How Different is an Eigenvalue Problem from an Ordinary Differential Equation

Question

I have been thinking about how different an eigenvalue problem such as that of the Sturm-Liouville Equation (SL) with that of a second-order linear differential equation. It doesn't seem to be the case to be the difference between a boundary value problem vs. an initial value problem, as we may also have BVPs that do not involve eigenvalues, in which case solutions might not exist in abundance as opposed to the case with IVPs. Perhaps, the eigenvalue parameter is there to allow us to speak of to which values of the parameter there exist a solution.

However, that brought me to the question as to what happens when no boundary values are imposed, such as in the case of the famous Simple Harmonic Oscillator (SHO) SL problem in Quantum Mechanics (QM), where the wavefunction, i.e. the solution of the "eigenvalue" problem, is defined throughout the entire real line without any restriction - except for the fact that they approach the $x$-axis asymptotically, which actually follows from square-integrability. I put the word "eigenvalue" in quotation marks because I am now questioning it. Let me explain.

The SHO problem is the seeking of the solution $\psi$ to the "eigenvalue" problem $\displaystyle-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}m\omega^2x^2\psi=E\psi$. We can clean up the equation by absorbing constants, but we will keep them. It is well-known in QM that solutions to this eigenvalue problem are products of Hermite polynomials with a Gaussian and the eigenvalues $E$ form a discrete set and are given by $E_n=\left(n+\frac{1}{2}\right)\hbar\omega$. Note that no boundary conditions are imposed although, of course, we are seeking for square-integrable functions. Herein lies the question:

If we rewrite the SHO equation as $\displaystyle-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\left(\frac{1}{2}m\omega^2x^2-E\right)\psi=0$, it does look to me like a legit 2nd order variable-coefficient linear ODE that has a solution for any $E$. Specifically, $E$ does not have to assume specific values that form a discrete set.

What am I missing here? That not all $E$s give square-integrable solutions? How is the SHO an eigenvalue problem and not a mundane differential equation problem?

Appreciate your insights.

Answer 1

Your problem has always the zero function as solution. So the real question to ask is how to get solutions apart from that. So in effect you are asking for a singular situation that only isolated values of $E$ will satisfy.

You can transform this into an ordinary boundary value problem by

• (zeroth reducing the infinite interval to some finite interval $[x_0,x_2]$)

• first adding $E$ as component with equation $\frac{dE}{dx}=0$, and

• then add the square integral as fourth component with equation $\frac{dI(x)}{dx}=|ψ(x)|^2$.

Now you have 4 state components and thus 4 slots for boundary conditions, which are $$ψ(x_0)=ψ(x_1)=0~\text{ and }~I(x_0)=0, ~~I(x_1)=1.$$

As with many BVP there will not be a unique solution, if you employ a numerical solver, the initial guess has to be close to some eigenfunction to get predictable results.

Answer 2

From the point-of-view of Quantum Mechanics, a valid solution of the harmonic oscillator equation must be square integrable in $x$, because the square of a valid solution is a probability distribution. That does not mean that the solution cannot be built from non-square-integrable solutions of $H\psi_{\lambda}=\lambda \psi_{\lambda}$ through a Fourier transform type of integral. But it would mean that there would not be exact energy levels for the oscillator if there were no $L^2$ eigenfunction solutions of $H\psi_{\lambda}=\lambda\psi_{\lambda}$ for some real $\lambda$.

When looking at the solutions of $H\psi= \lambda\psi$, where $\lambda$ is a separation constant that is also constant of energy, there's no reason to assume a priori that there will be $L^2(\mathbb{R})$ solutions for a given real $\lambda$. You let the chips fall where they may, and you find out that there is a discrete, countably infinite set of $\lambda_0 < \lambda_1 < \lambda_2 < \cdots < \lambda_n \rightarrow\infty$ for which there are $L^2(\mathbb{R})$ solutions and, furthermore, the null space of $\mathcal{N}(H-\lambda_n I)$ is one-dimensional for all $n=0,1,2,3,\cdots$. You can prove there is no continuous spectrum based solely on the fact that the normalized eigenfunctions $\{\varphi_n \}$ of $H$ form a complete orthonormal basis, which can be proved directly. The conclusion is that the Harmonic Oscillator has an infinite, discrete set of possible energy levels, with the only cluster point of energy being $\infty$. There is no continuous range of spectrum corresponding to unbound states. This seems reasonable from a Physics point-of-view, and is verified with Mathematics.

To follow this up: there is always one non-trivial solution of $H\psi_{\lambda}=\lambda\psi_{\lambda}$ that is square integrable on $(-\infty,0]$ and there is another non-trivial solution, say $\phi_{\lambda}$, that is square integrable on $[0,\infty)$. $\lambda$ is an eigenvalue iff $\psi_{\lambda}$ and $\phi_{\lambda}$ are linearly dependent and, in that case, $\psi_{\lambda}$ is square integrable on $\mathbb{R}$. So the eigenvalue equation is determined by finding $\lambda$ such that $\{ \psi_{\lambda},\phi_{\lambda} \}$ is a dependent set of eigenfunctions, which becomes a Wronskian condition. This is similar to the formalism that works for determining eigenvalues and eigenfunctions in cases where there is a left endpoint condition and a right endpoint condition to be satisfied in order to have an eigenfunction. That's why it is common to find that eigenvalues are zeros of a holomorphic function.