Dirac electron in graphene with magnetic fields arising from first-order intertwining operators

where the intertwining operators $L_{0}^{\mp}$, which satisfy the relations $H^{\pm}L_{0}^{\mp}=L_{0}^{\mp}H^{\mp}$, are given by:

with $W_{0}(x)$ being the superpotential, which allows to express the potentials (5) in compact form as follows:

$$V^{\pm}(x)=W_{0}^{2}(x)\pm W_{0}^{\prime}(x).$$

(9)

The action of the intertwining operators $L_{0}^{\pm}$ on the eigenstates of the Hamiltonians (4) become:

$$\psi^{+}_{n}(x)=\frac{L_{0}^{-}\psi^{-}_{n+1}(x)}{\sqrt{{\cal{E}}_{n+1}^{-}}},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \psi^{-}_{n+1}(x)=\frac{L_{0}^{+}\psi^{+}_{n}(x)}{\sqrt{{\cal{E}}_{n}^{+}}},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \psi_{0}^{-}(x)\sim e^{-\int W_{0}(x)dx},$$

(10)

where the operator $L_{0}^{-}$ annihilates the ground state of $H^{-}$, $L_{0}^{-}\psi_{0}^{-}(x)=0$, which implies that $W_{0}(x)=-{\psi^{-}_{0}(x)}^{\prime}/\psi^{-}_{0}(x)$. The energy levels of $H^{\pm}$ turn out to be:

These expressions indicate that the eigenfunctions and eigenvalues of the problem can be found through the operators $L_{0}^{\pm}$, which simplifies the calculations since this involves just first-order derivatives.
Note that the magnetic field amplitude $B_{0}(x)$ is connected with the $y$ component of the vector potential, with the superpotential $W_{0}(x)$ and with the ground state of $H^{-}$ as follows:

In addition, for some special forms of $B_{0}(x)$ the two SUSY partner potentials $V^{\pm}(x)$ become shape invariant [23, 24, 27, 28, 25, 26, 30, 31, 32], which implies that $H^{\pm}$ are exactly solvable so that the eigenfunctions and eigenvalues are known explicitly. Therefore, from a shape invariance point of view the problem is essentially solved in these cases. However, it is still feasible to generalize the method (see also [29, 36]), and to iterate it as many times as we wish, which will be shown in the following sections.

so that $\tilde{V}_{0}(x)\equiv V^{-}(x)-\epsilon_{1}$, where $\epsilon_{1}\leq{\cal{E}}_{0}^{-}=0$. The second step is to build a new Hamiltonian $H_{1}$ departing from $\tilde{H}_{0}$ through the intertwining relation:

where $H_{1}$ and $L_{1}^{\pm}$ are given by:

Equation (14) implies that the Hamiltonians $\tilde{H}_{0}$ and $H_{1}$ are factorized as $\tilde{H}_{0}=L_{1}^{-}L_{1}^{+}$ and $H_{1}=L_{1}^{+}L_{1}^{-}$, which leads to the following Riccati equation and relation between potentials [37]:

Let us suppose now that $W_{1}(x,\epsilon_{1})=u_{1}^{(0)^{\prime}}/u_{1}^{(0)}$; thus Eq. (16) is transformed into:

which is the stationary Schrödinger equation associated to $\tilde{H}_{0}$ for null factorization energy. A nodeless solution to this equation constitutes the information required to build the potential $V_{1}(x,\epsilon_{1})$ [37] (see also equation (17)). Moreover, the magnetic field giving place to $V_{1}(x,\epsilon_{1})$ is calculated as follows:

with $\epsilon_{1}\leq{\cal{E}}_{0}^{-}=0$. A diagram of these energies can be seen in figure 1. The unknown eigenfunctions associated to the previous energies are given by:

where the eigenfunctions $\psi_{n}^{-}(x)$ of $H^{-}$, and consequently those of $\tilde{H}_{0}$, are assumed to be known. In addition, the ground state of $H_{1}$ fulfills the condition $L_{1}^{-}\psi_{0}^{(1)}(x)=0$.

such that $\epsilon_{2}<\epsilon_{1}$. Thus, the ground state energy of $\tilde{H}_{1}$ is $\tilde{\cal{E}}_{0}^{(1)}=-\epsilon_{2}\geq 0$ (see figure 1), and we denote $\tilde{V}_{1}(x,\epsilon_{1})\equiv V_{1}(x,\epsilon_{1})-\epsilon_{2}$. The intertwining relation reads:

with the new Hamiltonian $H_{2}$ and intertwining operators $L_{2}^{\pm}$ being given by:

The Hamiltonians $\tilde{H}_{1}$, $H_{2}$ are expressed in terms of the intertwining operators as $\tilde{H}_{1}=L_{2}^{-}L_{2}^{+}$ and $H_{2}=L_{2}^{+}L_{2}^{-}$, which leads to a pair of equations similar to (16, 17):

Once again we pass from the Riccati equation (25) to the associated Schrödinger equation through the change:

leading to:

The intertwining operator $L_{1}^{+}$ helps to find $u_{2}^{(1)}$, by connecting it with the corresponding Schrödinger solution $u_{2}^{(0)}$ for $\tilde{H}_{0}$:

where $\mathsf{W}[u_{1}^{(0)},u_{2}^{(0)}]$ denotes the Wronskian of $u_{1}^{(0)}$ and $u_{2}^{(0)}$ while $u_{2}^{(0)}$ fulfills:

By combining Eqs. (26) and (29) it is obtained:

By means of Eqs. (27) and (29) the second-order magnetic field now can be found:

The eigenvalues of $\tilde{H}_{1}$ and $H_{2}$ are related through:

$$\begin{split}&\tilde{\cal{E}}_{n}^{(1)}={\cal{E}}_{n}^{(1)}-\epsilon_{2},\\ &{\cal{E}}_{0}^{(2)}=0,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ {\cal{E}}_{n+1}^{(2)}=\tilde{\cal{E}}_{n}^{(1)}.\end{split}$$

(33)

The unknown eigenfunctions, associated with the last eigenvalues, are:

where $n=0,1,2\dots$

where $\epsilon_{k}<\epsilon_{k-1}<\dots\epsilon_{2}<\epsilon_{1}\leq 0$. The intertwining relation for the $k$th step is:

while the Hamiltonians $H_{k}$, $\tilde{H}_{k-1}$ and intertwining operator $L^{\pm}_{k}$ are expressed as:

The factorizations involved are now $\tilde{H}_{k-1}=L_{k}^{-}L_{k}^{+}$ and $H_{k}=L_{k}^{+}L_{k}^{-}$, while the equations resulting from (36) and (37) become:

If the superpotential is expressed as $W_{k}(x,\epsilon_{k})=u^{(k-1)^{\prime}}_{k}/u^{(k-1)}_{k}$, then it is obtained the following Schrödinger equation for zero factorization energy:

The intertwining operators supply the relation between $u^{(k-1)}_{k}$ and its predecessor, i.e., $u^{(k-1)}_{k}\propto L^{+}_{k-1}u^{(k-2)}_{k}$, where $u^{(k-2)}_{k}$ fulfills the Schrödinger equation:

By iterating these expressions for lower indices it turns out that $u^{(k-1)}_{k}\propto L^{+}_{k-1}L^{+}_{k-2}...L^{+}_{1}u^{(0)}_{k}$, where

which allows to know all the superpotentials as well as the new potential:

The $kth$-order magnetic field reads now:

A useful form of the previous expression is also given by:

The eigenfunctions associated to the last eigenvalues are given by:

$$\begin{split}&\psi_{0}^{(k)}(x)\sim e^{-\int W_{k}(x,\epsilon_{k})dx},\\ &\psi_{1}^{(k)}(x)=\textstyle{\frac{1}{\sqrt{{\cal{E}}_{0}^{(k-1)}-\epsilon_{k}}}}L_{k}^{+}\psi^{(k-1)}_{0}(x),\\ &\vdots\\ &\psi_{k-1}^{(k)}(x)=\textstyle{\frac{1}{\sqrt{({\cal{E}}_{k-2}^{(k-1)}-\epsilon_{k})...({\cal{E}}_{0}^{(1)}-\epsilon_{2})}}}L_{k}^{+}...L_{2}^{+}\psi^{(1)}_{0}(x),\\ &\psi_{k+n}^{(k)}(x)=\textstyle{\frac{1}{\sqrt{({\cal{E}}_{k+n-1}^{(k-1)}-\epsilon_{k})...({\cal{E}}_{n}^{-}-\epsilon_{1})}}}L_{k}^{+}...L_{1}^{+}\psi^{-}_{n}(x).\end{split}$$

(47)

whose eigenvalues and eigenfunctions are given by:

$${\cal{E}}_{n}^{-}=\omega n,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \psi^{-}_{n}(x)=N_{n}e^{-\frac{\omega}{4}(x+\frac{2k}{\omega})^{2}}H_{n}{\textstyle[\sqrt{\frac{\omega}{2}}(x+\frac{2k}{\omega})]},$$

(49)

where $n=0,1,2\dots$, the normalization factor is $N_{n}=\sqrt{\frac{1}{2^{n}n!}(\frac{\omega}{2\pi})^{\frac{1}{2}}}$ and $H_{n}$ are the Hermite polinomials. The first step of the method is to move up the energy of $H^{-}$ as follows:

with $\epsilon_{1}\leq{\cal E}_{0}^{-}=0$. From the intertwining relation (14) it is obtained equation (16). The change $W_{1}(x,\epsilon_{1})=u_{1}^{(0)^{\prime}}/u_{1}^{(0)}$ in the last equation leads to the Schrödinger equation (18), whose general solution is:

with $a=-\epsilon_{1}/2\omega$. Note that, in order to avoid singularities in the potential $V_{1}(x,\epsilon_{1})$ the parameter $\nu_{1}$ must be restricted to the interval $(-1,1)$. In particular, for the parameters $\epsilon_{1}=-\omega/5$ and $\nu_{1}=0$ the superpotential becomes:

and the new potential reads:

In the variable $\zeta=\sqrt{\omega/2}(x+2k/\omega)$ it is obtained that:

which in a more compact form reads as:

For this case the magnetic field is given by:

Some graphs of the potential $V_{1}(x,\epsilon_{1})$ and associated magnetic field $B_{1}(x,\epsilon_{1})$ are shown in figure 2. Since the two lowest energy levels are very close, the form of $V_{1}(x,\epsilon_{1})$ resembles the famous Mexican hat, which illustrates the spontaneous symmetry breaking in field theory. In our case the Mexican hat can be modified by changing $\epsilon_{1}$, and its existence means that the Dirac electron can pass from the ground to the first excited state with a greater probability than for any two other states.

The eigenenergies of the system in our example are:

$$E_{0}^{(1)}=0,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ E_{n+1}^{(1)}=\hbar v_{F}\sqrt{\omega\left(n+\frac{1}{5}\right)},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ n=0,1,\dots$$

(57)

The corresponding eigenfunctions, taking into account (2), are given by:

With these expressions it is customary to calculate the probability density and probability current. The probability density for the excited states of graphene is $\rho_{n+1}(x)=|\psi_{n}^{(0)}(x)|^{2}+|\psi_{n+1}^{(1)}(x)|^{2}$ while for the ground state is $\rho_{0}(x)=|\psi_{0}^{(1)}(x)|^{2}$. The probability currents are $j_{n+1}(x)=ev_{F}\psi_{n}^{(0)}(x)\psi_{n+1}^{(1)}(x)$ and $j_{0}(x)=0$ respectively [23]. Some graphs for the probability density and probability current are shown in figure 3.

It is seen that the greater probability appears around the points where a minimum of the potential exists. Moreover, since the potential is symmetric with respect to the point $-2k/\omega$, then the probability density has also the same symmetry.