Why Kink can not tunnel to vacuum, and is topologically stable

Click For Summary
SUMMARY

The discussion centers on the inability of the kink solution, represented by the function φ(x) = tanh(x/ξ), to tunnel into the vacuum states +v or -v due to an infinite energy barrier. The energy density is defined as E(x) = (gv^4/2)sech^4(x/ξ), which integrates to a finite value over all space. However, the transition from a soliton configuration to the vacuum results in infinite energy for any intermediate state, confirming the impossibility of tunneling. This conclusion is supported by the boundary conditions and the nature of spontaneous symmetry breaking.

PREREQUISITES
  • Understanding of soliton solutions in field theory
  • Familiarity with spontaneous symmetry breaking concepts
  • Knowledge of energy density calculations in field configurations
  • Basic proficiency in mathematical functions such as tanh and sech
NEXT STEPS
  • Study the implications of spontaneous symmetry breaking in quantum field theory
  • Explore soliton dynamics and stability in various field theories
  • Investigate the mathematical properties of energy density in field configurations
  • Learn about tunneling phenomena in quantum mechanics and field theory
USEFUL FOR

The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, soliton dynamics, and the study of spontaneous symmetry breaking phenomena.

PRB147
Messages
122
Reaction score
0
Why the kink (\phi(x)=tanh(\frac{x}{\xi})),
can not tunnel into vacuum +vor -v (Spontaneous symmetry breaking vacuum).

From the boundary condition(x\rightarrow \pm\infty, \phi(x)\rightarrow \pm v),
it is self-evident.

but the book states:
Due to the infinite high energy barrier, the kink can not tunnel into the vacuum.
where is the infinite high energy barrier?
The energy density is E(x)=\frac{gv^4}{2}sech^4 (\frac{x}{\xi}),
whose integration over all space is finite.

where is the infinite high energy barrier?
 
Physics news on Phys.org
Let's write down a possible field history in which a soliton at ##t=0## goes to the vacuum at ##t=1##:

\phi(x,t) = (1-t)v\tanh(x/\xi) + tv

The starting configurationi at ##t=0## and the ending configuration at ##t=1## both have finite total energy. But If you calculate the energy of the field configuration for any ##0 < t < 1## you should find that it is infinite. This infinite energy barrier should appear for any possible path that goes from a solition configuration to the vacuum.
 
  • Like
Likes   Reactions: DEvens and fzero
thank you The_Duck for your clear answer.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K