Two ways to form $SU(2)$ singlets? I am trying to reconcile the two ways of forming $SU(2)$ singlets out of a pair of doublets.
Method (1):
If $v=\begin{pmatrix}v^1\\ v^2\end{pmatrix}$ and $w=\begin{pmatrix}w^1\\ w^2\end{pmatrix}$ are two $SU(2)$ doublets, then I can form a singlet by taking the antisymmetric combination:
$$(v\otimes w)_\text{singlet}=\epsilon_{ij}v^iw^j=v^1w^2-v^2w^1$$
Method (2):  Using the same objects from method 1, I can form a singlet by simply taking the conjugate-transpose of one of them, say $v^*=\begin{pmatrix}v_1^*& v_2^*\end{pmatrix}$, and contracting them straight-up:
$$(v\otimes w)_\text{singlet}=(v^*)_i w^i=v_1^*w^1+v_2^*w^2.$$
So are these two different ways to form singlets?  Is there a way to understand this from a more general point of view, say $SU(3)$?  Also, in elementary quantum mechanics, why is it that when we form spin-singlets out of wavefunctions using method (1), and not using method (2)?
--
As a note to myself and others, $(v^*)_i$ and $\epsilon_{ij}v^j$ are two different objects.  But, both transform as in the same way; namely as $\bar{\bf{2}}$.
 A: I think confusion arises because the metric tensor is hidden in $SU(2)$, so I'll work in $SU(p,q)$ and then specialize to $SU(2)$. The inequivalent defining reps of the general linear group $GL(m,C)$ carried on the four vector spaces $V_{m},\tilde{V}_{m}, V^{*}_{m}, \tilde{V}^{*}_{m}$ transform vectors as follows.
$$
v'^{a}=\,[D(g)]^{a}_{\ \ b}v^{b}\\
v'_{a}=\,[D(g^{-T})]_{a}^{\ \ b}v_{b}\\
v'_{\bar{a}}=[D(g^{*})]_{\bar{a}}^{\ \ \bar{b}}v_{\bar{b}}\\
v'^{\bar{a}}=[D(g^{-\dagger})]^{\bar{a}}_{\ \ \bar{b}}v^{\bar{b}}  
$$
$SU(p,q)$ with $p+q=m$ is defined as the subgroup under which the Hermitian metric tensor $I^{\bar{a}}_{\ \ b}$ transforms trivially. The metric tensor can be used to change ordinary indices into barred indices.
$$
v^{\bar{a}}=I^{\bar{a}}_{\ \ b}v^{b}
$$
The metric tensor is invariant under $SU(p,q)$ so it commutes with the group matrices and hence the rep $D(g)$ is equivalent to the rep $D(g^{-\dagger})$. A physical quantity represented by states $v^{a}$ is the same kind of physical quantity as represented by states $v^{\bar{a}}$.
Now specialize to $m=2$ so the group is $SU(2)$ or $SU(1,1)$. The Levi-Civita tensor $\epsilon_{ab}$ can be used to change contravariant indices into covariant indices.
$$
v_{a}=\epsilon_{ab}v^{b}
$$
The Levi-Civita tensor is also invariant under the group and hence the rep $D(g^{-T})$ is equivalent to the rep $D(g)$.  
The singlet $\epsilon_{ab}v^{a}w^{b}=v^{a}w_{a}$. These are two equivalent 1-d irreps carried on $V_{2}\otimes V_{2}$ and $V_{2}\otimes\tilde{V}_{2}$. They represent the same kind of physical quantity: a spin 0 state. So, if one has two spin half states $\psi^{a}$ and $\phi^{a}$, the spin zero state is $\epsilon_{ab}\psi^{a}\phi^{b}$ or, equivalently $\psi^{a}\phi_{a}$. In the second form one needs a state $\phi_{a}$ which is the same kind of physical quantity (a spin half) as $\phi_{\bar{a}}$. A $\phi_{\bar{a}}$ transforms as $D(g^{*})$ so if one has a $\phi^{a}$ then the natural way to get something transforming as $D(g^{*})$ is to use $(\phi^{a})^{*}$. In this way, a spin zero is also $\psi^{a}(\phi^{a})^{*}$.
A: OP's two methods are isomorphic. In general one is only interested in classifying representations modulo isomorphism. The point is that for the Lie group $SU(2)$, the spinor representation ${\bf 2}$ and the complex conjugate spinor representation $\bar{\bf 2}$ are equivalent representations ${\bf 2}\cong \bar{\bf 2}$. The equivalence is precisely given by multiplying with the epsilon symbol.
In contrast, the representation ${\bf 3}$ and complex conjugate representation $\bar{\bf 3}$ of the Lie group $SU(3)$ are not equivalent.
