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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 ($\phi=e^{i\omega t}f_{\omega}(x_1,x_2)$):

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

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.

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.

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

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

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.

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 ($\phi=e^{i\omega t}f_{\omega}(x_1,x_2)$):

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

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.

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

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

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.

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

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

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.