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I am studying quarks $u$, $d$ and $s$. I know they can be represented by 3-vectors:

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We also have the antiquarks that are represented by the same 3-vectors:

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The difference between them is that when acting with ladder operators one has to use the Gell-Mann matrices $\lambda_i$ for the quarks and the antifundamental representation $\bar{\lambda}_i$ when acting on antiquarks.

Writing all this in Dirac notation we could say that $$ \hat{\Lambda}_i |u\rangle \sim \lambda_i u \hspace{40pt} \hat{\Lambda}_i |\bar{u}\rangle \sim \bar{\lambda}_i \bar{u} $$ so $\hat{\Lambda}_i$ is the abstract operator that acts over the elements of the Hilbert space $ |u\rangle$ and $|\bar{u}\rangle$.

My question is: because $\hat{\Lambda}_i$ can act over $ |u\rangle$ and $|\bar{u}\rangle$ then I am tempted to say that they both live in the same Hilbert space. However if that were the case and if the 3-vector representation is correct, then they are the same state... So $ |u\rangle$ and $|\bar{u}\rangle$ must be different but are represented by the same 3-vectors... How does this work? Does $ |u\rangle$ and $|\bar{u}\rangle$ live in two different Hilbert spaces?

Also, a meson is composed by a quark and an antiquark. Lets consider the meson $\bar{u} u$. If I were to represent this meson in Dirac notation, would it be $ |u\rangle \otimes |\bar{u}\rangle$?

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The quarks and antiquarks transform by conjugate representations, respectively $(1,0)$ and $(0,1)$ in the Dynkin notation. In particular, the eigenvalues of the diagonal operators in the $(0,1)$ irrep are the negative of those in the $(1,0)$ since, if $e^{i\theta \Lambda}$ is diagonal (and hermitian) in $(1,0)$, then $$ (e^{i\theta \Lambda})^* = e^{-i\theta \Lambda} $$ in $(0,1)$, illustrating how eigenvalues in an irrep are the negatives of the eigenvalues in its conjugate irrep.

Clearly, since no similarity transformation can change the eigenvalues, the irrep $(1,0)$ and $(0,1)$ must be distinct.

In particular, the standard Gell-Mann matrices act in the defining $(1,0)$ irrep, but the matrix elements of the generators in the $(0,1)$ will have sign differences from those in $(1,0)$ and thus will NOT be the standard Gell-Mann matrices (even though they will be $3\times 3$ matrices since $(0,1)$ is of dimension $3$).

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