# $SU(2)$ Invariant Lagrangian

Consider two arbitrary scalar multiplets $$\Phi$$ and $$\Psi$$ invariant under $$SU(2)\times U(1)$$. When writing the potential for this model, in addition to usual terms like $$\Phi^\dagger \Phi + (\Phi^\dagger \Phi)^2$$, I often see in the literature, less usual terms like: $$\Phi^\dagger T^a \Phi \ \Psi^\dagger t^a \Psi + \Phi^{c\dagger} T^a \Phi \ \Psi^\dagger t^a \Psi^c$$ where $$T^a$$ and $$t^a$$ are SU(2) generators in different representations, and charge conjugate representation is defined as $$\Phi^c \equiv C \Phi^*$$, with $$C$$ being the anti-symmetric charge conjugation matrix (like $$\epsilon$$ matrix in 2-dimensional representation).

See for an example eqn (1) in this paper.

I am wondering why the above terms are invariant under an $$SU(2)$$ transformation?

Any helps or comments would be appreciated.

The idea of the $$C$$ matrix is that if $$\Phi$$ transforms like $$\Phi\rightarrow U\Phi$$, where $$U$$ is some representation of SU(2), then also $$\Phi^c\rightarrow U\Phi^c$$. You can work it out yourself using the $$\epsilon$$ matrix for the case of the fundamental representation.

Now taking the complex conjugate we have $$\Phi^\dagger\rightarrow \Phi^\dagger U^{-1}$$ (similarly for $$\Phi^c$$), and so $$\Phi^\dagger \tau^a \Phi\rightarrow \Phi^\dagger\left( U^{-1}\tau^a U\right)\Phi=R(U)^a_{\,b}\Phi^\dagger\tau^b\Phi$$ where $$R(U)$$ is the adjoint representation of SU(2).

So the terms as you wrote them are not $$SU(2)$$ invariant, they are in the adjoint representation. But if you look in the paper they are contracted with another factor in the adjoint representation (with index $$a$$), and so the entire term with both factors is invariant.

• Thanks for your answer. Can you please a bit elaborate on why $U^{-1} t^a U=R(U)^a_b \ t^b$. – Ramtin Dec 20 '18 at 12:11

They are suppressing the SU(2) indices. If this is your first pass you have every right to be confused. I will be very explicit with my indices so you can see the underlying structure.

Theorem: The product of two representations of $$\phi, \psi$$ of SU(2) is invariant iff their indices are contracted with an invariant tensor (of the same representation) of SU(2).

That's a bit of a mouthful, but the following should clarify. There are two invariant tensors of the 2 rep of $$SU(2)$$ (1) $$\epsilon_{ab}$$ and (2) $$\delta_{\bar{a} b}$$. The bar denotes an index that transforms under the complex conjugate representation.

Now, let's write out your two vectors $$\Phi$$ , and $$\Phi^\dagger$$ explicitly with indices. The would look like

$$\Phi \sim \Phi_a \qquad \Phi^\dagger \sim \Phi^\dagger_\bar{a}$$

Here is the issue I have with the question: As the question is phrased, it is simply not true in general. As mentioned in the first paragraph under the "theorem", the only two invariant contractions of two SU(2) doublets are

$$\Phi^\dagger_\bar{a} \delta_{\bar{a} b} \Phi_b \sim \Phi^\dagger \Phi$$

and

$$\Phi_a\Phi_b \epsilon_{ab} \sim \Phi \Phi$$.

The latter is referred to as the singlet representation, where on the right I have written the "suppressed index" notation of each contraction.

Possible resolution: My guess is that the paper is assuming that $$\Phi$$ transforms in the Adjoint representation of SU(2) (also referred to as the 3). If this is the case, then indeed the contraction between the 3 and another 3 given by

$$\Phi^\dagger_{i} T^a_{ij} \Phi_j$$

where $$T^a$$ is a generator of the fundamental of SU(2), is indeed an invariant of the adjoint representation if SU(2). If you look at the gauge-covariant derivative of QCD this is exactly the same thing (except with SU(3)).

Conclusion: The two different invariants you are talking about are for two different representations of SU(2). Indeed, almost whenever you see $$\Phi^\dagger T^a \Phi$$ they are almost always talking about the adjoint of SU(3) (the 8 of SU(3)) and not the 3 of SU(2). But the point is that the two invariants you wrote correspond to two different representations of SU(2) (the fundamental and the adjoint) and so can never appear in the same lagrangian (unless you furnish your reps with both indices), you will only see them in different theories.