Why can non-differentiable solutions to the Schrödinger equation be ignored?

Question

To clarify the question, let's consider the particle in a box (infinite potential $V$ outside [0,1], potential 0 inside [0,1]). (But the problems illustrated here also apply to particles in a non-infinite well.)

The differential equation for the time-independent Schrödinger equation is then: $$ H\psi = E\psi, \quad\text{where}\quad H\psi := -i\psi'' + V\psi $$ (I am omitting constants like $\hbar$ and $m$, and $\psi''$ is the twice-derivative on the position $x$.) Here $\psi$ is supposed to be an element of $L^2([0,1])$. ($L^2(\mathbb R)$ for non-infinite wells.)

In other words, we are looking for eigenvectors of $H$.

Then all accounts I read solve this differential equation / eigenvalue problem, get solutions $$\psi(x) = A\sin(k\pi x)+B\cos(k\pi x).$$ Then they argue that $\psi(0)=\psi(1)=0$ (typically referring to continuity, this part seems vague to me). Hence $B=0$ and $k\in\mathbb N$. So all solutions / eigenvectors are $\psi(x) = \sin(k\pi x)$ for $k\in\mathbb N$. The corresponding eigenvalues are $\pi^2k^2$.

However, I would claim that there are additional solutions: For example, $\psi(x) := \sin(2\pi x)$ on $[0,1/2]$ and $:=0$ outside. (I.e., we take the solution for $k:=2$, and set it to $0$ on $[1/2,1]$. This still is a $\pi^2k^2$-eigenvalue of $H$.

Note: one might complain that $H$ is not defined on this $\psi$ because $\psi$ is not differentiable at $1/2$. But since we are in $L^2([0,1])$, countably many undefined points don't matter, so $H$ is well-defined on $\psi$.

We can even have non-continuous $\psi$ that are additional eigenvalues of $H$. (E.g., $\psi(x) := \sin(\pi x)$ on $[0,1/2]$ with eigenvalue $\pi^2$.) Or we can piece together several shifted solutions $\sin(k\pi x + \delta)$. Even worse, we now have (even continuous) solutions for any real eigenvalue $E\geq \pi^2$, namely $\sin(\sqrt E x)$ on $[0,\pi/\sqrt E]$.

Now something is definitely wrong. Not only does this contradict all accounts I have read, but it is also mathematically inconsistent: If we have eigenvectors for all $E\geq \pi^2$, we get an uncountably large set of mutually orthogonal vectors. That in turns implies that the Hilbert space we are in ($L^2([0,1])$) is not separable. But $L^2([0,1])$ is known to be separable.

Question: Where is the flaw in my reasoning? Can we discount all non-differentiable solutions? Why? Or if not, how is the mathematical contradiction resolved, at least?

Answers I thought of:

• One can argue physically and say that we are simply not interested in non-differentiable solutions to the Schrödinger equation. (I.e., add this as an additional assumption in the physical model.) This is unsatisfactory to me because:
  • There remains the contradiction of purely mathematical nature. I.e., even if we say the I solved something different from what the physics wants, I still have solved a mathematically well-defined eigenvector problem and gotten a contradiction to the separability of $L^2$.

Answer 1

This is an issue of mathematical nature, however it has physical implications so that it deserves to be tackled.

Spectral theory as used in QM applies to so-called normal operators. In the concrete case at hand selfadjoint operators.

The spectral decomposition holds for this type of operators.

Selfadjoint operators cannot be differential operators, since the adjoint of a differential operator (of order $\geq 1$) is never differential.

If starting from symmetric differential operators, we have to look for selfadjoint extensions to apply the spectral machinery.

The fortunate case is when the initial operator is essentially selfadjoint, i.e., it admits a unique selfadjoint extension.

In general, the eigenvectors of selfadjoint extensions are not smooth functions, just because the actual operator whose they are eigenvectors is not a differential operator!

Nevertheless, there are several regularity conditions the must satisfy. In particular, if we start with a pure Laplacian on a bounded set with suitable boundary conditions which made it symmetric and essentially selfadjoint, the eigenfunctions of the selfadjoint extension are smooth.

Let us come to the concrete case.

One is looking here for the spectrum of the unique selfadjoint extension in $L^2([0,1])$ of the Laplace operator on $[0,1]$ initially defined on the domain $D$ of the smooth functions vanishing at the boundary.

(Alternatively you should fix another domain where the differential operator is essentially selfadjoint, but the properties you are discussing are referred to $D$).

Using the heat kernel theory as well as results of elliptic regularity, it is possible to prove that, in fact, there is only one such selfadjoint extension and that the following facts hold for the unique selfadjoint extension.

1. The spectrum is a pure point spectrum.

2. The eigenvectors are smooth functions in $(0,1)$.

The second point already rejects your supposition.

A more direct approach is the following one. On the mentioned domain $D$ the operator is symmetric. On the other hand, the standard countable Hilbert basis of $\sin$ functions is a family of eigenvectors of the said operator. I indicate them by $|n\rangle$. Hence this symmetric operator admits a set of analytic vectors whose span is dense. At this juncture, Nelson’s theorem immediately implies that the operator is essentially selfadjoint on that domain. A selfadjoint operator which extends the Laplacian can be constructed by summing over the orthogonal projectors of the said basis (in the strong operator topology) $$A = \sum_{n=1}^{+\infty}(n\pi)^2 |n\rangle\langle n|$$ Therefore that is the selfadjoint extension we are looking for. We conclude that the unique selfadjoint extension has the spectrum of $A$, namely, a point spectrum with eigenvectors given by the said family of smooth functions. No further eigenvectors, eigenvalues are permitted, and there is no continuous part of the spectrum.

The formal eigenfunctions you found are not eigenvectors of the initial symmetric operator. But also via direct inspection you see that they are not orthogonal to the standard eigenvectors which form a Hilbert basis, so that they are generally infinite linear combinations of them. Furthermore the discontinuous nature of them (and their first derivative) should be treated with the due care to avoid pathologies.

Answer 2

This is a response to supplement Valter Moretti's answer and to elaborate on my comments on his answer. The key idea is as follows. We're interested in operators of the form $H_0 = -\Delta\psi+V\psi$, where $V\in C^\infty$ is a smooth potential, and $H_0: C_c^\infty((-1,1))\to L^2([-1,1])$. I should note that essential self-adjointness does not even come into the picture. All we need to know is that the true self-adjoint operator is some hermitian extension $H:\mathcal{D}(H)\to L^2([-1,1])$, where $C_c^\infty((-1,1))\subset \mathcal{D}(H)$, and $H$ agrees with $H_0$ on $C_c^\infty((-1,1))$. We will denote this relation as $H_0\subset H$. Then notice: suppose that $\phi\in\mathcal{D}(H)$, $\psi \in C_c^\infty$. Then $$\langle {\psi, H \phi\rangle} = \langle{ H\psi, \phi\rangle}$$ because $H$ is hermitian and $\mathcal{D}(H_0)\subset\mathcal{D}(H)$. But then, $$=\langle H_0 \psi, \phi\rangle$$ because $H_0\subset H$. We conclude then that on any possibly non-smooth vector $\phi$, we need the intergration-by parts identity $$\int (H\phi)(x)\psi(x)dx = \int\phi(x)(-\Delta+V) \psi(x) dx$$

I leave it to you to show that the way you evaluated the Laplacian does not satisfy this identity.

The distribution $T(\phi) = \int \phi(x)\Delta\psi(x)$ is known as the distributional derivative.

Awesome fact: We have this wonderful family of theorems known as elliptic regularity. What it says is that if $P$ is an elliptic differential operator with smooth coefficients, and if $T$ is a distribution satisfying $PT=0$, then actually $T$ was a smooth function. Applying this to $(H-E)T=0$ then implies that the eigenfunctions of a hermitian extension of $H_0$ must all be infinitely differentiable.