Since we are on a planar system (2D system) the massless Dirac equation reads
$$\vec{\alpha}\cdot(\vec{p}-e\vec{A})\psi_E=E\psi_E$$

Here Dirac matrices are Pauli matrices ($\alpha^1=-\sigma^2$ , $\alpha^2=\sigma^1$)

To find zero-energy modes we write the wave function as $\psi_0= \begin{pmatrix}
u\\
v
\end{pmatrix}$, and choose Coulomb gauge for $\vec{A}$ assumed to be single valued and well behaved at the origin,

$$A^i=\epsilon^{ij}\partial_ja$$
$$B=-\nabla^2a$$

Then the Dirac equation reduces to pair,

(1)
$$(\partial_x + i\partial_y)u - e(\partial_x+i\partial_y)au=0$$
$$(\partial_x - i\partial_y)v + e(\partial_x-i\partial_y)av=0$$

with solutions

$$u=\exp(ea)f(x+iy)$$
$$v=\exp(-ea)g(x-iy)$$

For example if we apply a constant magnetic field with 2 possible Coulomb gauges
$$a_0^I=-{1\over 4} r^2B$$
$$a_0^{II}=-{1\over 2}x^2B$$

For $a_0^I$ case zero-energy states may be written as
$$\psi_{0(n)}^I=e^{-eBr^2/4}(x+iy)^n$$

For $a_0^{II}$ case zero-energy states may be written as
$$\psi_{0(k)}^{II}=e^{-eBx^2/2}e^{k(x+iy)}$$

**My questions are:**

How can I solve the differential equations given in (1) to obtain $u$ and $v$ solutions given above ? Where do $f(x+iy)$ and $g(x-iy)$ arise from ?

The author sets $f(x+iy) = (x+iy)^n$ and $g(x-iy) = e^{k(x+iy)}$. How can I determine these functions ?

**Reference**

R. Jackiw, Phys. Rev. D 29, 2375 (1984).

This post imported from StackExchange Physics at 2014-04-13 14:46 (UCT), posted by SE-user ucoskun