2j+1 d representation for Poincaré group

In summary, one writes down a Lorentz boost for a particle state in an inertial coordinate frame by writing down the commutation relations between the operators in that frame, and then completing the picture by applying the generators.
  • #1
terra
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 realized 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?
 
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  • #2
what does the state [itex]|jm>[/itex] represent (what is j and m)?
 
  • #3
terra said:
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
terra said:
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 realized 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.
 
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FAQ: 2j+1 d representation for Poincaré group

1. What is the significance of the "2j+1" in the 2j+1 d representation for Poincaré group?

The "2j+1" in the representation refers to the total angular momentum of a particle. This value determines the number of possible states a particle can have in the representation.

2. How is the "2j+1" value related to the spin of a particle in the 2j+1 d representation?

In the 2j+1 d representation, the "2j+1" value corresponds to the spin of a particle. For example, a particle with spin 1/2 would have a 2j+1 value of 2, while a particle with spin 1 would have a 2j+1 value of 3.

3. What is the difference between the 2j+1 d representation and other representations of the Poincaré group?

The 2j+1 d representation is a specific type of representation of the Poincaré group that is used for particles with spin. It is different from other representations, such as the Wigner's little group representation, which is used for particles with no spin.

4. How does the 2j+1 d representation for Poincaré group impact our understanding of particle physics?

The 2j+1 d representation is important in particle physics as it helps us understand the behavior of particles with spin. It allows us to mathematically describe and predict the properties and interactions of particles with spin, such as electrons and protons.

5. Are there any limitations to using the 2j+1 d representation for Poincaré group?

While the 2j+1 d representation is useful for describing particles with spin, it has limitations. It does not take into account other properties of particles, such as electric charge or mass. Therefore, it is often used in conjunction with other representations to fully understand the behavior of particles.

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