3
$\begingroup$

Consider the following theory in $1+1$ dimensions

$$\mathcal L = \frac12(\partial\phi)^2 - \frac\lambda4 (\phi^2 - v^2)^2 \,,$$

which exhibits a $\mathbb Z_2 = \{0,1\}$ symmetry, $\phi \to -\phi$, broken spontaneously by the classical vacua, $\phi = \pm v$. Naively, it appears to me that the vacuum manifold has four topologically distinct sectors, namely the two vacua above and the kink and the antikink solutions.

However, since the symmetry breaks like $\mathbb Z_2 \to \{1\}$ say, the vacuum manifold is given by $\frac{\mathbb Z_2}{\{1\}} = \mathbb Z_2$ which has two disjoint components. Therefore, there must be only two topologically distinct sectors. That means some of the above vacua can tunnel into each other.

Since the kink or the antikink cannot tunnel into the flat vacua $\phi = \pm v$, does that mean the kink and the antikink can tunnel into each other as do the flat vacua?

$\endgroup$
3
$\begingroup$

However, since the symmetry breaks like $\mathbb Z_2 \to \{1\}$ say, the vacuum manifold is given by $\frac{\mathbb Z_2}{\{1\}} = \mathbb Z_2$ which has two disjoint components. Therefore, there must be only two topologically distinct sectors.

This is missing an hypothesis. Say $G$ is the symmetry group; $G_0$ is the isotropy group of $X$, the set of all ground states.

Then $G/G_0$ is isomorphic to $X$ if $G$ acts transitively on X . In you example $G=\mathbb Z_2$ does not act transitively on $\{\text{kinks, other vacua }\}$ because there's no $g \in \mathbb Z_2$ that connects kinks to the other vacua. You can identify two sets that satisfy this requirement, namely $\{\text{kinks}\}$ and $\{\text{other vacua}\}$ which as expected are both comprised of 2 ($=\# \mathbb{Z_2}/ \mathbb{1} $)elements each.

edit:added the following.

proposition: Let $G$ be a group that acts transitively on $X$; $x\in X$ and $G_0$ the isotropy group of $x$. Then the map $$ \begin{align} \pi :& G/G_0 \to X \\ &gG_0 \mapsto gx \end{align}$$ is bijective.

proof: by the invariance of $x$ under $G_0$ we see that the map is well-defined, meaning it does not depend on the representative $g$ chosen for the class $gG_0$.

The map is injective: suppose that $\pi(g)=\pi(h)$. This implies $h^{-1}g \in G_0$, so $hG_0=gG_0$.

Surjectivity: let $x\neq y \in X $. By transitivity $\exists g \in G$ such that $gx=y$. Then $\pi(g)=y$

$\endgroup$
  • $\begingroup$ Why do you call $G_0$ the "isotropy" group? Do you mean the leftover symmetry group? Also, could you kindly point me to a reference for this hypothesis? Thanks. $\endgroup$ – Nanashi No Gombe Oct 9 '18 at 6:57
  • $\begingroup$ @NanashiNoGombe the name "isotropy" group is standard terminology as far as I know. Yes I mean the subgroup of G that leaves a vacuum invariant. Since I couldn't find a reference I added the proof to the post $\endgroup$ – tbt Oct 9 '18 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.