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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...

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  • $\begingroup$ Regarding "phenomonological restrictions", in quantum information (or maybe better mathematical physics?) one has to limit the energy in the system, in order to get well-behaved quantities such as channel capacities. (One typically bounds sth. like the average energy, rather than the maximum energy, and the bound itself is arbitrary, and the e.g. capacities one obtains will depend on the bound on chooses. It is basically capturing the idea that you are willing to invest a certain amount of energy into your system; depending on that amount, there is a rate at which you can transmit ... $\endgroup$ Commented Jun 13 at 8:02
  • $\begingroup$ ... information, and so forth. This goes back to the work of Holevo (and probably others) in the 70ies, before it was called quantum information. $\endgroup$ Commented Jun 13 at 8:02
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    $\begingroup$ The Sturm–Liouville theorem states quite generally that for this class of problems (Hamiltonians etc.) there is a minimum, but no maximum, eigenvalue. $\endgroup$
    – Roman
    Commented Jun 13 at 14:47
  • $\begingroup$ I am not sure about this but does string theory with $g_s=0$ count given that such systems have an upper limit of temperature called the Hagedorn temperature and $E_{\text{Hagedorn}}=k_B T_{\text{Hagedorn}}$? $\endgroup$
    – Sanjana
    Commented yesterday

4 Answers 4

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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.


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

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  • $\begingroup$ Hi, I have a quick question: Taking $H$ the Hamiltonian of the quantum harmonic oscillator, then $-H$ is bounded above, no? Where does your argument fail there? Or more generally, which assumption is crucial that it works (you also write that "boundedness from above is not possible for every choice of $V$)? Is it related to the domain? $\endgroup$ Commented Jun 12 at 6:53
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    $\begingroup$ Yes but that way you reversed the sign of the kinetic energy! That is not physical. My argument relies upon the fact that the kinetic energy is unbounded form above... $\endgroup$ Commented Jun 12 at 7:00
  • $\begingroup$ Ah, I see! Thanks $\endgroup$ Commented Jun 12 at 7:00
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    $\begingroup$ Your argument is nice but relies on the possibility of taking $k\rightarrow \infty$. In a discrete space, there would be a UV cutoff I believe though (This would either be in a lattice, or by the existance of a Planck length that bounds the wavelength from below). Infinite dimensional can still be achieved by the spatial domain being infinitely large. Not completely sure what to get from there because still higher bands for the same k may appear perhaps. $\endgroup$
    – Wouter
    Commented Jun 13 at 8:31
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    $\begingroup$ Nice answer. Regarding $k$, when $\psi(x)$ is real $k$ is indeed the expectation value of momentum in the state $\psi_k$. $\endgroup$
    – lcv
    Commented Jun 13 at 14:00
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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.

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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.

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  • $\begingroup$ But the hydrogen atom, as shown by the answer of Valter Moretti, is not bounded from above (only its discrete spectrum is). $\endgroup$
    – lcv
    Commented Jun 13 at 6:41
  • $\begingroup$ The hydrogen atom still has an infinite number of states in a finite energy window, so this argument still applies. $\endgroup$
    – 11zaq
    Commented Jun 13 at 14:37
  • $\begingroup$ Barring your explicit example, I cannot understand well why the singularity of $\rho(E)$ at $E_0$ is non-integrable in any cases... $\endgroup$ Commented Jun 14 at 6:08
  • $\begingroup$ Supposed that the microstate counting is analytic in $E^{-1}$ with leading order divergence $\Omega \sim E^{-\alpha}$ for $\alpha >0$. Then $\rho \sim E^{-\alpha-1}$ and so $Z \sim \Gamma(-\alpha)$, which diverges if $\alpha$ is an integer, which it will be if $\Omega$ is analytic in $E$. For non-integer $\alpha$, the original integral doesn't converge but maybe $\Gamma(-\alpha)$ is a useful definition of a regulated value in some cases. I'm not totally sure, it would probably depend on the setting. $\endgroup$
    – 11zaq
    Commented Jun 14 at 8:20
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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.

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    $\begingroup$ But what's stopping the electron to go to energies> $0$ eV? There are scattering solutions to the H-atom problem, right? $\endgroup$
    – Sanjana
    Commented Jun 12 at 1:40
  • $\begingroup$ Sure. But equally nothing stops us from declaring that the set of bound orbitals is our physical system. $\endgroup$ Commented Jun 12 at 1:51
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    $\begingroup$ Well, but then any self-adjoint (bounded above) operator is an example. Take e.g. $H=|\psi\rangle\langle\psi|$. Is that physical? Maybe. $\endgroup$ Commented Jun 12 at 5:48
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    $\begingroup$ This answer is just wrong. It tuncates the spectrum at some arbitrarily chosen point (here, 0, but any other point would do, e.g. we could cut the energy just above the ground state and would only have a single energy for the system. And this works for any systen.) $\endgroup$ Commented Jun 13 at 7:59
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    $\begingroup$ It truncates the spectrum at a place where people commonly do it for physically well motivated reasons. There is no way to avoid doing this. If you included the scattering states as well, there would still be electron-positron pairs that you are neglecting. Every Hamiltonian is an effective Hamiltonian. Although perhaps instead of "physical examples" the OP was more concerned with finding a bounded operator that cannot be obtained by restricting some unbounded operator that acts on a larger Hilbert space. $\endgroup$ Commented Jun 13 at 11:15

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