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The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bounded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large, and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may look for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily close to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x =1\:.$$ This constraint prevents us from finding functions with $$-\langle \psi|H\psi\rangle$$ arbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is instable as it is unbounded below. Fortunately for us, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.

The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bounded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large, and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may look for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily close to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x =1\:.$$ This constraint prevents us from finding functions with $$-\langle \psi|H\psi\rangle$$ arbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is unstable as it is unbounded below. Fortunately for us, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.

The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bounded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large, and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may look for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily close to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x <+\infty\:.$$ This constraint prevents us from finding functions with $$-\langle \psi|H\psi\rangle$$ arbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is instable as it is unbounded below. Fortunately for us, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.

The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bounded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large, and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may look for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily close to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x =1\:.$$ This constraint prevents us from finding functions with $$-\langle \psi|H\psi\rangle$$ arbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is instable as it is unbounded below. Fortunately for us, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.

The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bouded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large. and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may looks for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily cose to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x <+\infty\:.$$ This constraint prevents for finding functions with $$-\langle \psi|H\psi\rangle$$ arnbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is instable as it is unbounded below. Fortunatley for as, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.

The issue can be tackled from a completely mathematical perspective. What you are saying about $\phi^3$ is true at classical level. The classical Hamiltonian functional ${\cal H}(\phi,\partial\phi)$ is not bounded below because we can find field configurations $(\phi_0, \partial \phi_0)$ with $-{\cal H}(\phi_0, \partial \phi_0)$ arbitrarily large, and this gives rise to several pathologies of various nature.

The issue with the hydrogen atom is mathematically different. There you may look for wavefunctions $\psi$ such that $\langle \psi|H\psi\rangle$ is arbitrarily close to $-\infty$.

However, differently from the (classical) Klein-Gordon $\phi^3$, here you also have the constraint $$\langle \psi|\psi\rangle = \int_{\mathbb{R}^3} |\psi(\vec{x})|^2 d^3x <+\infty\:.$$ This constraint prevents us from finding functions with $$-\langle \psi|H\psi\rangle$$ arbitrarily large. The proof is the (well known) computation of the spectrum of the hydrogen atom.

(This is a sort of miracle, and it is one of the reasons why we are here. Since, classically speaking, the hydrogen atom and any system subjected to the attractive Coulomb force is instable as it is unbounded below. Fortunately for us, our universe is quantum.)

One can also consider the second quantization of $\phi$. In that case, after normal ordering renormalization one sees that it is however possible to find states $\Psi$ of the quantum field where $-\langle \Psi| \hat{\cal H} \Psi\rangle$ is arbitrarily large, in spite of the constraint $\langle \Psi| \Psi\rangle=1$, due to the nature of $\hat{\cal H}$ written in terms of creation and annihilation operators. Maybe there is a straightforward physical reason beyond the evident mathematical difference in the formalisation of the two systems (first and second quantization, respectively) but I do not know, sorry.