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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?

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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 $H$. $\Box$

See also e.g. this related Phys.SE post.

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  • $\begingroup$ 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, $(1-t)\phi_0+t\phi_1$? $\endgroup$
    – Sven2009
    Jul 13, 2020 at 5:03
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    $\begingroup$ 1. No. 2. It wouldn't satisfy condition (2). $\endgroup$
    – Qmechanic
    Jul 13, 2020 at 8:24
  • $\begingroup$ 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)? $\endgroup$
    – Sven2009
    Jul 13, 2020 at 11:21
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    $\begingroup$ Because the system cannot access infinite-energy configurations -- not even via quantum tunnelling. $\endgroup$
    – Qmechanic
    Jul 13, 2020 at 11:41

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