I think my problem is that I misunderstand what "transforms under" really means.

Let's take $SU(3)$, for the $\mathbf{3}$ with Dynkin indices $(1,0)$, a state transforms like : $ψ→gψ$. For the $\mathbf{\bar{3}}$ with Dynkin indices $(0,1)$, a state transforms like : $ϕ→ϕg^{−1}$. And for the adjoint representation $(1,1)$: $\mathcal{O}→g\mathcal{O}g^{−1}$.

But then, if I take $SU(2)$, because the $\mathbf{2}$ and $\mathbf{\bar{2}}$ are equivalent, they should transform in the same way? And what about a representation labeled by $(2,0)$ in $SU(3)$? Should a state in this representation transforms like $\Psi→g^2\Psi$?


1 Answer 1


For the case $SU(n)$, $n>2$ the matrix and its inverse are not related by a similarity transform, so the representation where one acts with $g$ and with $g^{-1}$ are not isomorphic. For $SU(2)$ you can check that $$g^{-1}=EgE^{-1} $$ where $$E=\begin{vmatrix} 0 & 1 \\ -1 &0 \end{vmatrix}$$ This means there is no reason to act with $g^{-1}$.

Secondly, it is better to write indices such that the element from representation space carries $g=g^{a}{}_{b}$ and $\psi=\psi^a$ for $(1,0)$ and $\psi=\psi_b$ for $(0,1)$. So your $(2,0)$ corresponds to a symmetric rank-two tensor $\psi=\psi^{ab}=\psi^{ba}$ and $g$ hits every index. Such simple notation as $g^2\psi$ becomes misleading.

  • $\begingroup$ Hm, isn’t it $(g^{-1})^{\mathsf{T}} = E g E^{-1}$ rather than $g^{-1} = E g E^{-1}$? $\endgroup$
    – Socob
    Jun 27, 2017 at 14:07

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