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

[PRB147]

Why the kink ($$\phi(x)=tanh(\frac{x}{\xi})$$),
can not tunnel into vacuum $$+v$$or $$-v$$ (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?

 

[The_Duck]

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.

 

[PRB147]

thank you The_Duck for your clear answer.