Lecture notes on the Skyrmion Makoto Oka Advanced Science Research - - PDF document

Skyrmion Lecture（2019）Oka

1

Lecture notes on the Skyrmion

Makoto Oka Advanced Science Research Center (ASRC), Japan Atomic Energy Agency (JAEA) (September 6, 2019)

1 Lagrangian

1.1 Chiral symmetry and Skyrme Lagrangian

Skyrmion is a classical solution of the equation of motion for the non-linear pion field theory. The pion fields πa(x) (a = 1, 2, 3) form an isospin triplet, I = 1, whose three charge states are given by π± = 1 √ 2(π1 ∓ iπ2) (1) π0 = π3 (2) The π+ and π− are related by the charge conjugation (conjugation of particle and anti-particle) and have exactly the same mass, m(π±) = 139.57 MeV/c2, while π0 has a different mass m(π0) = 134.98 MeV/c2 due to the isospin symmetry breaking and electromagnetic interaction. We construct an effective Lagrangian in terms of the non-linear form of the pion fields defined by U ≡ exp

• i 2

Fπ ⃗ τ · ⃗ π

• = 1 + i 2

Fπ ⃗ τ · ⃗ π − 2 F 2

π

(⃗ π)2 + . . . (3) (⃗ τ · ⃗ π) = √ 2(τ+π+ + τ−π−) + τ3π0 π± ≡ 1 √ 2(π1 ∓ iπ2), τ+ ≡ 1

• τ− ≡

1

• where ⃗

τ = (τ1, τ2, τ3) are the Pauli matrices for the isospin. As the (⃗ τ ·⃗ π) is an Hermitian matrix, U = U(x) is a field in the form of a 2×2 Unitary matrix with unit determinant, det U = 1, a member of SU(2) group.

Skyrmion Lecture（2019）Oka

2 Assume (for a while) that the masses of the up and down quarks are exactly zero, then the QCD Lagrangian is invariant under the chiral SU(2)R×SU(2)L symmetry (chiral limit). On the other hand, the QCD dynamics breaks the chiral symmetry spontaneously, that is the QCD vacuum at low temperature and density breaks the chiral symmetry by acquiring non- zero quark condensates, ⟨¯ uu⟩ ̸= 0 and ⟨ ¯ dd⟩ ̸= 0. As a result, the pions emerges as massless Nambu-Goldstone bosons. In terms of the nonlinear representation of the pion field, Eq. (3), the chiral SU(2)R×SU(2)L transformation can be expressed as U − → gLUg†

R,

gR ∈ SU(2)R, gL ∈ SU(2)L, (4) where gR and gL are an SU(2) matrix for the right- or left-component of the quark field and they are independent transforms. One sees that the vacuum, ⃗ π = 0 or U = 1, is not invariant under the transformation (4). On the other hand, it is invariant for the transform satisfying gR = gL (and g†

LgL = 1), which represents the remaining symmetry, SU(2)V , or equivalently the

isospin SU(2) symmetry. (Here we assume mu = md so that the strong interaction is isospin invariant.) In order to express the Lagrangian in a compact way, we define Lµ ≡ U †(x)∂µU(x) = i 2 Fπ ⃗ τ · ∂µ⃗ π(x) + . . . (5) ∂µ ≡ ∂ ∂xµ , where x = (t, ⃗ x) denotes the 4-dim. coordinates and µ = 0, 1, 2, 3. We choose the metric tensor as gµν = diag(1, −1, −1, −1). (We omit the sum for repeated Greek or Roman indices.) Skyrme proposed a chiral invariant Lagrangian, L(x) = −F 2

π

16 Tr[LµLµ] + 1 32e2Tr[Lµ, Lν]2 (6) for the Skyrmion. Here Fπ is a constant determined by the weak decay rate of the pion through its coupling to the axial-vector current. Experimental value of the constant is Fπ ≃ 186MeV. On the other hand, e is a dimensionless constant and a free parameter in this model. It is easy to see that Eq. (6) is invariant under the global (independent of x) chiral SU(2)R×SU(2)L transform Eq. (4) (See P2.) In the real world, the u and d quarks are not massless, but have finite masses, mu ∼ 2 MeV, md ∼ 5 MeV, and therefore the chiral symmetry is not exact. As the masses of light quarks

Skyrmion Lecture（2019）Oka

3 are much smaller than the QCD scale parameter ΛQCD ∼ 200 MeV, the chiral symmetry remains valid approximately. The pions also have finite masses, which must be zero if the chiral symmetry is exact (chiral limit).1 In the effective Lagrangian for the nonlinear pion field, chiral symmetry breaking can be intro- duced as the mass of the pions, L(x) = −F 2

π

16 Tr[LµLµ] + 1 32e2Tr[Lµ, Lν]2 + m2

πF 2 π

16 Tr[U + U † − 2] (7) P1 (a) Using the expansion Eq. (3), show that the 1st term of Eq. (7) is equivalent to (1/2)∂µ⃗ π· ∂µ⃗ π. (b) Show that the 3rd term is to −(1/2)m2

π(⃗

π)2. P2 (a) Prove that Eq. (6) is invariant under the chiral transform Eq. (4). (b) Show that the third term of Eq. (7) is not chiral invariant. P3 (a) Calculate the Noether current for the chiral transform. (b) Calculate the 4-divergence of the axial-vector current for the Lagrangian Eq. (7). (Hint: We have two independent Noether current, JR and JL, respectively for SU(2)R and SU(2)L

• transform. The axial-vector current is defined by JR − JL.)

1.2 Equation of Motion

The Euler-Lagrange equation for the above Lagrangian is obtained from stability against vari- ation of U(x) → Ueiϕ, given, for the massless pion, by Lµ → e−iϕ(Lµ + i∂µϕ)eiϕ (8) δL = −F 2

π

8 Tr[Lµi∂µϕ] + 1 8e2Tr[[Lµ, Lν][Lµ, i∂νϕ]] = F 2

π

8 Tr[∂µLµiϕ] − 1 8e2Tr[∂ν([Lµ, Lν]Lµ)iϕ] = 0 (9) F 2

π

8 ∂µLµ − 1 8e2∂ν([Lµ, Lν]Lµ) = 0. (10)

1The third flavor, strangeness, can be included in the chiral symmetry, as SU(3)R×SU(3)L, but the mass of

the strange quark (∼ 100 MeV/c2) and thus the Nf = 3 chiral symmetry is broken significantly. Accordingly, the kaon, the counter part of the pion in the strange sector, has a larger mass. In this note, I do not discuss the Nf = 3 case.

Skyrmion Lecture（2019）Oka

4 For the finite pion mass, we obtain F 2

π

8 ∂µLµ − 1 8e2∂ν([Lµ, Lν]Lµ) + m2

πF 2 π

16 (U − U †) = 0 (11)

1.3 Baryon number

The Skyrme Lagrangian has a special topological quantity, assigned as the baryon number. One sees that the following current is conserved. Bµ ≡ 1 24π2ϵµναβTr[LνLαLβ] = − 1 24π2ϵµναβTr[U †∂νU∂αU †∂βU] (12) The corresponding conserved charge is assigned to the baryon number, B0 = − 1 24π2ϵijkTr[LiLjLk] = 1 24π2ϵijkTr[U †∇iU∇jU †∇kU] (13) Now, we use L†

i = −Li and Tr[Li] = 0, then by using the notation, Li = iτaLa i (La i is real,

i, a = 1, 2, 3), we obtain B0 = − 1 12π2ϵijkϵabcLa

i Lb jLc k = − 1

2π2det[La

i ]

(14) B = − 1 12π2

• d3x ϵijkϵabcLa

i Lb jLc k = − 1

2π2

• d3x det[La

i ]

(15) The most amazing fact is that the above B can take only integer values for any static (time- independent) U(⃗ x), which satisfies the condition U(|⃗ x| → ∞) = 1 at the infinity of |⃗ x|. For such a U(⃗ x), the Euclidean three-dimensional space can be topologically equivalent to the compactified 3-dimensional sphere S3 as we can regard the infinity as a single point. Then U(⃗ x) is regarded topologically as a mapping from S3 to SU(2) ∼ S3. This mapping has non- trivial topology designated by the homotopy group, π3(SU(2)) = π3(S3) = Z. B is the winding number of the mapping and takes integers. It signifies how many times the mapping covers the target space. For continuous mappings, the number of the covering cannot jump from an integer n to another integer n′ ̸= n when the mapping function is modified continuously. Thus the baryon number B is stable (invariant) under continuous variation of the pion fields. P4 Prove that ∂µBµ = 0. This is a relation satisfied regardless of the Equation of motion.

Skyrmion Lecture（2019）Oka

5

1.4 Energy for the static U

The total energy of a static U(⃗ x) (L0 = 0) is given by Est[U] = −F 2

π

16

• d3x Tr[L2

i ] −

1 32e2

• d3x Tr[Li, Lj]2

−m2

πF 2 π

16

• d3x Tr[U + U † − 2]

(16) In the case of mπ = 0, the lower limit of the energy is given in terms of the baryon number. By replacing the variable by the dimensionless one x → (Fπe)x, Est[U] = Fπ e 1 8

• d3x
• (La

i )2 + (ϵijkϵabcLb jLc k)2

= Fπ e 1 8

• d3x
• (La

i ∓ ϵijkϵabcLb jLc k)2 ± 2ϵijkϵabcLa i Lb jLc k

• ≥ Fπ

e 1 4

• d3x ϵijkϵabcLa

i Lb jLc k

• = 3π2 |B| Fπ

e (17) where the equality may hold when La

i = ±ϵijkϵabcLb jLc k is satisfied.

2 Hedgehog Solution

Hedgehog solution is a static (time-independent) solution of the equation of motion with the form of πa(x) = (Fπ/2)ˆ raF(r), where ˆ r is the unit vector pointing outwards from the origin. Then we have U(⃗ r) = ei⃗

τ·ˆ rF(r) = cos F(r) + i⃗

τ · ˆ r sin F(r) (18) Lj

i = ˆ

riˆ rjF ′ + (δij − ˆ riˆ rj)sc r + ϵijkˆ rk s2 r (19) where s = sin F(r), c = cos F(r), and F ′ = dF(r) dr . Making a variable transform into dimensionless quantities, r → x ≡ (Fπe)r, µ ≡ mπ Fπe, we

• btain

1 4

• F ′′ + 2

xF ′ − 2sc x2

• +

2s2

x2 F ′′ + 2sc x2 (F ′)2 − 2s3c x4

• − 1

4µ2s = 0 (20)

Skyrmion Lecture（2019）Oka

6 The hedgehog is a solution in which the direction of the isospin of the pion fields and the radial vector from the origin are entangled. This structure is protected by the topological invariance, and therefore the baryon number is conserved. P5 For Li ≡ U †∇iU = iτjLj

i, prove Eq. (19).

2.1 Energy and Baryon number

The energy of the hedgehog solution, that is the mass of the Skyrmion MB, is given by (for the case of mπ = 0) MB = EH = Fπ e 1 8(E2 + E4) (21) E2 =

• d3x
• (F ′)2 + 2s2

x2

• (22)

E4 =

• d3x 4s2

x2

• 2(F ′)2 + s2

x2

• (23)

In order to have the energy not diverging, we need boundary conditions, sin F(x = 0) = 0 and sin F(x → ∞) = 0, or equivalently, F(0) = nπ and F(∞) = 0 without loosing generality. If we scale F(x) into F(x/a), then the energy reads EH(a) = Fπ e 1 8

• E2a + E4

a

• .

(24) According to the variational principle, the solution should take the extremum at a = 1. ∂EH(a) ∂a

• a=1

= 0 − → E2 = E4 (25) This is the Virial theorem for the Skyrmion. One sees that the Hedgehog solution minimize the static energy. The E2 term represents the kinetic energy of the pion field that gives outward pressure, while the E4 term is a potential that tends to compress the system. Their balance forms the static solution. The baryon number of this solution is given by B0 = − 1 2π2 s2 x2F ′ (26)

Skyrmion Lecture（2019）Oka

7 B = − 1 2π2

• d3x sin2 F dF

dx = − 2 π

F(∞)

F(0)

sin2 F dF = 1 π[F(0) − F(∞)] = n (27) Thus we conclude that the boundary values of the profile function F determine the baryon number of the hedgehog solution. This observation is consistent with the topological nature of the baryon number, that is, a continuous variation of the profile function does not change the baryon number as the baryon number is fixed only by the end values of the function.

2.2 Boundary conditions

The finite energy solution of Eq. (20) with the baryon number B should satisfy the boundary conditions F(x)

x→0

− → Bπ − kx (28) F(x)

x→∞

− → α x2(1 + µx)e−µx (29) wher k and α are parameters determined so that these conditions are simultaneously satisfied.

• Eq. (29) shows that the hedgehog solurion behaves at large x as

π(x) = Fπ 2 F(x) ∝ e−µx x (30) and this form indicates that the baryon solution contains a pionic tale at large x. Suppose that two Skyrmions of B = 1 are placed with a distance r apart, then we can prove that the total energy of the system as a function of r is given by E(r) = 2MB + V (r) (31) V (r) = g2

πBB

4π e−mπr r (32) gπBB = 4π 3 MBFπα. (33) This shows that the Skyrmions are interacting with each other by a pion exchange Yukawa interaction at large distances. The gπBB designates the coupling constant of the pion to the Skyrmion.

Skyrmion Lecture（2019）Oka

8

3 Quantization of Rotating Skyrmion

3.1 Nc counting

Witten argued that the Skyrmion is the baryon in the large Nc → ∞ QCD. Its mass is O(Nc) and its size is O(1). In order for the Skyrmion to satisfy these relations, we need to have the energy scale Fπ e to be O(Nc), and the length scale 1 Fπe to be O(1), and thus Fπ ∼ O(

• Nc)

(34) e ∼ O(1/

• Nc)

(35) Under these counting conditions, the pion-pion interaction behaves ∝ 1/F 4

πe2 ∼ O(1/Nc), which

is consistent with the ’t Hooft large Nc counting for mesons.

3.2 Rotation of Hedgehog

Spin ⃗ S and isospin ⃗ I are not individually good quantum numbers, but are entangled in the hedgehog, as is seen from the form (⃗ τ · ˆ r). However, the grandspin, defined by ⃗ K = ⃗ S + ⃗ I, is

• conserved. The hedgehog is invariant (spherical) under the grandspin rotation, K = 0.

The Skyrme Lagrangian is invariant under the spatial and isospin rotations independently. Therefore, the hedgehog rotated in the coordinate space or isospin space or both independently is also a classical solution with the same energy. Choose an arbitrary A ∈ SU(2) and rotate the hedgehog UH(⃗ r) in the isospin space as U A(⃗ r) ≡ AUH(⃗ r)A†. One can prove that both the baryon number and energy are invariant: B[U A] = B[UH], E[U A] = E[UH]. A simple calculation shows us that A is equivalently a spatial rotation in the opposite direction. U A(⃗ r) = AUH(⃗ r)A† = A exp(iτiˆ riF(r))A† = exp(iAτiA†ˆ riF(r)) = exp(iτiR−1

ij ˆ

rjF(r)) (36) where R−1

ij ≡ 1 2Tr[τjAτiA†] denotes an SO(3) rotational matrix.

Skyrmion Lecture（2019）Oka

9

3.3 Quantization of Rotation

The above rotation forms a zero mode (excitation without energy excess), and can be quantized as a collective coordinate, which results in the low energy excitation modes of the Skyrmion. Introducing A(t) = a0(t) + i⃗ τ · ⃗ a(t) (a2

0 + ⃗

a2 = 1) to the Lagrangian and take the term with the second derivatives in time, we obtain Lt =

• d3xF 2

π

16 2Tr[−A†∂tAA†∂tA + UHA†∂tAU †

HA†∂tA + . . .

= J Tr[∂tA∂tA†] = 2J (˙ a2

0 + ˙

⃗ a

2)

(37) J ≡

• d3x

F 2

π

6 sin2 F + 2 3e2 sin2 F

• F ′2 + sin2 F

r2

• ,

(38) where J is the moment of inertia for the rotation. Correspondingly, the spin ⃗ J and the isospin ⃗ Ioperators are given by Ji = −iJ Tr[τiA†∂tA] = 2J Ki (39) Ii = −iJ Tr[τiA∂tA†] = −2J RijKj (40) Ki ≡ 1 2Tr[τiA†∂tA] = a0 ˙ ⃗ a − ˙ a0⃗ a + ⃗ a × ˙ ⃗ a (41) and the Lagrangian is shown as L = Lt − EH(static) = Lt − MB (42) Lt = 2J ⃗ K2 = 1 2J ⃗ J2 = 1 2J ⃗ I2 (43) This results in the energy spectrum of the rotating Skyrmion as E = MB + I(I + 1) 2J ; J = I (44) The large Nc counting rule for the moment of inertia, J ∼ F 2

π(Fπe)−3 ∼ O(Nc), indicates that

the rotational energy is of the order O(1/Nc), two orders down from MB. Thus the Skyrmion baryon spectrum consists of a rotational band, tower of I = J states, such as I = J = 0, 1, 2, . . .. However, we note that baryons need to have half-integer spins and isospins for SU(2) flavor. By extending the Skyrme Lagrangian to Nf = 3, namely for the SU(3) flavor

Skyrmion Lecture（2019）Oka

10 symmetry, it was proven that for odd Nc, the B = 1 Skyrmion should have a half-integer spin, i.e., J = 1/2, 3/2, 5/2, . . .. The lowest-mass non-strange baryon is the nucleon I = J = 1/2, while the first excited state corresponds to the ∆ (3,3) resonance with I = J = 3/2. Similarly, it is shown that the lowest J = 1/2 states belong to the SU(3) octet and the next excited states with J = 3/2 form a 10 representation.