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2j+1 d representation for Poincaré group

  1. Jun 18, 2015 #1
    I want to learn how to write down a particle state in some inertial coordinate frame starting from the state ##| j m \rangle ##, in which the particle is in a rest frame.
    I know how to rotate this state in the rest frame, but how does one write down a Lorentz boost for it? Note that I am not looking to boost a 4-vector but a vector of length ##2j+1##, so I need a different representation.
    On one hand, I know that formally the boost is of the form ##e^{-i \alpha \bar{p} \cdot \bar{K}} ##, where K is the generator for boosts, but it's not helpful as ##K## is never written down.
    I'm probably somehow searching with wrong terminology as the representation I'm looking for is elusive.
    Edit: I just realised that this might be possible by simply looking at the commutation relations ## K## have to satisfy, since the operators ## J## can be deduced for each ## j##, I think. In any case, There probably wouldn't be an easy form to write down how they operate on ##| j m \rangle## if I try that. Shouldn't it be possible to write the operation as some sort of rotation?
     
    Last edited: Jun 18, 2015
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  3. Jun 18, 2015 #2

    ChrisVer

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    what does the state [itex]|jm>[/itex] represent (what is j and m)?
     
  4. Jun 18, 2015 #3

    strangerep

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    IF your j,m just refer to the usual quantum numbers for angular momentum, then,... it's hard to answer the question in isolation since these only relate to the ordinary rotational degrees of freedom.

    More generally, one constructs unitary irreducible representations of the Poincare group. This procedure (worked out by Wigner a long time ago) involves choosing a "standard" momentum, (corresponding to the rest frame, in the case of a massive field), finding the usual SO(3) reps in that frame, and then completing the picture by boosting to arbitrary momenta. The procedure is explained in Weinberg vol-1, also Maggiore and various other QFT books. Also Sexl & Urbantke.
     
  5. Jun 19, 2015 #4

    samalkhaiat

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    The Lorentz algebra consists of (basically) two commuting [itex]so(3)[/itex] algebras:
    [tex][J_{i} , J_{j}] = i \epsilon_{ijk} J_{k}, \ \ [K_{i} , K_{j}] = i\epsilon_{ijk}K_{k}, \ \ [J,K]=0.[/tex]
    It follows from this algebra that a finite dimensional irreducible representation space [itex]\mathcal{V}^{j j'}[/itex] can be spanned by a set of [itex](2j+1)(2j'+1)[/itex] basis vectors [itex]|jm;j'm'\rangle[/itex] where [itex]m \in [-j,j][/itex], [itex]m' \in [-j' ,j'][/itex]. The action of the generators [itex]\vec{J}[/itex] and [itex]\vec{K}[/itex] on these states can be obtained from the following
    [tex](J_{1} \pm i J_{2}) |j,m;j'm' \rangle = \sqrt{(j \mp m)(j \pm m +1)} | j, m \pm 1;j'm' \rangle[/tex]
    [tex]J_{3}|jm ; j'm' \rangle = m |jm ; j' m' \rangle ,[/tex]
    and similar ones for the components of the [itex]\vec{K}[/itex] generator.
     
    Last edited: Jun 19, 2015
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