In quantum mechanics we are told that

  1. the wavefunctions live in Hilbert space.

  2. the wavefunctions are continuous.

It recently came to my notice that in Mathematics, there is a theorem which says that the $\textit{continuous functions}$, for example $C[0,1]$ do not form a Hilbert space (see this). Now I wonder whether this is in contradiction with 1 and 2 or not?

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    $\begingroup$ Requiring stationary wavefunctions to be continuous is one thing. Requiring that any Cauchy sequence of them converge to another continuous function is quite another. $\endgroup$ Jul 23 at 11:25
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    $\begingroup$ Related/possible duplicate: physics.stackexchange.com/q/19667/50583, in which Qmechanic explains why solutions to 1D Schrödinger equations tend to be continuous even if wavefunctions in general are not, and its linked questions. $\endgroup$
    – ACuriousMind
    Jul 23 at 12:46

3 Answers 3


Item 2 is simply false if wavefunction is understood as a generic state vector of the system.

This is compatible with the answer NO to your question.

With the above interpretation wavefunctions are not "functions" in usual sense, since they are elements of $L^2(\mathbb{R}, dx)$ ($\mathbb{R}$ can be replaced by an interval $[a,b]$). That Hilbert space includes objects which are functions "up to arbitrary re-definition on zero-measure sets". For instance The set of (signed) rational numbers in $\mathbb{R}$ has zero measure. You see that, as a consequence, $L^2(\mathbb{R}, dx)$ contains true monsters.

On the other hand several wavefunctions must belong to the domain of some operators, usually differential operators, so it is convenient to deal (when requested and possible) with "monsters" which can be made differentiable (up to the requested order) by redefining them on zero measure set. That is the reason why one usually does not see monsters. However they are necessary for guaranteeing completeness of $L^2(\mathbb{R}, dx)$ as a Hilbert space. Completeness, in turn, guarantees the validity of some mathematical machinery of great relevance in Quantum Theory, as the spectral decomposition of selfadjoint operators.

There is nothing horrible for instance, dealing with a particle in an infinite potential box in $[0,1]$ to consider a discontinuous initial (unnormalized) state $\psi(x) = 0$ for $x\in [0,1/2]$ and $\psi(x)= 1$ for $x\in (1/2,1]$.

This state is not a mathematical chimera: it can be prepared in the practice by removing at $t=0$ a separator wall placed at $x=1/2$ after having inserted a particle in the portion $[1/2,1]$.

That function is discontinuous but its evolution is well defined. It is sufficient to decompose the function on the basis of the Hamiltonian in the box made of smooth functions $\psi_n(x)$ with eigenvalue $E_n$ (we can find then in every elementary textbook on QM) $$\psi(x) = \sum_{n=1}^{+\infty} c_n \psi_n(x)$$

That series converges in the $L^2$ sense, as it does, since every $L^2$ function can be expanded that way with respect to a Hilbert basis as $\{\psi_n\}_{n=1,2,\ldots}$. This is a basic fact of spectral decomposition. Our function $\psi$ is $L^2$ trivailly.

Its time evolution is $$\psi(t,x) = \sum_{n=1}^{+\infty} c_n e^{-it E_n/\hbar}\psi_n(x)\:.$$

We can equivalently rewrite this identity as $$\psi(t,x) = e^{-itH/\hbar}\psi(x)$$ where $H$ is the Himiltonian operator. The evolutor $e^{-itH/\hbar}$ is defined in whole Hilbert space. If $\psi$ belongs to the domain of $H$, and this is not the case for our $\psi$ we can differentiate in the $L^2$ topology obtaining the Schroedinger equation $$i\frac{d}{dt} \psi(t,x) = H \psi(t,x)\:.$$ (that is not the whole story because the topology used in computing the derivative is not the pointwise one, but it is possible to prove that, if the function is sufficiently regular the Schroedinger equation above turns out to be the standard one in the sense of PDE)

If for wavefunction we intend a solution of the Schroedinger equation viewed as a classic PDE, then we should add some regularity hypotheses to the functions. However, that is a too restrictive viewpoint in my opinion, because it suggests that only these "wavefunctions" have time evolution (solutions of the Schroedinger equation). Instead, every state has time evolution regardless its regularity as function as I pointed out above as an example.

  • $\begingroup$ "This state is not a mathematical chimera: it can be prepared in the practice by..." -- the "separator wall" construct is unnecessary since a wavefunction with $\psi = 1$ on $[0, 1]$ and $\psi = 0$ elsewhere (outside the box) is already discontinuous (at $x = 0$ and $x = 1$). Having the particle start in a smaller box $[1/2, 1]$ doesn't fundamentally change anything. The larger issue is that a discontinuous wavefunction (whether with support on $[1/2, 1]$ or $[0, 1]$) has infinite expectation value of kinetic energy and so cannot be prepared in practice. $\endgroup$
    – nanoman
    Jul 24 at 5:01
  • $\begingroup$ @nanoman Unfortunately I have contracted covid and it is difficult to continue this discussion. I partially agree with your remark, but a separate discussion would be necessary to analyze your general argument in realistic situations. It is time to rest for me now. $\endgroup$ Jul 24 at 8:24
  • $\begingroup$ @ValterMoretti Best wishes for a quick recovery! At some point I would be interested to see your thoughts on whether having a finite $\langle E\rangle$ is, in fact, an absolute prerequisite for a physically reasonable state. I have never been fully convinced that this is so. $\endgroup$
    – J. Murray
    Jul 25 at 17:11
  • $\begingroup$ Thanks! Today I feel I bit better. Why only the expectation value of the energy? Also the other physical quantities as momentum and angular momentum should be finite along this idea, no? So there should be a subspace, the intersection of the domains of 7 operators at least (why not also the pure kinetic energy?) which contains the physical states. All physically meaningful procedures should leave this subspace invariant...That is impossible in the general case because the time evolution itself does not, barring trivial potential functions. $\endgroup$ Jul 25 at 17:51
  • $\begingroup$ So, though the argument is in principle relevant, in my view it leads to consequences even more difficult to accept. However I do not have a definite answer. $\endgroup$ Jul 25 at 17:53

Item 2 is false. The Hilbert space $\mathcal H=L^2(\mathbb R)$ consists of all square integrable functions, including discontinuous ones. Even though the Hamilton operator $H=-\Delta+V$ is a differential operator, which isn't defined on all of $\mathcal H$, the time evolution operator $e^{-i t H}$ is a densely defined bounded linear operator and can thus be extended to all of $\mathcal H$ by the BLT theorem (bounded linear transformation theorem), so the time evolution of all square integrable functions is well defined, even if they are discontinuous.

(As Valter pointed out in the comments, one needs to take a bit more care when defining $e^{-i t H}$. In order to apply the measurable functional calculus, $H$ needs to be self-adjoint. The differential operator version of $H$ is only symmetric, but possesses a unique self-adjoint extension if one starts out with a suitable initial domain of definition, such as the space of Schwartz functions.)

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    $\begingroup$ I disagree on a point. The Hamiltonian you use to construct the evolutor, is not $H$, but its closure. The domain of that operator is not made of regular functions since it is not a differential operator but the adjoint of a differential operator $\endgroup$ Jul 23 at 12:37
  • $\begingroup$ You are correct, I was being a bit sloppy here $\endgroup$
    – Nullstein
    Jul 23 at 12:38
  • $\begingroup$ I was also sloppy in my answer assuming to be more understandable. As a result I produced a flame. $\endgroup$ Jul 23 at 12:39
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    $\begingroup$ I am very disappointed that your answer gets downvoted ☹ $\endgroup$
    – Nullstein
    Jul 23 at 12:46
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    $\begingroup$ Dear Nullstein, great answer. Just a general hint: If you want that a user gets notified when you write a comment, use @ followed by the username. $\endgroup$ Jul 23 at 12:48

Some say that item 1 is also false. Basically a rigged Hilbert space is more appropriate. See e.g. https://arxiv.org/abs/quant-ph/0502053 or https://doi.org/10.1007/3-540-088431-1.

  • $\begingroup$ Rigged Hilbert spaces are Hilbert spaces too, just equipped with additional structure. They are in fact optional and unnecessary to do computations, but they justify the physicist's way of doing computations, i.e. Dirac notation. $\endgroup$
    – Nullstein
    Jul 24 at 22:42

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