About the traditional explanation of the continuity of the first derivative of a 1D wavefunction

Question

I would like to receive some clarifications about the traditional explanation of the continuity of the first derivative of a 1D wavefunction (E.g. see the very clear answer by @ZeroTheHero Continuity of wave function derivative, but there are other similar answers in PSE, some by @Qmechanic, e.g. Continuity & smoothness of wave function).

Before introducing the issue, some premises are necessary. As some of PSE friends know, I am a mathematical physicist, but I studied theoretical physics, so that I am able to well understand and also to use the various arguments and theoretical procedures proper of theoretical physics.

The approaches of mathematical physics and theoretical physics are quite distinct. Roughly speaking, TP uses the mathematical machinery as a slave, MP as a guide.

• Theoretical physics uses basic, but crucial, facts of the physical phenomenology to interpret more complex facts by a suitable theoretical ``dilatation'' of the setup: The theoretical description of the two-slits experiment encompasses almost all the explanations of quantum phenomena of a single constituent (e.g., physics of entanglement is excluded).

• Mathematical physics instead constructs a fine dictionary to translate the physical world to the mathematical universe, one works out the objects of the latter, and she/he comes back to physics through the inverse use of the dictionary.

However, when physics works well, then mathematics works well as well or, in some cases, a new mathematics pops out capable to explain new physics.

Let us come to the issue. Let us focus on the Schrödinger equation $$H_0 \psi_E = E \psi_E$$ on the real line $x\in \mathbb{R}$ with a continuous potential $V=V(x)$, that can be singular at a point $x_0$. We are looking for eigenfunctions (I suppose properly normalized) $\Psi_E$ of the Hamiltonian $$H_0 := -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x)\:.$$

A basic fact one already learns in elementary courses of QM is that, if the singularity of $V$ is sufficiently tamed -- it is discontinuous at $x_0$ but the limits $V(x_0^-)$ and $V(x_0^+)$ exist and are finite -- then $\psi_E$ must be everywhere twice differentiable, excluding $x_0$, where it has continuous first derivative.

When I was a physics student, so adopting the perspective of the TP, I got crazy to understand the physical reasons of that continuity result.

I know the reason as a mathematical physicist (*), but I am here interested in the traditional answer of TP one can find also in textbooks written by very outstanding theoretical physicists.

One of the traditional explanations in TP of the continuity requirement (I know another slightly different explanation based on the fluid dynamics interpretation of the wavefunction) is something like this.

For some constant $c\neq 0$ (it trivially arises out of the Schroedinger equation), $$c \int_{x_0-\epsilon'}^{x_0+\epsilon} (V(x)-E) \psi(x) dx = \int_{x_0-\epsilon'}^{x_0+\epsilon} \frac{d^2 \psi}{dx^2} dx = \frac{d\psi}{dx}(x_0+\epsilon)- \frac{d\psi}{dx}(x_0-\epsilon')\:.$$

The leftmost hand side is finite if $\psi$ is continuous and $V$ has a finite step singularity. Hence, the limit for $\epsilon, \epsilon' \to 0^+$ yields the continuity of $\frac{d\psi}{dx}$ at $x_0$.

I do not understand well this argument because I do not understand well the hypotheses and the physical reasons behind them.

It seems to me that we are assuming here that

• (1) $\psi$ is continuous around $x_0$;
• (2) $\frac{d^2\psi}{dx^2}$ is integrable around $x_0$.

I cannot see any cogent physical reason to assume (1) and (2). One could assume directly from scratch that $\frac{d\psi}{dx}$ is continuous.

Why are the two hypotheses above more physically preferable than the direct assumption of the continuity of $\psi'$?

However, even clarifying that fact,

continuity of $\psi'$ seems to me as physically arbitrary as the couple of hypotheses above.

An "at end of the day" physical reason would be that, assuming continuity of $\frac{d\psi}{dx}$ or, equivalently, the two requirments above, there exist a basis of eigenfunctions of $H_0$ (with the eigenvalue equation suitably interpreted at the discontinuity of $V$). But this is exactly the mathematical physics reason (*): we are actually looking for the eigenfunctions of $H^\dagger$ which is selfadjoint.

Maybe, I am missing some physical subtle point in my view and I am asking you for some illumination.

---

(*) A basic requirement on observables is that they are self-adjoint operators, as this assures that they have a spectral decomposition which is one of the basic conceptual tools of QM. $H_0$ is not selfadjoint because, if it were selfadjoint, then $H_0=H_0^\dagger$, but the latter is never a differential operator if $H_0$ is. As is known, the "true" observable is $H_0^\dagger$ because, on a certain domain $D(H^\dagger_0)$ it holds $H_0^\dagger = (H_0^\dagger)^\dagger$. Notice that $H_0$ completely determines the observable $H_0^\dagger$ in that way, as it is expected from physics. The vectors in that domain are some functions in $C^2(\mathbb{R}\setminus \{x_0\})\cap C^1(\mathbb{R})$ as first established by H. Weyl. So, when requiring that the Hamiltonian is selfadjoint, we, in fact, have that the relevant eigenfunctions have continuous derivative where $V$ is discontinuous. That is because they are eigenvectors of $H_0^\dagger$ rather than $H_0$ itself, which would require more regularity. The result extends to $\mathbb{R}^n$ in various ways.

Answer 1

This "explanation" is probably too naive, but for some physicists, the derivative of the $\theta$-function is the $\delta$-function, $\theta'(x) = \delta(x)$. This means that the derivative of the function $\psi(x)$, which is discontinuous at some point $x_0$, contains the term $A\delta(x-x_0)$, where $A = \psi(x_0+0) - \psi(x_0-0)$. Therefore, if the potential $V(x)$ does not contain $\delta$-terms, then the solution of the stationary Schrodinger equation, $\psi(x)$, and its derivative, $\psi'(x)$, must be continuous functions. Otherwise, $\delta$ and $\delta'$-terms will appear from $-\frac{\hbar^2}{2m}\psi''(x)$ and will not be reduced by $(V(x)-E)\psi(x)$.

The same naive logic works well in the case when the potential is $\delta$-functions, $V(x) = U_0\delta(x/a)$. The wave function $\psi(x)$ is continuous at $x=0$, so we can replace $V(x)\psi(x)$ with $\psi(0)U_0\delta(x/a)$. The derivative $\psi'(x)$ is discontinuous at $x = 0$, so $-\frac{\hbar^2}{2m}\psi''(x)$ produces the term $A\delta(x)$, where $A = -\frac{\hbar^2}{2m}(\psi'(+0)-\psi'(-0))$. The requirement that two $\delta$-terms reduce each other leads to a well-known condition $$ -\frac{\hbar^2}{2m}(\psi'(+0)-\psi'(-0)) + U_0a\psi(0) = 0. $$

Answer 2

The question of the properties of $\psi$ were a subject of debate shortly after the original work of Schrödinger. He assumed that $\psi$ was continuous, single-valued, and that $\psi'$ always existed.

A good source for the history is

Jammer, M., 1989. The conceptual development of quantum mechanics, 2nd ed.

Since $\hat p= i\hbar \frac{d}{dx}$ in the position representation, the (original) argument is that $\psi'$ must exist everywhere else momentum would not be defined everywhere, which seems physically implausible. This is apparently as good at it gets in terms of justification.

Another line of argument, proposed by Langer and Rosen (the latter of the Einstein-Podolsky-Rosen fame) in Phys.Rev. 37 (1931) p. 658, argues that the fundamental quantity is a variational so "the function $\psi$ is determined by the conditions that make" the integral $$ J=\int \left(\frac{h^2}{4\pi^2}T(q,\frac{\partial \psi}{\partial q})+V\psi^2\right)d\tau $$
a minimum. Both authors make it clear that

A rough working rule we may demand of the function is that it be integrable in the square and that it be finite and continuous wherever the potential energy is finite.

Jaffe in Zeitschrift für Physik, vol 66 (1930) pp.770-774 (the paper is in German and so I only have an approximate appreciation) insisted that only physical arguments should determine $\psi$ and conceded that the function must be continuously differentiable throughout configuration space and must have a second derivative except possibly at points where the potential is discontinuous.

Of course there are situations in classical physics where the change in momentum is not continuous: a solid marble hitting perfectly elastically an infinitely hard wall for instance. In this example, the momentum changes from $+p$ to $-p$ discontinuously. In this case one can legitimately wonder what is the momentum "at the wall" and the physics answer is often to take it to be $0$ (as the average of the values on either side).

This infinitely hard wall is an idealization, and one can "tame" the discontinuity by taking the limit of a model where the change in momentum is continuous (some sort of impulse calculation which yields the change in momentum without worrying too much about the details of the force causing this change). It is in this spirit that some limit procedure is used to handle the discontinuity of $\psi'$ at those points where the potential has infinite discontinuities (edge of an infinite well, $\delta$-well or $\delta$-barrier, etc)

Kennard in Nature vol. 127 (1931) pp.892-893 proposes that the solutions of the Schrödinger equation must form a complete set, irrespective of issues about derivatives of this function.

It should be noted that once Born realized that the physically meaningful quantity was not $\psi$ but rather $\vert \psi\vert^2$, the argument moved from properties of $\psi$ to properties of $\vert \psi\vert^2$. The discussion was dominated by the issue of single-valuedness, especially after the introduction of spin state, but the arguments to justify the existence of $\psi'$ seems to have remained unchanged. In fact, continuity needs to be discussed using $\vert \psi\vert^2$ rather than $\psi$ itself. As a simple example, the radial wavefunction of hydrogen $R_{10}(r)\sim e^{-r/a_0}$ does not go to $0$ as $r\to 0$, but the probability density in spherical, which is $r^2 \vert R_{10}(r)\vert^2$, does.

The issue of single-valuedness was studied by Pandres in J.Math.Phys vol. 3 (1962) pp.305-308. He claims that single-particle $\psi$ must be single-valued else there are issues with the N-particle wave functions. I did not have time to carefully read this paper.