# SU(2) Invariant Lagrangian

• A
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 the usual terms like ##\Phi^\dagger \Phi + (\Phi^\dagger \Phi)^2##, I often see in the literature, less usual terms like:
$$\Phi^\dagger T^a \Phi \ \Psi^\dagger t^a \Psi$$
where ##T^a## and ##t^a## are SU(2) generators in different representations.
See for an example eqn (4) in this paper

I am wondering why the above term is invariant under an SU(2) transformation?

Any helps or comments would be appreciated.

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PS: I was told that after a transformation of the form ##\Phi \to U \Phi## and similarly ##\Psi \to u \Psi##, we get:
$$\Phi ^\dagger T^a \Phi \to \Phi ^\dagger U^{-1} T^a U \Phi = ad(U)_b^a \ \Phi ^\dagger T^b \Phi$$
and similarly, for the term involving ##\Psi##. where ##ad(U)## is the adjoint representation of SU(2).

But I cannot work out how to eliminate the adjoint representations in the above expression to get back to the original term in the Lagrangian.

samalkhaiat
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 the usual terms like ##\Phi^\dagger \Phi + (\Phi^\dagger \Phi)^2##, I often see in the literature, less usual terms like:
$$\Phi^\dagger T^a \Phi \ \Psi^\dagger t^a \Psi$$
where ##T^a## and ##t^a## are SU(2) generators in different representations.
This is the scalar product of the two adjoint vectors

$$V_{a} \equiv \Psi^{\dagger}T_{a}\Psi, \ \ \mbox{and} \ \ v_{a} \equiv \psi^{\dagger}t_{a}\psi$$

Now, if $$\Psi \to U(\alpha) \Psi = e^{-i\alpha^{a}T_{a}}\Psi,$$$$\psi \to u(\alpha)\psi = e^{-i\alpha^{a}t_{a}}\psi,$$ then $$V_{a}v_{a} \to \Psi^{\dagger} \left( U^{-1}T_{a}U \right)\Psi \ \psi^{\dagger}\left( u^{-1}t_{a}u\right) \psi.$$ Now, if you expand $U^{-1}, U, u^{-1}$ and $u$ and use the algebra, you find $$U^{-1}(\alpha)T_{a}U(\alpha) = D_{ab}(\alpha)T_{b},$$$$u^{-1}(\alpha)t_{a}u(\alpha) = D_{ac}(\alpha)t_{c},$$ where $$D_{bc}(\alpha) = \left( e^{-i\alpha^{a}J_{a}}\right)_{bc} = D_{cb}(- \alpha),$$ and $$(J_{a})_{bc} = - i \epsilon_{abc}.$$ So, if you substitute these, you find $$V_{a}v_{a} \to D_{ab}(\alpha)D_{ac}(\alpha)V_{b}v_{c} = \left( D(- \alpha)D(\alpha)\right)_{bc} V_{b}v_{c} = \delta_{bc}V_{b}v_{c} = V_{c}v_{c}.$$

odietrich and Ramtin123