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The solution just found represents a baryon in its ground energy state. We would like to quantize the solution and to describe also the excitations. The baryon is very heavy, its mass being given in first approximation as 8π²κ ∝ λNc. Thus a non relativistic description is perfectly suitable (this is a general feature of baryons on large Nc QCD obviously). The program we would like to follow is:

i) Introduce another modulus so far neglected: the gauge group orientation modulus (analogous to g in (3.2.31)).

ii) Write down the quantized solution allowing the moduli to be time dependent.

iii) Find the metric in the moduli space and the Lagrangian of the moduli.

iv) Write down and solve the Schr¨odinger equation.

The first point is achieved by a so called “wrong” gauge transformation. The same idea will be used in Section 6.3.3 in a little more detail, so we postpone the discussion. The transformation is

A₀ 7−→ A⁰₀ = 0 ,

A_M 7−→ A⁰_M = V AMV⁻¹− iV ∂_MV⁻¹ . (5.3.1) In this section primed quantities are computed after the transformation by V and unprimed quantities are computed in the classical solution. The function V (x) has a fixed boundary behaviour

z→±∞lim V(x) = a , (5.3.2)

where a is a SU (2) matrix corresponding to the gauge group orientation modulus.

Let us regard all moduli as time dependent quantities

X^M(t) = { ~X(t), Z(t)} , ρ(t) , a(t) . (5.3.3) This procedure is not completely harmless, we must be careful so that the equa-tions of motion remain satisfied. The only equation to be taken care of is the Gauss Law constraint

D_MF_0M + N_c

64π²κε_{M N P Q}Fb_{M N}F_{P Q} = 0 . (5.3.4) We want to solve it in the gauge A0 = 0; moreover in our solution the second term vanishes reducing the equation to DMF_0M = 0. The idea is to make use of

the arbitrariness of V to solve this constraint. Calling³ Φ = −iV⁻¹V˙ and using Thus Φ must cancel all the other terms. The most natural thing to do is to decompose Φ as a sum, each term taking care of one specific piece: Φ = ΦX +

But we have already solved these equations! In Section 3.2 we described the zero modes of the solution. In order to define a metric in terms of them we had to impose a gauge fixing DM(δαAN) = 0 and we used a function Λ to enforce this gauge condition. Here we have simply Λα = − ˙X^αΦα for α = X, ρ, SU (2). There is no need to redo the computation, the result is exactly the same

ΦX = − ˙X_NΛN = − ˙X^NA_N , Φρ = − ˙X_ρΛρ= 0 ,

Φ^a_{SU (2)} = Λ^a = f (ξ)gτ^a 2 g⁻¹ ,

(5.3.7)

where f (ξ) and g are defined in (5.2.2). We have 3 independent solutions for Φ^a, the correct one is obtained by fixing the boundary value

ΦSU (2) = −if (ξ)g(a⁻¹˙a)g⁻¹ (5.3.8) This ends points i) and ii). Point iii) is easy because we already have the metric of the moduli space

ds² = gαβdX^αdX^β = M₀

2 δ_{M N}dX^MdX^N + M0 dρ²+ ρ²δ_IJda^Ida^J

, (5.3.9) where M0 = 8π²κ is the leading order expression for the baryon mass. The potential (the full baryon mass MB) is given by

V(X^α) = M0

3The dot is a time derivative

The Lagrangian is readily written as the standard kinetic term in curved space

where we have switched to cartesian coordinates for (ρ, a) defining y^I as ρa^I. Finally point iv) is standard quantum mechanics, some terms are easier and some a little more complicated. First of all the Hamiltonian is easily computed as a Legendre transform

H = M0+ HX + HZ+ Hy , (5.3.12) where the single terms are

H_X = − 1

They respectively describe a free particle in 3 dimensions, a simple harmonic oscillator in 1 dimension and an harmonic oscillator in 4 dimensions with an extra centrifugal energy. The solution of the first two is immediate:

Ψ( ~X) = 1 The last one is not too complicated if one realizes that 1/ρ² is a centrifugal term, hence it is sufficient to switch to spherical coordinates in R⁴. The Laplacian decomposes as

The eigenstates of ∇²_S3 are the spherical harmonics, which are defined in S³ as

Y<sup>(`)</sup>= CI<sub>1</sub>···I`a^I1· · · a<sup>I`</sup> , C ∈ 1 · · · ` , (5.3.17) where CI₁···I` is a totally symmetric and traceless tensor (as the Young tableaux says).

The tensor C is determined in the following way: if one multiplies Y^{(`)</sup> by ρ<sup>`} the result is an harmonic function (∇²= 0). The eigenvalue equation for Y^(`) is

∇²_S3Y^{(`)</sup>= −`(` + 2)Y<sup>(`)}. (5.3.18) Here we will use these eigenfunctions to solve the problem, but the following isomorphism is very useful to gain a physical interpretation of the result:

SO(4) ∼= SU (2) ⊗ SU (2)/Z2. (5.3.19) The irreducible representation `, given by the single row tableaux, is mapped into the symmetric combination <sup>`</sup>₂,^`₂
. The two quantum numbers are interpreted as the spin and the isospin of the baryon. This picture enables us to describe only baryonic states with equal spin and isospin: this is consistent with the Skyrme model where this con-straint comes directly form the “hedgehog ansatz ” used to solve the equations.

Here we try to give a brief explanation of this physical interpretation. Recall the transformation of the Skyrmion described in Section 3.6: a left rotation is an isospin rotation while a right rotation is a global rotation. Using the isomorphism SO(4) ∼= SU (2) ⊗ SU (2) we can interpret the two factors as left and right rotations. In fact the isomorphism is manifest in the formalism of quaternions. Let h ∈ H be

h = a + ib + jc + kc . (5.3.20)

If we regard h as a four dimensional vector, a general SO(4) rotation can be parametrized by a couple of quaternions of modulus 1 q₁ and q₂

h → q₁h q₂. (5.3.21)

This is indeed a left/right rotation because the set of quaternions of modulus 1 is mapped in SU (2) by identifying

i = iσ³, j = iσ², k = iσ¹, 1 =12. (5.3.22) We can also understand the reason of the “/Z2”: multiplying q₁ and q₂ by −1 has no effect on (5.3.21).

The obvious ansatz for Ψ(y^I) is

Ψ(y^I) = R(ρ)Y^(`)(a^I) . (5.3.23) The centrifugal term modifies the angular momentum

`(` + 2) → `(` + 2) + N_c²M₀²

80π⁴κ² ≡ ˜`(˜`+ 2) . (5.3.24)

Thus, upon substituting ` → ˜` we end up with a regular harmonic oscillator in 4 dimensions in spherical coordinates. The solution is

R(ρ) = e^−M0√6ρ2ρ^`˜₁F₁

n_ρ, ˜`+ 2,p2/3M₀ρ²

, n_ρ ∈ N . (5.3.25) The function1F₁(a, b, c) is the Confluent Hypergeometric Function. The energies are We can also give some explicit examples of the angular dependence: in general the spherical harmonics Y^(`)(a^I) will be polynomials in the a^I. The degree must be odd if we want to describe baryons because, since they are fermions, the wavefunction must be parity odd:

Ψ(−y^I) = −Ψ(y^I) . (5.3.27)

In particular the states with spin s3 and isospin I3, |s3, I₃i are

|s₃, I₃i = | ↑, ↑i ∝ a1 + ia2 , | ↑, ↓i ∝ a₄+ ia3 ,

| ↓, ↑i ∝ a₄− ia₃ , | ↓, ↓i ∝ a₁− ia₂ . (5.3.28) In the general case one should apply these operators to the states to compute their quantum numbers It can be easily checked that they both form a SU (2) algebra, moreover they are related to the SO(4) generators of rotations MIJ in a simple way

M_ab = Ia+ Ja , M_4a= Ia− J_a. (5.3.30) A baryon is a state |B, si in the Hilbert space defined by this Hamiltonian. As said above s is the (iso)spin of the baryon, equal to `/2. The quantum numbers n_ρ and nZ describe excited baryons and/or resonances, the case nρ = nZ = 0 corresponds to the neutron (I3 =↓) and the proton (I3 =↑).