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I was going through this paper on QCD chaos, where in Appendix B (page 10), for equation B12: $$\frac{\mathcal{S}}{\mathcal{T}}= \int \mathrm{d}t\sum _{n=0,1} \left(\dot{c}_n^2-c_n^2 \omega _n^2\right)+11.3c_0^3+21.5 c_0 c_1^2+10.7c_0 \dot{c}_0^2+3.32 c_0 \dot{c}_1^2+6.64\dot{c}_0 c_1 \dot{c}_1.\tag{B12}$$

The authors state that:

The action (B12) describes the motion of the string in a trapping potential with an unstable equilibrium point on its edge, as explained above. One problem of this action is that the kinetic term of the Hamiltonian constructed from this action is not positive definite in some regions within the trapping potential, hence the time evolution there is not well-posed.

My question is how to verify that kinetic term is not positive definite for some regions and why is it important?

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  • $\begingroup$ The verification could be done by look at the eigenvalues of the Hamiltonian, if the eigenvalues are all positive then the Hamiltonian is positive definite. As to why it matters, sometimes having systems with negative energy isn't really physically meaningful. $\endgroup$
    – Ali
    Commented Jun 2 at 15:00
  • $\begingroup$ Thanks for the reply. The first eigenvalue $\omega_0^2$ is negative(-1.40), but that is bound to happen since the system is unstable and shows chaos. Why does the author state that "kinetic term of Hamiltonian" explicitly? Does it carry any greater significance? $\endgroup$
    – codebpr
    Commented Jun 3 at 4:36

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