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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)?

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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 $\bar{\bf{2}}$. –  QuantumDot Feb 28 '13 at 2:55
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2 Answers 2

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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.

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Just to clarify, I have a spinor $\psi$ in representation $\bf{2}$. Then is $\psi^*=\epsilon\psi$ an equality? or is this taken to mean that only the LHS and RHS transform in the same way (i.e. $\bar{\bf 2}$). –  QuantumDot Feb 27 '13 at 20:27
    
Well, if the metric to raise and lower spinor indices is $\delta_{ij}$, then you can get from an object that transform as a spinor, to an object that transforms as a complex conjugate spinor, by multiplying with the epsilon symbol. This works for any sign convention of epsilon. –  Qmechanic Feb 27 '13 at 20:36
    
Oh this is something new! I thought that the metric necessarily had to be $\epsilon$. I can chose it to be $\delta$? Anyway, what if my metric to raise and lower indicies is in fact $\epsilon_{ij}$; what then would be the appropriate thing to multiply by? Thanks! –  QuantumDot Feb 27 '13 at 21:56
    
The epsilon symbol is not a metric. A (real) metric $\eta_{ij}$ is by definition symmetric. For the Lie group $U(2)$, the metric is positive definite. A $2\times 2$ matrix $U\in U(2)$ satisfies $U^{\dagger}\eta U= \eta$. So a change in the choice of metric changes the possible choice of group element $U$. Any sane person would chose $\eta_{ij}=\delta_{ij}$. –  Qmechanic Feb 27 '13 at 22:02
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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 $\psi^{a}(\phi^{a})^{*}$.

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This answer together with @Qmechanic 's result has clarified the confusion! This answer, in particular is invaluable. Would you know where I could find info on the representation theory of $GL(m,C)$ as you mention in this answer? –  QuantumDot Feb 28 '13 at 0:56
    
@QuantumDot : I learnt this from Wu-Ki Tung's book "Group Theory in Physics". –  Stephen Blake Feb 28 '13 at 1:15
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