Dirac equation

The one-dimensional Dirac equation for a particle of rest mass m₀ has a matrix form

$$\frac{d{\bf Z}(x)}{dx} \left(\begin{array}{cc} 0 & m_0-(V(x)-E) \\ m_0+(V(x)-E) & 0 \end{array}\right) {\bf Z}(x).$$ = (10)

which is structurally somewhat different from the Schrödinger equation because of the symmetric term depending on the rest mass, and the antisymmetric presence of the kinetic energy in two different places.

The substitutions

$$\kappa = [m_0^2-(V-E)^2]^{1/2}, \sigma = \left(\frac{m_0-(V-E)}{m_0+(V-E)}\right)^{1/2}$$ (11)

lead to the same form of solution matrix as before,

$${\bf Z}(x) \left(\begin{array}{cc} \cosh\ \kappa x & \sigma\ \sinh\ \kappa x... ...(1/\sigma)\ \sinh\ \kappa x & \cosh\ \kappa x \end{array} \right) {\bf Z}(x_0),$$ = (12)

but still only for a constant energy difference E - V. Otherwise the derivative of V(x) would have to be taken into account, introducing additional terms into the coefficient matrix.

When the rest mass is zero, the coefficient matrix is a scalar multiple of the unit antisymmetric matrix, so the use of the angle

 

$$\varphi \int_{x_0}^x(V(t)-E)dt$$ = (13)

gives simply

 

$${\bf Z}(x) \left(\begin{array}{cc} \cos\ \varphi & -\sin\ \varphi \\ \sin \varphi & \cos\ \varphi \end{array} \right) {\bf Z}(x_0),$$ = (14)

Irrespective of either m₀ or E, if V is very large yet extending over a small interval such that $V\times(x-x_0)=P$, the result is this same rotation matrix, now running through the angle P. An interesting consequence is that strengths differing by $2\pi$ make identical changes to the solution.

The fact that a concentrated potential produces a finite rotation rather than a shear is somewhat disconcerting, since it implies a discontinuity in both components of a solution vector. But mainly, this illustrates the fact that not all potentials give rise to continuous derivatives, nor even to continuous wave functions.

Following the Kronig-Penney precedent of compounding the delta-function potential with a plane wave, we would get

$${ \left(\begin{array}{cc} \cos\ P & -\sin\ P \\ \sin\ P & \cos\ ... ...appa L \\ (1/\sigma)\ \sinh\ \kappa L & \cosh\ \kappa L \end{array}\right) = }$$

$$\left(\begin{array}{cc} \cos(P)\cosh(\kappa L) - (1/\sigma)\sin(P... ... L) & \cos(P)\cosh(\kappa L) + \sigma \sin(P)\sinh(\kappa L) \end{array}\right)$$ (15)

with the dispersion relation

$$\cosh(\theta) \cos(P) \cosh(\kappa L) + \frac{1}{2}(\sigma - \frac{1}{\sigma}) \sin(P) \sinh(\kappa L).$$ = (16)

This is the spherical cosine law for a unit hyperboloid with an angle between edges of the triangle of $\cosh(\gamma) = \frac{1}{2}(\sigma - 1/\sigma)$. We don't see such clear bumps on the cosine curves as for the Schrödinger equation, but the result is very similar.