I don't have a real question, this is more confusion about multiplets and irreducible representations fo $\text{SU(2)}$. I am looking for an epiphany.
I know that for each irreducible representation of the group symmetry I have a degenerate multiplet of N elements, with N the dimension of the space on which the irreducible representation acts.
For example: given the irreducible representation of $\text{SO}(3)$ labelled with $\ell=1$ I get three degenerate states (given $[H,L_i]=[H,L^2]=0$ where $L^2$ is the Casimir operator $L^2=L_1^2+L_2^2+L_3^2$ wich commutes with the generators by definition). In this example the irreducible (and fundamental) representation acts on the space $\mathbb{R}^3$: N=3 and this means there are three degenerate states (the three projections of the orbital angular momentum).
If we talk about $\text{SO(N)}$ no problem. I now have a problem with $\text{SU(2)}$: I got that its fundamental representation is given by the $2\times 2$ unitary matrices and the generators are proportional to the Pauli matrices. As long as this group represents a symmetry, I have degenerate multiplets like in the previous case: this time the space on which the fundamental representation acts is $\mathbb{C}^2$. I know from physcs that in this case the multiplet is a spin doublet (up and down). Writing down the functions I have $$\psi(x)=\left(\genfrac{}{}{0pt}{}{\phi_1+i\phi_2}{\phi_3+i\phi_4}\right)$$ and each $\phi_i$ represents a degree of freedom. So we have 4 degree of freedom but just a spin doublet.
What I thought is that when you're acting on a complex space you have to consider that for every particle you have its antiparticle (e.g. the complex scalar field): in this case I would have to consider an "anti-doublet" (and this would make sense, for example both electron and positron have spin). So am I wrong if I say that for $j=1$ I will have a degenerate triplet and an "anti-triplet"? This means I would get six particles degenerate in energy.
My doubts arose considering the Spontaneous Symmetry Breaking for the irreducible representation of $\text{SU}(2)$: since there are 3 broken generators I should have three massless and spinless Goldstone bosons (3 degrees of freedom) and this obviously requires more than just a doublet.