What is the physical reason (i.e. without any maths) that the derivative of a wavefunction (except with infinite potentials) has to be continuous?

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I know that in the classical wave case it is because otherwise an infinitesimal mass would feel an infinite acceleration, but since since the wavefunction is not 'physical' I don't think a direct analogy can be made.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/19667/2451 , physics.stackexchange.com/q/1067/2451 and links therein. $\endgroup$ – Qmechanic May 27 '15 at 18:43
  • $\begingroup$ Well, I find a wavefunction quite physical already. It describes (mechanical) waves on a string, (sound) waves through air, (electromagnetic) waves of field distortion... and then also some quantum stuff. $\endgroup$ – Steeven May 27 '15 at 19:50
  • $\begingroup$ To OP: Please substantiate the claim (i.e. give a reference) that the derivative of a wavefunction should be continuous. (Given that the space of wavefunctions is commonly said to be $L^2(\mathbb{R}^3)$, this strikes me as evidently false, since not all functions in $L^2$ are even differentiable.) $\endgroup$ – ACuriousMind May 28 '15 at 13:04
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    $\begingroup$ @ACuriousMind: V.S. Mathur & S. Singh, Concepts in QM, p. 89 $\endgroup$ – Quantum spaghettification May 28 '15 at 13:17
  • $\begingroup$ Mathur & Singh are talking about TISE not TDSE. That are very different situations. For TISE, see my Phys.SE answer here. $\endgroup$ – Qmechanic May 31 '15 at 17:27

There is no need, nor mathematically nor physically, for the wavefunction or its (spatial) derivative to be continuous.

In fact, the space of wavefunctions is usually considered to be an Hilbert space (and there are very poignant physical and mathematical motivations, the first is that any algebra of physical observable that satisfies reasonable assumptions is represented isomorphically as operators on some Hilbert space).

The aforementioned Hilbert space is usually of infinite dimension, and to (irreductibly) represent the finite dimensional canonical commutation relations (a very natural physical requirement) it must be isomorphic to $L^2(\mathbb{R}^d)$. And the functions of $L^2(\mathbb{R}^d)$ can easily be non-continuous and non-differentiable.

With respect to time instead, there is a continuity requirement of the map of wavefunctions $\psi(t):\mathbb{R}\times \mathscr{H}\to\mathscr{H}$ that associates to an initial wavefunction its time evolution. That requirement is, roughly speaking, related to a suitable existence and uniqueness theorem for the solution to the Schrödinger equation, i.e. of the quantum dynamics that is described by the equation $$i\partial_t \psi(t)=H\psi(t)\; .$$ As a matter of fact, since $H$ is required to be a (real-numerical-value) observable, then it should be a self-adjoint operator. That is very nice, since the self-adjoint operators are in 1-1 correspondence with strongly continuous unitary groups of operators; and the group $e^{-itH}$ corresponding to $H$ solves in a unique fashion the Schrödinger equation: i.e. $$\psi(t)=e^{-itH}\psi_0\; ,\; \psi_0\in D(H)$$ is the unique solution in $\mathscr{H}$ of the Cauchy problem $$\left\{\begin{aligned}&i\partial_t \psi(t)=H\psi(t)\\&\psi(0)=\psi_0\end{aligned}\right .\; ;$$ and can be extended to a non-differentiable solution when $\psi_0$ is a generic vector of $\mathscr{H}$. This unique solution is also always continuous in time, i.e. $\psi(\cdot)\in C^0(\mathbb{R},\mathscr{H})$; however as I said it is only differentiable on $D(H)$, that is usually a densely defined Hilbert subspace of $\mathscr{H}$.

Comment (about the referred book). Mathematically speaking, there is a weaker notion of derivative that make the point raised by the authors incorrect. It is possible to define derivatives in a distributional sense, and this does not require continuous functions (it is possible to define a derivative for distributions). Also, exploiting the properties of the Fourier transform there is a quite natural notion of functions of $L^2$ that admit square-integrable derivatives, and they form Hilbert spaces that are called the Sobolev spaces. Again no smoothness at all is required, and this framework is ideal for quantum mechanics. In addition, the existence of solutions to the elliptic equation they cite (the time-independent Schrödinger equation) can be studied (and it is quite customary) on these function spaces with "weak" derivatives.

  • $\begingroup$ Nice answer, though your first argument is not entirely convincing: compactly supported smooth functions are dense in $L^2$ so shouldn't sufficiently well behaved operators on both spaces coincide? From this one could argue that going to the completion makes the math easier, but it doesn't necessarily have a direct physical meaning. $\endgroup$ – doetoe May 28 '15 at 14:27
  • $\begingroup$ @doetoe In my opinion the algebraic point of view in this context is quite enlightening. You may say it is a mathematical point, but in my opinion grasps the physical difference between classical and quantum systems. There is a natural realization of abelian algebras (i.e. classical observables) as $C_0^\infty$ functions, that makes classical states probability distributions. However, there is not for non-abelian algebras (i.e. quantum observables): the latter are naturally realized as operators in Hilbert spaces, with density matrices as (normal) states... $\endgroup$ – yuggib May 28 '15 at 14:40
  • $\begingroup$ And even if we could imagine to restrict the quantum Hilbert space to only the $C_0^\infty$ functions (if that Hilbert space can be written as $L^2$), then you would lose something maybe not in describing observables (that however may map you outside of $C_0^\infty$) but in describing the possible states: in some sense you are forbidding general wavefunctions that are discontinuous in space (and from a physical standpoint I see no reason that the quantum (well also classical) probability distribution of finding a particle outside or inside a box should change smoothly w.r.t. space). $\endgroup$ – yuggib May 28 '15 at 14:44
  • $\begingroup$ I wasn't disputing your conclusion, only the argument. Good point about mapping outside the smooth functions though, you can only restrict the domain, not the codomain, so the math would become particularly messy. $\endgroup$ – doetoe May 28 '15 at 14:56
  • $\begingroup$ @doetoe Ok, I see your point. Anyways as a realistic example any Coulomb potential would map you outside of the $C_0^\infty$ states, for it has a singularity in $x=y$ ($x$ and $y$ being the coordinates corresponding to two particles). So there seems to be, in my opinion, also a physical reason to consider more general wavefunctions than just $C_0^\infty$. $\endgroup$ – yuggib May 28 '15 at 15:12

For the time-dependent Schrödinger equation there might be one. The spatial derivative of the wave function is connected to a "flow of probability" associated with the squared absolute value of the wave function which gives a probability density.

You can view this probability density as a fluid with mass conservation (probability conservation). A discontinuous derivative of the wave function would cause a discontinuous current. And in case of an ordinary fluid this would be rather strange physically, since you are forced to uphold the conservation laws.


I'd say that a good physical reason for the first derivative being continuous is for the probability current to be continuous too, since it's constructed from $\psi(x,t)$ and $ \nabla \psi(x,t)$.


It started out smooth, and it can't go from smooth to not smooth.

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    $\begingroup$ The wave function is typically not assumed to be smooth (i.e. $\mathcal C^\infty$) $\endgroup$ – Danu May 27 '15 at 18:54
  • $\begingroup$ I didn't mean smooth in the strict mathematical sense. I meant C^however high it has to be to stay smooth. $\endgroup$ – DanielLC May 28 '15 at 0:23
  • $\begingroup$ so what exactly would that be, according to you? $\endgroup$ – Danu May 28 '15 at 5:05
  • $\begingroup$ I don't know. I didn't bother to work it out. Maybe it is C^infinity. $\endgroup$ – DanielLC May 28 '15 at 7:50
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    $\begingroup$ it is usually only taken to be (a slightly weakened version of) $\mathcal C^2$. Given that you currently appear unsure of your own argument, I'd consider either improving it or deleting the answer. -1 $\endgroup$ – Danu May 28 '15 at 8:03

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