How to find the conserved quantities of $\phi^4$ model?

Question

Consider \begin{equation}\label{1} \partial^2_t\phi-\partial^2_x\phi=\phi -\phi^3,\: \ (x,t) \in \Bbb{R}\times \Bbb{R} \tag{1} \end{equation} the $\phi^4$ model, often used in quantum field theory and other areas of physics. It can be found here, Ref. [1].

How to calculate the conserved quantities for this model?

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[1] Michał Kowalczyk, Yvan Martel and Claudio Muñoz: Kink dynamics in the $\phi^4$ model: Asymptotic stability for odd perturbations in the energy space, J. Amer. Math. Soc. 30 (2017), 769-798. doi:10.1090/jams/870

Answer 1

A Lagrangian density for the system in 1+1D spacetime is $$ {\cal L}~=~\frac{1}{2}(\partial_t\phi)^2 -\frac{1}{2}(\partial_x\phi)^2-{\cal V}, \qquad {\cal V}~= ~\frac{1}{4}\phi^4-\frac{1}{2}\phi^2. $$ We can now calculate the conserved quantities as Noether charges for the symmetries of the action functional. Poincare symmetry with Lie algebra $iso(1,1)$ is generated by 2 spacetime translations and 1 boost. The corresponding 3 Noether charges are energy, momentum & boost, $$\begin{align} H~=~& \int_{\mathbb{R}} \!\mathrm{d}x~{\cal H}, \qquad {\cal H}~=~\frac{1}{2}(\partial_t\phi)^2 +\frac{1}{2}(\partial_x\phi)^2+{\cal V},\cr P~=~& \int_{\mathbb{R}} \!\mathrm{d}x~{\cal P}, \qquad {\cal P}~=~\partial_t\phi~\partial_x\phi,\cr B~=~& \int_{\mathbb{R}} \!\mathrm{d}x~{\cal B}, \qquad {\cal B}~=~t{\cal P}-x{\cal H}. \end{align}$$

Answer 2

Define the energy $ E(t) = \frac12\! \int_{\Bbb R} {\phi_t}^2 + {\phi_x}^2 \,\text d x $ as usually done for wave equations. Thus, \begin{aligned} \frac{\text d}{\text d t}E(t) &= \int_{\Bbb R} {\phi_t}\phi_{tt} + {\phi_x}\phi_{xt} \,\text d x \\ &= \int_{\Bbb R} {\phi_t}(\phi_{xx} + \phi - \phi^3) + {\phi_x}\phi_{tx} \,\text d x \\ &= \int_{\Bbb R} ({\phi_t}\phi_{x})_x + {\phi_t} (\phi - \phi^3) \,\text d x \\ &= \int_{\Bbb R} \big(\tfrac12\phi^2 - \tfrac14\phi^4\big)_t \,\text d x \\ &= -\frac{\text d}{\text d t}D(t) \end{aligned} with $D(t) = \frac14\! \int_{\Bbb R} \phi^2(\phi^2-2) \,\text d x$. Therefore, the energy $E +D$ is conserved. Note that the energy found in the article is also conserved. Indeed, it is of the form $E+C$ with $C(t) = \frac14\! \int_{\Bbb R} (1-\phi^2)^2 \,\text d x$; the densities corresponding to $C$ and $D$ differ only by a constant.