)~\sigma _3+V(x)~ \right] \varPsi (x,y)= & {} 0, \end{aligned}$$
(9)

where \(\sigma _j\), \(j=1,2,3\), are the Pauli matrices, m and V represent the position-dependent mass function and the potential, respectively, and \(\varPsi \) denotes the two-component solution. The principal idea of our construction is to convert our Dirac equation (9) into Schrödinger form (1), apply the Darboux transformation described in Sect. 2 , and afterward reinstate Dirac form of the resulting transformed equation. Hence, the first step consists in decoupling (9) and afterward convert it to a second-order equation that matches the form (1).

3.1 Decoupling the Dirac equation

In order to decouple the Dirac equation (9), we make use of the fact that both potential and position-dependent mass do not depend on the variable y. We set

$$\begin{aligned} \varPsi (x,y)= & {} \exp (i~k_y~y)~\left[ \varPsi _1(x), \varPsi _2(x)\right] . \end{aligned}$$
(10)

Here, the constant \(k_y\) stands for free motion in y-direction. We now relate the component functions \(\varPsi _1\) and \(\varPsi _2\) to each other as follows

$$\begin{aligned} \varPsi _2(x)= & {} i~\frac{k_y~\varPsi _1(x)-\varPsi _1'(x)}{m(x)-V(x)}. \end{aligned}$$
(11)

Upon substituting this setting along with (10), the second component of the Dirac equation (9) is satisfied, while the first component takes the form

$$\begin{aligned}&\varPsi _1''(x)+\frac{V'(x)-m'(x)}{m(x)-V(x)}~\varPsi _1'(x)+ \left[ m(x)-V(x)\right] ^{-1} \Bigg \{ -m(x)^3+k_y^2~V(x) \\&\quad + m(x)^2~V(x)-V(x)^3+m(x)\left[ V(x)^2-k_y^2 \right] +k_y \left[ m'(x)-V'(x) \right] \Bigg \} ~\varPsi _1(x) =0. \end{aligned}$$

In the next step, we gauge away the first-derivative term by defining

$$\begin{aligned} \varPsi _1(x)= & {} \sqrt{m(x)-V(x)}~\psi _0(x), \end{aligned}$$
(12)

where the function \(\psi _0\) is a solution of

$$\begin{aligned}&\psi _0''(x) - \Bigg \{ k_y^2+\frac{V'(x)-m'(x)}{m(x)-V(x)}~k_y+\frac{3}{4} \left[ \frac{m'(x)-V'(x)}{m(x)-V(x)}\right] ^2+m(x)^2-V(x)^2 \nonumber \\&\quad +\frac{V''(x)-m''(x)}{m(x)-V(x)} \Bigg \}~ \psi _0(x) = 0. \end{aligned}$$
(13)

We observe that this linear second-order equation matches the form (1), provided we make the following definitions

$$\begin{aligned} E= & {} k_y \end{aligned}$$
(14)
$$\begin{aligned} V_0(x)= & {} \frac{V'(x)-m'(x)}{m(x)-V(x)} \end{aligned}$$
(15)
$$\begin{aligned} u_0(x)= & {} \frac{3}{4} \left[ \frac{m'(x)-V'(x)}{m(x)-V(x)}\right] ^2+m(x)^2-V(x)^2 +\frac{V''(x)-m''(x)}{m(x)-V(x)}. \end{aligned}$$
(16)

Due to Eqs. (1) and (13) matching, we are now in position to perform a Darboux transformation.

3.2 The Darboux transformation

We will now apply our Darboux transformation (4) to the solution \(\psi _0\) of Eq. (13). While the transformed solution \(\psi _n\) is given by (4), the transformed potential terms \(V_n\), \(U_n\) can be found by inserting the above settings (15), (16) into (7) and (8), respectively. This yields

$$\begin{aligned} V_n(x)= & {} \frac{V'(x)-m'(x)}{m(x)-V(x)} + \frac{\mathrm{d}}{\mathrm{d}x}~\log \left[ \frac{{\hat{W}}_{n-1}(x)}{W_{v_0,\ldots ,v_{n-1}}(x)} \right] \end{aligned}$$
(17)
$$\begin{aligned} U_n(x)= & {} \frac{3}{4} \left[ \frac{m'(x)-V'(x)}{m(x)-V(x)}\right] ^2+m(x)^2-V(x)^2 +\frac{V''(x)-m''(x)}{m(x)-V(x)} \nonumber \\&-\frac{n}{2}~\frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{V'(x)-m'(x)}{m(x)-V(x)} \right] +\frac{V'(x)-m'(x)}{2~[m(x)-V(x)]} ~\frac{\mathrm{d}}{\mathrm{d}x}~\log \left[ \frac{{\hat{W}}_{n-1}(x)}{W_{v_0,\ldots ,v_{n-1}}(x)} \right] \nonumber \\&+\frac{3}{4} \left[ \frac{{\hat{W}}'_{n-1}(x)}{{\hat{W}}_{n-1}(x)}\right] ^2+ \frac{3}{4} \left[ \frac{W_{v_0,\ldots ,v_{n-1}}'(x)}{W_{v_0,\ldots ,v_{n-1}}(x)} \right] ^2\nonumber \\&- \frac{{\hat{W}}_{n-1}'(x)~W_{v_0,\ldots ,v_{n-1}}'(x)}{2~{\hat{W}}_{n-1}(x)~W_{v_0,\ldots ,v_{n-1}}(x)}-\frac{{\hat{W}}_{n-1}''(x)}{2~{\hat{W}}_{n-1}(x)} \nonumber \\&-\frac{W_{v_0,\ldots ,v_{n-1}}''(x)}{2~W_{v_0,\ldots ,v_{n-1}}(x)}. \end{aligned}$$
(18)

Thus, the function (4) with \(\psi _0\) from (13) is a solution of the transformed equation (2) for the settings (14)–(16). Now that we have generated the transformed linear second-order equation (2), the remaining task is to cast the latter equation in Dirac form.

3.3 Reinstating Dirac form: matching conditions

In order to revert the decoupling procedure for the Dirac equation, we need to match the transformed equation (2) for (17) and (18) with the general shape of (13). More precisely, we require the transformed equation (2) to read

$$\begin{aligned}&\psi _n''(x) - \Bigg \{ k_y^2+\frac{U'(x)-M'(x)}{M(x)-U(x)}~k_y+\frac{3}{4} \left[ \frac{M'(x)-U'(x)}{M(x)-U(x)}\right] ^2+M(x)^2-U(x)^2 \nonumber \\&\quad +\frac{U''(x)-M''(x)}{M(x)-U(x)} \Bigg \}~ \psi _n(x) = 0, \end{aligned}$$
(19)

introducing a transformed Dirac potential U and a transformed position-dependent mass function M. We will now proceed by matching the coefficients pertaining to the \(k_y\)-powers. Since the term \(k_y^2\) is already matching, we continue with the coefficient of \(k_y\) in (19), using its explicit form (17). This results in the equation

$$\begin{aligned} \frac{U'(x)-M'(x)}{M(x)-U(x)}= & {} \frac{V'(x)-m'(x)}{m(x)-V(x)} + \frac{\mathrm{d}}{\mathrm{d}x}~\log \left[ \frac{{\hat{W}}_{n-1}(x)}{W_{v_0,\ldots ,v_{n-1}}(x)} \right] . \end{aligned}$$

We can solve this constraint for the transformed mass function M by means of logarithmic integration. The result can be simplified as to remove all integrations in the following way:

$$\begin{aligned} M(x)= & {} U(x)-\exp \left\{ \int \limits ^x -\frac{V'(t)-m'(t)}{m(t)-V(t)} - \frac{\mathrm{d}}{\mathrm{d}t}~\log \left[ \frac{{\hat{W}}_{n-1}(t)}{W_{v_0,\ldots ,v_{n-1}}(t)} \right] \mathrm{d}t \right\} \nonumber \\= & {} U(x)-\exp \left[ \int \limits ^ x-\frac{V'(t)-m'(t)}{m(t)-V(t)}~\mathrm{d}t \right] \exp \left\{ -\int \limits ^x \frac{\mathrm{d}}{\mathrm{d}t}~\log \left[ \frac{{\hat{W}}_{n-1}(t)}{W_{v_0,\ldots ,v_{n-1}}(t)}\right] ~\mathrm{d}t \right\} \nonumber \\= & {} U(x) - \big [V(x)-m(x) \big ]~\frac{W_{v_0,\ldots ,v_{n-1}}(x)}{{\hat{W}}_{n-1}(x)}. \end{aligned}$$
(20)

It remains to match the terms that do not depend on \(k_y\). This gives a condition on the transformed Dirac potential U in the form

$$\begin{aligned}&\frac{3}{4} \left[ \frac{M'(x)-U'(x)}{M(x)-U(x)}\right] ^2+M(x)^2-U(x)^2 +\frac{U''(x)-M''(x)}{M(x)-U(x)} \nonumber \\&\quad = \frac{3}{4} \left[ \frac{m'(x)-V'(x)}{m(x)-V(x)}\right] ^2+m(x)^2-V(x)^2 +\frac{V''(x)-m''(x)}{m(x)-V(x)}+ U_n(x)-U_0(x), \nonumber \\ \end{aligned}$$
(21)

where the function \(U_n\) and the transformed mass M are displayed in (18) and (20), respectively. We omit to include the explicit form of these functions, as the resulting expressions would be very large. We can solve Eq. (21) with respect to the potential U, which gives the lengthy result

$$\begin{aligned} U(x)= & {} \sqrt{V(x)-m(x)}~ \Bigg \{ \frac{W_{v_0,\ldots ,v_{n-1}}(x)}{2~{\hat{W}}_{n-1}(x)} +\frac{{\hat{W}}_{n-1}(x)}{2~W_{v_0,\ldots ,v_{n-1}}(x)} ~\Big [V(x)+m(x) \Big ] \Bigg \} \nonumber \\&+ \sqrt{\frac{1}{V(x)-m(x)}}~\Bigg \{-\frac{[U_2(x)-U_0(x)]~{\hat{W}}_{n-1}}{2~W_{v_0,\ldots ,v_{n-1}}(x)} +\frac{3~{\hat{W}}_{n-1}(x)~[W_{v_0,\ldots ,v_{n-1}}'(x)]^2}{8~W_{v_0,\ldots ,v_{n-1}}(x)^3} \nonumber \\&-\frac{W_{v_0,\ldots ,v_{n-1}}'(x)~{\hat{W}}_{n-1}'(x)}{4~W_{v_0,\ldots ,v_{n-1}}(x)^2}- \frac{[{\hat{W}}_{n-1}'(x)]^2}{8~W_{v_0,\ldots ,v_{n-1}}(x)~{\hat{W}}_{n-1}(x)} \nonumber \\&- \frac{{\hat{W}}_{n-1}(x)~W_{v_0,\ldots ,v_{n-1}}''(x)}{4~W_{v_0,\ldots ,v_{n-1}}(x)^2}+ \frac{{\hat{W}}_{n-1}''(x)}{4~W_{v_0,\ldots ,v_{n-1}}(x)} \Bigg \} \nonumber \\&+ \Big [V(x)-m(x) \Big ]^{-\frac{3}{2}}~ \Bigg \{\Big [V'(x)-m'(x)\Big ]~\frac{{\hat{W}}_{n-1}(x)~W_{v_0,\ldots ,v_{n-1}}'(x)}{8~W_{v_0,\ldots ,v_{n-1}}(x)^2} \nonumber \\&- \Big [V'(x)-m'(x)\Big ]~\frac{{\hat{W}}_{n-1}'(x)}{8~W_{v_0,\ldots ,v_{n-1}}(x)} \nonumber \\&+\Big [V''(x)-m''(x) \Big ]~\frac{{\hat{W}}_{n-1}(x)}{8~W_{v_0,\ldots ,v_{n-1}}(x)} \Bigg \} \nonumber \\&+ \Big [V(x)-m(x) \Big ]^{-\frac{5}{2}}~\Bigg \{ \frac{7~{\hat{W}}_{n-1}(x)}{32~W_{v_0,\ldots ,v_{n-1}}(x)} ~\Bigg [ -[m'(x)]^2+2~m'(x)~V'(x) \nonumber \\&-{\hat{W}}_{n-1}(x)~[V'(x)]^2 \Bigg ] \Bigg \}. \end{aligned}$$
(22)

In summary, if we choose the transformed mass M and the transformed Dirac potential U as in (20) and (22), then equation (19) matches the form (2), where the potential terms \(V_n\) and \(U_n\) are given by (17) and (18), respectively. We are now ready to convert our transformed linear second-order equation (2) for (17) and (18) to Dirac form.

3.4 Reinstating Dirac form: employing initial quantities

We obtain our desired Dirac form by reverting the decoupling process that was performed in Sect. 3.1 . To this end, let us first state the transformed Dirac equation. It reads

$$\begin{aligned} -i~\sigma _1~\frac{\partial }{\partial x}~\varPhi (x,y)-i~\sigma _2~\frac{\partial }{\partial y}~\varPhi (x,y)+\left[ M(x)~\sigma _3+U(x)~ \right] \varPhi (x,y)= & {} 0, \end{aligned}$$
(23)

where the mass M and the potential U are given in (20) and (22), respectively. The two-component solution of (23) can be constructed in a similar way as its counterpart \(\varPsi \). We just need to rewrite our results (10), (11), (12), where we replace initial quantities by their transformed partners. We find

$$\begin{aligned} \varPhi (x,y)= & {} \exp (i~k_y~y)~\left[ \varPhi _1(x), \varPhi _2(x)\right] . \end{aligned}$$
(24)

The component functions \(\varPhi _1\) and \(\varPhi _2\) are interrelated by means of

$$\begin{aligned} \varPhi _2(x)= & {} i~\frac{k_y~\varPhi _1(x)-\varPhi _1'(x)}{M(x)-U(x)}, \end{aligned}$$
(25)

and the first component \(\varPhi _1\) of (24) is given by

$$\begin{aligned} \varPhi _1(x)= & {} \sqrt{M(x)-U(x)}~\psi _n(x) \nonumber \\= & {} \sqrt{M(x)-U(x)}~\frac{W_{v_0,\ldots ,v_{n-1},\psi _0}(x)}{\sqrt{{\hat{W}}_{n-1}(x)~W_{v_0,\ldots ,v_{n-1}}(x)}}. \end{aligned}$$
(26)

At this point, there is one more task remaining: we observe that the transformed quantities (20), (22), (25), (26) are expressed in terms of solutions \(\psi _0\) and \(h_j\), \(j=0,\ldots ,n-1\), to the Schrödinger-type equation (1), see Sect. 2 for their definition. This is not desirable, since we are aiming at constructing a Darboux transformation between Dirac equations without the use of any intermediate result or equation. For this reason, we must now establish a connection between solutions of (13) and corresponding solutions of the initial Dirac equation (9). The first of these connections is given by (12). Inversion gives

$$\begin{aligned} \psi _0(x)= & {} \sqrt{\frac{1}{m(x)-V(x)}}~\varPsi _1(x). \end{aligned}$$
(27)

Next, we need to consider the functions \(v_j\), \(j=0,\ldots ,n-1\), that are defined in (3). We will now relate these functions to auxiliary solutions \(u_j\), \(j=0,\ldots ,n-1\), of the Dirac equation (9) that pertain to \(k_y\)-values \(\lambda _j\), \(j=0,\ldots ,n-1\), respectively. Upon taking into account (3) and (12), we obtain

$$\begin{aligned} v_j= & {} \sqrt{\frac{1}{m(x)-V(x)}}~\exp \left[ (E-\lambda _j)~x \right] u_j(x),~~~j=0,\ldots ,n-1. \end{aligned}$$
(28)

For the sake of brevity, let us now introduce the abbreviation

$$\begin{aligned} w_{v_0,\ldots ,v_j}(x)= & {} \exp \left[ (E-\lambda _j)~x \right] u_j(x),~~~j=0,\ldots ,n-1. \end{aligned}$$
(29)

We can now combine (27)–(29) in order to rewrite the Wronskians that appear in the transformed quantities (20), (22), (25), (26). Starting out with the Wronskian \(W_{v_0,\ldots ,v_{n-1}}\) of the functions \(v_j\), \(j=0,\ldots ,n-1\), we obtain

$$\begin{aligned} W_{v_0,\ldots ,v_{n-1}}(x)= & {} \left[ m(x)-V(x)\right] ^{-\frac{n}{2}} W_{w_0,\ldots ,w_{n-1}}(x). \end{aligned}$$
(30)

Here, we made use of the fact that a common factor in each entry of the matrix associated with the Wronskian can be pulled out like a constant [19]. A similar argumentation leads to

$$\begin{aligned} W_{v_0,\ldots ,v_{n-1},\psi _0}(x)= & {} \left[ m(x)-V(x)\right] ^{-\frac{n+1}{2}} W_{w_0,\ldots ,w_{n-1},\varPsi _1}(x). \end{aligned}$$
(31)

In the final step, we proceed to rewrite the quantity \({\hat{W}}_j\), as defined in (6). To this end, we note that the latter quantity can be stated in explicit form as an actual Wronskian [15]. To summarize, we have

$$\begin{aligned} {\hat{W}}_{n-1}(x)= & {} \frac{(-2)^{n}}{F(x)}~W_{v_0,\ldots ,v_{n-1},F}(x), \end{aligned}$$
(32)

where the function F is given by

$$\begin{aligned} F(x)= & {} \exp \left( \frac{1}{2}~\int \limits ^x V_0(t)+2~k_y~\mathrm{d}t \right) . \end{aligned}$$
(33)

Upon implementation of (28) and (29), we can cast (32) in the form

$$\begin{aligned} {\hat{W}}_{n-1}(x)= & {} \frac{(-2)^{n}}{F(x)} \left[ m(x)-V(x)\right] ^{-\frac{n+1}{2}} W_{w_0,\ldots ,w_{n-1}, \sqrt{m-V} F}(x). \end{aligned}$$
(34)

We are now ready to state the Darboux transformation that connects the initial Dirac equation (9) with its transformed counterpart (23). Substitution of (30), (31), (34) into (26) gives the result

$$\begin{aligned} \varPhi _1(x)= & {} \left( -\frac{1}{2} \right) ^\frac{n}{2} \frac{\sqrt{F(x)~[M(x)-U(x)]}}{\left[ m(x)-V(x) \right] ^\frac{1}{4}}~ \frac{W_{w_0,\ldots ,w_{n-1},\varPsi _1}(x)}{W_{w_0,\ldots ,w_{n-1}, \sqrt{m-V} F}(x) ~W_{w_0,\ldots ,w_{n-1}}(x) }. \nonumber \\ \end{aligned}$$
(35)

This is the first component of the solution to the transformed Dirac equation (23). The second component can be obtained by plugging (35) into (25). We do not show the explicit form of the resulting expression due to its length. Next, let us state the mass (20) that enters in the transformed equation (23). Since this mass is expressed in terms of a Wronskian ratio, it is convenient to determine the latter ratio first. Combination of (30) and (34) gives

$$\begin{aligned} \frac{W_{v_0,\ldots ,v_{n-1}}(x)}{{\hat{W}}_{n-1}(x)}= & {} \left( -\frac{1}{2} \right) ^n F(x)~ \sqrt{m(x)-V(x)}~\frac{W_{w_0,\ldots ,w_{n-1}}(x)}{W_{w_0,\ldots ,w_{n-1}, \sqrt{m-V} F}(x)}. \end{aligned}$$
(36)

Upon plugging this expression into the mass function (20), we arrive at

$$\begin{aligned} M(x)= & {} U(x) + \left( -\frac{1}{2} \right) ^n F(x)~ \big [m(x)-V(x) \big ]^\frac{3}{2}~ \frac{W_{w_0,\ldots ,w_{n-1}}(x)}{W_{w_0,\ldots ,w_{n-1}, \sqrt{m-V} F}(x)}. \end{aligned}$$
(37)

Now that we have rewritten the mass function in terms of quantities related to the initial Dirac equation, we can perform the same process with the transformed Dirac potential U by substituting the Wronskian ratio (36) into (22). Since the resulting form of the latter potential will become very long and involved, we omit to show it here.

4 Application: hyperbolic potential and mass

We will now demonstrate how the Darboux transformation between Dirac equations with position-dependent mass is applied. To this end, let us start by considering a particular case of our initial equation (9). We choose the potential and the mass functions as follows

$$\begin{aligned} V(x)= & {} 20~\text{ sech }^2(x)+\frac{1}{2} \qquad \qquad \qquad m(x) = 20~\text{ sech }^2(x). \end{aligned}$$
(38)

Let us mention here that the primary reason for choosing these settings is to keep subsequent calculations as simple and transparent as possible. In general, even Darboux transformations of first order result in expressions that are not manageable due to their length. In the following, we will perform a second-order transformation, where excessively long expressions that result from it will be stated in “Appendix A.” Now, upon implementing the settings (38), the Dirac equation (9) is taken at zero energy and admits solutions of bound-state type, characterized by particular values of the parameter \(k_y\). We define these values as

$$\begin{aligned} k_y= & {} \frac{1}{2}~\sqrt{1+4~n_{k_y}^2}, \end{aligned}$$
(39)

where \(n_{k_y}\) stands for a nonnegative integer. A particular solution to the Dirac equation for the settings (38) and (39) is given by (10) with components

$$\begin{aligned} \varPsi _1(x)= & {} \frac{i}{\sqrt{2}}~P_4^{n_{k_y}}\left[ \tanh (x) \right] \end{aligned}$$
(40)
$$\begin{aligned} \varPsi _2(x)= & {} -\sqrt{2}~(n_{k_y}-5)~P_5^{n_{k_y}}\left[ \tanh (x)\right] + \sqrt{\frac{1}{2}}~\left[ \sqrt{1+4~n_{k_y}}\right. \nonumber \\&\left. -10~\tanh (x)\right] P_4^{n_{k_y}}\left[ \tanh (x) \right] . \end{aligned}$$
(41)

Here, P stands for the associated Legendre functions of the first kind [1]. Note that the second component can be obtained from its counterpart by means of (11). Figure 1 shows normalized probability densities \(|\varPsi (x,0)|^2\) for several values of \(n_{k_y}\).

Fig. 1
figure 1

Graph of the initial probability density \(|\varPsi (x,0)|^2\) for \(n_{k_y}=4\) (black curve), \(n_{k_y}=3\) (gray curve), and \(n_{k_y}=2\) (dashed curve)

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We will now apply a second-order Darboux transformation to our Dirac equation (9) for the settings (38). To this end, we need to identify two auxiliary solutions of the latter equation. We take the first components of these solutions from (40) as

$$\begin{aligned} u_0(x)= & {} \frac{i}{\sqrt{2}}~P_4^{0}\left[ \tanh (x) \right] \qquad \qquad \qquad u_1(x) = \frac{i}{\sqrt{2}}~P_4^{1}\left[ \tanh (x) \right] , \end{aligned}$$

while the second components of the auxiliary solutions are irrelevant here. In accordance with (29), we define

$$\begin{aligned} w_0(x)= & {} \exp \left[ \left( E-\frac{1}{2} \right) x \right] u_0(x) \qquad \qquad w_1(x) = \exp \left[ \left( E-\frac{\sqrt{5}}{2} \right) x \right] u_1(x), \end{aligned}$$
(42)

note that the values for \(\lambda _0\) and \(\lambda _1\) in (29) are obtained from (39) for \(n_{k_y}=0\) and \(n_{k_y}=1\), respectively. We are now ready to apply our Darboux transformation that is determined by the transformed solution \(\varPhi _1\), the mass function M, and the associated potential U, as given in (35), (20), and (22), respectively. Let us first consider the difference between the transformed mass M and the potential U. Substitution of (42) into (20) gives

$$\begin{aligned} M(x) -U(x)= & {} -\left\{ 2~\cosh ^2(x) \Bigg [2~\cosh (x)+36~\cosh (3 x)-4~\cosh (5 x)+2~\cosh (7 x) \right. \nonumber \\&+ (\sqrt{5}-1)~\Big [-329~\sinh (x)+ 138~\sinh (3x)-22~\sinh (5x)\nonumber \\&\left. +\sinh (7x) \Big ]\Bigg ] \right\} \nonumber \\&\times \Bigg \{ 6~(2873+14~\sqrt{5})~\cosh (x)+10~(-1159+14~\sqrt{5})~\cosh (3x) \nonumber \\&+ 30~(191+2~\sqrt{5})\cosh (5x)+(-2+4~\sqrt{5})~ \cosh (9x)-2~\Bigg [ -4020 \nonumber \\&+ 3860~\sqrt{5}+5~(567-575~\sqrt{5})~\cosh (2x)-14~(-9+\sqrt{5})~\cosh (4x) \nonumber \\&+(-9+\sqrt{5})~\cosh (6x) \Bigg ] \Bigg [ -6~\sinh (x)+\sinh (3x) \Bigg ] \Bigg \}^{-1}. \end{aligned}$$
(43)
Fig. 2
figure 2

Graphs of the function \(M-U\), as given in (43)

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Figure 2 shows two graphs of this expression. We observe that away from zero, both transformed mass and potential differ by a constant amount, where close to zero they are almost equal. Let us now state the results of our Darboux transformation. Starting out with the transformed mass (20), it turns out that after substitution of (22), (33), (38), and (42), we obtain a very long and involved expression. After having this expression simplified by Mathematica (Wolfram Research Inc.), the result is shown in “Appendix A.” Similarly, we use the latter result for finding the explicit form of the transformed Dirac potential (22) by means of (43), see “Appendix A.” Figure 3 shows graphs of the initial mass and potential, along with their transformed counterparts. Note that the graphs are scaled in vertical direction, as the transformed quantities take much larger values as compared to their initial partners. It remains to determine the solution (35) for the present case. We follow the same route as for the transformed mass and potential functions by stating the explicit form of the latter solution in “Appendix A.” The second component of the solution can be obtained through (25); we omit to present its explicit form. Figure 4 shows normalized probability densities \(|\varPhi (x,0)|^2\) for several values of the parameter \(n_{k_y}\).

Fig. 3
figure 3

Left plot: vertically scaled graph of initial mass (38) (dashed curve) and transformed mass (20) (solid curve). Right plot: vertically scaled graph of initial potential (38) (dashed curve) and transformed potential (22) (solid curve)

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Fig. 4
figure 4

Graphs of the probability density \(|\varPhi (x,0)|^2\) for \(n_{k_y}=4\) (black curve), \(n_{k_y}=3\) (gray curve), and \(n_{k_y}=2\) (dashed curve)

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Fig. 5
figure 5

The transformed mass function (20), calculated by Mathematica (Wolfram Research Inc.)

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Inspection of the figure indicates that the probability densities pertain to solutions of bound-state type.

5 Concluding remarks

We have constructed arbitrary-order Darboux transformation for Dirac equations with position-dependent mass. Even though applications typically result in large expressions for the transformed quantities, our method can be easily implemented by means of symbolic calculators. In particular, no symbolic integration is required except for the typically simple integral in (33). Our method has a technical shortcoming that concerns the transformed position-dependent mass function (37). In order to be physically meaningful, this function must be nonnegative. However, due to its complicated form and involved dependence on parameters, we do not have a condition for nonnegativity. A similarly important task is to obtain regularity constraints for the transformed potential terms (7) and (8), as they exist in the conventional Darboux transformation (SUSY formalism). Finally, we point out that our method can be extended in a straightforward manner to work with non-diagonal matrix potentials.