We will now turn our attention to where the CSL can be realized in QCD. Firstly, there is not enough energy to excite any particles if µB ≤ mN, where mN is the nucleon mass. In that case, we have a QCD vacuum with φ = 0. Consequently, by eq. (8.5), the energy density is zero. However, if we obtain the soliton energy in the CSL ground state by combining eqs. (8.23) and (8.30) we find
E
S = 4mπfπ2
k− 1 k
K(k)<0, (8.32)
where we have used that K(k) > 0 and 0 ≤ k < 1. As a result, the CSL ground state is energetically more favorable than the QCD vacuum when µB ≤ mN. In other words, nature will prefer the CSL over the QCD vacuum if µB ≤ mN and µBH ≥(µBH)CSL is fulfilled.
Finally, nuclear matter is a substance consisting of an infinite number of nucle-ons. It is distributed with uniform density over an infinite volume. Nucleons have baryon number one, causing nuclear matter to have energy mN −µB per baryon number. The soliton energy per baryon number in the CSL ground state is given by eqs. (8.23), (8.19) and (8.30), and reads
E
NB = 8πmπfπ2
H (k−1/k)K(k))<0. (8.33)
The CSL ground state is therefore energetically more favorable than nuclear matter whenµB ≈mN. The binding energy of nucleons reduces the total energy of nucleons
implying thatµBcan be slightly less thanmN. We conclude that the CSL is realized when the condition in eq. (8.31) is fulfilled.
In order to determine the validity of the derivative expansion in ChPT, we can obtain the momentum scale associated with the exact solution in eq. (8.14). This scale is given by the maximum value of
∂zφ(¯z) = 2mπ
k dn(¯z, k) = µH
8πfπ2E(k)dn(¯z, k), (8.34) where we have inserted the value ofk using eq. (8.30). This has the maximal value
pCSL = µH
8πfπ2. (8.35)
When µBH = (µBH)CSL, which satisfies eq. (8.31), we have a maximal value of pCSL = 2mπ ≈280 MeV 4πfπ ≈1156 MeV. (8.36) We can see that this satisfies the constraint on ChPT given in section 4.3 when mπ ≈ 140 MeV and fπ ≈92 MeV. The ChPT Lagrangian in eq. (8.2) is of second order in the derivative expansion, meaning that it includes all corrections up to the order (pCSL/4πfπ)2. Any higher order corrections can be neglected as long as pCSL 4πfπ.
Chapter 9
Excitation spectrum in a chiral soliton lattice
In the previous chapter, we observed how the CSL leads to a modulation of the background in the direction of the external magnetic field. We will in this chapter see how this gives rise to a phonon of the soliton lattice. The rest of this thesis will focus on a finite isospin density instead of a finite baryon density. This amounts to letting µB → µI/2 in the previous chapter. First, we will derive the dispersion relation of the phonon in a fixed external magnetic field. This was done for a CSL with finite baryon density in [23]. Next, we introduce dynamical electromagnetic fields, which alter the excitation spectrum in presence of the CSL background. The dispersion relations are obtained by solving the equations of motion linearized around the CSL background. Closed analytic solutions are found in the chiral limit and for pion-like excitations propagating along the direction of the external magnetic field. In other cases, we provide approximate solutions to the dispersion relations. We will use the same method as in [47] where the excitation spectrum is derived in presence of a baryon chemical potential. It is interesting to note that the system has three gapless modes in the absence of pion-photon coupling. These are the phonon of the soliton lattice and the two photon polarizations. However, turning on the coupling will give rise to two gapped excitations and only one gapless mode. This happens even though the Higgs mechanism for photons is absent and the fact that spontaneous breaking of spatial translations occurs.
9.1 Phonons
The dispersion relation of the phonon of the soliton lattice is derived from the Lagrangian in presence of a fixed external magnetic fieldH and an isospin chemical potential µI. The full Lagrangian has the form
L =LChPT+LWZW = fπ2 4
Tr(DµΣ†DµΣ) + 2m2πReTrΣ + µI
8π2H·∇φ, (9.1) where the ChPT Lagrangian is given by eq. (4.16). Furthermore, the ordinary derivative has been promoted to a covariant derivative reading
DµΣ≡∂µΣ−i[δµ0µII3−Qµ,Σ] = ∂µΣ− 1
2i(δµ0µI−Aµ) [τ3,Σ] =∂µΣ, (9.2)
whereQµ ≡Aµτ3/2. The commutator is zero since the CSL consists solely of neutral pions, implying that Σ =eiτ3φ. The last term of the Lagrangian is the Wess-Zumino-Witten term given by eq. (6.37). Following similar arguments as in sections 8.1 and 8.2 results in an identical equation of motion
∂z2φ =m2πsinφ, (9.3)
with the CSL solution cos
φ(¯z) 2
= sn(¯z, k) ⇔ φ(¯z) = 2am(¯z, k)−π, (9.4) where ¯z = zmkπ. We will in the following see how this solution gives rise to a phonon of the soliton lattice. First, we note that SSB leads to broken generators of the broken symmetry. If the Noether charge densities of the broken generators are linearly dependent, it will lead to redundant Goldstone bosons [48]. The Noether charge densities for translation T0i and rotation R0i are related by
R0i =εijkxjT0k. (9.5)
The CSL spontaneously breaks continuous translational symmetry down to a dis-crete one in the z-direction. In addition, it spontaneously breaks continuous ro-tational symmetry down to roro-tational symmetry around the z-axis. The Noether charge densities associated with the generators of these broken symmetries are re-lated through eq. (9.5). Consequently, we expect exactly one Goldstone boson of the CSL. This is the phonon of the soliton lattice. The dispersion relation of the phonon can be obtained by examining the linear perturbations of the CSL solution.
We start out by introducing linear perturbations, π(x), of the neutral pion field φ such thatφ →φ+π(x), where x is a four-vector. From eq. (9.3), we conclude that fluctuations with full spacetime dependence yield the linearized equation of motion
2+m2πcosφ
π = 0. (9.6)
Observing that the CSL background only depends on thez-coordinate, we can carry out a Fourier transform in x,y and t, resulting in
−∂z2¯+ 2k2sn2(¯z, k)
π = k2 m2π
m2π+ω2− p2x+p2y
π, (9.7)
where we used eq. (9.4) and the fact that cos(2am(¯z, k)) = 1−2sn2(¯z, k). The left-hand side is known as the Lam´e operator with n = 1 [49]. The general form of the Lam´e equation is
−∂z2¯+n(n+ 1)k2sn2(¯z, k)
ψ(¯z) =Aψ(¯z). (9.8) Having n = 1 gives the solution
ψ(¯z) = H(¯z+σ, k)
Θ(¯z, k) e−¯zZ(σ,k), (9.9) where H, Θ and Z are the Jacobi’s eta, theta and zeta functions, respectively.
Moreover, σ is a complex parameter related to A by
A= 1 +k2cn2(σ, k). (9.10)
We can cast eq. (9.9) into a Bloch form reading The two first factors are together periodic in ¯z with period 2K(k), meaning that the period of z is 2kK(k)/mπ = `. The last factor has the form eipzz = eik¯zpz/mπ. Hence, the crystal momentum associated with the phonon of the soliton lattice is
pz = mπ
The solution in eq. (9.11) is only physically acceptable when it is bounded. Hence, Z(σ, k) has to be purely imaginary. This happens for Re(σ) = 0 or Re(σ) =K(k).
The Lam´e equation with n = 1 thus has two solutions corresponding to the two bands
σv=K(k) +iκ,
σc=iκ, (9.13)
where σv corresponds to the ”valence band” with the phonon excitation, while σc corresponds to the ”conduction band”.1 The next step in finding the dispersion relation of the phonon is to identifyA in eq. (9.8). We start out by observing that the minimum of the valence band appears atκ =K(k0), where k0 ≡√
1−k2 is the complementary elliptic modulus. We shift κ to the minimum by δκ ≡ κ−K(k0).
This allows us to show that
cn(σv, k) =−ik0
k cn(δκ, k0), (9.14)
where we have exploited the periodicity of the Jacobi elliptic functions given by eq.
(16.8.2) in [43]. Thus, eq. (9.10) becomes
A= 1−k02cn2(δκ, k0). (9.15) A purely longitudinal motion where px =py = 0 will from eqs. (9.7) and (9.8) give
ω band. A relation between κ and the crystal momentum pz is found by exploiting the identity
dZ(σ, k)
dσ = dn2(σ, k)− E(k)
K(k). (9.17)
Again, we rewrite the Jacobi elliptic function at the valence band in terms of δκ, yielding
dn(σv, k) =−k0sn(δκ, k0), (9.18) which vanishes to all orders whenδκ= 0. Combining eqs. (9.12) and (9.17) provides the relation
dpz
dκ = mπE(k)
kK(k) (9.19)
1A study of the band structure resulting from the Lam´e equation can be found in [50].
at the bottom of the valence band. Keeping a purely longitudinal motion and using eqs. (9.16) and (9.19) gives the phonon group velocity
cph = dω
dpz = dω/dκ dpz/dκ =√
1−k2K(k)
E(k). (9.20)
The group velocity rapidly increases towards the speed of light as the magnetic field increases above the critical magnetic field for the formation of the CSL. Restor-ing transverse motion on the CSL background gives by eq. (9.7) the full phonon dispersion relation where the crystal momentum pz is measured from the bottom of the valence band.
The neutral pion fluctuations of the CSL are represented by this phonon.2