Imagine we want to solve the equations $$ i \hbar \frac{\partial}{\partial t} \left| \Psi \right> = \hat{H}\left| \Psi \right> $$ where $$\hat{H} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac{1}{2} m \omega^2 x^2.$$

Denote $\left| \Psi_n \right>$ each eigenstate of the equation. Imagine initial condition $$ \left| \Psi^0 \right> = \frac{\pi}{\sqrt{6}}\sum_{n=0}^\infty \frac{1}{n+1} \left| \Psi_n \right> $$ Then, $\hat{H}\left| \Psi^0 \right>$ does not converge. What is, in a mathematically rigorous manner, $\hat{H}$?

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    $\begingroup$ What do you mean with $What is, mathematically rigurous, 𝐻̂?" at all? It is an operator. The Hamiltonian of the quantum harmonic oscillator is not everywhere defined. $\endgroup$ Commented May 7 at 20:03
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    $\begingroup$ Why would you close this question? There is nothing wrong with it: Jorge, to his surprise, found out that the HO Hamiltonian is unbounded and got confused. Even seasoned people get confused by the subtleties of functional analysis and @TobiasFunke has fixed their confusion. So now at least two people have learned from Jorge's question, this is the best use of this stack exchange. $\endgroup$
    – hyportnex
    Commented May 7 at 20:50

3 Answers 3


$\hat{H}$ is an unbounded operator. It is a fact from Functional Analysis that Hermitian unbounded operators cannot be defined on the entire Hilbert space, only in subspaces of it: this result is known as the Hellinger–Toeplitz theorem. Hence, it is not surprising that there are vectors for which $\hat{H} |\psi\rangle$ is not defined.

$\hat{H}$ is not alone in this sense. Position and momentum are unbounded too (and thus it is not surprising that the sum of their squares is unbounded).

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    $\begingroup$ +1, but your second sentence is wrong. What is true, however, is that self-adjoint unbounded operators cannot be everywhere defined (Hellinger-Toeplitz). $\endgroup$ Commented May 7 at 20:22
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    $\begingroup$ @TobiasFünke I wasn't aware of that restriction! Thanks for pointing it out. I corrected the answer and added the Wikipedia reference to the Hellinger-Toeplitz theorem (I preferred the term Hermitian, though, since it is more common in physics and the domain subtleties are already addressed in the theorem) $\endgroup$ Commented May 7 at 20:26
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    $\begingroup$ Yes. However, IIRC, the claim I made (that there always exists an everywhere defined unbounded operator on any Hilbert space) requires the axiom of choice. $\endgroup$ Commented May 7 at 20:36
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    $\begingroup$ @Jorge It tells us, in particular, that the right-hand side should be well-defined. This restricts the possible states to those which are in the domain of the Hamiltonian $\endgroup$ Commented May 7 at 21:01
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    $\begingroup$ The entire field of functional analysis relies heavily on the axiom of choice, starting with the Hahn-Banach Theorem. To mis-quote the subtitle of Dr. Strangelove, one should stop worrying and learn to love the axiom of choice. $\endgroup$
    – Lee Mosher
    Commented May 8 at 16:02

The state $$ \lvert \Psi^0 \rangle = \frac{\sqrt{6}}{\pi}\sum_{n=0}^\infty \frac{1}{n+1} \lvert \Psi_n \rangle $$ is in the Hilbert space, as it is square integrable. However, it is not a (standard) solution to the time-dependent Schrodinger equation, exactly because, as OP notices, $\hat{H}\lvert \Psi^0 \rangle$ is not square-integrable and hence isn't in the Hilbert space.

However, if we proceed naively and just assume that we can time-evolve this state in the standard way, which is to say $$ \lvert \Psi^0(t) \rangle = \frac{\sqrt{6}}{\pi}\sum_{n=0}^\infty \frac{1}{n+1} e^{-i\hbar\omega(n+1/2)t}\lvert \Psi_n \rangle\,, $$ then we get a (seemingly) perfectly-well defined (time-dependent) quantum state, because this state is square-integrable and it evolves in the "standard" way.

This is what is called (I believe!) a weak solution to the partial differential equation (or it is at least related to the same idea). The idea is that while the derivatives of the function might exist, we can transform the differential equation into an (almost-equivalent) integral equation, and it turns out that more functions satisfy the integral equation than satisfy the differential equation.

In fact, these weak solutions can be non-differentiable. The same sort of thing shows up in the context of the wave equation. One can solve for the normal modes of a string with fixed ends, but it turns out that you can expand functions that don't satisfy the boundary conditions as a linear combination of these normal modes, and the expansion works. The function then satisfies the wave equation in only the weak sense.

So, physically speaking, this might be the way of resolving things. Look for weak solutions to the time-dependent Schrodinger equation rather than "strong" solutions, and perhaps we can call these perfectly well-defined quantum states. Whether this is relevant physically is another matter.

See also Sobolev space.


An aspect complementary to the already existing answer: the problem with the initial state $$|\psi(0)\rangle = \frac{\sqrt{6}}{\pi} \sum\limits_{n=0}^\infty \frac{1}{n+1} |\phi_n\rangle, \quad \langle \phi_m |\phi_n\rangle= \delta_{mn}, \quad H |\phi_n\rangle = \hbar \omega (n+1/2) |\phi_n\rangle,$$ (note the correct normalization ensuring $\langle \psi(0) |\psi(0)\rangle=1$) is not its time evolution. $|\psi(t)\rangle$ is perfectly well defined by the unitary transformation $$|\psi(t)\rangle = e^{-iHt /\hbar} |\psi(0)\rangle = \frac{\sqrt{6}}{\pi} \sum\limits_{n=0}^\infty \frac{e^{-i\omega(n+1/2)t}}{n+1}| \phi_n \rangle,$$ in spite of the fact that $|\psi(0)\rangle $ (and, more generally, $|\psi(t)\rangle$) is not in the domain of $H$.

The real problem with the initial state $|\psi(0)\rangle$ is the fact that it is not even in the domain of $\sqrt{H}$, such that the expectation value of $H$ in this state does not exist (diverges). In other words, $|\psi(0)\rangle$ is sick because it corresponds to a state with infinite energy.

  • $\begingroup$ Can you please explain the relevance of $\sqrt H$? $\endgroup$
    – Ghoster
    Commented May 8 at 5:23
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    $\begingroup$ @Ghoster For the quadratic form $\langle \psi |H \psi \rangle \to \langle \sqrt{H} \psi | \sqrt{H} \psi \rangle$ to be well defined, it is not necessary that $\psi \in D(H)$ but $\psi \in D(\sqrt{H})$ would be sufficient. But even this weaker condition is not fulfilled in the present example as $\sum\limits_{n=0}^\infty \frac{n+1/2}{(n+1)^2}$ still diverges. $\endgroup$
    – Hyperon
    Commented May 8 at 5:40

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