For a Hamiltonian that includes a $^3$ term, one can make the Hamiltonian tend to negative infinity by letting
approach negative infinity (depending on the sign in front of the term), meaning that the Hamiltonian has no ground state. Is there a more general method to determine the lower bound of any Hamiltonian?
Is there a more general method to determine the lower bound of any Hamiltonian?
Generally, in quantum mechanics, you have to solve the time-independent Schrodinger equation: $$ \hat H |\psi_n\rangle = E_n |\psi_n\rangle\;, $$ and then the smallest $E_n$ is the greatest lower bound.
E.g., for the hydrogen atom, -13.6 eV is the greatest lower bound.
In classical theory, you can try to minimize the Hamiltonian as a function of its variables, via differential calculus (or by inspection).
E.g., for a classical harmonic oscillator, the classical Hamiltonian is $$ H = p^2/2m + kx^2/2\tag{1} $$ and $ \frac{\partial H}{\partial p} = 0 $ and $ \frac{\partial H}{\partial x} = 0 $ tells you that an extremum is $p=x=0$ at which point $H=0$, which is the greatest lower bound. But, of course, you could tell that just by looking at Eq. (1) since all the terms are manifestly non-negative (since $m$ and $k$ are positive).
In classical field theory, you can do the same
E.g., for a classical KG field: $$ H = \int d^3x \frac{1}{2}\left(\pi^2 + |\nabla \phi|^2 + m^2\phi^2\right) $$ the greatest lower bound is zero.
