I am trying to understand the mass acquisition of particles in the Standard Model based on the book 'Fundamentals of Neutrino Physics and Astrophysics' by C. Giunti, and several doubts have arisen regarding the Higgs Mechanism. In particular:

1. In the book, it is mentioned that to have a neutral vacuum, the Higgs VEV must be in the form of $$\langle\Phi\rangle=\frac{1}{\sqrt{2}}\binom{0}{v}, \quad v \equiv \sqrt{-\frac{\mu^2}{\lambda}} .$$ I understand that indeed only the uncharged part of the doublet can be non-zero, but how do we know it is $$1/\sqrt{2}~ v$$ and not any other value?

2. Then it states that the symmetry $$SU(2)_L\times U(1)_Y$$ is spontaneously broken by the VEV because \begin{aligned} & I_1\langle\Phi\rangle=\frac{\tau_1}{2}\langle\Phi\rangle=\frac{1}{2 \sqrt{2}}\binom{v}{0} \neq 0, \\ & I_2\langle\Phi\rangle=\frac{\tau_2}{2}\langle\Phi\rangle=-\frac{i}{2 \sqrt{2}}\binom{v}{0} \neq 0, \\ & I_3\langle\Phi\rangle=\frac{\tau_3}{2}\langle\Phi\rangle=-\frac{1}{2 \sqrt{2}}\binom{0}{v} \neq 0, \\ & Y\langle\Phi\rangle=\langle\Phi\rangle \neq 0, \end{aligned} but I fail to see the relationship between these, wouldn't this happen with any other doublet of the Higgs?"

• See e.g. Secs. 4.3.2-4.3.4 here.
– J.G.
Commented Apr 27 at 18:28

1. That is a definition of the value $$v$$. The $$\frac{1}{\sqrt{2}}$$ is merely a convention. You could also have picked other conventions. That is only a particular choice and a definition of $$v$$.

2. Other conventions for the vacuum expectation value would lead to the same results. This particular convention is not preferred in any way.

Remember: spontaneous symmetry breaking means you have a multitude of vacua, all of which are equally good. There is no preferred vacua among them. However, physically there must be one true vacuum, which is the one about which we expand to define particles and so on. We convention that this vacuum is such that we get the VEV you wrote down: $$\langle \Phi \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ v \end{pmatrix}.$$ This is merely a choice of coordinates in the "vacuum space", so to speak. It is similar to when you define spherical coordinates about the $$z$$ axis: there is nothing special about the $$z$$ axis, but you must make a choice, so you pick your coordinates so that they are convenient.

Nothing forbids you from picking other options, other conventions, other definitions of $$v$$, and so on. This choice is made by the author of the book you're following only because they find it convenient.

To digress a bit more about point 2, what is relevant is that all of the infinitely many vacua you could have chosen break the symmetry. Think about the Mexican hat potential: whenever you pick a minimum of the potential, you are breaking the rotational symmetry it once had. Breaking the symmetry is not a property of your particular choice of vacuum, but it must happen. After all, the symmetry of the theory is spontaneously broken.

• Maybe I should add I'm a bit rusty on SSB, so let me know if anything seems odd Commented Apr 27 at 19:14
1. You were presumably taught that the vacuum state must minimize the potential $$V(\Phi)= \lambda [(\Phi^\dagger \Phi-v^2/2)^2 -v^4/a ]$$ in its v.e.v., so this is the value your complex 2-vector (spinor) has been arranged/dialed to take here.

2. You have been taught that for any of the four components $$\phi^i$$ of $$\Phi$$, the ones for which $$\langle [Q,\phi^i]\rangle \neq 0$$ for some Q are Goldstone bosons for a SSB symmetry, but the ones for which $$\langle [Q,\phi^i]\rangle = 0$$ are not, so the Q's leaving $$\langle \Phi \rangle$$ invariant are not broken. You chose the only non-vanishing component of $$\langle \Phi \rangle$$ to be the real part of of the second component. So, you are checking that $$I_{1,2,3}$$ & Y are broken, but the electric charge, $$I_3-Y/2$$ isn't, as required.

• Recall the definition of SSB is $$e^{\theta_i I_i}\langle \Phi\rangle\neq \langle \Phi\rangle$$, that is, an isorotation of this particular v.e.v. does not act like the identity on it; and similarly for the hypercharge.

Had you aligned the v.e.v. in any other direction, you'd have effectively rotated your representation basis, and the action of these generators on all states of the theory would be counter-rotated accordingly.

• So, like $I_i <\Phi> \neq 0 \Rightarrow e^{I_i \theta_i} <\Phi> \neq 0$ the vacuum Higgs doublet is not invariant under the transformation, it is broken? Commented Apr 28 at 10:43
• Indeed, an SU(2) rotation does not leave the vacuum invariant, the very definition of SSB. So the r.h.side of your second formula here should be $\langle \Phi \rangle$, not 0. Commented Apr 28 at 10:47
• Excuse me, I am afraid I do no understand what you said Commented Apr 28 at 10:58
• I added an explanatory paragraph. For the v.e.v. of the Higgs field to be invariant under an isorotation, the exponent (generators) must act trivially on the vacuum, but here it does not! Commented Apr 28 at 11:02