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

Why the kink $$\phi(x)=v\tanh(\frac{x}{\xi}) ,$$ 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 $$E(x)=\frac{gv^4}{2}{\rm sech}^4 (\frac{x}{\xi}),$$ whose integration over all space is finite.

Where is the infinite high energy barrier?

• There is information missing here: What theory are you considering, i.e. what is the Lagrangian, what is the $\xi$ in the kink, and how is $v$ determined? – ACuriousMind Aug 18 '15 at 0:56
• Which book states? – Qmechanic Aug 18 '15 at 17:52

Here we assume that OP's question asks about $\phi^4$-theory in 1+1D, where the lagrangian density reads

$$\tag{1} {\cal L}~=~\frac{1}{2}\dot\phi^2 -{\cal U}, \qquad {\cal U}~:=~ \frac{1}{2} \phi^{\prime 2} + {\cal V},\qquad \phi \in C^1(\mathbb{R}^2),$$

where the $\phi^4$-potential density

$$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$

has two minimum points at $\phi=\pm v$, i.e. a double-well. In eq. (1) the dot (prime) means differentiation wrt. $t$ ($x$), respectively. We rephrase OP's question as follows:

Prove that there don't exist finite energy homotopies $\phi:\mathbb{R}^2\to \mathbb{R}$ between the following 4 topological sectors: the kink, the antikink, and the two vacuum solutions $\phi=\pm v$.

Here the kink has limits $$\tag{3} \lim_{x\to \pm \infty}\phi(x)~=~\pm v,$$ and the antikink has limits $$\tag{4} \lim_{x\to \pm \infty}\phi(x)~=~\mp v.$$

Sketched indirect proof: Assume that a homotopy $\phi$ exists. To be concrete, say, between the kink and the left vacuum solution $\phi=-v$. So the homotopy $\phi$ has to change valley for positive $x$. Since this is Phys.SE rather than Math.SE, we are for simplicity going to assume that for arbitrary instants $t\in\mathbb{R}$, the limits

$$\tag{5} f_+(t)~:=~\lim_{x\to \infty}\phi(x,t) \quad\text{and}\quad f_-(t)~:=~\lim_{x\to -\infty}\phi(x,t) ,$$

exist. Then to have finite potential energy

$$\tag{6} V(t)~:=~ \int_{\mathbb{R}}\! dx~{\cal V}(\phi(x,t))~<~\infty,$$

it follows that the two functions $f_{\pm}(t)$ can only take the values $\pm v$. Intuitively $\phi$ is then topologically confined to the two potential valleys for sufficiently large $x$. It follows that there exists a sufficiently large constant $K$ such that $\forall x\geq K$ the function $t \mapsto \phi(x,t)$ cannot be continuous in $t$. Contradiction.

References:

1. S. Coleman, Aspects of Symmetry, 1985; Section 6.3.1.

2. R. Rajaraman, Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, 1987; Sections 2.3-2.4.

The energy density of the state $\pm v$ is going to be something like $\propto μ^4$, if you are using the basic $\varphi^4$ theory. While the energy of the domain wall is finite, the energy of the vacuum state is not, and so the transition to the vacuum state iver all space will be infinite.

• Thank you very much for the great answers, sorry for the delayed response, I just recovered from a serious illness. – Qft_Phys Dec 18 '16 at 6:17