Solving a time-dependent Dirac Equation in the Plane How would one go about solving the following time dependent Dirac equation: $$\frac{\partial u(\mathbf{x},t)}{\partial t}=\mathcal{D}u(\mathbf{x},t)$$ where $\mathcal{D}=-i(\sigma_x\partial_x+\sigma_y\partial_y)$ is the standard Dirac operator associated with a spin-1/2 particle in the plane and $\sigma_x$ and $\sigma_y$ are Pauli matrices: $$\sigma_x=\left(\begin{matrix}0&1\\1&0\end{matrix}\right),\quad\text{and}\quad\sigma_y=\left(\begin{matrix}0&-i\\i&0\end{matrix}\right).$$
How can I solve this given the initial condition of $u(\mathbf{x},0)=\delta(\mathbf{x})$?
 A: I do not understand what sort of solution you need. The solution in compact form  is 
$$u (t,\vec{x})= e^{t {\cal D}}\delta (\vec{x})\:$$
Passing to Fourier transform, this espression can be expanded as a linear combination of some  $dk $ integrals of Bessel functions $J_n(|k| |x|)$, with $n=0,1$ I think multiplied by I and the two Pauli matrices...
Let us assume that the solution is a distribution parametrized in $t$ so that it admits Fourier transform $\hat{u}(t,\vec{k})$ and
$$u(t,\vec{x})= \frac{1}{2\pi}\int _{\mathbb R}e^{i\vec{k}\cdot \vec{x}} \hat{u}(t,\vec{k})   d^2k\:.$$ Inserting it in the differential equation we find
$$\int_{\mathbb R} e^{i\vec{k} \cdot \vec{x}} \left(\partial_t \hat{u}(t,k) +   i \vec{k}\cdot \vec{\sigma} \hat{u}(t,k) \right) d^2k =0\:.$$
Hence $$\partial_t \hat{u}(t,k) =- i \vec{k}\cdot \vec{\sigma} \hat{u}(t,k)$$ which has solution
$$ \hat{u}(t,\vec{k}) = e^{- i \vec{k}\cdot \vec{\sigma}} \hat{u}(0,\vec{k})$$
that is, if $k:= |\vec{k}|$, 
$$ \hat{u}(t,\vec{k}) = \left(\cos k I -i \sin k \:\frac{\vec{k}}{k} \cdot \vec{\sigma} \right)\:\: \hat{u}(0,\vec{k})$$
Since $u(0,\vec{x})= \delta(\vec{x}) e$
(where $e= (1,1)^t$ because $\cal D$ acts on two components spinors and I interpret your $u(0,\vec{x})= \delta(\vec{x})$ as if the delta stays in each component of the spinor)
 we have $\hat{u}(0,\vec{k})= \frac{1}{2\pi} e$. Summing up
$$ \hat{u}(t,\vec{k}) = \frac{1}{2\pi}\left(\cos k I -i \sin k \:\frac{\vec{k}}{k} \cdot \vec{\sigma} \right)\:\: e$$
Finally
$$u(t,\vec{x}) = \frac{1}{(2\pi)^4}\int_{\mathbb R} e^{ i \vec{k}\cdot \vec{x}}\cos k \: d^2k\:e  -i \frac{1}{(2\pi)^4}\int_{\mathbb R} e^{ i \vec{k}\cdot \vec{x}}\sin k \:\frac{\vec{k}}{k} \cdot \vec{\sigma} d^2k e$$
Noticing that $d^2k = dk d\theta$ and $\vec{k}\cdot \vec{x} = kx \cos \theta$, the $\theta$ integration can be computed obtaining Bessel functions  like $J_0(kx)$, but  the final result remains quite messy...
