6
$\begingroup$

Let $\xi$ be a spinor. If $(\theta ,\phi)$ are the parameters of a rotation and pure Lorentz transformation, then how can we prove that the transformation rule for $\xi$ can be written as $$\xi ~\rightarrow~ \exp\left(\ i \frac{\bf{\sigma}}{2}\cdot \theta +\frac{\bf{\sigma}}{2} \cdot \phi\right) \xi,$$ where $\sigma$ are the Pauli matrices?

$\endgroup$
  • $\begingroup$ Are you comfortable with pure rotations, i.e. do you still have trouble if I set $\phi=0$? $\endgroup$ – Michael Brown Jan 2 '13 at 23:41
  • 1
    $\begingroup$ It is not clear what the question is asking for. Do you mean to ask why a spinor transforms that way? $\endgroup$ – Siva Apr 30 '13 at 23:07
  • $\begingroup$ @unlimited-dreamer, I suppose you've offered a bounty because you're not satisfied with the current answer. You've got to explain what you're looking for. Else, this question could be downvoted for being poorly framed. $\endgroup$ – Siva May 4 '13 at 6:15
  • $\begingroup$ my question isn't fuzzy i think. I just wanted to get the equation$\xi ~\rightarrow~ \exp\left(\ i \frac{\bf{\sigma}}{2}\cdot \theta +\frac{\bf{\sigma}}{2} \cdot \phi\right) \xi~?$. any help would be approciated. Thanx. $\endgroup$ – Unlimited Dreamer May 4 '13 at 9:58
5
+50
$\begingroup$

You can write an infinitesimal transformation, with generator $J$, as $$ R(\delta\theta) = 1 + iJ\delta\theta $$ A finite transformation is a succession of $N\to\infty$ infinitesimal transformations, $$ R(\theta) = (1 + iJ\theta/N)^N = e^{iJ\theta} $$

The rotations $O(3)$ are isomorphic to $SU(2)$, with generators $J = \sigma/2$. The Lorentz transformations are similar to rotations, but with hyperbolic functions rather than trigonometric functions; $\sinh=\gamma\beta$ and $\cosh\phi=\gamma$, because the boosts satisfy $\gamma^2-\gamma^2\beta^2=1$. You can find that the Lorentz generators are $K = \pm i\sigma/2$.

Putting this together, for the negative solution, you find $$ R(\theta, \phi) =\exp\left(\ i \frac{\bf{\sigma}}{2}\cdot \theta +\frac{\bf{\sigma}}{2} \cdot \phi\right) $$ This is the right-handed $(1/2,0)$ solution. (The alternative positive solution is the left-handed $(0,1/2)$ solution.)

$\endgroup$
4
$\begingroup$

You are giving the lorentz transformation of a left-handed Weyl spinor. A very detailed derivation of those formulas is given in [1] for example.

In short the appearence of the two pauli matrices stems from the fact that the Lie algebra $\mathfrak{so(3,1)}$ of the lorentz group is the same as $\mathfrak{su(2)\times\mathfrak{su(2)}}$. So the rotations are generated by $\mathfrak{su(2)}$ as well as the boost (however note the extra factor $i$).

[1] Maggiore, Michele A Modern Introduction to Quantum Field Theory, 2005

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.