WITTEN’S PROBLEMS (Spring Term), Set One – N^{o}. 1 (solution by D. Freed)

Problem:

(1) The strong interactions are described to good approximation by a theory which has a symmetry group G = SU(2)×SU(2) spontaneously broken down to a diagonal subgroup H ≅ SU(2). There are no massless particles except the Goldstone bosons, which, to the extent G is a valid symmetry, are exactly massless. The theory, to good approximation, has a G-invariant Lagrangian L.

However, G is not an exact symmetry of nature. It is broken by perturbations that transform in the (2,2) representation of G (that is, the tensor product of the two dimensional representations of the two SU(2)’s.) Among the operators of the G-invariant theory are two “multiplets” of operators O_{i} and Õ_{j}, i, j = 1 … 4, each transforming as (2,2). The Langrangian is, more accurately,

(1)  L̃ = L + ∑_{i=1}^{4} (ε_{i}O_{i} + ε̃_{i}Õ_{i}) .

The ε_{i} and ε̃_{j} are, of course, all real. (That there are precisely two possible multiplets O and Õ of perturbations is explained by the underlying gauge theory, but for the present exercise we simply assume it.)

For a generic choice of the ε_{i} and ε̃_{i} the exact symmetry group of the model is only H′ = U(1), and this is, in fact, the situation in nature.

Treat the effective theory to first order in ε and ε̃ (which are, in fact, in nature small and comparable to each other).

(a) Show that, in first order, you would expect the degeneracy of the vacuum to be lifted, and there to appear a unique vacuum with mass gap. In particular, you expect the H′ symmetry to be unbroken. (Abstractly, H′ can be embedded in H; the point that is slightly non-trivial is that the perturbation picks a vacuum, of the unperturbed theory, that is invariant under an H ≅ SU(2) that actually does contain H′. H could of course have been conjugated to a group that does not contain H′.) In particular, there is no exact symmetry that is spontaneously broken, and no exact Goldstone boson; the former Goldstone bosons all get mass.

(b) More surprisingly, show that to first order in ε and ε̃, the three Goldstone bosons – the one that is invariant under H′ and the two that transform in a non-trivial representation of H′ – all get equal masses. (This is also observed in nature, to high precision, the particles being respectively the neutral and charged pions.)

(c) The three Goldstone bosons will not have exactly equal masses. In what order in ε and ε̃ will the degeneracy be lifted and how would you describe this lifting in the effective Lagrangian?

Solution:

Since the only massless particles in the unbroken (ε = ε̃ = 0) theory are the Goldstone bosons, in the infrared it flows to a σ-model into the homogeneous space G/H. Topologically G/H is a 3-sphere. Now with ε, ε̃ nonzero we expect a perturbation of the σ-model Langrangian, and we are asked to contemplate first a perturbation linear in ε, ε̃. The most relevant perturbation involves no derivative of the field in the σ-model, so is simply a potential function V on G/H. Now G double covers SO_{4}, and we are given that the operators O, Õ in the problem transform as vectors under SO_{4}. Thus if we transform ε, ε̃ as vectors then the perturbed theory retains the full G invariance. For fixed ε, ε̃ the group G is broken to the subgroup which fixes ε, ε̃, which for generic vectors ε, ε̃ is isomorphic to SO_{2}. (This is called “H′” in the problem.) Now the potential V must be SO_{4}-invariant if we transform ε, ε̃, and it follows easily that it has the form

V_{ε,ε̃}(x) = (λε + λ̃ε̃) · x.

Here we write x ∈ G/H ≃ S^{3} ⊂ ℝ^{4}, the ‘·’ denotes the dot product in ℝ^{4}, and λ, λ̃ are real constants. For fixed ε, ε̃ the potential V_{ε,ε̃} is the restriction to S^{3} of a linear function in ℝ^{4}. Thus it has a unique minimum on S^{3}. This is the unique vacuum of the perturbed theory (question (a)). Furthermore, it is clear that the Hessian of V_{ε,ε̃} at the minimum is scalar; that is, its eigenvalues – the masses of the pseudo Goldstone bosons – are all equal (question (b)). Observe that the potential is invariant under an S0_{3} subgroup of SO_{4} (which fixes the vector λε + λ̃ε̃), an accidental symmetry of the infrared theory. (We only expected SO_{2} invariance.) Finally, a perturbation quadratic in ε, ε̃ — for example V_{ε,ε̃}(x) = (ε · x) (ε̃· x) — can break this SO_{3} invariance and so destroy the equality of the Goldstone boson masses (question (c)).

Spring Term QFT Problems, Set One – N^{o}. 2

(a) In this problem we work classically. Then the minimum occurs when the gauge fields are zero and ϕ = ϕ_{0} is a constant with |ϕ_{0}|^{2} = v^{2}. Locally we can identify the SU_{2} × U_{1} in the problem with U_{2}, which then acts on ϕ via the two dimensional representation. Fixing the gauge fields and ϕ breaks the gauge symmetry down to the global U_{1} gauge transformations which fix ϕ. (A nonzero vector in ℂ^{2} is fixed by U_{1} ⊂ U_{2}.)

(b), (c) We expand the action around the minimum up to quadratic terms. Write A, iB for the SU_{2} and U_{1} connections and set

A = ( \begin{matrix}iA_{3} & A_{1} + iA_{2} \\ −A_{1} + iA_{2} & −iA_{3} \end{matrix} ) .

Here A_{1}, A_{2}, A_{3}, B are all real. Choose ϕ_{0} = \binom{0}{v} and write

ϕ = \binom{0}{v} + \binom{ϕ_{1}}{ϕ_{2}} .

Then up to quadratic terms we have

|F_{B}|^{2} ∼ |dB|^{2},
|F_{A}|^{2} ∼ |dA_{1}|^{2} + |dA_{2}|^{2} + |dA_{3}|^{2},
|Dϕ|^{2} ∼ |( \begin{matrix}dϕ_{1} + (A_{1} + iA_{2})v \\ dϕ_{2} + iBv − iA_{3}v \end{matrix} )|^{2},
∼ |dϕ_{1}|^{2} + |dϕ_{2}|^{2} + v^{2}A_{1}^{2} + v_{2}A_{2}^{2} + v^{2}(B − A_{3})^{2},
\frac{λ}{8} (|ϕ|^{2} − v^{2}) ∼ \frac{λ}{8} |ϕ_{1}|^{2} + \frac{λ}{8} |ϕ_{2}|^{2} + 2v Re ϕ_{2}.

Now rescale A → e_{2}A, B → e_{1}B so that after introducing a gauge fixing term (as in Problem Set Six, N^{o}. 2 in the fall), the quadratic Langrangian is (putting a coefficient 1/2 in front of |Dϕ|^{2}):

L_{quad} = ∫ d_{x}^{4} (\frac{1}{2} |dϕ_{1}|^{2} + \frac{λ}{8} |ϕ_{1}|^{2}) + (\frac{1}{2} |dϕ_{2}|^{2} + 2v Re ϕ_{2} + \frac{λ}{8} |ϕ_{2}|^{2})
+ (\frac{1}{2}(ΔA_{1}, A_{1}) + \frac{1}{2} v^{2}e_{2}^{2}A_{1}^{2}) + (\frac{1}{2} (ΔA_{2}, A_{2}) + \frac{1}{2} v^{2}e_{2}^{2}A_{2}^{2})
+ (\frac{1}{2}(ΔA_{3}, A_{3}) + \frac{1}{2}(ΔB, B) + \frac{1}{2} v^{2} (e_{1}B − e_{2}A_{3})^{2}) .

From this we see that there are two massive complex scalars (ϕ_{1}, ϕ_{2}); two massive vectors (A_{1}, A_{2}) with mass ve_{2}; a massless vector and a massive vector with mass v\sqrt{e_{1}^{2} + e_{2}^{2}}. To prove the last statement observe that in L_{quad} the gauge fields A_{3}, B have a mass matrix with eigenvalues 0 and v\sqrt{e_{1}^{2} + e_{2}^{2}}.

The Lie algebra of the unbroken U_{1} is the subalgebra where A_{1} = A_{2} = 0 and e_{2}A_{3} = e_{1}B. Thus ϕ_{1} has charge 2, ϕ_{2} is uncharged, A_{3} and B are uncharged, and A_{1}, A_{2} have charge 2.

Note: If we had followed the normalization indicated in the problem, we would have replaced A, B by \frac{A}{2}, \frac{B}{2} , and then the charge of ϕ_{1}, A_{1}, A_{2} would be 1 rather than 2. However, we wanted to avoid the denominators in the computation.