2.2 Vacua

The combined system of the vector multiplet and the hypermultiplets has the Lagrangian which is the sum of (2.1.11) and (2.1.14). The supersymmetric vacua are given by the following conditions.

First, the variation of the D auxiliary fields gives

1 g2[Φ, Φ] + (QiQi Q̃iQ̃i)traceless = 0, (2.2.1)

where X|traceless for an N × N matrix is defined by

X|traceless = X 1 NtrX. (2.2.2)

We use the convention that a scalar is multiplied by a unit matrix when necessary.

Second, the variation of the F auxiliary field of Φ gives

QiQ̃itraceless = 0 (2.2.3)

and the F auxiliary fields of Qi, Q̃i give

ΦQi + μijQj = 0,Q̃iΦ + μjiQ̃j = 0 (2.2.4)

for all i. The total scalar potential is a weighted sum of absolute values squared of (2.2.1), (2.2.3) and (2.2.4).

So far we only used the supersymmetry condition with respect to the 𝒩=1 supersymmetry manifest in the superfield notation. By massaging the cross terms between the first term and the second term of (2.2.1) and combining them with the squares of (2.2.4), we can re-write the total scalar potential as a weighted sum of the following objects. First, we have one term purely of Φ:

[Φ, Φ] = 0. (2.2.5)

Second, we have terms purely of Q and Q̃: one is

(QiQi Q̃iQ̃i)traceless = 0 (2.2.6)

and another is (2.2.3). Finally, we have terms mixing Φ and Q, which are (2.2.4) together with

ΦQi + μijQj = 0,Q̃iΦ + μjiQ̃j = 0. (2.2.7)

Note that (2.2.5) and (2.2.6) are the SU(2)R singlet and triplet parts of the equation (2.2.1), respectively. Furthermore, the equation (2.2.6) together with the real and the imaginary parts of the equation (2.2.3) form the triplet of SU(2)R. Finally, the equations (2.2.4) and (2.2.7) transform as a doublet of SU(2)R.

Let us summarize. We first demanded that one 𝒩=1 sub-supersymmetry is unbroken in (2.2.1), (2.2.3) and (2.2.4). We found the equations satisfied are automatically SU(2)R invariant, and therefore we see that all the 𝒩=2 supersymmetry is automatically unbroken.

One easy way to have supersymmetry is to demand (2.2.5) and set Q = Q̃ = 0. This subspace of the supersymmetric vacuum moduli is called the Coulomb branch, since there usually remain a number of Abelian gauge fields in the infrared.

Another extreme is just to demand (2.2.6) and (2.2.3), and set Φ = 0. This is called the Higgs branch. Some people in the field reserve the word the Higgs branch for the branch where the gauge group is completely broken, but theoretically the Higgs branch as defined here behaves more uniformly under various operations.

The branches with when both the hypermultiplet scalars Q, Q̃ and the vector multiplet scalars Φ are nonzero are called the mixed branches.

From (2.2.5) we see that Φ can be diagonalized in the supersymmetric vacua. For definiteness let G = SU(2). Then Φ = diag(a,a). When a0 this breaks the gauge group to U(1). As there is a Coulomb field remaining in the infrared, these vacua are called the Coulomb branch. Let us compute the mass of the resulting W-bosons. From

1 g2tr|DμΦ|2 = 1 g2tr(μΦ + [Aμ, Φ])2 (2.2.8)

we have a term

1 g2tr[Aμ,Φ]2 (2.2.9)

in the Lagrangian, which gives a mass to the vector field. Writing

Aμ = A0W+W A 0 μ, (2.2.10)

we find

0Wμ+ 0 0 , a 0 0 a = 2a 0Wμ+ 0 0 . (2.2.11)

The kinetic term in our convention is trFμνFμν(2g2), and therefore this gives the mass

MW = |2a|. (2.2.12)

The mass terms of the fields Qi, Q̃i for fixed i are

Q̃iΦQi+μiQ̃iQi = (Q̃1i,Q̃2i) a + μi 0 0 a + μi Q1i Q2i . (2.2.13)

Therefore we have

MQi,1 = |a + μ|,MQi,2 = |a + μ|. (2.2.14)

We studied the classical mass of the monopole in this model in (1.3.12) when 𝜃 = 0. In general, this is given by

Mmonopole = |2τa|. (2.2.15)

Classically, there is a general inequality for the mass of a particle

M |na + m(2τa) + ifiμi| (2.2.16)

where n, m, fi are the electric, magnetic and flavor charges of the particle. Here the i-th flavor charges are associated to the symmetry

Qi eiφiQi,Q̃i eiφiQ̃i. (2.2.17)

This inequality, called the Bogomolnyi-Prasad-Sommerfield (BPS) bound, persists in the quantum system, once quantum corrections are taken into account to a and 2τa. Let us study this point next.