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My lecture notes state that we need to classify all finite-dimensional irreducible representations of the proper, orthochronous Lorentz group in order to formulate a QFT for particles with non-zero spin.

This is done by characterising the Lorentz algebra by the eigenvalues $a (a + 1)$ and $b (b + 1)$ of the square of the operators $$ \vec{A} = \frac{1}{2} (\vec{J} + i \vec{K}) \\ \vec{B} = \frac{1}{2} (\vec{J} - i \vec{K}) , $$ where $\vec{J}$ is the generator of rotation and $\vec{K}$ the generator of boosts.

The corresponding representation of the Lorentz group is then obtained by taking the exponential map of particular operators like $\frac{\vec{\sigma}}{2}, 0$ for $a = \frac{1}{2}, b = 0$.

Can $\vec{A}^2$ an $\vec{B}^2$ be understood as the Casimirs of the Lie algebra or do they have something in common with the concept (I am missing some understanding here)?

How can I guarantee that taking the exponential map of an irreducible representation of the Lie algebra gives me an irreducible representation in the corresponding Lie group?

Qmechanic
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rgba
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1 Answers1

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  1. Finite-dimensional irreducible

    • (i) representations of the double cover $Spin^+(1,3,\mathbb{R})\cong SL(2,\mathbb{C})$ of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$,

    • (ii) representations of the corresponding Lie algebra $so(1,3;\mathbb{R})$,

    • (iii) projective representations of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$,

    are all labelled by two non-negative half-integers $$(a,b)~\in~\frac{1}{2}\mathbb{N}_0 \times\frac{1}{2}\mathbb{N}_0.$$

    See also e.g. this Phys.SE post and links therein.

  2. If $a+b~\in~\mathbb{N}_0$ is an integer, it is also a group representation of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$ itself.

  3. $A_i$ and $B_i$, $i\in\{1,2,3\},$ are the $3+3=6$ generators of the complexified Lie algebra $$so(1,3;\mathbb{C})~\cong~sl(2,\mathbb{C})_A\oplus sl(2,\mathbb{C})_B,$$ with quadratic Casimirs $\vec{A}^2$ and $\vec{B}^2$.

  4. The exponential map $\exp: so(1,3;\mathbb{R})\to SO^+(1,3;\mathbb{R})$ for the restricted Lorentz group is surjective, cf. e.g. this Phys.SE post.

Qmechanic
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