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?

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

1 Answer 1


As your text (for which there can be no substitute!) should stress, the SU(2) transformation mostly scrambles the spinor components of your complex doublet, whereas the U(1) of hyper charge simply alters their phase, in lockstep. So the sombrero potential you are looking at breaks both SU(2) and this U(1). (Indeed, you may think of SU(2)~O(3) and U(1)~O(2), but this sets you up for confusions, unless you are computationally adept.)

$$ \delta \Phi=-i(\beta+ \vec \alpha\cdot \vec \tau/2 )\Phi, $$ 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, $H^0$.)

Consider a one-generation lepton doublet to illustrate these statements and their consequences.


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