Consider Dirac equation in 2D with Hamiltonian given by (arb. units) \begin{equation} H=-i \begin{pmatrix} 0&\partial_x-i\partial_y\\ \partial_x+i\partial_y & 0\\ \end{pmatrix}. \end{equation} The equation to solve, is simply \begin{equation} H\psi=E\psi, \ \psi=\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \end{equation} I expand the wave function in some basis of spatial-localized functions \begin{equation} \psi(x,y)=\sum_n^N \begin{pmatrix} c^1_n \\ c^2_n \end{pmatrix} \phi_n(x_n,y_n)+\sum_m^M \begin{pmatrix} b^1_m \\ b^2_m \end{pmatrix} \phi_m(x_n,y_n) \end{equation} Let the functions indexed by $m$ lie on the boundary. After plugging this form to the equation I obtain the matrix equation \begin{equation} H_D\psi_D=ES_D\psi_D, \end{equation} $H_D $ is the matrix $2(N+M) \times 2(N+M) $ (size of the problem). On the right hand side I also have overlap matrix due to non orthogonality of my basis functions. The last thing to do is imposing the boundary condition. The question is, how to impose the boundary condition of the form \begin{equation} \psi_2=\alpha\psi_1, \end{equation} where $\alpha$ is some complex number. I know how to impose Dirichlet boundary condition, but I have hard time with this one. I will be glad for any ideas or any refs. Is there any name for this type of boundary ?? Greetings.
1 Answer
The general boundary condition for the Dirac equation is a local linear restriction on the components of the spinor wave function at the boundary, $$\psi=M\psi,\;\;M=\begin{pmatrix} n_z&n_x-in_y\\ n_x+in_y&-n_z \end{pmatrix}$$ with ${\mathbf n}=(n_x,n_y,n_z)$ a unit vector. (Check that $M^2=\mathbb{1}$.)
If you are interested in the generalization of the Dirichlet boundary condition, then you will want to ensure that zero current flows through the boundary. This further restricts $M$ to vectors ${\mathbf n}$ that satisfy $({\mathbf n},{\mathbf n}_B)=0$, where ${\mathbf n}_B$ is a unit vector in the $x$-$y$ plane perpendicular to the boundary.
All of this was worked out in the context of graphene, see arXiv:0710.2723. In that context the different boundary conditions have names: $n_x=0=n_y$ is called the "zigzag" boundary condition and $n_z=0$ is called the "mass" boundary condition. There is also something called the "armchair" boundary condition, which you need if you put the Dirac equation on a lattice.
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$\begingroup$ Thank you very much for your response, paper is great, very explanatory. But now I am wondering how to actually use this condition. I have an equation discretized on the $N+M$ points, this gives me a problem of size $2(N+M) \times 2(N+M)$ because of the spinor. $N$ is the number of nodes inside the domain and $M$ nodes lie on the boundary. The boundary condition should lower the number of degrees of freedom, but now it's not very clear, how to put them into the matrix equation. $\endgroup$– drszdrszCommented Mar 25, 2017 at 21:01
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$\begingroup$ it may help to think about it in a one-dimensional setting first; $\psi(x)=M\psi(x)$ at $x=0$ (the boundary) tells you that $\psi(x)$ is an eigenfunction of $M$ with eigenvalue $+1$, and then $\psi(x)$ for $x>0$ follows upon integration, $\psi(x)=\exp(iEx\sigma_x)\psi(0)$ ($\sigma_x$ is a Pauli matrix). But be advised that you cannot discretize the Dirac equation without running into the fermion doubling problem, see arXiv:0810.4786 $\endgroup$ Commented Mar 25, 2017 at 21:10
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$\begingroup$ Hmm, I am not sure If I see what you mean. How did you get that relation? And how to use it in the context of Finite Element Method or Finite Difference Method ? Thank you for the warning, I see there is a problem with simple discretization. $\endgroup$– drszdrszCommented Mar 28, 2017 at 10:38
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$\begingroup$ for translational invariance in $y$ the equation $H\psi=E\psi$ is a first order differential equation in $x$, multiply both sides with the Pauli matrix $\sigma_x$ and integrate with respect to $x$; if there is a $y$-dependence you have coupled differential equations for each transverse mode, which you solve via a transfer matrix, see arXiv:0707.0886; for solution via finite differences you have to proceed differently, it is typically much more cumbersome and I would not recommend it. $\endgroup$ Commented Mar 28, 2017 at 13:58
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$\begingroup$ Ok, I see the problem with finite differences is well known and problematic enough. I will stick to the Finite Element Method. For now I am still struggling with this boundary. Thank you for all the hints, papers you suggested are very interesting sure will be helpful. $\endgroup$– drszdrszCommented Mar 29, 2017 at 20:11