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The problem considered here is, as the title says, to construct a Lagrangian that is invariant with respect to $SU(2)$ transformations. I will present first the context and the reason of my questioning and then form the question in the end.

The problem physically is to build the simplest phenomenological Lagrangian of the last century for NN scattering via a pion interaction with isospin invariance. We deal here only with isospin invariance and we do not care with Lorentz symmetry.

We categorize the three pions as a pion triplet in $SU(2)$ adjoint representation, that is a vector space with three dimensions( we will hereafter write the representations as Dsomething). Also we categorize the proton and neutron as a doublet of $SU(2)$ $D_{1/2}$ representation, that is a 2d space with $p$ and $n$ as base vectors.

In my understanding ( which may be wrong and then should be corrected) the construction of the invariant Lagrangian follows from building inside the two representation spaces $D_1$ and $D_{1/2} $ invariants from both the doublet and the triplet.

In the D1/2 representation we have as an invariant the term $ \psi ^T \psi$, where $\psi$ stands for the doublet and T for the dagger symbol.

My problems comes when I have to mix this term with the triplet $ \phi$ of the pions in the D1 repr. If I just take the term $\psi ^T \psi $ and put it as it is in the D1 rep and multiply it by $ \phi $ I surely don' t have an invariant( taking that the quantity \psi ^T \psi remains in the same form?).

What I have seen in practice is that one should take the term $\Phi = \tau _k \phi ^k $, and then construct the invariant as $L_{int} = \psi ^T \Phi \psi = \psi ^T \phi _k \tau ^k \psi .$ $\tau $ are the Pauli matrices in 2x2 form as generators of the group SU(2).

Question:

What I don't get is how this term is invariant by construction in both representation spaces. This is because the term $\Phi$ is a term that as I see it belongs, not to the D1 space but at the space of 2x2 matrices where the generators are in 2x2 form and work as basis vectors. What I think is troubling( and most probably for the wrong reasons?) is that this way my invariant doesn't seem to belong to the two representation spaces.

So, how is this prescription justified? Must one, for the construction of invariants in both spaces work on the space of the direct product of the two of them or must she find the irreducible representations and then construct invariant in there somehow? Or should one's work be based on the papers:

Papers -

  1. S. Coleman, J. Wess and B. Zumino, Phys. Rev. 177, 2239 (1969).

  2. C. G. Callan, S. Coleman, J. Wess and B. Zumino, Phys. Rev. 177, 2247 (1969).

Or in the end should one find how to represent the objects of the D1/2 representation to the D1 representation and then construct the invariant. It is obvious that somewhere in the road I got lost. Just please, if this is only a problem of finding the irreducible representations of the direct product of the D1 and D1/2 , clarify how then the interaction should be constructed and I will work on the problem.

Note: One can find such Lagrangians and further discussion in papers like

  1. http://www.sciencedirect.com/science/article/pii/0370157374900088

  2. http://www.sciencedirect.com/science/article/pii/S0370157387800029

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    $\begingroup$ Can anyone assume why the post got downvoted? If there is any point that' s not clear I would happily make comments or editions. Thanks. $\endgroup$ – Constantine Black Jan 13 '17 at 15:46
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Let's account for isospin through the dimensionalities of the multiplets, which best count states, so the isospinor fermion is a doublet, 2, and the isovector scalar is a triplet, 3, the adjoint.

So, since $ {\bf 2}\otimes {\bf 2}\otimes {\bf 3}= {\bf 5}\oplus {\bf 3}\oplus {\bf 3}\oplus {\bf 1}$, and you are only interested in the singlet of the Lagrangean interaction term, you have to see how it is ivariant.

The generic SU(2) isorotation in the defining (doublet, $D_{1/2}$) representation is $U=\exp{i{\boldsymbol \theta}\cdot {\boldsymbol \tau}}$, manifestly unitary, $U^\dagger U=1\!\!1$.

Under this generic rotation, $$ \psi \mapsto U\psi ,\\ \psi^\dagger \mapsto \psi^\dagger U^\dagger, \\ {\boldsymbol \phi}\cdot {\boldsymbol\tau} \mapsto U{\boldsymbol \phi}\cdot {\boldsymbol\tau} U^\dagger= {\boldsymbol \phi}\cdot (U{\boldsymbol \tau} U^\dagger)= ({\boldsymbol \phi} -2{\boldsymbol\theta}\times{\boldsymbol \phi+... })\cdot {\boldsymbol \tau}= \\ = {\boldsymbol \phi}\cdot \bigl ( \boldsymbol{\tau} ~ \cos (2\theta)+ \boldsymbol{ \hat{\theta} }\times \boldsymbol{\tau} ~\sin (2\theta)+ \boldsymbol{\hat{\theta}} ~ \boldsymbol{\hat{\theta}} \cdot \boldsymbol{\tau} ~ (1-\cos (2\theta))\bigr ) ~. $$ The last term in parenthesis specifies for you how the triplet of τ s (Pauli vector) transform into each other through the similarity transformation, i.e. how the triplet, $D_1$, adjoint indices transform in the symmetric (triplet) part of the $ {\bf 2}\otimes {\bf 2}.$ (This is but the celebrated [Rodrigues' rotation formula].) This procedure of rotating an adjoint vector by a double operation on the fundamental indices of the Pauli vector saturating it, instead, generalizes to all groups, so, in SU(3) of QCD, for example, where the 8x8 matrices would be huge and messy, but the 3x3 "Gell-Mann vector" is sparse and manageable.

So, manifestly, $$\psi^\dagger U^\dagger U{\boldsymbol \phi}\cdot {\boldsymbol\tau } U^\dagger U\psi = \psi^\dagger {\boldsymbol \phi}\cdot {\boldsymbol\tau} \psi ~,$$ an invariant, alright. It really is no different than composing angular momenta of two spinors with a vector and extracting the singlet.

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  • $\begingroup$ Thank you Professor, I appreciate your response and help. Minor questions: you write 2x2x3 and not 2x3 for the direct product of representations; is that because we have two iso-fermions in the scattering or because you are using both $\psi$ and $\psi$ dagger? $\endgroup$ – Constantine Black Jan 14 '17 at 10:39
  • $\begingroup$ Also, can we say that the transformation of $\phi \cdot t $ is a rotation of a vector by showing that the absolute value of $\phi $ remains the same; but in what vector space does this rotation takes place? Why does that rotation gives us the transformation of the adjoint vector in the fundamental representation( or have I misunderstood)? And just for being more precise: how and what are now the term in both representation spaces that are now invariant; or do we have the invariants now in another space? Again, thank you and if there are any references from where I could study please inform me. $\endgroup$ – Constantine Black Jan 14 '17 at 10:45
  • $\begingroup$ Two isospinors, each in its own 2d space, combine to a 3d isovector, $\psi^\dagger \tau^i \psi$, which is dotted to a fixed magnitude isovector $\phi^i$ to yield the isosinglet. All isorotations are in the internal isospin space, but are formally identical to space rotations. You learn about the rotation group in basic Quantum Mechanics books, where you must do all angular momentum exercises to understand basic group representation theory. $\endgroup$ – Cosmas Zachos Jan 14 '17 at 12:44
  • $\begingroup$ Can I say that the similarity transformation of $\tau _k $implies a change of basis in the space of the algebra since the algebra is isomorphic to the adjoint representation by definition of the last as a map to the algebra? So we change basis from the triplet of the φ to the basis of τ .Thank you. $\endgroup$ – Constantine Black Jan 15 '17 at 11:04
  • $\begingroup$ Up to a point, you might. But, this change of basis is a rotation, the inverse isorotation of φ you saw worked out. $\endgroup$ – Cosmas Zachos Jan 15 '17 at 14:34

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