The Harmonic Oscillator as a Self-Adjoint Operator

The harmonic oscillator has an explicit complete eigenbasis, making its unbounded domains visible as weighted sequence spaces. The canonical oscillator treatment develops the energy and wavefunction derivations. This page uses those results to establish which infinite expansions belong to the Hamiltonian domain, why finite Hermite sums form a core, and where ladder manipulations require qualifications.

Required background. Self-Adjoint Operators supplies the domain criterion. Test Function Spaces connects Schwartz regularity to oscillator powers.

The Hermite basis defines a maximal diagonal domain

For $m,\omega>0$, introduce $y=x\sqrt{m\omega/\hbar}$. The dimensionless oscillator is

$$L=-\frac{d^2}{dy^2}+y^2,\qquad H=\frac{\hbar\omega}{2}L.$$

The normalized Hermite functions $h_n$ form a complete orthonormal basis and satisfy $Lh_n=(2n+1)h_n$. Completeness is a substantive Hermite-function theorem, not a consequence of a ladder formula alone.

For $\psi=\sum_{n\geq0}c_nh_n$, define

$$D(L)=\left\{\psi:\sum_{n\geq0}(2n+1)^2|c_n|^2<\infty\right\}.$$

The diagonal operator on this domain is self-adjoint because it is multiplication by a real sequence under the unitary Hermite transform. Its spectrum consists precisely of the displayed eigenvalues, each simple.

The associated form domain has the weaker weight $\sum_n(2n+1)|c_n|^2<\infty$. Thus finite mean oscillator energy does not imply membership in the operator domain. Arbitrary square-summable coefficients still define states and unitary evolution.

Differential regularity agrees with the spectral domain

The maximal differential realization has

$$D(L)=H^2(\mathbb R)\cap D(y^2),\qquad Q(L)=H^1(\mathbb R)\cap D(y).$$

An estimate explaining the operator-domain statement follows first on Schwartz functions. Integration by parts gives

$$\|L\psi\|^2 =\|\psi''\|^2+\|y^2\psi\|^2 +2\|y\psi'\|^2-2\|\psi\|^2.$$

Therefore the graph norm of $L$ controls both $\psi''$ and $y^2\psi$. The reverse bound follows from the triangle inequality. Closure, together with smooth cutoff and mollification approximation in these norms, identifies the differential domain with the weighted Hermite domain.

The form identity is simpler: $\langle\psi,L\psi\rangle=\|\psi'\|^2+\|y\psi\|^2$ on the core, followed by form closure. It controls one derivative and one position weight, not two.

These distinctions explain why requiring a wavefunction merely to decay at infinity does not specify the operator domain. Both derivatives and weighted integrability matter.

Finite Hermite sums are an operator core

Let $\Pi_N$ project onto $h_0,\ldots,h_N$. For $\psi\in D(L)$,

$$\|\psi-\Pi_N\psi\|^2 +\|L(\psi-\Pi_N\psi)\|^2 =\sum_{n>N}\bigl[1+(2n+1)^2\bigr]|c_n|^2 \longrightarrow0.$$

Thus finite Hermite sums are dense in the graph norm. The symmetric differential operator on Schwartz space contains this core and is contained in the maximal realization, so it is essentially self-adjoint. Smooth compactly supported functions also form a core by the graph-norm approximation above.

The resolvent is compact because its diagonal coefficients $(2n+1-z)^{-1}$ tend to zero. Finite-rank resolvent cutoffs converge in operator norm. This compactness is stronger than having a list of known eigenfunctions; it excludes an omitted continuous spectral component once completeness is established.

Ladder operators require weighted domains

In the Hermite basis the closed annihilation and creation operators act by

$$ah_n=\sqrt n\,h_{n-1},\qquad a^*h_n=\sqrt{n+1}\,h_{n+1}.$$

Both have domain $\{\psi:\sum_n n|c_n|^2<\infty\}$; adding the zeroth norm makes the weights $n$ and $n+1$ equivalent. Their products $a^*a=N$ and $aa^*=N+I$ have domain $\sum_n n^2|c_n|^2<\infty$.

Hence $[a,a^*]\psi=\psi$ is an operator identity on $D(N)$, including finite Hermite sums. It is not an invitation to apply either product to an arbitrary normalized state. The formal differential identities become valid after their product domains are specified.

The same issue persists under computation: compressing a ladder operator to finitely many levels changes its commutator at the upper boundary.

Exercise: calculate the finite-level commutator defect

On $\operatorname{span}\{h_0,\ldots,h_N\}$, put $a_N=\Pi_Na\Pi_N$. Compute $[a_N,a_N^*]$.

Solution

For $n<N$, the ordinary ladder actions give $[a_N,a_N^*]h_n=h_n$. At the upper level, $a_N^*h_N=0$ while $a_N^*a_Nh_N=Nh_N$, so the commutator is $-N$ there. Thus

$$[a_N,a_N^*]=I_{N+1}-(N+1)|h_N\rangle\langle h_N|.$$

Its trace is zero, as every finite commutator’s must be. Expectations approximate the canonical value only when the top-level population multiplied by $N+1$ is controlled; small population alone without its scaling is not a uniform bound.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — oscillator eigenfunctions and unbounded ladder operators.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Section 8.3 — Hermite completeness and the oscillator realization.