# Conserved current from the three $SU(2)$ transformations

We are asked to show that the following Lagrangian is invariant under the three $$SU(2)$$ transformations $$\Phi \rightarrow \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi}$$, where $$\Phi$$ is a doublet complex scalar field$$\Phi =( \phi_1 \phi_2)^{T}$$ The given Lagrangian is

$$\mathcal{L} = \partial_\mu\Phi^{\dagger}\partial_\mu\Phi-m^2\Phi^{\dagger}\Phi$$

I have re-written the Lagrangian as

$$\mathcal{L} = g^{\mu\nu}\partial_\mu{\phi_i}^\dagger\partial_\nu{\phi_i}-m^2\phi^{\dagger}_i\phi_i$$

Where $$i = 1,2$$.

I derived $$\delta\mathcal{L}=0$$ using the fact that $$\vec{\sigma}_j = \vec{\sigma}_j^\dagger$$. My problem is finding $$\delta\phi_i$$ and $$\delta\phi_i^\dagger$$. How can I expand the exponential? And also, will I get 3 conserved currents for each complex field?

How would I calculate $$\alpha_j(j^\mu)_i = \frac{\partial\mathcal{L}}{\partial({\partial_\mu\phi_i})}\delta\phi_i+\frac{\partial\mathcal{L}}{\partial({\partial_\mu\phi^\dagger_i})}\delta\phi^\dagger_i$$

Edit 1:

Attempting to calculate $$\delta\phi_i$$ I expand the exponential to linear order since the parameter $$\alpha_j \ll 1$$. Hence $$\delta\phi_i= I + i\alpha_j\sigma^j$$. Assuming this is true, how does that summation look like? Would each j have a different charge?

• That was very helpful! I have edited the question. – fielder Oct 14 '19 at 20:21
• I'm sorry, I shouldn't have used $\alpha$ for both the transformation parameter and the index of the fields. Does this edit make my question valid now ? – fielder Oct 14 '19 at 21:57
• Yes. ..."..."........... – Cosmas Zachos Oct 26 '19 at 10:34

You might be misunderstanding the notation. $$\delta \Phi = \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi} -\Phi +O(\alpha^2)= \frac{i}{2}\alpha_j\sigma^j \Phi,$$ so that $$\delta \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} = \frac{i}{2} \begin{pmatrix} \alpha_3&\alpha_1-i\alpha_2\\ \alpha_1+i\alpha_2&-\alpha_3 \end{pmatrix} \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} ,$$ which you may perform to get $$\delta\phi_j$$.
Now $$\frac{i}{2} \alpha_j ~ (\partial^\mu \Phi^\dagger \sigma^j \Phi - \Phi^\dagger \sigma^j \partial^\mu \Phi)$$ represents three currents, an SU(2) adjoint triplet, the coefficients of the three parameters $$\alpha_j$$, normally omitted. Did you compute them? Indeed, integrating their zero components produces three isocharges, closing into an su(2) Lie algebra.