# Discreteness of set of energy eigenvalues

Given some potential $V$, we have the eigenvalue problem

$$-\frac{\hbar^2}{2m}\Delta \psi + V\psi = E\psi$$

with the boundary condition

$$\lim_{|x|\rightarrow \infty} \psi(x) = 0$$

If we wish to seek solutions for

$$E < \lim_{|x|\rightarrow \infty}V(x),$$

I find textbooks like Sakurai and others assert that (without proof) set of all such eigenvalues $E$ that admits nontivial solutions is discrete, which gives us liberty to index them with natural numbers (principal quantum numbers). Can anyone help me with a little elaboration of this mathematical fact ? As a student of mathematics, I know that compact operators have discrete spectrums, so for the operator $T : L^2 \rightarrow L^2$given as

$$T(\psi) = -\frac{\hbar^2}{2m}\Delta \psi + V\psi.$$

Are there methods to show that $T$ is compact? Or possibly other ways to establish discreteness of its eigenspectrum?

• T is not a bounded operator and therefore not compact. But it's resolvent might be compact. You will find the answer certainly in one of the Reed-Simon books. May 23, 2013 at 17:09
• There might be some fancy proofs, but the simplest way to see this is through Sturm-Liouville theory, en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory. Schrödinger's equation is a special case of Sturm-Liouville (plug in $p(x) = 1$). Then assuming some conditions on $V(x)$ and the boundary conditions, there is a theorem (see Wiki) that tells you about the spectrum and the eigenfunctions. May 23, 2013 at 19:10
• @smiley06: Can you quote exactly where this is mentioned in Sakurai? Page no. and line no.? Also you need to clarify what you mean by 'non-trivial'.
– guru
Jul 30, 2013 at 9:09
• @guru Last and first lines of page 99-100 of 'Modern Quantum Mechanics' by J.J Sakurai. And $\psi = 0$ is called a trivial solution. Jul 30, 2013 at 10:15

Consider the Sturm-Liouville regular problem in self-adjoint form $$(A-\zeta)\,v=f$$ whose explicit solution is $$v=\int_a^bG(x,s)\,f(s)\,ds$$ where the Green function $$G(x,s)=\begin{cases}\frac{\varphi_b(x)\,\varphi_a(s)}{W(s)},&a\leq{s}\leq{x}\\[0.1in]\frac{\varphi_a(x)\,\varphi_b(s)}{W(s)},&x\leq{s}\leq{b}\end{cases}$$ is built by the solutions $\varphi_a$, $\varphi_b$ of the equation $A\,\varphi=0$ which satisfy respective boundary conditions at $x=a,b$. It's a long way to get here, but you can verify this. The resolvent associated to $A$ is given by $$R(A,\zeta)\,f=\int_a^bG(x,s)\,f(s)\,ds$$ Now, showing that an operator $K$, whose kernel $k(x,s)$ satisfies $$\int_a^b\int_a^b\left|k(x,s)\right|^2ds\,dx<\infty\hspace{0.5in}(*)$$ is compact (meaning the functions $k$ are of square-integrable), may be used to show that if the Green function $G$ of the resolvent operator $R$ is continuous, then $G$ satisfies equation $(*)$ on a finite interval $[a,b]$, meaning that $R$ is a compact self-adjoint operator for regular Sturm-Liouville problems. You could just do this last thing in general, and just show that with the operator $$\hat{H}\equiv{T}=-\frac{\hbar^2}{2m}\nabla^2+V$$ (the Hamiltonian), the eigenvalue problem $\hat{H}\psi=E\psi$ (known as Schrödinger stationary -or time independent- equation) can be reduced to a regular Sturm-Liouville problem.

All of this because, complementing what you said, the eigenvalues of the problem $T\varphi_k=\lambda_k\varphi_k$ for a compact symmetric operator $T$ on a Hilbert space $H$ with interior product $\langle\cdot\mid\cdot\rangle_H$ are a bounded countably infinite set that converges to zero, $\displaystyle{\lim_{k\to\infty}\lambda_k=0}$. Other (pretty relevant) properties are that the multiplicity of each eigenvalue $\lambda_k$ is finite, and that the set of all eigenfunctions $\varphi_k$ define a complete basis of the space $H$, so that any element $f$ of $H$ can be expanded as $\displaystyle{f=\sum_{k=1}^\infty{f_k}\,\varphi_k}$.

Update

Please note the relevance of the condition $\boxed{\displaystyle{E<\lim_{|x|\rightarrow \infty}V(x)}}$. Consider this next image as an illustration (I've taken it directly from the web, so take the potential $V=U$); as the math is pretty straightforward, I'll go a little more with physical picture this time

with the condition above, the results found are valid for the equation $\hat{H}\psi=E_i\psi$ with $i=0,1$. Indeed, both equations may presumably be reduced to a regular S-L problem because they would be ones with a finite or penetrable potential well, where we can know the energy spectrum just from $\psi$ inside the potential well, i.e. for a finite interval, whereas considering $i=2$ the interval for $\psi$ would inevitably be infinite and it is evident that $\hat{H}\psi=E_i\psi$ couldn't be reduced to a regular S-L problem and thus the energy spectrum presumably wouldn't be discrete. Feel free to consider any potential you like.

This corresponds to a beautiful analogy with classical mechanics when we find closed or open orbits, which here are bounded and unbounded states (hope I got the translations right) respectively. Think of a massive nucleus and an electron that interact by means of an atractive Columb potential. If the condition above is satisfied, the system may be a hydrogen-type atom that would have a discrete energy spectrum, as known. Although if someone shot the electron against the nucleus from very far away and with enough kinetic energy, the total energy would be positive and the nucleus would deflect the electron without capturing it and without changing it's energy, which evidently is free to take any value. The simplest case for a continuous energy spectrum in QM is that of the free particle, i.e. $V(\mathbf{r})=0,\,\forall\mathbf{r}\in\mathbb{R}^3$; there the normalization condition $\int\psi^*_n\psi_m\,dx=\delta_{nm}$ is not valid and may be generalized using the Dirac delta, also wavefunctions strictly belong to a Banach space.

• Unfortunately it is still problematic that the interval is not finite. Even if $V(x) > E$ for sufficiently large $x$, the particle is not constrained to a finite region; it can tunnel. The way to resolve this is to abandon the Sturm-Liouville formualtion altogether. Instead, you want to look at minimax principles for the Schrodinger operator in order to find the structure of the whole spectrum, discrete and continuous, which, as you mention, will correspond to the bound and unbound states. Jul 30, 2013 at 7:36
• Maybe I understand where the trouble is. In cases like this, we say the particle is confined, even when the potential is penetrable. For $E_{1,0}$, the wavefunction $\psi$ is not the same "inside" and "outside" the potential, though it has the same energy and satisfies continuity conditions at the potential boundaries. Consider $E_1$, and let it intersect $V$ at $x=x_{11},x_{12}$ with $x_{11}<x_{12}$, then $\psi=\left\{\begin{array}{ll}\psi_1,&x<x_{11}\\\psi_2,&x_{11}<x<x_{12}\\\psi_3,&x>x_{12}\end{array}\right.$.
– user24999
Jul 30, 2013 at 8:09
• From $\hat{H}\psi_2=E_1\psi_2$, it presumably follows that $E_1$ takes discrete values as I mention in my answer. That's what I meant by saying that the interval becomes finite.
– user24999
Jul 30, 2013 at 8:13
• I agree with your physical interpretation, but I'm only commenting that it is impossible to mathematically show that the set of bound states is discrete using the Sturm-Liouville strategy, because there is no way to reduce any aspect of the problem to the S-L approach. If you try to restrict the problem to a PDE with boundary conditions (at the finite boundaries above) that are just continuity conditions, then in the weak formulation the problem fails to be self-adjoint, when you use integration by parts. Jul 30, 2013 at 19:04
• I did not explicitly show how it is reduced (I intentionally kept saying the word presumably) to a regular S-L problem, it might be a rather laborious task. Now, continuity conditions on $\psi_2$ are simply nonhomogeneous Neumann and Dirichlet BCs, so they can be reduced to homogeneous BCs by means of e.g. interpolation of a new function that satisfy the original conditions, so there's always $f,g$ such that $\left(f\dot{g}-\dot{f}{g}\right)\big|_{x_{11}}^{x_{12}}=0$, and therefore the problem might be treated as I show in the first part of the answer.
– user24999
Jul 30, 2013 at 23:42