A The kinetic term of the Hamiltonian is not positive definite

[codebpr]

TL;DR Summary

Unable to comprehend why the kinetic term of the Hamiltonian constructed from the action of perturbative string motion is not positive definite

I am trying to reproduce the results from this paper. On page 10 of the paper, they have an equation:
$$ \frac{S}{T}=\int dt\sum _{n=0,1} (\dot{c_n}{}^2-c_n^2 \omega _n^2)+11.3 c_0^3+21.5 c_0 c_1^2+10.7 c_0 \dot{c_0}{}^2+3.32 c_0 \dot{c_1}{}^2+6.64 \dot{c_0} c_1 \dot{c_1} \tag{B12} $$
where they make a statement that the action is problematic because the kinetic term of the Hamiltonian constructed from this action is not positive definite in some regions within the trapping potential, so they apply a variable change to solve this problem.

I am unable to comprehend the reason for it and why the variable change was needed in the first place. Any clear explanation for my doubt would be truly beneficial!

 

[PeterDonis]

Moderator's note: Thread moved to Beyond the Standard Model forum.

 

[Demystifier]

Since you read a very specialized paper on string theory, I presume that you know some basics of field theory, in particular something about spontaneous symmetry breaking. To remind you, a simple field theory with spontaneous symmetry breaking is described by an action containing a potential of the form
$$V(\phi)=\mu^2\phi^2+\lambda \phi^4$$
where $\lambda>0$, but $\mu^2<0$. Since $\mu^2$ is negative, the "point" $\phi=0$ is a local maximum, not a local minimum, of the the potential. Hence the system is unstable around $\phi=0$, so $\phi$ is not a convenient variable for perturbative calculations with small field. Hence one finds a true mimimum value $v$ of the potential and introduces a new field variable
$$\phi'=\phi-v$$
which is naturally small near the minimum of the potential so is convenient for perturbative calculations. I presume that you already know all this, but this is just a quick reminder. If you are not familiar with this, take any QFT textbook and read a chapter on spontaneous symmetry breaking.

I think that what you have here in your string example is something very similar, except that now $\phi$ is called $c_0$ and $\mu^2$ is called $\omega_0^2$ which is negative (they say it's $\omega_0^2=-1.40$). Note that all the terms in the Lagrangian that don't contain the time derivatives are the $-V$, and by "kinetic term" the authors seem to mean all the terms which are quadratic in $c$'s or $\dot{c}$'s. The "kinetic term" in this sense is not positive definite because $\omega_0^2$ is negative. However, I'm also confused because I don't see how (B.14) has a positive kinetic term. I suspect that they made some error in their calculation.

 

[codebpr]

Thank you for the illuminating reply. I had to brush up on my field theory basics and I finally got to understand your point. Now I get why the kinetic term was not positive definite but I also had the same confusion as yours, on rechecking the calculations I found that instead of -7.57 in the modified action it should be +2.64 which makes the time evolution well-posed. Just a small doubt is still lingering in the back of my mind, why they chose that specific variable change? I was going through a related paper where they analyze the magnetic field on the chaotic dynamics, they also use the same variable change arguing that "in some regions of the potential the kinetic term is negative"(page10).