# Lagrangian for a $SU(2)$ doublet field

My professor asked me to find the lagrangian density for a complex scalar field and a lagrangian density for a field that is a doublet of $SU(2)$.

I started from the well established lagrangian density for a scalar field and I found for the complex scalar field:

$$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^\dagger)(\partial^\mu\phi)-m^2\phi^\dagger\phi - \lambda(\phi^\dagger\phi)^2$$

Its symmetry is $U(1)$ I believe.

I had taken only an introductory course on group theory so I know not much about $SU(2)$ and $SU(2)\times SU(2)$. So it is kind of cumbersome to write formal arguments to construct the Lagrangian for the $SU(2)$ doublet. Anyway, I tried and my efforts took me to consider the field $\phi_i= (\phi_1\phi_2)^T$

I know that are some arguments of isomorphism between $U(1)$ and $SU(2)$, so I just wonder if I could just start from the complex scalar lagrangian density and write:

first: $$(\partial_\mu\phi^\dagger_i)(\partial^\mu\phi_i) = (\partial_\mu\phi^\dagger_1)(\partial^\mu\phi_1) + (\partial_\mu\phi^\dagger_2)(\partial^\mu\phi_2)$$

second: $$\phi^\dagger_i\phi_i = \phi^\dagger_1\phi_1 + \phi^\dagger_2\phi_2$$

Third: $$(\phi^\dagger_i\phi_i)^2 = (\phi^\dagger_1\phi_1 + \phi^\dagger_2\phi_2)^2 = (\phi^\dagger_1\phi_1)^2 + (\phi^\dagger_2\phi_2)^2 + 2\phi^\dagger_1\phi_1\phi^\dagger_2\phi_2$$

So... I just put it all in the lagrangian above and I believe that the answer is right although I am not sure about the formal aspects of what I am really doing. My professor didn't asked for formality anyway, because we will start the formalism after, what I really want here is to know if I am reasoning fairly.

It is actually not so complicated. The key is that the single complex field is converted into a doublet and this is compactly represented by simply adding an index to the field $$\phi \rightarrow \phi^a .$$ Here $a$ is an index that denotes the two elements of the doublet.
To maintain the SU(2) symmetry, these doublets need to be contracted onto each other, similar to the way that the complex field is always multiplied with its complex conjugate to maintain the U(1) symmetry. Hence, $$\phi^{\dagger} \phi \rightarrow \phi^{a\dagger} \phi^a .$$ The reason why this maintains the symmetry is because of the way these fields transform $$\phi^a \rightarrow U^{ab}\phi^b ~~~~~~~ \phi^{a\dagger} \rightarrow\phi^{c\dagger} [U^{\dagger}]^{ca} ,$$ where $U$ and $U^{\dagger}$ are the unitary matrices that are elements of SU(2). Hence we have that $$[U^{\dagger}]^{ca} U^{ab} = \delta^{cb} .$$ It thus follows that $$\phi^{c\dagger} [U^{\dagger}]^{ca} U^{ab}\phi^b = \phi^{a\dagger} \phi^a .$$ By implication, one can convert the Lagrangian for a complex scalar field with U(1) symmetry into one with SU(2) symmetry, simply by adding indices to all the fields.