# Stability and topological charge of kink (anti-kink) solutions (soliton)

I am reading the book << Gauge theory of elementary particle physics >>. In chapter 15, it presents a model having finite-energy solution. First, we have a $$1+1D$$ spacetime model $$\begin{equation} \mathcal{L}=\int dx [\frac{1}{2}(\partial_0 \phi)^2-\frac{1}{2}(\partial_x \phi)^2-V(\phi)] \end{equation}$$ where $$\begin{equation} V(\phi)=\frac{\lambda}{2}(\phi^2-a^2)^2,~~~a^2=\mu^2/\lambda. \end{equation}$$ Now we have four static (time-independent) solutions to equation of motion, and they will give finite energy: ground-state configuration: $$\pm a$$ and kink (anti-kink) solutions: $$\pm a \tanh (\mu x)$$. The so-called topological charge is given by $$\begin{equation} Q=\int_{-\infty}^{+\infty} \partial_x \phi dx=n(2a), \end{equation}$$ where $$n=0$$ for two ground states, $$n=\pm1$$ for kink (anti-kink).

The book claims there is no transition between kink (anti-kink) and ground states and they are stable. So how can I see there is no transition between kink $$a\tanh \mu x$$ and ground state $$+a$$? Is this because of time-independence?

Further, the book gives an intuitive explanation: converting kink to ground state needs penetrating barrier around $$\phi=0$$ and will take infinite amount of energy. How to prove this point mathematically?

Here is one approach:

1. Let us for simplicity only consider static (i.e. time-independent) configurations. The time-dependent case is left to the reader.

2. The stationary solutions are the 2 ground states, the kink and the antikink. (The 2 latter have a moduli parameter.)

3. The 2 ground states are obviously locally stable. That the kink and the antikink are locally stable follows from the Bogomol'nyi-Prasad-Sommerfield (BPS) rewriting, cf. e.g. this related Math.SE post.

4. To prove that there doesn't exists a continuous homotopy $$H:\mathbb{R}\times [0,1]\to\mathbb{R}$$ from 2 different stationary solutions $$H(x,\lambda\!=\!0)~=~\phi_0(x)\qquad\text{and}\qquad H(x,\lambda\!=\!1)~=~\phi_1(x),\tag{1}$$ such that $$\forall\lambda\in[0,1]:~~ H(\cdot,\lambda)\text{ has finite energy}.\tag{2}$$ we will make the additional technical assumption that $$\forall\lambda\in[0,1]:~~ \lim_{x\to\infty} H(x,\lambda)\qquad\text{and}\qquad \lim_{x\to-\infty} H(x,\lambda) \qquad\text{ exist}.\tag{3}$$ It is then not difficult to show that the limits must in fact be $$\pm a$$. This in turn clashes with the continuity of the homotopy $$H$$. $$\Box$$

• Thank you very much. 1. It seems $H(x,\lambda)$ must be the solution of equation of motion for any $\lambda$? 2. Otherwise we can construct arbitary homotopy, for instance, $(1-t)\phi_0+t\phi_1$? Jul 13, 2020 at 5:03
• I am a little confused. If the $H(x,\lambda)$ is not the solution of EOM, then why do we require it to have finite energy, i.e., eq(2)? Also in the last link, eq(6)? Jul 13, 2020 at 11:21