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Rotations

Consider a rotation through an angle $\phi$ about the $z$-axis:


\begin{displaymath}
\left( \begin{array}{c}
{x^0}^\prime \\ {x^1}^\prime \\ {x^2...
...begin{array}{c}
x^0 \\ x^1 \\ x^2 \\ x^3
\end{array} \right) .
\end{displaymath} (5.140)

The matrix $I_n$ has all zero elements accept for $I^1_{\ 2} =
-I^2_{\ 1} = 1$, or $I^{12} = -I^{21} = -1$. Thus


\begin{displaymath}
\sigma_{\mu\nu}I^{\mu\nu} = 2\sigma_{12}I^{12} = -2i\gamma_1...
... \sigma_3 & 0 \\ 0 & \sigma_3 \end{array}\right) = -2\Sigma_3,
\end{displaymath} (5.141)

Therefore


\begin{displaymath}
S = \exp\left(-\frac{i}{4}\phi\sigma_{\mu\nu}I^{\mu\nu}_n\right) =
\exp\left(\frac{i}{2}\phi\Sigma_3 \right) .
\end{displaymath} (5.142)

For a rotation about an arbitrary axis $\hat{n}$, we write


\begin{displaymath}
S = \exp\left(\frac{i}{2}\phi\vec{\Sigma}\cdot\hat{n} \right) .
\end{displaymath} (5.143)

The appearance of the half-angle is an expression of the double-valuedness of the spinor law of rotation; it takes a rotation of $4\pi$ to return $\psi(x)$ to its original value. This is a characteristic of 1/2 integer spin. Therefore physical observables in the Dirac theory must be bilinear, or an even power in $\psi(x)$. By using the rotation operator upon the solution for the Dirac particle at rest and polarized in the $z$-direction, it is possible to form states of any arbitrary direction. Since $\vec{\Sigma}$ is hermitian, $S^\dagger=S^{-1}$ for spatial rotations. We shall henceforth call the wave function of the Dirac theory a spinor.


next up previous contents index
Next: Lorentz Boosts Up: Proof of Covariance Previous: Proof of Covariance
Douglas M. Gingrich (gingrich@ ualberta.ca)
2004-03-18