Transformation of solutions of the Dirac equation

[Gene Naden]

I am working through "Lessons on Particle Physics." The link is https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf. I am on page 21, equation (1.5.50), which is
$S(\Lambda)=1-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}$.
I would like some motivation for this equation. I wonder what the $\omega$'s are. When I derived the equations that came after, the $\omega$'s dropped out. I get the general idea that this is the transformation of the solution of the equation, corresponding to a Lorentz transformation of the coordinates.

 

[stevendaryl]

I am working through "Lessons on Particle Physics." The link is https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf. I am on page 21, equation (1.5.50), which is
$S(\Lambda)=1-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}$.
I would like some motivation for this equation. I wonder what the $\omega$'s are. When I derived the equations that came after, the $\omega$'s dropped out. I get the general idea that this is the transformation of the solution of the equation, corresponding to a Lorentz transformation of the coordinates.

\omega_{\mu \nu} are just the parameters of the Lorentz transformation.

An infinitesimal change of coordinates from one inertial coordinate system to another (with the same origin) can be characterized by two 3-D vectors:

\vec{R}: a rotation
\vec{B}: a "boost" (change of velocity)

These 6 components can be combined into a single antisymmetric tensor \omega_{\mu \nu} as follows:

1. \omega_{0j} = B_j
2. \omega_{xy} = R_z
3. \omega_{yz} = R_x
4. \omega_{zx} = R_y

(and then use \omega_{\mu \nu} = -\omega_{\nu \mu} to get the other components)

 

[Gene Naden]

OK, thanks. So $S(\Lambda)=1-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}$ is reasonable in that it reduces to unity if the $\omega_{\mu\nu}=0$. And I suppose that the difference from unity is linear in $\omega$. I am still vaguely unsatisfied about this equation but maybe the best thing into accept it for now and move forward.

 

[George Jones]

OK, thanks. So $S(\Lambda)=1-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}$ is reasonable in that it reduces to unity if the $\omega_{\mu\nu}=0$. And I suppose that the difference from unity is linear in $\omega$.

No, this is an approximation that is useful when the $\omega_\mu \nu$ are very small.

The exact expression for $S \left( \Lambda \right)$ is given by (1.5.54). For any square matrix $X$, the exponential $e^X$ is defined by the series
$$e^X = 1 + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \dots$$
which is convergent for all $X$.

Using this series expression for (1.5.54) shows that (1.5.50) is a two-term approximation to (1.5.54) that is "valid" for small $\omega_\mu \nu$.

The relationship between $S \left( \Lambda \right)$ and $\Sigma^{\mu\nu}$ is that of a representation of the Lorentz Lie Group and the corresponding representation of the Lorentz Lie algebra (derivative at the identity).

 

[Gene Naden]

Thank you; the authors go on to develop the exponential for rotations and boosts.