- #1
LAHLH
- 409
- 1
Hi,
I just wondered if someone could check my understanding is correct on this topic. I understand that to find the irreps of a group we can find the irreps of the associated Lie algebra, i.e. in the case of SO(3) find irreducible matrices satisfying the comm relations [tex] \left[X_i,X_j\right]=i\epsilon_{ijk}X_k[/tex].
I of course recognise that these are the commutation relations that the angular momentum operators satisfy in standard QM. Thus if we know what the eigenvectors/eigenvalues are for the angular momentum operators, we can choose to work in the basis of eigenvectors, thus diagonalising, say, [tex] X_3[/tex] as [tex] \langle jm'\mid X_3\mid jm\rangle=m\delta_{m'm}[/tex]. What I don't really understand however is how this is an irrep of [tex] X_3[/tex] matrix, since clearly if this matrix is diagonal then it is reducible further?
Thanks
I just wondered if someone could check my understanding is correct on this topic. I understand that to find the irreps of a group we can find the irreps of the associated Lie algebra, i.e. in the case of SO(3) find irreducible matrices satisfying the comm relations [tex] \left[X_i,X_j\right]=i\epsilon_{ijk}X_k[/tex].
I of course recognise that these are the commutation relations that the angular momentum operators satisfy in standard QM. Thus if we know what the eigenvectors/eigenvalues are for the angular momentum operators, we can choose to work in the basis of eigenvectors, thus diagonalising, say, [tex] X_3[/tex] as [tex] \langle jm'\mid X_3\mid jm\rangle=m\delta_{m'm}[/tex]. What I don't really understand however is how this is an irrep of [tex] X_3[/tex] matrix, since clearly if this matrix is diagonal then it is reducible further?
Thanks