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Let the entities $\psi^i$ transform as the fundamental representation of $\mathrm{SU}(n)$, denoted by ${\bf n}$: $$ \psi^{\prime i}=U^{i}_{~j}\psi^j, $$ where, of course, $U$ represents $n\times n$ unitary matrices with unit determinant. Taking complex conjugates on both sides, we get, $$ \left(\psi^{\prime i}\right)^{*}=\left(U^{i}_{~j}\right)^{*}(\psi^j)^{*}. $$ Introducing the notation $$ \psi_i=\left(\psi^i\right)^{*}, \quad U_{i}^{~j}=\left(U^{i}_{~j}\right)^{*},\tag{1} $$ we have $$ \psi^\prime_i=U_{i}^{~j}\psi_j. $$ Therefore, by definition, $\psi_i=(\psi^i)^{*}$, which transform as $\bar{\bf n}$.

On the other hand, if I am not wrong, for the group $\mathrm{SU}(2)$, $\psi_i$ is also defined to be equal to $$\psi_i=\varepsilon_{ij}\psi^j.\tag{2}$$

How can we have two different definitions [in (1) and in (2)] for the same object $\psi_i$?

Emilio Pisanty
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Solidification
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1 Answers1

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OP's question is closely related to the fact that

are inequivalent representations, except for $n=2$ due to the existence of the invariant 2D Levi-Civita symbol $\epsilon$, cf. the following identity $$(\rho(g)^{-1})^T~=~\epsilon \rho(g)\epsilon^{-1},\qquad g~\in~SU(2).\tag{3}$$

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$^1$Note that the dual/contragredient/transposed representation (2) is equivalent to the complex conjugate representation $\bar{\rho}$ due to unitarity.

Qmechanic
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