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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?

Community
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Eweler
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3 Answers3

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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.

Cosmas Zachos
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    How did you find that $U$ matrix? – Charlie Sep 20 '18 at 19:28
  • @Charlie I tweaked the last paragraph to reflect the method. – Cosmas Zachos Sep 20 '18 at 23:29
  • Hello professor, in your $L_3$ form, how do we identify the hypercharge $Y$? In $T_3$ form, we can obtain $Q=T_3 + \frac{Y}{2}$, but now $L_3$ don't has the diagonal components, how do we know the $Q$ or $Y$ of a triplet? Thanks! – Daren Jan 05 '23 at 12:34
  • The hypercharge Y can always be proportional to the identity, so identical in the two forms, since a change of basis cannot affect it. So, check that Q has the same eigenvalues in both bases. It's just that the charge eigenstates for the triplet, the 3 Ws, are not simple components as before, but the three eigenvectors of $L_3$; find them! – Cosmas Zachos Jan 05 '23 at 14:15
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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.

Valter Moretti
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  • Ah, so does the difference in convention come from the presence/absence of a factor of $i$ in the exponentiation of the Lie algebra? – Eweler Sep 14 '16 at 01:46
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    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 $. – Valter Moretti Sep 14 '16 at 05:36
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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.

Abdelmalek Abdesselam
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