For a Hamiltonian that includes a $𝜙^3$ term, one can make the Hamiltonian tend to negative infinity by letting 𝜙 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?
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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.
