Consider a doublet of complex scalar fields $\mathbf{\Phi}$. The $SU(2)$ transformation is $\textrm{exp}\left(it^{i}\tau_{i}\right)$ with generators $\tau_{i}=\frac{\sigma_{i}}{2}$, whilst the $U(1)$ transformation is $\textrm{exp}\left(\frac{1}{2}i\alpha I\right)$. Hence we may write an $SU(2)\times U(1)$ transformation as $$\mathbf{\Phi}\rightarrow\textrm{exp}\left(\frac{1}{2}i\alpha I\right)\textrm{exp}\left(it^{i}\tau_{i}\right){\mathbf{\Phi}}\approx\left(I+\frac{1}{2}i\alpha I+it^{i}\tau_{i}\right)\mathbf{\Phi}$$ where the last equality applies in the infinitesimal case.

I have been looking at vacuum solutions $\mathbf{v}=\left(0,v_{0}\right)^{T}$ for Lagrangians that are invariant under this transformation. In order to determine the Goldstone modes, one needs to find the generators of the group that do not map the vacuum to zero. The result of this process is to find that $$\tau_{1}\mathbf{v}\neq0$$ $$\tau_{2}\mathbf{v}\neq0$$ $$(\tau_{3}-\frac{1}{2}I)\mathbf{v}\neq0$$ $$(\tau_{3}+\frac{1}{2}I)\mathbf{v}=0$$ and so there are three massless bosons.

However I am unable to understand why the generators of $SU(2)\times U(1)$ have been chosen as the specific combinations $\tau_{1}$, $\tau_{2}$, $(\tau_{3}-\frac{1}{2}I)$ and $(\tau_{3}+\frac{1}{2}I)$.

Edit (due to comment below): There is free choice of Lie algebra basis, but I don't see how that is an answer. In the case of $SU(2)$ symmetry breaking I have found that the number of broken generators is basis independent, and I would like to think that this would hold generally. So, surely there must be some additional constraint here.

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    $\begingroup$ I'm not sure what exactly you want to know - you give the reason in your question already, namely "one needs to find the generators of the group that do not map the vacuum to zero" (and therefore implicitly those that map it to zero, too, and that's what makes $\tau_3 + \frac{1}{2}I$ a unique choice. $\endgroup$ – ACuriousMind Apr 1 '17 at 18:53
  • $\begingroup$ @ACuriousMind This does clear up my confusion somewhat, since I now realize it's just a matter of choice rather than there being one possible set of generators. Suppose I chose the generators as $\tau_{1}$, $\tau_{2}$, $\tau_{3}$ and $\frac{1}{2}I$ instead though - it seems weird to me that whether or not you get massless excitations depends on the choice of Lie algebra basis... $\endgroup$ – klgklm Apr 1 '17 at 19:03
  • $\begingroup$ The choice of generators, $\tau_1$, $\tau_2$, $(\tau_3-\frac{1}{2})$, and $(\tau_3+\frac{1}{2})$ preserve the Hermitian property of the generators. This is typically useful when going to the unitary gauge. The ladder operators that you mentioned would lead to non-unitary transformation. $\endgroup$ – T-Ray Apr 1 '17 at 19:33
  • $\begingroup$ @T-Ray Wouldn't $(\tau_{1}-\frac{1}{2})$, $(\tau_{1}+\frac{1}{2})$, $\tau_{2}$ and $\tau_{3}$ also preserve this (but not have any Goldstone modes)? $\endgroup$ – klgklm Apr 1 '17 at 19:38
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    $\begingroup$ Your choices are Hermitian, sure. But typically the motivation is to find unbroken Hermitian generators. It appears that you can always break your broken and unbroken generators into broken generators. But I suspect that when you go to the unitary gauge and expand your potential, you'll find that the field corresponding to the $(\tau_3+\frac{1}{2})$ combination will still appear as a Goldstone mode in the Lagrangian. You could give this a try and let us know how it goes. $\endgroup$ – T-Ray Apr 1 '17 at 20:01

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