Representation of Angular Momentum Operator in the (j,j')

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Hello All,

I'm trying to understand how the (j,j') representation of the Lorentz group. Following Ryder, I can see why we define A=J+iK and B=J-iK, which each form an SU(2) group. So it's clear to me what the rep of these generators is when acting on a state (j,j'): [tex]Rep(A)\otimes1+1\otimes Rep(B)[/tex]. Where Rep(A) and Rep(B) are the appropriate j and j' reps.

My question is this: given the rep [tex](j,j')\oplus(j',j)[/tex], what is the induced rep on the generators? For example how do I act with A or J on this state?

Thanks a whole bunch
 

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  • #2
fzero
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If you have a representation [tex]r[/tex] and associated generators [tex]R[/tex], then the generators for the representation [tex]r\oplus r'[/tex] are

[tex]\begin{pmatrix} R & 0 \\ 0& R'\end{pmatrix}.[/tex]

As a sort of converse, if a representation [tex]\tilde{r}[/tex] is reducible to a sum [tex]r\oplus r'[/tex], then the generators [tex]\tilde{R}[/tex] are block diagonal, as above.
 
  • #3
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Thanks for the quick reply. But I didn't understand. In the [tex](1/2,1)\oplus(1,1/2)[/tex] of Lorentz, does the operator A (the left SU(2)) look like this

[tex]
\begin{pmatrix}
A_{2\times2} & 0 & 0 & 0 \\
0 & 1_{3\times3} & 0 & 0 \\
0 & 0 & A_{3\times3} & 0 \\
0 & 0 & 0 & 1_{2\times2}
\end{pmatrix}
[/tex]
 
  • #4
fzero
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Thanks for the quick reply. But I didn't understand. In the [tex](1/2,1)\oplus(1,1/2)[/tex] of Lorentz, does the operator A (the left SU(2)) look like this

[tex]
\begin{pmatrix}
A_{2\times2} & 0 & 0 & 0 \\
0 & 1_{3\times3} & 0 & 0 \\
0 & 0 & A_{3\times3} & 0 \\
0 & 0 & 0 & 1_{2\times2}
\end{pmatrix}
[/tex]
No, there's separate blocks for the right and left-generators. The block decomposition is only for the sums, not the tensor products. So you'd have

[tex]
\begin{pmatrix}
A_{2\times2} & 0 \\
0 & A_{3\times3}
\end{pmatrix}
[/tex]
 
  • #5
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Sorry I still don't understand. Each (j,j') is a (2j+1)X(2j'+1) dimensional vector space. So in the case of [tex](1/2,1)\oplus(1,1/2)[/tex] it should be a twelve dimensional vector space. What you wrote is five dimensional. Maybe the answer is [tex] A_{2\times2}\otimes 1_{3\times3} \oplus A_{3\times3}\otimes 1_{2\times2} [/tex]? This would give

[tex] \begin{pmatrix}
\left( A_{2\times2}\otimes 1_{3\times3} \right)_{6\times 6} & 0 \\
0 & \left( A_{3\times3}\otimes 1_{2\times2} \right)_{6\times 6}
\end{pmatrix}
[/tex]

which is a twelve dim operator like I'd expect. Does this make any sense?

Thank!!
 
  • #6
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Sorry I still don't understand. Each (j,j') is a (2j+1)X(2j'+1) dimensional vector space. So in the case of [tex](1/2,1)\oplus(1,1/2)[/tex] it should be a twelve dimensional vector space. What you wrote is five dimensional. Maybe the answer is [tex] A_{2\times2}\otimes 1_{3\times3} \oplus A_{3\times3}\otimes 1_{2\times2} [/tex]? This would give

[tex] \begin{pmatrix}
\left( A_{2\times2}\otimes 1_{3\times3} \right)_{6\times 6} & 0 \\
0 & \left( A_{3\times3}\otimes 1_{2\times2} \right)_{6\times 6}
\end{pmatrix}
[/tex]

which is a twelve dim operator like I'd expect. Does this make any sense?

Thank!!
This is right.
 

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