Matrix differential equations

All in all, the Dirac harmonic oscillator is a good place to study resonances, especially if much smaller masses are considered than the ones which characterize the actual experience of physicists. Of course, that makes a comparison with experiment correspondingly difficult.

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

$$\frac{d{\bf Z}(x)}{dx} \left(\begin{array}{cc} 0 & m_0+(V(x)-E) \\ m_0-(V(x)-E) & 0 \end{array}\right) {\bf Z}(x).$$ = (1)

The substitutions

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

lead to a convenient solution matrix,

$${\bf Z}(x) \left(\begin{array}{cc} {\rm cos\ } k\ x & \sigma\ {\rm sin\ } k\ x \\ (1/\sigma)\ {\rm sin\ }k\ x & {\rm cos\ }k\ x \end{array} \right),$$ = (2)

but only for a constant energy difference E - V, and for real k. Formally the result holds for immaginary k but in that case, a classically forbidden region, it is preferable to redefine both k and $\sigma$ to make them real and then use hyperbolic functions.

There are other approximations which are useful; for example when the mass is small enough to be neglected, the coefficient matrix is antisymmetric. Setting

 

$$\varphi(x,x_0) \int_{x_0}^x(v(t)-E)dt$$ = (3)

the solution becomes

 

$${\bf Z}(x) \left(\begin{array}{cc} {\rm cos\ } \varphi(x,x_0) & - {\rm sin\ ... ... }\varphi(x,x_0) & {\rm cos\ } \varphi(x,x_0) \end{array} \right) {\bf Z}(x_0),$$ = (4)

which is a pure rotation in the phase plane.

There is a standard procedure for correcting an approximate coefficient matrix in a system of differential equations. Suppose that the full coefficient M is split into a sum

M = A + B, (5)

wherein A is the convenient approximation and B is a correction, not necessarily small. All kinds of splittings are possible: symmetric and antisymmetric parts, averages and deviations, large and small components, to mention three.

Supposing that the auxiliary equation

$$\frac{d{\bf U}(x)}{dx} A(x) {\bf U}(x)$$ = (6)

has already been solved subject to the initial condition $U(x_0)={\bf I}$, and that we intend to write ${\bf Z}(x) = {\bf U}(x){\bf V}(x)$, we find that ${\bf V}(x)$ must solve the differential equation

$$\frac{d{\bf V}(x)}{dx} {\bf U}(x)^{-1}B(x){\bf U}(x)\ {\bf V}(x)$$ = (7)

subject to the same initial condition as ${\bf Z}$.