2j+1 d representation for Poincaré group

  • #1
27
2
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:

Answers and Replies

  • #2
ChrisVer
Gold Member
3,352
452
what does the state [itex]|jm>[/itex] represent (what is j and m)?
 
  • #3
strangerep
Science Advisor
3,162
996
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? [...]
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.
 
  • #4
samalkhaiat
Science Advisor
Insights Author
1,714
985
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?
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:

Related Threads on 2j+1 d representation for Poincaré group

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
12
Views
6K
  • Last Post
Replies
10
Views
4K
Replies
3
Views
1K
Replies
9
Views
3K
Replies
8
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
4K
Replies
7
Views
1K
Top