$\text{SU}(2)$ irreducible representation and degenerate multiplets 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.
 A: Generally, if your particle transforms in the representation $V$, then the anti-particle transforms in the conjugate representation $\bar{V}$, which is the same as the dual representation for groups like $\mathrm{SU}(n)$ and $\mathrm{SO}(n)$ because we consider only unitary representations of them.
As for the symmetry breaking: The generators always transform in the adjoint representation, since by definition the generators are just a basis of the Lie algebra. The adjoint of $\mathrm{SU}(2)$ is, naturally, the triplet, and indeed the Higgs fields that usually occur in spontaneous symmetry breaking also often is taken to be in the adjoint.
You should be careful of your usage of the definite article - there is no unique "the irreducible representation" - all of the spin representations of $\mathrm{SU}(2)$ labeled by half-integers $j$ on $\mathbb{C}^{2j+1}$ are irreducible. Nothing a priori forces the physical fields to transform only in the fundamental or adjoint - in principle, one can imagine fields/particles transforming in arbitrary irreducible representations. However, it turns out that we simply don't observe any fundamental particles transforming in such representations, so their use in model building is rare.
