# Explicit solutions to 2-d Dirac Equation

The 2-d Dirac equation without any constants is represented usually as $$i*dt (\phi) = D (\phi)$$ where $D = m\sigma_2-i\sigma_1dx-i\sigma_3dy$. Where can I find explicit closed form solutions to this equation? If there are none, are there perturbation-like non-exact solutions?

Thanks alot.

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Just try a plane wave solution (as in 4d, where the problem is more difficult but still solved in all textbooks). –  Vibert Sep 10 '13 at 18:42
Is there specific solution? I'm wondering how to solve out the 2-vectors though. –  Xingyou Sep 10 '13 at 18:50
@Xingyou I think this link has what you want: sciencedirect.com/science/article/pii/S0375960104001136 –  dj_mummy Sep 10 '13 at 19:06
Sorry, but I don't think that gives a (2+1) solution, but rather only talks about (1+1) in certain curved manifolds. –  Xingyou Sep 11 '13 at 2:50

## 1 Answer

For the Dirac equation $D \psi=0$, we may use the following matrix $D$:

$$D = \begin{pmatrix} m+\partial_y&\partial_x - \partial_t\\ \partial_x + \partial_t& m-\partial_y\end{pmatrix} \tag{1}$$

A general solution of the Dirac equation is $\psi = \tilde D \phi$, where $\phi(x,y,t)$ is any 2-row vector, such as $(\partial_t^2-\partial_x^2-\partial_y^2+m^2) \phi = 0$, and $\tilde D$ is the following matrix:

$$\tilde D = \begin{pmatrix} m-\partial_y&\ -(\partial_x - \partial_t)\\ -(\partial_x + \partial_t)& m+\partial_y\end{pmatrix} \tag{2}$$

This is because $D \tilde D = (\partial_t^2-\partial_x^2-\partial_y^2+m^2) \mathbb{Id}$

A general expression for $\phi$ could be obtained with plane waves :

$$\phi(x,y,t) = \int dp_x dp_y f(p) e^{\large i(p_x x + p_y y - \sqrt{p_x^2+p_y^2+m^2}t) }\tag{3}$$ where $f(p)$ is any 2-row vector. The expression for $\psi = \tilde D \phi$ is straightforward.

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What do the p_x and p_y operators represent? Momentum? Thank you. –  Xingyou Sep 12 '13 at 17:00
$p_x$ and $p_y$ are momenta, and $p_0 = \sqrt{p_x^2+p_y^2+m^2}$ is the energy, but these are not operators, these are usual real commutative quantities. The momentum operators $P_i$ are defined as $-i \frac{\partial}{\partial x^i}$ (in $\hbar=1$ units) if they apply to a function of space-time like $\phi(x, y,t)$. Applyied to a function of momenta, you have obviously : $P_i f(p) = p_i f(p)$ –  Trimok Sep 12 '13 at 17:37
What is the domain of momentums we're integrating over? Could you give an example? –  Xingyou Sep 12 '13 at 18:30
@Xingyou : All real values, from $-\infty$ to $+\infty$. And you may choose any function $f(p_x,p_y)$. –  Trimok Sep 12 '13 at 18:33
@Xingyou : Simple example is a pure wave plane solution like : $\phi_p(x,y,t) =\begin{pmatrix} 1\\0\end{pmatrix}e^{\large i(p_x x + p_y y - \sqrt{p_x^2+p_y^2+m^2}t) }$ –  Trimok Sep 12 '13 at 18:36