I recently learned about an example where an operator $O$ (not Hermitian) acted on a wavefunction $|\psi \rangle$ and carried the wavefunction out of Hilbert space, i.e. $O|\psi \rangle$ is not in Hilbert space.

Can a Hermitian operator $A$ carry a wavefunction out of Hibert space? Up till now, I always did calculations in QM assuming that $A|\psi \rangle$ is still in Hilbert space.

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    $\begingroup$ It is not reccomended to name $|\psi\rangle$ by "wavefunction", but by its true name, a ket. $\endgroup$
    – DanielC
    Sep 6, 2019 at 21:22

2 Answers 2


Formally self-adjoint, but unbounded, operators can easily take a normalizable state (i.e a state in the Hilbert space) and make it unnormalizable and therefore no longer in the space. This can lead to all sorts of aparent paradoxes. For example consider the operator $p^4= \partial_x^4$ on infinite square well $[0,1]$. Let it act on the wavefunction $$ \psi(x)=\sqrt{30} x(1-x)= \sum_{n=0,{\rm odd}}^\infty \frac{\sqrt{960}}{n^3\pi^3} \phi_n(x) $$ where $\phi_n(x)= \sqrt{2}\sin n\pi x$ are the normalized eigenfunctions of $p^2=-\partial_x^2$. This wavefunction statisfies the usual boundary condition of a "particle in a box," and is normalized.

Clearly differentiaing $\psi(x)$ four times gives zero, but acting four times on the eigenfunction expansion gives $$ p^4 \psi= \sum_{n=0}^\infty \sqrt{960}n\pi \phi_n(x) $$ which has $||\psi||^2 \propto \sum_{n=1}^\infty n^2 = \infty$, so the resultant state is no longer in the Hilbert space.

For this reason, and others, unbounded operators are not allowed to act on all states in the Hilbert space, but instead have a domain which is at best a dense linear subspace of the Hilbert space, and always includes the restriction that the action of the operator on any state remains in the Hilbert space. In the example above $\psi$ is in the domain of $p^2$ as $p^2\psi$ is normalizable, but it is not in the domain of $p^4$. Therefore $p^4\ne (p^2)^2$ because the output of $p^2$ no longer in the domain of the second $p^2$ factor.

The notion of a "Hermitian" operator, although popular in intro QM courses, is unsafe as "Hermitian", only impies that $p$ and $p^\dagger$ are given by the same differential operator.The better notion is self-adjoint because it requires that $p$ and $p^\dagger$ are not only given by then same formuala, but also that they also have the same domain. Any quantum obsservable has to be self-adjoint. There are many operators that are hermitian, but do not have a complete set of eigenfunctions, and therefore have no meaning in Quantum mechanics.

  • $\begingroup$ To add a bit to the domain part, here is the formal definition. Let A be an operator in a Hilbert space. Then the maximal domain of A, denoted $D_A$ is defined as: $ D_A :=\{\psi\in\mathcal{H}| ||A\psi||_{\mathcal{H}} <\infty \}$. $\endgroup$
    – DanielC
    Sep 6, 2019 at 16:15
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    $\begingroup$ It is false that, for instance, $A^4\neq A^2 \circ A^2$. $\endgroup$ Sep 6, 2019 at 20:50
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    $\begingroup$ The composition of operators win non-maximal domain is defined on standard domains, in particular $D(AB):=\{x\in D(B) \:|\: Bx \in D(B)\}$. With this definition the composition is associative and the identity you claim to be false is instead true. Similarly $D(A+B):= D(A)\cap D(B) $. With this well known definition also the sum of operators is associative. All functional calculus of selfadjoint operstors automatically embodies this formalism and results. $\endgroup$ Sep 6, 2019 at 20:55
  • $\begingroup$ One show that the self-adjointness domain of a linear operator is always its maximal domain, so, in case of unbounded operators, one needs to relax the self-adjointness condition to at most essential self-adjointness, in order to make sense of sum and composition of unbounded operators. $\endgroup$
    – DanielC
    Sep 6, 2019 at 21:19
  • $\begingroup$ @DanielC I disagree. Sum and composition of operators are well defined as I wrote independently from selfadjointness issues. In fact, what I wrote also applies to operators which are not symmetric and also in pure Banach spaces where notions like selfadjointness do not apply. $\endgroup$ Sep 7, 2019 at 5:20

The answer is negative from scratch. An operator $A:D(A) \to H$ with $D(A)$ a linear subspace of $H$ is said Hermitian when $\langle Ax|y\rangle= \langle x| Ay\rangle$ if $x\in D(A)$. It is clear that $Ax \in H$ in any cases if $x$ is an element of the Hilbert space such that $Ax$ is defined. Also the Hermiticity requirement is quite irrelevant in this discussion.

This sort of issues just arises from some confusion about basic mathematical tools used in physics. I admit that sometimes, or also often, physical motivations play a crucual propulsive role in developing the formalism, but once the basic formalism is established, similar issues (discussed as they deserve) should be eventually viewed as nonsensical.

There are however many cases where apparently operators bring their argument outside the Hilbert space as clearly illustrated in the other answer by Mike Stone which I interpret as a very good practical discussion about the reason why this apparent paradox pops out in the standard manipulation of the mathematical formalism of physicists. (It is however false that $p^4 \neq p^2 p^2$ when properly using standard mathematical definitions regarding compositions of operators).

However in a strict sense, referring to the mathematical formalism of Newtonian physics, an issue similar to that raised here could be "is there a triangle with four edges?".


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