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

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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?
 
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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.
 
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Likes DEvens and fzero
thank you The_Duck for your clear answer.
 
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