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Consider a theory with a $\phi^6$-scalar potential: $$ \mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-\phi^2(\phi^2-1)^2. $$

I solved its equation of motion but found that the general form of its solution, $$ \phi(x) = \frac{e^x}{\sqrt{e^{2x}-C_1}}-\frac{e^{-x}}{\sqrt{e^{-2x}-C_2}}, $$ is not a soliton. Why isn't this solution to the equation of motion a soliton?

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  • $\begingroup$ Why should it be a soliton? Also, I only see $\phi^4$, not $\phi^6$ in your Lagrangian. $\endgroup$ – ACuriousMind May 8 '15 at 18:44
  • $\begingroup$ I just want to understand why it is not. Plus I just fixed the Lagrangian for it to actually be a $\phi^6$ potential. $\endgroup$ – NSERC Protester May 8 '15 at 18:45
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Why do you say it is not a soliton solution?

I did not verify your answer but assuming its true, then it seems like a soliton to me.

Your vacuum consists of $\phi=0,\pm 1$. This solution obviously interpolates between the two vacua $\phi(+\infty)\rightarrow +1$ and $\phi(-\infty)\rightarrow -1$, and is therefore topologically stable. Moreover the region where potential energy is stored in this solution is localized.

Try to play with $C_1$ and $C_2$ to visually see the location and dimensions of the kink.

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  • $\begingroup$ More generally, one has $$ \mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-(\phi^2-\epsilon)(\phi^2-1)^2. $$ If $\epsilon\neq 0$ then it's not a soliton anymore $\endgroup$ – NSERC Protester May 9 '15 at 18:29
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Just to complement Ali Moh's answer:

We can define a topological current in the same way we do for the $\phi^4$ kink, $$J_\mu=C\epsilon_{\mu\nu}\partial^\nu\phi(t,x),$$ where $C$ is a normalization constant and $\epsilon_{01}= -1$. The topological charge then is $$Q=\int_{-\infty}^\infty J_t dx=C\int_{-\infty}^\infty\partial_x\phi dx=C\left[\phi(\infty,t)-\phi(-\infty,t)\right].$$ Using the solution given in the question (I am also assuming it is true) and setting $C=1/2$ we get $$Q=1.$$ Notice that the fact the solution interpolates two distinct vacua is enough for a conserved topological charge. Inded, $\partial_\mu J^\mu=0$ implies $$\frac{dQ}{dt}=\int_{-\infty}^\infty \partial_x J_x dx=J_x(\infty,t)-J_x(-\infty,t)=C\left[\frac{\partial\phi}{\partial t}(-\infty,t)-\frac{\partial\phi}{\partial t}(\infty,t)\right]=0,$$ since asymptotically $\phi\rightarrow\pm 1$ .

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