# Hamiltonian positive definite and vacuum state

For the self-interating $$\phi^4$$ Lagrangian density $$\mathcal L=\frac{1}{2}\partial_\mu\phi \partial^\mu\phi-\frac{m^2}{2}\phi^2-\frac{\lambda}{4!}\phi^4,$$ the corresponding Hamiltonian is $$H=\int d^3x (\frac{1}{2}\dot{\phi}^2+\frac{1}{2}\nabla\phi\cdot\nabla\phi+\frac{m^2}{2}\phi^2+\frac{\lambda}{4!}\phi^4).$$

I read that since every term in the Hamiltonian is postivie definite, the Hamiltonian will be bounded from below leading to a meaninful vacuum state.

Does this mean that there will be a ground vacuum state $$|0\rangle$$ with energy eigenvalue that is positive? Why does a positive Hamiltonian operator guarantee the existence of a ground state? Also, why does a positive Hamiltonian mean positive energy eigenvalues?

• Positive-definite means by definition that all eigenvalues are positive. Feb 20, 2021 at 10:06
• @NDewolf Given this Hamiltonian, how can we know that all its eigenvalues are positive? I can only see that the Hamiltonian when integrated is positive. Feb 22, 2021 at 6:08

The operator $$- \frac{d^2}{dx^2}$$ is positive but it does not have a (normalisable) ground state.