# What's the difference between the global transformations of $SU(2)$ and $U(1)$?

I am studying the spontaneously broken global non-Abelian symmetry. Suppose we have an $$SU(2)$$ doublet of bosons $$\Phi = (\phi^+, \phi^0)^T$$, with Lagrangian density $$\mathcal L = (\partial_\mu\Phi^\dagger)(\partial^\mu\Phi)+\mu^2\Phi^\dagger\Phi-\frac{\lambda}{4}(\Phi^\dagger\Phi)^2$$

This theory has $$SU(2)\times U(1)$$ symmetry. For global $$SU(2)$$ transformations, we have $$\Phi\rightarrow \Phi' = \exp(-i\vec\alpha\cdot\vec\tau/2)\Phi$$ where $$\vec\alpha = (\alpha_1,\alpha_2, \alpha_3 )$$; While for global $$U(1)$$ symmetry, we have $$\Phi\rightarrow \Phi' = \exp(-i\beta)\Phi$$ My question is what's the difference between these two transformations? Is it right to say for $$SU(2)$$ the Lagrangian is invariant under 3-dimensional rotation, but for $$U(1)$$ I can imagine there is only one axis of rotation? After the spontaneous symmetry breaking, do we still have the transformation for the unbroken subgroup?

• The 3 is indeed the number of generators but Pauli matrices are not numbers. Apr 13, 2023 at 0:48
• @Connor Behan Ah I see. Thank you!!
– IGY
Apr 13, 2023 at 1:10

$$\delta \Phi=-i(\beta+ \vec \alpha\cdot \vec \tau/2 )\Phi,$$ so for $$\langle \Phi\rangle= (0,v)^T$$, you have $$\langle \delta \Phi\rangle= -i{v\over 2} \begin{pmatrix} \alpha_1-i\alpha_2\\ 2\beta-\alpha_3 \end{pmatrix} ,$$ which cannot vanish for nonvanishing real angles, unless $$\beta=2\alpha_3$$, as your instructor must have computed for you.
That is, a linear combination of $$T_3$$ of su(2) and Y of u(1) escapes spontaneous breaking, and amounts to the generator of unbroken EM, $$Q= T_3+Y/2$$, a different U(1): the net number of independent generators broken is 3, not 4! The unbroken Q transformation, then, is $$\Phi'=\exp(-i(\beta +\alpha_3/2))\Phi$$, with the angle parameter which dropped out of $$\langle \delta \Phi\rangle$$. cf. (Check it for the Higgs, $$H^0$$.)