Domain walls intersection

Question

I was reading this article(On domain shapes and processes in supersymmetric theories). In the paragraph about domain walls intersection (paragraph $4$, page $7$) the authors say:

In a one-field theory it is known that an intersection of domain walls is unstable

They explain it as follows:

Indeed the configuration shown in Fig. 2a has translational zero modes, whose shape is given by the gradient of the field. Since there are directions in the plane, where the field approaches the same values at both innities, the component of the gradient in such direction necessarily has zero. Thus the zero mode cannot be the lowest in the spectrum and a negative mode exists, leading to a separation of the walls.

So, i have the theory of one scalar field in $2+1$ with mexican hat potential: $V=\frac{\lambda}{4}\left( \phi^{2}-v^{2}\right)^{2} $.

I consider configuration (domain walls intersection) with boundary conditions:

$$ \phi_{d}(+\infty, +\infty)=-v $$ $$ \phi_{d}(+\infty, -\infty)=v $$ $$ \phi_{d}(-\infty, +\infty)=v $$ $$ \phi_{d}(-\infty, -\infty)=-v $$ Here $\phi_{d}$ satisfies the stationary equation of motion (recall that $\dot{\phi}=0$):

$$ \Delta \phi_{d}-\frac{\partial V}{\partial \phi}(\phi_{d})=0. $$

I consider the small excitation over $\phi_d$: $\tilde{\phi}=\phi_{d}+\phi$. After linearizing, I obtain the following equation ($\phi=e^{i\omega t}f_{\omega}(x_1,x_2)$):

$$\left[-\Delta + \frac{\partial^{2}V}{\partial \phi^{2}}(\phi_{d})\right]f=\omega^{2}f \tag{1}$$.

I have found zero modes: its directional derivative : $f_{0}(x_{1},x_{2})=(\nabla\phi, \mathbf{n})$. It satisfies the equation $(1)$ with $\omega=0$.

Now I can consider the directions along "diagonals", at their ends the field takes the same value $v$ or $-v$ respectively (see figure 2 at page 10 in the file). So at this diagonals there is a point where $f_{0}=0$, as authors said.

But I cant understand the last sentence of their explanation: How the existance of such point (where $f_{0}=0$) implies that we also have negative modes? Can you explain me this? Thanks in advance.

PS. I will accept another approaches to show the existance of negative modes (and therefore instability of such configuration) too.

Answer 1

In one dimensional case you have the oscillation theorem: $n$-th level has $n$-zeroes. As a special case ground state has no zeroes. It doesn't generalize on the case of several dimensions however there is still a theorem that the ground state is non-degenerate and has no zeroes.

Thus the observation that the Goldstone mode vanishes somewhere means that it is not a ground state. Therefore there are some modes with lower (i.e. negative) eigenvalue, hence instability.

Answer 2

As I can't comment posts, let me make a few observations in this post. Hopefully they might be steps in the right direction.

1) If $f_0=0$, then $\tilde\phi=0$ meaning your perturbation is as big as $\phi_d$. For me this would invalidate the use of perturbation theory.

2) $\phi_d$ is supposedly smooth, however the boundary conditions impose it to vanish at least some points. Perturbation theory around those points is, at least, tricky (as far as I understand). Omitting this detail and looking at your equation (1), I think one could use this fact to show existence of negative values of $\omega^2$, i.e. instability.