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Suppose I want to write down an interaction term for an action for spin 1/2 fermions that is $SU(2)$-symmetric.

I start from the most naive general form of such an action: $$S_{int} ~=~ \int_{4321} \sum_{\alpha \beta \gamma \delta} \bar \psi(4)_\alpha \bar \psi(3)_\beta \psi(2)_\gamma \psi(1)_\delta V(4,3,2,1)_{\alpha \beta \gamma \delta}$$

where the indices $1$ to $4$ stand for momenta and frequencies of my fermions.

Now I want to find the form $V$ must have in order to be $SU(2)$ symmetric. By transforming the fermion fields and demanding that the action must stay invariant under that, I can show that $V$ must transform as $$V_{\alpha' \beta' \gamma' \delta'} ~=~ \sum_{\alpha\beta\gamma\delta} R^\dagger_{\alpha \alpha'} R^\dagger_{\beta \beta'} R_{\gamma \gamma'} R_{\delta \delta} V_{\alpha \beta \gamma \delta}$$ where $R \in SU(2)$.

Well, and now I'm stuck continuing from here. Using some handwaving I think I could argue that $V$ must preserve total spin and also total spin in $z$-direction I could probably argue that $V$ can only scatter triplets to triplets, singlets to singlets, and also can't change the $z$-component of the triplet, but I would rather use a more rigorous approach.

Which will probably involve irreducible representations? I could probably get to the singlet/triplet statement above by noting that $SU(2)$ will transform multiplets into the same multiplet, so the singlet would be invariant under $SU(2)$ and the triplets would somehow mix. But why is it appropriate then to look at an "ingoing" singlet or "ingoing" triplet formed by indices $\gamma$ and $\delta$ as opposed to forming such states with, e.g., indices $\alpha$ and $\gamma$?

ADDENDUM: Well, I guess I can also start with the spins in a different basis: Assuming that I can put the two "ingoing" and the two "outgoing" spins into either a singlet or one of three triplets, I guess I can write the action as $$S \sim \int_{1234} \sum_{jm j'm'} (\bar \psi(4) \bar \psi(3))_{jm} (\psi(2) \psi(1))_{j'm'} V(4,3,2,1)_{jm;j'm'}$$ Then I can first argue that due to conservation of total spin we require $j = j'$. And then for $V$ I can look at singlet-singlet scattering and triplet-triplet scattering separately: For $j = j' = 0$, $m$ must be $0$ and so $V$ is a scalar, invariant under $SU(2)$, But for $j = j' = 1$, the states with $m = 0, \pm 1$ transform into each other in some way, and thus I must work a bit harder to get the symmetry right. I'll think about this, but in the meantime I'm open for more suggestions.

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Is your action going to be renormalizable? In how many dimensions? It's worth noting that in 3+1D renormalizability limits you to no more than two fermion fields in a single term (because each has dimension 3/2, and you need to have dimension 4 total). –  David Z Mar 30 '12 at 21:31
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It's for a condensed matter system so I don't worry about renormalizability; I'll always assume that there is a "natural" cut-off –  Lagerbaer Mar 30 '12 at 22:10
    
Oh, OK. Ignore me then :-) –  David Z Mar 30 '12 at 23:21
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This is very much a quick and dirty answer, I havn't though too much about it. I might update my answer if I find time in the weekend, otherwise I hope others will give more precise answers.

$V_{\alpha\beta\gamma\delta}$ transforms reducibly under $SU(2)$, as tensor products of four spin $\frac 12$ representations. Using $\bf\frac 12\otimes\frac 12 = 0 \oplus 1$ and $\bf 1\otimes 1 = 0 \oplus 1 \oplus 2$, we find that $V_{\alpha\beta\gamma\delta}$ decomposes into these irreducible representations $$\mathbf{\frac 12\otimes\frac 12\otimes\frac 12\otimes\frac 12} = (2\times\mathbf{0})\oplus (3\times \mathbf{1}) \oplus \mathbf{2},$$ two singlets, three triplets and a spin 2 (5-dimensional representation). There is an action of the permutation group $S_4$ on the indices of $V_{\alpha\beta\gamma\delta}$, for $SU(N)$ it turns out that decomposing this tensor into irreducible representations of the permutation group also corresponds to irreducible representations of $SU(N)$. This can be done rather quickly using Young Tableau, you can find the details in most representation theory books for physicists (I don't have a relevant book here and don't remember the details. I might add the answer in the weekend). In the case of a tensor product $\bf 1\otimes 1 = 0\oplus 1\oplus 2$, we get the following decomposition

$$ T_{ij} = \delta_{ij}\frac{tr(T)}3 + \left(\frac{T_{ij}-T_{ji}}2\right) + \left(\frac{T_{ij}+T_{ji}}2 - \delta_{ij}\frac{tr(T)}3\right).$$ These three terms transform irreducibly as spin $0$, spin $1$ and spin $2$ representations of $SU(2)$, respectively. In terms for the permutation group, these are the trivial, anti-symmetric and trace less symmetric representations, respectively.

Something similar can be done for $V_{\alpha\beta\gamma\delta}$, by using Young Tableau or just by playing around with it.

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Hm, when you decompose $V$ into irreducible representations you combine all four spins into total spins of $0$, $1$ or $2$. But should I, given that two of the spins transform in one direction and the other two into the "opposite" direction, keep the spins whose operators have a "bar" on top separate from those without the bar, i.e. $(0 \oplus 1) \otimes (0 \oplus 1)$? –  Lagerbaer Mar 30 '12 at 22:48
    
Mh, I think I'm just a bit confused. Maybe for clarification: When my general operator $V_{\alpha\beta\gamma\delta}$ is decomposed as you describe, are then the two singlets the only parts that are invariant under rotation as they transform like scalars? Or am I getting things completely wrong? –  Lagerbaer Mar 31 '12 at 5:01
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You were right; the fact that I get two terms in the action that transform like singlets is related to the irreducible representations: One singlet is obtained by combining two singlets, the other is obtained by combining two triplets. –  Lagerbaer Apr 4 '12 at 18:21
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