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15.1SolitonsNow we change subject completely and talk about solitons. Solitons are localized solutions of classical fieldequations. By localized we mean that if it starts at some finite region in space it stays there. Originallythey were required to retain their form even after scattering.This is the soliton in mathematical senseoccurring in the solution in nonlinear differential equations.Let’s start with some classical field theory in 1 + 1 dimensions. The Lagrangian isL=12(∂μϕ)2-V(ϕ),V(ϕ) =-m22ϕ2+λ4ϕ4+λ4v4=λ4(ϕ2-v2)2(15.3)where we definedv=pm2/λ. The minima of the potential are atvand-v. We know from symmetrybreaking that there are two vacuum states and elementary excitation has mass√2m.If you are in avacuum state and are asked which vacuum you are in, there is no way to tell, because physics looks thesame.The classical field equation is∂2ϕ∂t2-∂2ϕ∂x2+∂V∂ϕ= 0(15.4)The static solution is just when the time derivative is zero, and it is equivalent to minimizing the potentialenergyU[ϕ(x)] =Zdx12ϕ02+V(ϕ)(15.5)67
Quantum Field Theory IIILecture 15Now we need to look for a solution.It had better have finite energy and go to one vacuum or anotherat infinity. Let’s consider a configuration which connects one vacuum to the other at different infinities.Now if we start from this configuration and do a variation continuously to lower the energy, we will hita minimum, which must be different from either vacuum solution.So we should have a solution whichconnects two vacuua and we call this a “kink”. Now mathematicians will say the space of functions is notcompact and we can’t use this argument, but it turns out that here it is fine.Let’s multiply the equation byϕ00 =-ϕ0ϕ00+ϕ0∂V∂ϕ=ddx-12ϕ02+V(15.6)So the term in brackets is independent ofx, and we can shift it to be zero, which we already did whenintroducingv. So we havedϕdx=±√2V ,dϕ√2V=dx(15.7)and we can integrate the equation. Let’s call the point whenϕ= 0 byx0, and integrate from therer2λZϕ0dϕ0v2-ϕ2=Zxx0dx0=x-x0(15.8)If we do the integral on the left then we will find thatϕ=vtanhm√2(x-x0)(15.9)The size of the kink is on the order of 1/mwhich can be seen from the equation directly. Soϕ0is onthe order ofm2v2∼m4/λ. The classical energy of this solution isEcl=Mcl∼1mm4λ∼m3λ=2√23m3λ=2√23m2λ!m(15.10)The term in the final bracket is dimensionless in 1 + 1 dimensions. In the limit of weak coupling we haveλ/m21, so this mass, or energy, is much much larger than the elementary excitations√2m.Now we can consider solutions of many kinks. But we need to have the above integral relation whichtells us that we can only go to vacuum atx→ ∞. But we can have time-dependent solution, which arekinks moving around.When they collide a kink will dissipate with an antikink.The number of kinksminus the number of antikinks will be conserved and we can define the topological currentJμ=12v