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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 $\Bbb R^n$)?

Cameron L. Williams
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    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 01 '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 L. Williams Feb 02 '18 at 00:20
  • 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 02 '18 at 00:36
  • @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 L. Williams Feb 02 '18 at 00:37
  • 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 02 '18 at 00:40
  • ... and now it occurs to me that I do not immediately know a "complete reducibility" result for repns of the Heisenberg Lie algebra... So, at least for the moment, I do not know that knowledge of the uniqueness of the irreducible is adequate. Probably someone else knows this immediately, since it should be an old question. – paul garrett Feb 02 '18 at 01:00
  • @paulgarrett Yeah that has been my main hang-up. I'm not sure if it is completely reducible but it's so hard to find hard analysis results because there is so much information about the Lie algebra (due in large part to the QM nature of it) that they get drowned out. I'll keep looking though. If I find anything, I'll be sure to ping you. :) Thanks a bunch. – Cameron L. Williams Feb 02 '18 at 01:22
  • I would indeed be interested to hear how this turns out. Thx! -pg – paul garrett Feb 02 '18 at 02:55
  • @paulgarrett I do not know if this has a full solution, but I think it is at least in the right direction. https://projecteuclid.org/euclid.cmp/1103839851 I have skimmed through it a couple of times, but now I'm digging into the finer details. – Cameron L. Williams Feb 02 '18 at 03:04
  • @paulgarrett I think the answer below might interest you. – Cameron L. Williams Feb 09 '18 at 21:54
  • Yes, indeed! Thanks for the ping. It's easier than I'd thought (modulo Stone's formula, which should count as standard). – paul garrett Feb 09 '18 at 22:19

1 Answers1

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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.

Disintegrating By Parts
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  • Wow this is such an obvious approach in retrospect haha. I'm still going over the finer details, but I love it! Once I go over the fine details and it all looks good, I will award you the bounty! – Cameron L. Williams Feb 09 '18 at 14:28
  • @CameronWilliams : It's always good to check what I do. Trust me on that. :) – Disintegrating By Parts Feb 09 '18 at 16:52
  • So just a couple of points of clarification. Going from $a^(a^a-\lambda) = (a^a - (\lambda+1))a^$ to $a^(a^a-\lambda)^{-1} = (a^a-(\lambda+1))^{-1}a^$ is the usual functional calculus extension, I assume (perhaps via a formal power series)? And secondly: the spectral theorem has so many variants that I get confused which formulations (easily) tell you what. The spectrum is clearly only ${0,1,2,\ldots}$. Unless I am mistaken, $E(\Bbb C)$ would be the identity operator, i.e. the projections onto the subspaces with eigenvalues $0,1,2,\ldots$ would span the space? – Cameron L. Williams Feb 09 '18 at 17:26
  • Furthermore, this does not guarantee that the eigenvalues $0,1,2,\ldots$ are realized without the added assumption that $\ker a \neq {0}$, right? This approach suggests to me that for positive operators with ladder operators (under modest assumptions), the spectrum is always the eigenvalues. – Cameron L. Williams Feb 09 '18 at 17:30
  • @CameronWilliams: Yes, the whole space decomposes as an sum of Eigenspaces. Moreover, $a: Eig_\lambda(a^\ast a) \to Eig_{\lambda-1}(a^\ast a)$ is injective for $\lambda=1,2,...$ because $\frac{1}{\lambda}a^\ast$ is a left-inverse. Therefore: If any eigenvectors exist at all (i.e. if $H\neq 0$), then there also exist at least as many elements of the kernel. And vice versa: Every element of the kernel gives you a sequence of eigenvectors with the required property. This is what the last sentence means: $H$ decomposes as direct sum of $\dim(\ker(a))$ many copies of the classical case. – Johannes Hahn Feb 09 '18 at 18:54
  • @JohannesHahn I just realized that it's guaranteed that the spectrum is nonempty, so that you can apply the lowering operator to then get the $0$ eigenvalue eigenfunctions(s), then build back up with the raising operator. – Cameron L. Williams Feb 09 '18 at 19:08
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    Thanks guys! I am rather disappointed that I couldn't find this proof anywhere or this result anywhere despite hours and hours of searching. Hopefully this will be very useful for others. – Cameron L. Williams Feb 09 '18 at 19:15
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    @cameronwilliams : I have tried to do this several times over the years and could not. So I was elated that I saw the missing piece this time. Who knows, maybe this is new. :) If it is, then publish it; I don't care, this is just a hobby for me. – Disintegrating By Parts Feb 09 '18 at 19:33
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    @DisintegratingByParts I have been trying for quite some time as well, though I was trying to do it straight from the spectrum formula - which was a nightmare. Spectral measures are so much more friendly. I don't like to take credit for someone else's work, but I might just reference this MSE post in a paper of mine though. :) – Cameron L. Williams Feb 09 '18 at 19:43
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    Yes, a quite pretty argument. This should be a fundamental example... e.g., of application of Stone's formula, but/and abstracting the "quantum harmonic oscillator" biz, obviously. – paul garrett Feb 09 '18 at 22:22
  • @paulgarrett : Thank you. – Disintegrating By Parts Feb 09 '18 at 23:00