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Let's denote a spinor $\xi$. If $(\theta ,\phi)$ are the parameters of a rotation and pure Lorentz transformation, then how $\xi$ could be written as $$\xi ~\rightarrow~ \exp\left(\ i \frac{\bf{\sigma}}{2}\cdot \theta +\frac{\bf{\sigma}}{2} \cdot \phi\right) \xi~?$$

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Are you comfortable with pure rotations, i.e. do you still have trouble if I set $\phi=0$? –  Michael Brown Jan 2 '13 at 23:41
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It is not clear what the question is asking for. Do you mean to ask why a spinor transforms that way? –  Siva Apr 30 '13 at 23:07
    
@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. –  Siva May 4 '13 at 6:15
    
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. –  Unlimited Dreamer May 4 '13 at 9:58

2 Answers 2

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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.)

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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

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