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

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  1. Aug 17, 2015 #1
    Why the kink ([tex]\phi(x)=tanh(\frac{x}{\xi})[/tex]),
    can not tunnel into vacuum [tex]+v [/tex]or [tex]-v[/tex] (Spontaneous symmetry breaking vacuum).

    From the boundary condition([tex]x\rightarrow \pm\infty, \phi(x)\rightarrow \pm v[/tex]),
    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 [tex]E(x)=\frac{gv^4}{2}sech^4 (\frac{x}{\xi})[/tex],
    whose integration over all space is finite.

    where is the infinite high energy barrier?
     
  2. jcsd
  3. Aug 17, 2015 #2
    Let's write down a possible field history in which a soliton at ##t=0## goes to the vacuum at ##t=1##:

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

    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.
     
  4. Aug 19, 2015 #3
    thank you The_Duck for your clear answer.
     
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