# Is the adjoint representation of $SU(2)$ the same as the triplet representation?

Is the triplet representation of $$SU(2)$$ the same as its adjoint representation? Where the convention for the adjoint representation used is the one used in particle physics, where the structure constants are real and antisymmetric:

$$\mathrm{ad}(t^b_G)_{ac} = i f^{abc}$$

I was under the impression that is was, but I see two different forms of the generators in the triplet representations used, one being just the real skew symmetric generators of the $$SO(3)$$ rotation group, which agrees with the adjoint representation, and the other being:

$$T^1 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \quad T^2 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{matrix}\right)$$

These two representations do not agree, I assume that my idea about the adjoint reperesentation of $$SU(2)$$ being its triplet representation is wrong, but why?

• Dec 20 '17 at 22:54

The hermitian generators $$T^1 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \quad T^2 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{matrix}\right)$$ may be unitarily transformed to the hermitian generators $$UT^jU^\dagger=L^{~j}$$, satisfying the very same Lie algebra, by the equivalence transformation $$U= \frac{1}{\sqrt{2}} \left(\begin{matrix} -1 & 0 & 1 \\ -i & 0 & -i \\ 0 & \sqrt{2} & 0\end{matrix}\right) ~,$$ so that $$L_{1} = i\left(\begin{matrix}0&0&0\\0&0&-1\\0&1&0\end{matrix}\right) , \quad L_{2} = i\left(\begin{matrix}0&0&1\\0&0&0\\-1&0&0\end{matrix}\right) , \quad L_{3} = i\left(\begin{matrix}0&-1&0\\1&0&0\\0&0&0\end{matrix}\right),$$ that is, i times the real antisymmetric basis of classical angular momentum.

Observe how U sends the evident eigenvectors of $$T_3$$ to the eigenvectors of $$L_3$$; this is, in fact, how it was constructed. The simple diagonal form of $$T_3$$ is a defining feature of this spherical basis.

• How did you find that $U$ matrix? Sep 20 '18 at 19:28
• @Charlie I tweaked the last paragraph to reflect the method. Sep 20 '18 at 23:29

It is just matter of a missed factor $i\sqrt{2}$ due to different conventions. The antiHermitian matrices $i\sqrt{2} T^k$ can be transformed into the real antisymmetric matrices $L_k$ (which therefore are also complex antiHermitian) by means of a suitable unitary matrix $U$, $$L_k = U i\sqrt{2}\:T^kU^\dagger\quad k=1,2,3\:.$$ This is because both triples of matrices are irreducible representations of the Lie algebra of $SU (2)$ with the same value of the Casimir operator $\sum_k (\sqrt{2} T^k)^2 = -\sum_k (L_k)^2 =2I$ (so that $2= j(j+1)$ with $j=1$ which is the spin of the representation). As is known, up to unitary equivalences there is only one unitary irreducible representation of $SU (2)$ for every value of the spin, essentially due to Peter-Weyl theorem.

• Ah, so does the difference in convention come from the presence/absence of a factor of $i$ in the exponentiation of the Lie algebra? Sep 14 '16 at 1:46
• Yes, properly speaking representations of Lie algebras sould be made of anti selfadjoint operators. Physicists are interested in observables and therefore use selfadjoint operators instead...the difference is a factor $i$. Sep 14 '16 at 5:36

Let $V=\mathbb{C}^2$. Then the irreps are given by the symmetric powers $V_k:={\rm Sym}^k(V)$ with $k=0,1,\ldots$. The irrep $V_k$ has dimension $k+1$. I suppose triplet means of dimension three. So it is $V_2$ which lives inside $V\otimes V$ in the same way the adjoint rep lives inside $V\otimes V^{\ast}$ (2 by 2 matrices) where $V^{\ast}$ denotes the dual. Now a very special fact about $SU(2)$ is that $V$ and $V^{\ast}$, as representations, are the same. That's why adjoint and triplet are the same.