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What does it physically mean, if a model in quantum mechanics or in quantum field theory has no ground state?

So I am talking about a Hamiltonian $H$ such that $\sigma(H)$ is bounded from below. So what is the physical difference, if $H$ has an eigenstate at the bottom of the spectrum, compared to no eigenstate existing at the bottom of the spectrum?

I am asking this because many authors in mathematical quantum theory care about proving the existence of a ground state for a given model, but I don't understand why that is so important.

Edit:

The only answer i can come up with by myself is the following. If the lower bound $E_g$ is an eigenvalue with eigenvector $f_g(0,x)$, we know according the Schrödinger equation that the system once being in the state $f_g(0,x)$ at a later point it will be in the state $f_g(t,x)=e^{iE_gt}f(0,x)$. Thus up to a complex phase, it will remain in that state.

However if the lower bound of the spectrum $E_g$ is no eigenvalue, we might measure the system at energy $E_g$ but it does not have to remain in this state. In this sense it is kind of unstable.

What does it physically mean, if a model in quantum mechanics or in quantum field theory has no ground state?

So I am talking about a Hamiltonian $H$ such that $\sigma(H)$ is bounded from below. So what is the physical difference, if $H$ has an eigenstate at the bottom of the spectrum, compared to no eigenstate existing at the bottom of the spectrum?

I am asking this because many authors in mathematical quantum theory care about proving the existence of a ground state for a given model, but I don't understand why that is so important.

The only answer i can come up with by myself is the following. If the lower bound $E_g$ is an eigenvalue with eigenvector $\psi_g(0,x)$, we know according the Schrödinger equation that the system once being in the state $\psi_g(0,x)$ at a later point it will be in the state $\psi_g(t,x)=e^{-iE_gt}\psi(0,x)$. Thus up to a complex phase, it will remain in that state.

However if the lower bound of the spectrum $E_g$ is no eigenvalue, we might measure the system at energy $E_g$ but it does not have to remain in this state. In this sense it is kind of unstable.

What does it physically mean, if a quantum mechanical system has no ground state?

So I am talking about a Hamiltonian $H$ such that $\sigma(H)$ is bounded from below. So what is the physical difference, if $H$ has an eigenstate at the bottom of the spectrum, compared to no eigenstate existing at the bottom of the spectrum?

I am asking this because many authors in mathematical quantum theory care about proving the existence of a ground state for a given model, but I don't understand why that is so important.

Edit:

The only answer i can come up with by myself is the following. If the lower bound $E_g$ is an eigenvalue with eigenvector $f_g(0,x)$, we know according the Schrödinger equation that the system once being in the state $f_g(0,x)$ at a later point it will be in the state $f_g(t,x)=e^{iE_gt}f(0,x)$. Thus up to a complex phase, it will remain in that state.

However if the lower bound of the spectrum $E_g$ is no eigenvalue, we might measure the system at energy $E_g$ but it does not have to remain in this state. In this sense it is kind of unstable.

What does it physically mean, if a model in quantum mechanics or in quantum field theory has no ground state?

So I am talking about a Hamiltonian $H$ such that $\sigma(H)$ is bounded from below. So what is the physical difference, if $H$ has an eigenstate at the bottom of the spectrum, compared to no eigenstate existing at the bottom of the spectrum?

I am asking this because many authors in mathematical quantum theory care about proving the existence of a ground state for a given model, but I don't understand why that is so important.

Edit:

The only answer i can come up with by myself is the following. If the lower bound $E_g$ is an eigenvalue with eigenvector $f_g(0,x)$, we know according the Schrödinger equation that the system once being in the state $f_g(0,x)$ at a later point it will be in the state $f_g(t,x)=e^{iE_gt}f(0,x)$. Thus up to a complex phase, it will remain in that state.

However if the lower bound of the spectrum $E_g$ is no eigenvalue, we might measure the system at energy $E_g$ but it does not have to remain in this state. In this sense it is kind of unstable.

What does it physically mean, if a quantum mechanical system has no ground state?

So I am talking about a Hamiltonian $H$ such that $\sigma(H)$ is bounded from below. So what is the physical difference, if $H$ has an eigenstate at the bottom of the spectrum, compared to no eigenstate existing at the bottom of the spectrum?

I am asking this because many authors in mathematical quantum theory care about proving the existence of a ground state for a given model, but I don't understand why that is so important.

Edit:

The only answer i can up with by myself is the following. If the lower bound $E_g$ is an eigenvalue with eigenvector $f_g(0,x)$, we know according the Schrödinger equation that the system once being in the state $f_g(0,x)$ at a later point it will be in the state $f_g(t,x)=e^{iE_gt}f(0,x)$. Thus up to a complex phase, it will remain in that state.

However if the lower bound of the spectrum $E_g$ is no eigenvalue, we might measure the system with in the energy $E_g$ but it does not have to remain in this state. In this sense it is kind of unstable.

What does it physically mean, if a quantum mechanical system has no ground state?

So I am talking about a Hamiltonian $H$ such that $\sigma(H)$ is bounded from below. So what is the physical difference, if $H$ has an eigenstate at the bottom of the spectrum, compared to no eigenstate existing at the bottom of the spectrum?

I am asking this because many authors in mathematical quantum theory care about proving the existence of a ground state for a given model, but I don't understand why that is so important.

Edit:

The only answer i can come up with by myself is the following. If the lower bound $E_g$ is an eigenvalue with eigenvector $f_g(0,x)$, we know according the Schrödinger equation that the system once being in the state $f_g(0,x)$ at a later point it will be in the state $f_g(t,x)=e^{iE_gt}f(0,x)$. Thus up to a complex phase, it will remain in that state.

However if the lower bound of the spectrum $E_g$ is no eigenvalue, we might measure the system at energy $E_g$ but it does not have to remain in this state. In this sense it is kind of unstable.