151 Solitons Now we change subject completely and talk about solitons Solitons

151 solitons now we change subject completely and

This preview shows page 67 - 69 out of 125 pages.

15.1 Solitons Now we change subject completely and talk about solitons. Solitons are localized solutions of classical field equations. By localized we mean that if it starts at some finite region in space it stays there. Originally they were required to retain their form even after scattering. This is the soliton in mathematical sense occurring in the solution in nonlinear differential equations. Let’s start with some classical field theory in 1 + 1 dimensions. The Lagrangian is L = 1 2 ( μ ϕ ) 2 - V ( ϕ ) , V ( ϕ ) = - m 2 2 ϕ 2 + λ 4 ϕ 4 + λ 4 v 4 = λ 4 ( ϕ 2 - v 2 ) 2 (15.3) where we defined v = p m 2 . The minima of the potential are at v and - v . We know from symmetry breaking that there are two vacuum states and elementary excitation has mass 2 m . If you are in a vacuum state and are asked which vacuum you are in, there is no way to tell, because physics looks the same. The classical field equation is 2 ϕ ∂t 2 - 2 ϕ ∂x 2 + ∂V ∂ϕ = 0 (15.4) The static solution is just when the time derivative is zero, and it is equivalent to minimizing the potential energy U [ ϕ ( x )] = Z dx 1 2 ϕ 0 2 + V ( ϕ ) (15.5) 67
Image of page 67
Quantum Field Theory III Lecture 15 Now we need to look for a solution. It had better have finite energy and go to one vacuum or another at 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 hit a minimum, which must be different from either vacuum solution. So we should have a solution which connects two vacuua and we call this a “kink”. Now mathematicians will say the space of functions is not compact and we can’t use this argument, but it turns out that here it is fine. Let’s multiply the equation by ϕ 0 0 = - ϕ 0 ϕ 00 + ϕ 0 ∂V ∂ϕ = d dx - 1 2 ϕ 0 2 + V (15.6) So the term in brackets is independent of x , and we can shift it to be zero, which we already did when introducing v . So we have dx = ± 2 V , 2 V = dx (15.7) and we can integrate the equation. Let’s call the point when ϕ = 0 by x 0 , and integrate from there r 2 λ Z ϕ 0 0 v 2 - ϕ 2 = Z x x 0 dx 0 = x - x 0 (15.8) If we do the integral on the left then we will find that ϕ = v tanh m 2 ( x - x 0 ) (15.9) The size of the kink is on the order of 1 /m which can be seen from the equation directly. So ϕ 0 is on the order of m 2 v 2 m 4 . The classical energy of this solution is E cl = M cl 1 m m 4 λ m 3 λ = 2 2 3 m 3 λ = 2 2 3 m 2 λ ! m (15.10) The term in the final bracket is dimensionless in 1 + 1 dimensions. In the limit of weak coupling we have λ/m 2 1, so this mass, or energy, is much much larger than the elementary excitations 2 m . Now we can consider solutions of many kinks. But we need to have the above integral relation which tells us that we can only go to vacuum at x → ∞ . But we can have time-dependent solution, which are kinks moving around. When they collide a kink will dissipate with an antikink. The number of kinks minus the number of antikinks will be conserved and we can define the topological current J μ = 1 2 v
Image of page 68
Image of page 69

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture