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 π a ( x ) ( a = 1 , 2 , 3) form an isospin triplet, I = 1, whose three charge states are given by π ± = 1 2( π 1 ∓ iπ 2 ) √ (1) π 0 = π 3 (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 /c 2 , while π 0 has a different mass m ( π 0 ) = 134 . 98 MeV /c 2 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 i 2 = 1 + i 2 π − 2 � � π ) 2 + . . . U ≡ exp ⃗ τ · ⃗ π ⃗ τ · ⃗ ( ⃗ (3) F 2 F π F π π √ 2( τ + π + + τ − π − ) + τ 3 π 0 ( ⃗ τ · ⃗ π ) = � 0 � 0 π ± ≡ 1 1 0 � � √ 2( π 1 ∓ iπ 2 ) , τ + ≡ τ − ≡ 0 0 1 0 where ⃗ τ = ( τ 1 , τ 2 , τ 3 ) 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- uu ⟩ ̸ = 0 and ⟨ ¯ zero quark condensates, ⟨ ¯ 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 → g L Ug † U − R , g R ∈ SU(2) R , g L ∈ SU(2) L , (4) where g R and g L 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 g R = g L (and g † L g L = 1), which represents the remaining symmetry, SU(2) V , or equivalently the isospin SU(2) symmetry. (Here we assume m u = m d so that the strong interaction is isospin invariant.) In order to express the Lagrangian in a compact way, we define L µ ≡ U † ( x ) ∂ µ U ( x ) = i 2 τ · ∂ µ ⃗ ⃗ π ( x ) + . . . (5) F π ∂ ∂ µ ≡ , ∂x µ where x = ( t, ⃗ x ) denotes the 4-dim. coordinates and µ = 0 , 1 , 2 , 3. We choose the metric tensor as g µν = 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 1 16 Tr[ L µ L µ ] + π 32 e 2 Tr[ L µ , L ν ] 2 (6) for the Skyrmion. Here F π 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 F π ≃ 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, m u ∼ 2 MeV, m d ∼ 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 Λ QCD ∼ 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 32 e 2 Tr[ L µ , L ν ] 2 + m 2 π F 2 1 Tr[ U + U † − 2] 16 Tr[ L µ L µ ] + π π (7) 16 P1 (a) Using the expansion Eq. (3), show that the 1st term of Eq. (7) is equivalent to (1 / 2) ∂ µ ⃗ π · ∂ µ ⃗ π . (b) Show that the 3rd term is to − (1 / 2) m 2 π ) 2 . π ( ⃗ 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, J R and J L , respectively for SU(2) R and SU(2) L transform. The axial-vector current is defined by J R − J L .) 1.2 Equation of Motion The Euler-Lagrange equation for the above Lagrangian is obtained from stability against vari- ation of U ( x ) → Ue iϕ , given, for the massless pion, by L µ → e − iϕ ( L µ + i∂ µ ϕ ) e iϕ (8) δ L = − F 2 8 Tr[ L µ i∂ µ ϕ ] + 1 π 8 e 2 Tr[[ L µ , L ν ][ L µ , i∂ ν ϕ ]] = F 2 8 Tr[ ∂ µ L µ iϕ ] − 1 π 8 e 2 Tr[ ∂ ν ([ L µ , L ν ] L µ ) iϕ ] = 0 (9) F 2 8 ∂ µ L µ − 1 π 8 e 2 ∂ ν ([ L µ , L ν ] L µ ) = 0 . (10) 1 The 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/c 2 ) and thus the N f = 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 N f = 3 case.

Skyrmion Lecture （ 2019 ） Oka 4 For the finite pion mass, we obtain F 2 8 e 2 ∂ ν ([ L µ , L ν ] L µ ) + m 2 π F 2 8 ∂ µ L µ − 1 π π ( U − U † ) = 0 (11) 16 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. 1 1 B µ ≡ 24 π 2 ϵ µναβ Tr[ L ν L α L β ] = − 24 π 2 ϵ µναβ Tr[ U † ∂ ν U∂ α U † ∂ β U ] (12) The corresponding conserved charge is assigned to the baryon number, 1 1 B 0 = − 24 π 2 ϵ ijk Tr[ U † ∇ i U ∇ j U † ∇ k U ] 24 π 2 ϵ ijk Tr[ L i L j L k ] = (13) Now, we use L † i = − L i and Tr[ L i ] = 0, then by using the notation, L i = iτ a L a i ( L a i is real, i, a = 1 , 2 , 3), we obtain 1 k = − 1 B 0 = − 12 π 2 ϵ ijk ϵ abc L a i L b j L c 2 π 2 det[ L a i ] (14) 1 k = − 1 � � d 3 x ϵ ijk ϵ abc L a i L b j L c d 3 x det[ L a B = − i ] (15) 12 π 2 2 π 2 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 S 3 as we can regard the infinity as a single point. Then x ) is regarded topologically as a mapping from S 3 to SU (2) ∼ S 3 . This mapping has non- U ( ⃗ trivial topology designated by the homotopy group, π 3 ( SU (2)) = π 3 ( S 3 ) = 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 ∂ µ B µ = 0. This is a relation satisfied regardless of the Equation of motion.

Skyrmion Lecture （ 2019 ） Oka 5 Energy for the static U 1.4 The total energy of a static U ( ⃗ x ) ( L 0 = 0) is given by E st [ U ] = − F 2 1 � � π d 3 x Tr[ L 2 d 3 x Tr[ L i , L j ] 2 i ] − 32 e 2 16 − m 2 π F 2 � d 3 x Tr[ U + U † − 2] π (16) 16 In the case of m π = 0, the lower limit of the energy is given in terms of the baryon number. By replacing the variable by the dimensionless one x → ( F π e ) x , E st [ U ] = F π 1 � i ) 2 + ( ϵ ijk ϵ abc L b � k ) 2 � d 3 x ( L a j L c e 8 = F π 1 � k ) 2 ± 2 ϵ ijk ϵ abc L a � � d 3 x ( L a i ∓ ϵ ijk ϵ abc L b j L c i L b j L c k e 8 ≥ F π 1 � = 3 π 2 | B | F π � � � d 3 x ϵ ijk ϵ abc L a i L b j L c � � (17) � � k e 4 e � where the equality may hold when L a i = ± ϵ ijk ϵ abc L b j L c k is satisfied. 2 Hedgehog Solution Hedgehog solution is a static (time-independent) solution of the equation of motion with the form of π a ( x ) = ( F π / 2)ˆ r a F ( r ), where ˆ r is the unit vector pointing outwards from the origin. Then we have rF ( r ) = cos F ( r ) + i⃗ τ · ˆ r ) = e i⃗ U ( ⃗ τ · ˆ r sin F ( r ) (18) s 2 r j ) sc r j F ′ + ( δ ij − ˆ L j i = ˆ r i ˆ r i ˆ r + ϵ ijk ˆ r k (19) r where s = sin F ( r ), c = cos F ( r ), and F ′ = dF ( r ) . dr Making a variable transform into dimensionless quantities, r → x ≡ ( F π e ) r , µ ≡ m π F π e , we obtain � 2 s 2 x 2 ( F ′ ) 2 − 2 s 3 c � 1 F ′′ + 2 xF ′ − 2 sc x 2 F ′′ + 2 sc − 1 � � 4 µ 2 s = 0 + (20) 4 x 2 x 4