I can see there can be multiple interpretations of this question. I will be interested in an answer to any of them.

"quantum systems" could include

  • Actual systems observed in experiment with quantum characteristics
  • Actual systems that must exist in nature with quantum characteristics
  • Conceptual models or scenarios within accepted quantum theory
  • Speculative quantum models beyond accepted theory

"no eigenstates" could be because

  • There is no Hamiltonian at all (I realise this stretches the grammar of the question slightly). For example maybe a system with no time aspect / no explicit time aspect? Or for any other reason.
  • There is a Hamiltonian, but it has no eigenstates. Whether or not this is allowed would be even more interesting for me.

Related question: Does the Hilbert space include states that are not solutions of the Hamiltonian?


1 Answer 1


This falls under "Conceptual models or scenarios within accepted quantum theory":

Free particle Hamiltonian

$$ \hat{H} = \frac{\hat{\mathbf p}^2}{2m} $$

has no proper eigenfunctions. It is true there are functions $\psi$ of $\mathbf r$ that obey

$$ (\hat{H}\psi) (\mathbf r) = E \psi(\mathbf r) $$

but these do not belong to any Hilbert space, since they are not normalizable ($\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$).

The normalizability is important in order to apply the Born interpretation to $|\psi|^2$. $\psi$ that is not normalizable can be used in expansions, but we cannot interpret $|\psi|^2$ as density of probability. That is why such functions are not admitted as proper description of state of a system; only their superposition that is normalizable can be admitted.

The function $\psi(\mathbf r) = Ce^{i\mathbf p_0\cdot\mathbf r/\hbar}$ is of similar status as the "delta function" $\psi(\mathbf r) = \delta(x-x_0)\delta(y-y_0)\delta(z-z_0)$ is. They are useful as tools to ease the work with the normalizable functions, but they can never be used as functions describing actual state.

In order to have existence of some proper Hamiltonian eigenfunctions (so they are members of some Hilbert space), the Hamiltonian has to contain, in addition to the kinetic terms, also some sufficiently well-behaved potential terms. For example, the Hamiltonian of harmonic oscillator does have proper eigenfunctions. But the Hamiltonian of a free particle does not.

  • $\begingroup$ Are there bases for a free particle, like maybe the set of all wave packets, which make a proper Hilbert space / "acceptable" quantum system? Or equivalently, does the Hamiltonian have some special status or is it just like any other observable and a quantum system perfectly "acceptable" without Hamiltonian eigenstates? I ask because it always seems when I hear people talk about free particles in infinite space they imply it is in some way fundamentally pathological and cannot be used properly. $\endgroup$
    – user183966
    Apr 23, 2018 at 19:20
  • 1
    $\begingroup$ Free particle in space is an idealization of a real situation where the particle is far from others so interaction can be neglected. This idealization can be described properly ($\psi$ function that can be normalized) or improperly ($\psi$ function that cannot be normalized, with unclear interpretation). In the proper description, one can choose any L2 basis to expand the $\psi$ function into linear combination of basis functions, like the eigenfunctions of the harmonic oscillator Hamiltonian (even if the actual Hamiltonian is that of free particle). $\endgroup$ Apr 24, 2018 at 10:29
  • $\begingroup$ Thank you very much this is extremely helpful to and gives me a new perspective on what is a quantum system and the role of the Hamiltonian. $\endgroup$
    – user183966
    Apr 24, 2018 at 14:07

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