Examples of systems with infinite dimensional Hilbert space, whose energy is bounded from above

Question

We often encounter (and love to!) deal with systems whose energy is bounded from below, for good reasons like stability, etc. But what about systems whose energy is bounded from above? In finite dimensional Hilbert spaces, this is obvious. But what about infinite dimensional Hilbert spaces? Is there any theoretical or even phenomenological restrictions on the upper bound to energy of a system with infinite dimensional Hilbert space? Mathematically, we are just looking for bounded operators acting on infinite dimensional Hilbert spaces.

With help from the hbar chatroom, I can think of one very simple example: If our Hamiltonian is the shift operator $\psi_i\mapsto \psi_{i+1}$, then energies are bounded both from above and below. But I am still posting this on the main site because I would prefer more "physical" examples. I would prefer "physical" examples but that doesn't mean I wouldn't want an answer full of mathematical examples only, or an example of some exotic theory which simply has an energy bound from above due to some theoretical principle/phenomenological reason...

Answer 1

In standard quantum mechanics, more precisely for systems described in $L^2(\mathbb{R}, dx)$ (and this generalizes to many dimensions) Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ initially defined in the space of smooth compactly supported wavefunctions.

Here, boundedness from above is not possible for any choice of $V$.

Here is a sketch of proof.

Boundedness from above means that (*) $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and all normalized vectors $\psi$ in the domain of the (selfadjoint extension of the) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |ik\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.

The result easily generalizes to many dimensions with $$H= -\sum_{i=1}^N\frac{\hbar^2}{2m_i}\Delta_{\vec{x}_i}+ V(\vec{x}_1, \ldots, \vec{x}_N)\:.$$ I suspect that also introducing a magnetic potential would not change the final result (I should check it however).

In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators of non-standard type.

The crucial fact in the sketch of proof above is that the kinetic energy is unbounded from above. Reversing the sign of $H$ is not physically permitted as it would correspond to a negative kinetic energy.

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(*) That is equivalent to $spect(H) < c< +\infty$, which can be given as an alternative definition of above boundedness.

Answer 2

As Valter Moretti's answer explained, the unboundedness of usual Hamiltonians comes from the quadratic kinetic energy. There are some situations where the quadratic kinetic energy may not be appropriate. A typical example of this is in condensed matter where you rather observe bands of energy. You can model a single band with tight binding-like approaches, and by construction the Hamiltonian will have a bounded spectrum.

The simplest example would be the 1D chain. Your infinite dimension Hilbert space can have the real space orthonormal basis $(|n\rangle)_{n\in\mathbb Z}$. With nearest neighbour hopping, you get the Hamiltonian: $$ \begin{align} H &= -t\sum_{n\in\mathbb Z}|n+1\rangle\langle n|+|n-1\rangle\langle n|\\ &= -t(S+S^{-1}) \\ &= -2t\cos(p) \end{align} $$ with $S$ you shift operator and $p$ the quasi momentum operator.

Answer 3

One point to make about the physical implications of being bounded above (and below) is that the Bolzano-Weierstrass Theorem shows that such a system must have a divergence in the density of states $$\rho(E) \sim \frac{d\Omega}{dE},$$ because $\Omega(E)$ will diverge. To see this, the BW theorem says that there exists at least one energy level $E_0$ such that an infinite number of states have energy $E_n$ arbitrarily close to $E_0$.

Because $E_0$ is bounded, this will also imply that the partition function $$Z = \int dE \rho(E) e^{-\beta E}$$ isn't well-defined because the Boltzmann suppression can't help cure the divergence at finite $E_0$. That's going to leave a big imprint on the physics, because it will mean that at least some subset of the thermodynamic quantities of interest will diverge as well. Maybe you can still make sense of some of the log derivatives of it if you regulate the integral though.

As Connor pointed out in his answer, the hydrogen atom is such an example. There, one must note that there are $n^2$ states at level $n$, so because $E\sim n^{-2}$, we have that $$\Omega(E) \sim E^{-1},$$ as well as $$\rho(E) \sim E^{-2}.$$ See this post for a discussion about the resolution/ interpretation to how to deal with this divergence for the hydrogen gas system.

Answer 4

The spectrum of a Hydrogen atom goes from the $-13.58 \text{eV}$ ground state to a free electron at $0 \text{eV}$. It does this in infinitely many discrete steps.