3.3 𝒩=1 pure Yang-Mills

3.3.1 Confinement and gaugino condensate

As an example of the application of what we learned in this section, let us consider the 𝒩=1 pure supersymmetric Yang-Mills theory with gauge group SU(N). The content of this section will not be used much in the rest of the lecture note.

This theory has just the vector multiplet, with the Lagrangian

L =d2𝜃 i 8π τtrWαWα + cc.,Wα = λα + Fαβ𝜃β + (3.3.1)

The one-loop running of the coupling is given by

E Eτ(E) = 3N, (3.3.2)

and therefore we define the dynamical scale Λ by the relation

Λ3N = e2πiτUV ΛUV 3N. (3.3.3)

We assign R-charge zero to the gauge field, and R-charge 1 to the gaugino λα. The phase rotation λα eiφλα is anomalous, and needs to be compensated by 𝜃 𝜃 + 2Nφ. The shift of 𝜃 by 2π is still a symmetry, therefore the discrete rotation

λα eπiNλα,𝜃 𝜃 + 2π (3.3.4)

is a symmetry generating 2N. Note that under this symmetry, Λ defined above has the transformation

Λ e2πi(3N)Λ. (3.3.5)

This theory is believed to confine, with nonzero gaugino condensate λαλα. What would be the value of this condensate? This should be of mass dimension 3 and of R-charge 2. The only candidate is

λαλα = cΛ3 (3.3.6)

for some constant c. The symmetry (3.3.5) acts in the same way on both sides by the multiplication by e2πiN. Assuming that the numerical constant c is non-zero, this 2N is further spontaneously broken to 2, generating N distinct solutions

λαλα = ce2πiNΛ3 (3.3.7)

where = 0, 1,,N 1. Unbroken 2 acts on the fermions by λα λα, which is a 360 rotation. This 2 symmetry is hard to break.

It is now generally believed that this theory has these N supersymmetric vacua and not more. For other gauge groups, the analysis proceeds in the same manner, by replacing N by the dual Coxeter number C(adj) of the gauge group under consideration. For example, we have N 2 vacua for the pure 𝒩=1 SO(N) gauge theory.

3.3.2 The theory in a box

It is instructive to recall another way to compute the number of vacua in the 𝒩=1 pure Yang-Mills theory with gauge group G, originally discussed in [39]. We put the system in a spatial box of size L × L × L with the periodic boundary condition in each direction. We keep the time direction as . By performing the Kaluza-Klein reduction along the three spatial directions, the system becomes supersymmetric quantum mechanics with infinite number of degrees of freedom.

The box still preserves the translation generators Pμ and the supertranslations Qα unbroken. We just use a linear combination 𝒬 of Qα and Qα, satisfying

H = P0 = {𝒬,𝒬}. (3.3.8)

We also have the fermion number operator (1)F such that

{(1)F,𝒬} = 0. (3.3.9)

Consider eigenstates of the Hamiltonian H, given by

H|E = E|E. (3.3.10)

In general, the multiplet structure under the algebra of 𝒬, 𝒬, H and (1)F is of the form

𝒬|E (𝒬𝒬𝒬𝒬)|E |E 𝒬|E (3.3.11)

involving four states. When 𝒬|E = 0 or 𝒬|E = 0, the multiplet only has two states. If 𝒬|E = 𝒬|E = 0, the multiplet has only one state, and E is automatically zero due to the equality

EEE = E|H|E = E|(𝒬𝒬 + 𝒬𝒬)|E = |𝒬|E|2 + |𝒬|E|2. (3.3.12)

We see that a bosonic state is always paired with a fermionic state unless E = 0.

This guarantees that the Witten index

treβH(1)F = trE=0(1)F (3.3.13)

is a robust quantity independent of the change in the size L of the box: when a perturbation makes a number of zero-energy states to non-zero energy E0, the states involved are necessarily composed of pairs of a fermionic state and a bosonic state. Thus it cannot change tr(1)F.

Therefore, we can compute the Witten index in the limit where the box size L is far smaller than the scale Λ1 set by the dynamics. Then the system is weakly coupled, and we can use perturbative analysis. To have almost zero energy, we need to have Fμν = 0 in the spatial directions, since magnetic fields contribute to the energy. Then the only low-energy degrees of freedom in the system are the holonomies

Ux,Uy,Uz SU(N), (3.3.14)

which commute with each other. Assuming that they can be simultaneously diagonalized, we have

Ux = diag(ei𝜃1x,,ei𝜃Nx), (3.3.15) Uy = diag(ei𝜃1y,,ei𝜃Ny), (3.3.16) Uz = diag(ei𝜃1z,,ei𝜃Nz). (3.3.17)

together with gaugino zero modes

λ1α = 1,,λNα = 1,λ1α = 2,,λNα = 2 (3.3.18)

with the condition that

i𝜃ix = i𝜃iy = i𝜃iz = 0, iλiα = 1 = iλiα = 2 = 0. (3.3.19)

The wavefunction of this truncated quantum system is given by a linear combination of states of the form

λi1α1λi2α2λiαψ(𝜃ix; 𝜃iy; 𝜃iz) (3.3.20)

which is invariant under the permutation acting on the index i = 1,N. To have zero energy, the wavefunction cannot have dependence on 𝜃ix,y,z anyway, since the derivatives with respect to them are the components of the electric field, and they contribute to the energy. Thus the only possible zero energy states are just invariant polynomials of λs. We find N states with the wavefunctions given by

1,S,S2,,SN 1 (3.3.21)

where S = iλiα = 1λiα = 2. They all have the same Grassmann parity, and contribute to the Witten index with the same sign. Thus we found N states in the limit of small box, too.

The construction so far, when applied to other groups, only gives 1 + rankG states. For example, let us consider for G = SO(N) for N > 4. Then the method explained so far only gives N2 + 1 states

1,S,S2,,SN2, (3.3.22)

and does not agree with C(adj) = N 2 when N 7. This conundrum was already pointed out in [39] and resolved later in the Appendix I of [40] by the same author.6 What was wrong was the assumption that three commuting matrices Ux,y,z can be simultaneously diagonalized as in (3.3.17). It is known that there is another component where they cannot be simultaneously diagonalized into the Cartan torus. For SO(7), an example is given by the triple

Ux(7) = diag(+ + + ), (3.3.23) Uy(7) = diag(+ + + ), (3.3.24) Uz(7) = diag( + + + ). (3.3.25)

These three matrices might look diagonal, but not in the same Cartan subgroup. This component adds one supersymmetric state. Then, in total, we have (72 + 1) + 1 = 5 = 7 2, reproducing C(adj).

For larger N, one can consider Ux,y,z given by the form

Ux = Ux(7) Ux,Uy = Uy(7) Uy,Uz = Uz(7) Uz, (3.3.26)

where Ux,y,z are in the Cartan subgroup of SO(N 7). Applying the analysis leading to (3.3.21) in both components, i.e. in the component where Ux,y,z are in the Cartan subgroup of SO(N), and in the component where Ux,y,z has the form (3.3.26), we find in total

(N2 + 1) + ((N 7)2 + 1) = N 2 (3.3.27)

zero-energy states, thus reproducing C(adj) states. This analysis has been extended to arbitrary gauge groups [4142].