What does the absence of a ground state physically mean?

Question

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.

Answer 1

Very interesting question. I don't believe there is any principle prohibiting a scenario where there is no ground state, but the Hamiltonian is bounded from below. It's just that for most systems there is a ground state, and showing the existence of the ground state is the easiest way to prove that the Hamiltonian is bounded from below, and also provides a nice way to build up all of the other states via operating on that ground state.

Answer 2

The free particle Hamiltonian has spectrum $\sigma(H_0) = [0,\infty)$, but $0$ is not an eigenvalue. More generally, any system that possesses only scattering states (For example $H = H_0+ V$ where $V$ is a positive bump or otherwise has some monotone falloff) will have positive spectrum with no bound states. In that sense there isn't anything particularly wrong with such systems. However, having a ground state has important implications for the dynamics of a theory, its phase structure, etc. For that reason it is useful for both qualitative and quantitative analysis of a model to understand whether or not it possesses such a state, and if it does, whether or not it is unique.

To elaborate on the point about dynamics, a theorem to look into is the RAGE theorem. Roughly speaking it states that for Schrödinger operators, a state that belongs to the continuous spectrum is 'scattering' in the sense that mass escapes any ball after sufficiently large time. On the other hand, eigenvectors are 'bound' , meaning their mass remains mostly localized in a bounded region. So if a system lacks any eigenstate, then it means that any and all states become delocalized given enough time.

Answer 3

I am not aware of any example where a Hamiltonian would be bounded from below but do not have a ground state.

On the other hand, one can easily come up with Hamiltonians not bounded from below and therefore not having a ground state. E.g., a particle in an electric field: $$ H=\frac{p^2}{2m} - eEx $$ In this case one would be typically interested not in the properties of the ground state, but in a flux generated by the electric field (in presence of interactions with, e.g., phonons, it can be stationary.)

Another example is the inverted parabolic potential, sometimes used to study tunneling problems: $$ H = \frac{p^2}{2m}-\frac{m\omega^2x^2}{2}. $$ Scattering solutions can be found for this potential in terms of parabolic cylinder functions.

Answer 4

Consider a Hamiltonian bounded from below. If there is no ground state, any state will not be the lowest energy state, which means that it could suffer a transition to a lower energy state. We can build a sequence of eigenstates with lower and lower energies, to see what happens near the bound.

If $B$ is the lower bound, for any given eigenstate associated to an energy $E_0$, there would be another eigenstate with energy $B < E_1 < E_0$ (otherwise, it would be the ground state). Following this construction, we can build a subsequence of Eigenvalues $E_0,E_1,...,E_k$ with associated eigenvectors and, for any number $N$, there will always be some eigenstate with energy

$$ | B - E_N | < \frac 1N $$

which means that at some point, the eigenstates start to be more and more closer with each other in this sequence. Now, physically, we couldn't measure energy with arbitrarily precision, with means that at some point, we couldn't realize any experiment to differ $E_N$ from $B$. In this case, $B$ (or a very closer number), would be considered an approximate lower eigenvalue.

Now, the question is if the eigenstates $\Phi_k\rangle$ have the same property, i.e., if for $m,n>N$, we have

$$ |||\Phi_m\rangle - |\Phi_n\rangle ||^2 < f(N) $$

For some bound $f(N)\rightarrow 0$ at large N. If this property doesn't hold, the system is transitioning between eigenstates with very similar energy, but very different states. It could be emerge as "degeneracy" of the "practical eigenvalue", some kind of degeneracy of the ground state.

If this property holds, for sufficient high $N$, we couldn't differentiate between different eigenstates, and we could say that the sequence converges for a non generate "practical ground state".

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A bit more rigour

Actually, what we showed is that we can build a sequence of eigenvalues with associated eigenvectors $\{E_k\}$ that is decreasing by construction and bounded by below. So it is convergent. The question is if the associtated eigenvectors also converges. Now, it can happen that

1. The eigenvectors does not converge
2. The eigenvectors coverge in an unphysical state

If they don't converge, there's nothing we can conclude mathematically. The second is the best situation, because if it converge to an unphysical state, it would be at least arbitrarily close to physical ones. Now, for the first possibility, we could only say that there is a sufficient condition to not have it.

Eigenstates associated with arbitrarily close eigenvalues are arbitrarily close of each other

This assumption is highly assumed in perturbation theory, but I think it could not be proved for all systems. If it holds, we can say that the sequence of eigenvectors above always converge.

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In conclusion, it seems that despite mathematically possible, a bounded Hamiltonian without ground state assumes arbitrary precision for measuring energies and to distinguish eigenstates, something that we couldn't achieve in practice. Any configuration like that would imply in a bounded sequence of eigenvalues and eigenvectors that, for sufficient high terms, we couldn't distinguish anymore and we could say that "converges" for some ground state

Answer 5

This is likely not what you are looking for, but it is an important point nonetheless. Consider a free, non-relativistic particle in 1 dimension (or more). The Hamiltonian $H = - \nabla^2$ has spectrum $\sigma(H) = [0,\infty)$, but no ground state at $0$. Note that, in the infinite-dimensional case, $H$ need not have an eigenvector associated with each element of the spectrum.

More interesting is the case, which I think you are concerned with, where we have a sequence $\psi_j$ of energy eigenstates with energy $E_j \rightarrow E_{\mathrm{inf}}$ converging to the greatest lower bound of $\sigma(H)$, but for which $E_{\mathrm{inf}}$ is not itself an energy. Such a system should be constructable, in a rather silly way, using a infinite "comb" of square wells, each with depth slightly lower than the previous (but lower-bounded depth). You'll notice in this case that transitions are disallowed in a rather silly way. I cannot think of a more "interesting" example of such a sequence of bound states, but I have a feeling that any such examples will look rather contrived.