# Does every representation of the harmonic oscillator Lie algebra necessarily admit a basis of eigenfunctions?

It is well-known in quantum mechanics that the harmonic oscillator Hamiltonian given by $\mathcal{H} = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 - \frac{1}{2}$ admits a basis of eigenfunctions on $L^2(\Bbb R,dx)$. There are many proofs of this ranging from hard analysis (brute force proving density) to Bargmann transform techniques to showing that the resolvent is compact. An important part of the analysis is that the Gaussian $e^{-\frac{x^2}{2}}$ is an eigenfunction with eigenvalue $0$.

Associated to this system is a very nice Lie algebra. Define the operator $a$ by

$$a = \frac{1}{\sqrt{2}}\left(\frac{d}{dx}+x\right)$$

and its (formal - I'm not going through the awful nitty gritty details of domains of definition, etc) adjoint

$$a^* = \frac{1}{\sqrt{2}}\left(-\frac{d}{dx}+x\right).$$

$\mathcal{H}$ has the following representation:

$$\mathcal{H} = a^*a.$$

Moreover, this triple of operators satisfies (on sufficiently nice functions):

$$[\mathcal{H},a^*] = a^*, \qquad [\mathcal{H},a] = -a, \qquad [a,a^*] = -1.$$

The quadruplet $\mathcal{H},a^*,a,1$ generates a Lie algebra. Because of these relations, applying $a^*$ repeatedly to the Gaussian leads to successive eigenfunctions (that they are in $L^2$ needs to be checked, but it is straightforward by induction) and indeed these eigenfunctions form a basis.

My question is this: does a representation of the Lie algebra

$$[a^*a, a^*] = a^*, \qquad [a^*a, a] = -a, \qquad [a,a^*] = -1$$

on an abstract separable Hilbert space $\mathfrak{H}$ (not necessarily $L^2(\Bbb R,dx)$) necessarily admit a basis of eigenfunctions for $a^*a$ assuming that $\ker a\neq \{0\}$? That is, if $\{\psi_i:i\in I\}$ denotes a basis for the kernel of $a$, does $\{(a^*)^m\psi_i:m\in\Bbb N_0,i\in I\}$ form a basis for $\mathfrak{H}$? Or is this somehow a happy accident of working on $\Bbb R$ (or, as we know, more generally $\Bbb R^n$)?

• Up to normalization, some version of this question is equivalent to asking for the Stone-vonNeumann theorem, which asserts the uniqueness of the isomorphism class of the irreducible repn of the Lie algebra of the Heisenberg group with given (non-trivial) unitary central character (thus proving that all models ... of a certain sort... of quantum mechanics are "the same"). In particular, if one such decomposes discretely, then all do... Is this the sort of thing you're asking about? – paul garrett Feb 1 '18 at 23:57
• @paulgarrett Hmm perhaps? I know that Stone-von Neumann would tell us that the $a^*$ and $a$ (or rather linear combinations thereof) are basically $x$ and $p$ on the irreducible representations. However, does that tell us anything necessarily about the eigenfunctions of $a^*a$ themselves? Would it tell us that on the irreps that they are harmonic oscillators? And hence within each irrep, we have a basis, then putting each basis together (by direct sum as per usual), we have a basis for the whole space? – Cameron Williams Feb 2 '18 at 0:20
• Well, perhaps one could take Hilbert direct integrals to thwart the discrete-spectrum property of irreducible repns of the Heisenberg (-Lie) algebra. In a simpler case, e.g., the circle group tends to discretely decompose things, but we can sabotage it if we try. Do you have more hypotheses? – paul garrett Feb 2 '18 at 0:36
• @paulgarrett Well if it's a separable Hilbert space (nonseparable spaces scare me hahah), we wouldn't end up with direct integrals, but instead direct sums, right? Which would keep the spectrum discrete still? – Cameron Williams Feb 2 '18 at 0:37
• Yes, separability would keep things saner. (Not-separable Hilbert spaces are not convivial... E.g., Dixmier's books about C^* algebras and von Neumann algebras would be 1/10-th the size if they did not attempt to accommodate not-necessarily-separable Hilbert spaces.) – paul garrett Feb 2 '18 at 0:40

Assume a unit $e$, and assume there is an underlying Hilbert space $H$. Then $$(a^*a)a^*-a^*(a^*a)=a^* \\ (a^*a-e)a^* = a^*(a^*a) \\ (a^*a-e-\lambda e)a^* =a^*(a^*a-\lambda e) \\ a^*(a^*a-\lambda e)^{-1}=(a^*a-(\lambda+1)e)^{-1}a^*$$ If $E$ is the spectral measure of $a^*a$, Stone's formula for real $r < s$ is $$\frac{1}{2}(E(r,s)+E[r,s]) \\ =s\mbox{-}\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{r}^{s}(a^*a-(u+i\epsilon)e)^{-1}-(a^*a-(u-i\epsilon)e)^{-1}du,$$ which leads to $$a^*E(r,s) = E(r+1,s+1)a^*.$$ Taking the adjoint yields $$a E(r,s) = E(r-1,s-1)a.$$ $E(r,s)H$ is in the domain of $a$ and $a^*$ for all finite $r,s$. The intervals can be open, closed, or half-open. $E$ is the spectral measure of the positive operator $a^*a$; hence $E(-\infty,0)=0$. Therefore $$aE[0,1)=E[-1,0)a=0, \\ a^{n}E[n-1,n)=0,\;\; n=1,2,3,\cdots.$$ Then $a^*aE[0,1)=0$ forces $E[0,1)=E\{0\}$, which is the classical case where the spectrum of $a^*a$ is $\{0,1,2,3,\cdots\}$ and $a^*,a$ act as ladder operators. In the classical case, one also has $\mbox{dim}E\{0\}=1$; in your case, there may be multiple identical and orthogonal copies of the classical one-dimensional space.