Is there anything physical that enforces the wave function to be $C^2$? Are weak solutions to the Schrödinger equation physical? I am reading the beginning chapters of Griffiths and he doesn't mention anything.


4 Answers 4


Here we want to give an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation (TISE) tend to be rather nice. First formally rewrite the differential form

$$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \tag{1}$$

into the integral form

$$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) .\tag{2}$$

In this answer, we will assume that the integral form (2) [rather than the oft-written differential form (1)] is the starting point.

There are various cases.

  1. Case $V \in {\cal L}^2_{\rm loc}(\mathbb{R})$ is a locally square integrable function. Assume the wavefunction $\psi \in {\cal L}^2_{\rm loc}(\mathbb{R})$ as well. Then the product $(V-E)\psi\in {\cal L}^1_{\rm loc}(\mathbb{R})$ due to Cauchy–Schwarz inequality. Then the integral $y\mapsto \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z)$ is continuous, and hence the wavefunction $\psi$ on the lhs. of eq. (2) is smooth $\psi\in C^{1}(\mathbb{R}).$

  2. Case $V \in C^{p}(\mathbb{R})$ for a non-negative integer $p\in\mathbb{N}_0$. Similar bootstrap argument shows that $\psi\in C^{p+2}(\mathbb{R}).$

The above two cases do not cover a couple of often-used mathematically idealized potentials $V(x)$, e.g.,

  1. the infinite wall $V(x)=\infty$ in some region. (The wavefunction must vanish $\psi(x)=0$ in this region.)

  2. or a Dirac delta distribution $V(x)=V_0\delta(x)$. See also here.

  • $\begingroup$ I do not understand well your answer. What hypotheses on $\psi$ are you assuming in (1) and to pass from (1) to (2)? In your first considered case, why not to iterate the argumen once more proving that $\psi$ is $C^2$? As it stands, without a clear procedure to pass to (2) from (1) the argument seems a bit tautological.... $\endgroup$ Jun 26, 2017 at 13:53
  • $\begingroup$ I tried to make the logical line of reasoning hopefully more clear in an update. Also the notion of weak solutions is not mentioned to keep technicalities at a minimum. $\endgroup$
    – Qmechanic
    Jun 26, 2017 at 14:40
  • 1
    $\begingroup$ Ok, assuming that the integral version is the correct one everything works. It should be equivalent to deal with the adjoint if the initially gived differential operator. Because its adjoint is selfadjoint and thus defines the Hamiltonian. The initially given operator is only essentially self-adjoint. This is the real reason for imposing continuity of the derivative: the elements of the domain of the adjoint operator satisfy this constraint... $\endgroup$ Jun 26, 2017 at 15:45
  • $\begingroup$ Hi @Valter Moretti: Thanks for the feedback. I have undeleted our previous comments, which might be relevant. $\endgroup$
    – Qmechanic
    May 27, 2023 at 11:10

Some of this was discussed elsewhere. See « significance of unbounded operators » https://physics.stackexchange.com/a/19569/6432 .

It is not true the wave function has to be continuous, it just has to be measurable (i.e., a limit of step functions almost everywhere). Naturally you might wonder what sense Schroedinger's equation makes if you apply it to a step function...but the answer is easier than worrying about distributional weak solutions. The point is that you can solve the time-dependent Schroedinger equation with the exponential $$e^{itH},$$ which is a family of unitary operators, and which is better behaved than the $H$ you have to use in Schroedinger's equation. The $H$ you have to use, for example $$-{\partial ^2\over\partial x^2} + \mathrm{other\ stuff}, $$ is unbounded. And non-differentiable functions are not in its domain. But plugging it in to the power series for exponential converges in norm anyway, and so the resulting operator, being bounded and even unitary on a dense domain of the Hilbert space, can be extended painlessly to the entire space, even step functions. So it makes more sense to say that the solution to Schroedinger's equation with a given initial condition $\psi_o$ is $$\psi_t (x) = e^{itH}\cdot \psi_o (x)$$ and there is no need to bring in distributional weak solutions. These considerations are called the Stone--von Neumann theorem.

But such functions are not very important and indeed it is possible to do all of Quantum Mechanics with smooth functions, especially if you take the attitude that, for example, a square well potential would also be unphysical and is really just a simplified approximation of a physical potential which smoothed off those square corners but had a formula that was unmanageable.... See Anthony Sudbery, Quantum Mechanics and the Particles of Nature, which since it is written by a mathematician, is careful about unimportant issues like this.

That family of operators I wrote down is called the time-evolution operators, and they are an example of a unitary group with one-parameter, time. It is easy to see that if $\psi_o$, the initial condition, the state of the quantum system at time $t=0$, is nice and smooth, then all the future states will be nice and smooth too. Furthermore, all the usual quantum observables have eigenstates which are nice and smooth, so if you perform a future measurement, you will get a function which is nice and smooth and its future time evolution will remain that way, until the next measurement, etc. until Doomsday.

That said, for all practical purposes you may assume all wave functions are smooth and that the only reason you study discontinuous ones is as convenient approximations.

The comment one sometimes hears is that a wave function that was not in the domain of the Hamiltonian would « have infinite energy » but this is nonsense. In Quantum Mechanics, you are not allowed to talk about a quantum system as having a definite value of an observable unless it is in an eigenstate of that observable. What you can ask is, what would be the expectation of that observable. If the wave function $\psi$ is discontinuous and not in the domain of the Hamiltonian, it cannot be an eigenstate, but if its energy is measured, the answer will always be finite. Yet, the expectation of its energy does not exist, or you could say, the expectation « is infinite ». Not the energy, its expectation. There is nothing very unphysical about this because expectation itself is not very directly physical: you cannot measure the expectation unless you make infinitely many measurements, and your estimated answer, even for this discontinuous function, will always be a finite expectation. It's just that those estimates are way inaccurate, the expectation really is infinite (like the Cauchy distribution in statistics).

But even for such a « bad » wavefunction, all the axioms of Quantum Mechanics apply: the probability that the energy, if measured, will be 7 erg, is calculated the usual way. But these bad wave functions never arise in elementary systems or exercises so most people think they are « unphysical ». And, as I said, if the initial condition is a « good » wave function, the system will never evolve outside of that. This, I think, is connected with the fact that in QM, all systems have a finite number of degrees of freedom: this would no longer be true for quantum systems with infinitely many degrees of freedom such as are studied in Statistical Mechanics.

  • 1
    $\begingroup$ Right, there's nothing wrong about step functions, delta-functions (the derivatives of the former), and others, and that's why physicists freely work with them and never mention artificial mathematical constraints. Still, some discontinuities may make the kinetic energy infinity, so they don't exist in the finite-energy spectrum. I would add that the most natural space of functions to consider is $L^2$, all square-integrable functions. They may be Fourier-transformed or converted to other (discrete...) bases. A subset also has a finite (expectation value of) energy. $\endgroup$ Jan 18, 2012 at 7:22
  • 1
    $\begingroup$ The delta-function is not in L2 and is problematic. In physical language, this connects to the uncertainty principle: the location of a particle is never known with perfect certainty. $\endgroup$
    – 5th decile
    May 5, 2021 at 14:45

The time-independent Schroedinger equation for the position-space wavefunction has the form $$\left(\frac{-\hbar^2}{2m}\nabla^2 +(V-E) \right)\Psi=0$$

Where $E$ is the energy of that particular eigenstate, and $V$ in general depends on the position. All physical wavefunctions must be in some superposition of states that satisfy this equation.

At least in nonrelativistic QM, the wavefunction is not allowed to have infinte energy. If the second derivative of the wavefunction does not exist or is infinite, it implies that either $V$ has some property that "cancels out" the discontinutiy (as in the infinite square well), or that the wavefunction is continuous and differentiable everywhere.

Generally, $\Psi$ must always be continuous, and any spatial derivative of $\Psi$ must exist unless $V$ is infinite at that point.


The solution of the Dirac equation for ground-state hydrogen atom (under point nuclei, fixed nuclei position) includes, as quoted in this link, $$ r^{\gamma-1},$$ where $$ \gamma = \sqrt{k^2 - Z^2 \alpha^2 } = \sqrt{ 1 - 1/137^2} < 1,$$ thus it diverges at the origin. Namely not continuous (not belong to $C^0$, thus cannot be $C^2$), but still within $L^2$ space.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Dec 27, 2021 at 4:30
  • $\begingroup$ @Community I am a bit confused. If a wavefunction is diverged, it does not belong to $C^0$ class (en.wikipedia.org/wiki/Smoothness), thus cannot be $C^2$.,thus no enforcement to let wavefunction be $C^2$. May I know which part is not clear or am I enirely wrong? And personally, I think provide full link address is better than proving a name of it (https.... vs this link, and full link address appears in the second answer already) $\endgroup$
    – AlphaF20
    Dec 27, 2021 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.