# Finding the vacuum which breaks a symmetry

I will start with an example. Consider a symmetry breaking pattern like $SU(4)\rightarrow Sp(4)$. We know that in $SU(4)$ there is the Standard Model (SM) symmetry $SU(2)_L\times U(1)_Y$ but depending on which vacuum we use to break this symmetry, in a case you can totally break the SM symmetry, with the vacuum : $$\Sigma_1 = \begin{pmatrix} 0& I_2 \\ -I_2 & 0 \end{pmatrix}$$ and in another case, these symmetry is preserved, with the vacuum $$\Sigma_2 = \begin{pmatrix} i\sigma_2& 0 \\ 0 & i\sigma_2 \end{pmatrix}$$ In the first case (with $\Sigma_1$), the generators corresponding to the SM symmetry are part of the broken generators so the SM symmetry is totally broken. In the second case ($\Sigma_2$), the SM generators are part of the unbroken generators then the SM symmetry is preserved. As you can read, I know the answers but not how to find them !

So, my questions are :

1. How is it possible in general (not only for the $SU(4)\rightarrow Sp(4)$ breaking pattern) to construct the vacuum that breaks the symmetry ?

2. Is it possible, when constructing the vacuum, to ensure that the vacuum will (or not) break a sub-symmetry like the SM symmetry in the previous example ?

• It's an interesting question. Maybe what you are looking for is the most general tensor that is invariant under the subgroup. The $Sp$ groups preserve precisely your $\Sigma_1$ and no other (Lin. Indep.) tensor. – Robin Ekman Apr 20 '14 at 13:51
• You're right for the case where you want to preserve the SM symmetry but not if you want to break this symmetry... – KoObO Apr 25 '14 at 14:24
• What I mean is that if you want to break $G$ to its subgroup $H$, any VEVs should be invariant tensors under $H$. E.g. when we break $SU(2)_L \times U(1)_Y$ to $U(1)_\text{EM}$, the Higgs VEV is an $SU(2)_L \times U(1)_Y$ tensor invariant under $U(1)_\text{EM}$ (i.e. it is electrically neutral). I haven't seen applications where determining the vacuum was important, just which symmetries were preserved by it. – Robin Ekman Apr 25 '14 at 17:07
• If a vacuum does not break the symmetry, then the unbroken generator must annihilate the vacuum. So the problem can be translated to finding null eigenvectors of the unbroken generators which are not null eigenvectors of the broken generators. I wonder if there is a representation theoretic argument from this point. – Siva Apr 26 '14 at 4:39
• Maybe It's a little above my head, but the potential you choose decides how the symmetry is broken. Are you asking how do you choose your potential? – kηives May 12 '15 at 1:14

2. Usually it is possible, but this is a question depending on the particular circumstances of G and H. If you are ambitious, you could run through the tables of Slansky's 1981 review article to reassure yourself. For your particular example above, the answer is "yes". SU(4) has 15 generators, Sp(4) has 10, and the SM has but 4. Preserving your symplectic metric $\Sigma_1$ preserves Sp(4), but you may check by writing down the generators that the alternative does not; nevertheless you may rewrite $\Sigma_2= 1\!\!\! 1 \otimes i\sigma_2$ with the 2x2 identity matrix on the left, preserving SU(2) of course; and the right group trivially preserves itself, a U(1), at the very least!
A scalar field transforming like a vector under SU(n) [or SO(n)] can break these symmetries down to SU(n-1) $\otimes$ U(1) [or SO(n-1)], since the group transformations in the subspace orthogonal to the direction of the VEV leave the vacuum invariant. There are many other examples in the literature.