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

## Main Question or Discussion Point

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'): $$Rep(A)\otimes1+1\otimes Rep(B)$$. Where Rep(A) and Rep(B) are the appropriate j and j' reps.

My question is this: given the rep $$(j,j')\oplus(j',j)$$, 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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fzero
Homework Helper
Gold Member
If you have a representation $$r$$ and associated generators $$R$$, then the generators for the representation $$r\oplus r'$$ are

$$\begin{pmatrix} R & 0 \\ 0& R'\end{pmatrix}.$$

As a sort of converse, if a representation $$\tilde{r}$$ is reducible to a sum $$r\oplus r'$$, then the generators $$\tilde{R}$$ are block diagonal, as above.

Thanks for the quick reply. But I didn't understand. In the $$(1/2,1)\oplus(1,1/2)$$ of Lorentz, does the operator A (the left SU(2)) look like this

$$\begin{pmatrix} A_{2\times2} & 0 & 0 & 0 \\ 0 & 1_{3\times3} & 0 & 0 \\ 0 & 0 & A_{3\times3} & 0 \\ 0 & 0 & 0 & 1_{2\times2} \end{pmatrix}$$

fzero
Homework Helper
Gold Member
Thanks for the quick reply. But I didn't understand. In the $$(1/2,1)\oplus(1,1/2)$$ of Lorentz, does the operator A (the left SU(2)) look like this

$$\begin{pmatrix} A_{2\times2} & 0 & 0 & 0 \\ 0 & 1_{3\times3} & 0 & 0 \\ 0 & 0 & A_{3\times3} & 0 \\ 0 & 0 & 0 & 1_{2\times2} \end{pmatrix}$$
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

$$\begin{pmatrix} A_{2\times2} & 0 \\ 0 & A_{3\times3} \end{pmatrix}$$

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

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

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

Thank!!

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

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

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

Thank!!
This is right.