# Representations of the rotation group

(I have already done a similar question, but I did not express myself very well and the question was a bit confusing, so let me try again. If you find the question repetitive, I apologize and you can remove it). Consider a Lie group with 3 generators $$J_{1}, J_{2}$$ and $$J_{3}$$ that satisfy $$[J_{a},J_{b}] = i\epsilon_{abc} J_{c}.$$ We can show that $$J^{2}$$ commutes with all these generators, so we can diagonalize $$J^{2}$$ and one of them togheter (we choose $$J_{3}$$. If we call $$|j,m \rangle$$ a eigenvector of $$J_{3}$$ and $$J^{2}$$, we find that $$J^{2}|j,m \rangle = j(j+1)|j,m \rangle,$$ $$J_{3}|j,m \rangle = m |j,m \rangle,$$ where $$j = 0, 1/2, 1, 3/2, ...$$ and $$m = -j, -j+1, ..., j$$. I have read that for each value of $$j$$ is associated a representtion of this group, given by the matrices with elements $$\langle j, m | J_{a} |j,m \rangle$$. In fact, for $$j = 1/2$$ we obtein the Pauli matrices, which are related to the generators of $$SU(2)$$. But how can I show that this matrices form a representation? It's not obvious to me.

• Possible duplicate by OP: physics.stackexchange.com/q/481141/2451 – Qmechanic May 24 at 2:40
• yes, it was to this question that I was referring to when I said that I had already asked a similar question – AlfredV May 24 at 12:12

## 1 Answer

Simply show that the matrices satisfy the commutation relations that define the algebra. The commutator is defined in terms of matrix multiplication. For matrices $$X,Y$$, we define $$[X,Y]=XY-YX$$.

• Why? To be a representation, shouldn't these matrices have the same multiplification rule of the elements of the group? That is, if A and B are elements of the group, and D(A) and D(B) the matrices associated with these elements, then we should have D(A)D(B) = D(AB). Show that these matrices satisfy the commutation relation is equivalent to this? – AlfredV May 24 at 12:10
• These are elements of the algebra, not the group. In any case for $x,y$ in the algebra there is usually no meaning for $xy$ so the requirement that $D(A)D(B)=D(AB)$ makes no sense. Our notion of multiplication on the algebra is commutation. – user26872 May 24 at 12:19
• I'm sorry, I am confused. What's the definition of representation? What's the relation of the algebra with the group? – AlfredV May 24 at 13:40
• – user26872 May 24 at 19:09
• Roughly, the algebra and group are related through exponentiation. – user26872 May 24 at 19:12