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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?

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    $\begingroup$ Positive-definite means by definition that all eigenvalues are positive. $\endgroup$
    – NDewolf
    Feb 20, 2021 at 10:06
  • $\begingroup$ @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. $\endgroup$
    – TaeNyFan
    Feb 22, 2021 at 6:08

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

In QFT, even the free Hamiltonian is not a well defined operator unless it is Wick ordered. It then has an isolated least eigenvalue. The interaction part is well defined in 1+1 dimensional space time if it is Wick ordered and has a space cut-off. The Hamiltonian with this interaction can be shown to be bounded below and have an isolated least eigenvalue. See, for example, Summers' review, page 6.

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