# Domain of simple quantum harmonic oscillator

When discussing the spectral theory of unbounded operators, one often starts with an operator defined on a densely defined subspace of your Hilbert space, and then proves that the operator is essentially-self adjoint which allows you to apply a spectral theory. In the case of the simple quantum harmonic oscillator one starts with the operator

$$L: -\partial_{xx} + x^2$$

defined on the dense domain of smooth compactly support functions, $C^{\infty}_0(\mathbb{R}) \subset L^2(\mathbb{R})$. Then one proceeds to show that $L$ is essentially self-adjoint which means that the closure of $L$ (a closed extension of $L$ to some larger domain) is self-adjoint.

My question is: What is the domain of this extension? My guess is that $L$ extends to $\bar{L}$ where the domain of $\bar{L}$ is given by:

$$D(\bar{L}) = \{ e^{-x^2/2}P(x) : P(x)\,\, \text{is a polynomial.}\, \}$$

Is this true? If not, then does anyone know explicitly the domain of $\bar{L}$? Any references illustrating this fact would be greatly appreciated.

As you may know there are eigenfunctions of $L$ written by means of the Hermite functions; they also forms a basis of $L^2(\mathbb{R})$. Let's call $\{h_n(x)\}_{n\in\mathbb{N}}$ such a basis. Also, as usual, denote by $$\langle h_n,\psi\rangle=\int_{\mathbb{R}}\bar{h}_n(x)\psi(x)dx$$ the scalar product between the $n$-th Hermite function and a generic function $\psi\in L^2(\mathbb{R})$. Then the domain of $L$ can be written as: $$D(L)=\Bigl\{\psi\in L^2(\mathbb{R})\, ,\, \sum_{n=0}^\infty n^2 \lvert\langle h_n,\psi\rangle\rvert^2<+\infty\Bigr\}\equiv \Bigl\{\psi\in L^2(\mathbb{R})\, ,\, n\langle h_n,\psi\rangle \in l^2\Bigr\}$$ where $l^2$ is the usual sequence space over $\mathbb{C}$.
• One cool thing about the Hermite functions is that they are analytic vectors for the position and momentum operators $\hat{X}$ and $\hat{P}$. This fact is used in a proof of the infinitesimal version of the Stone-von Neumann Theorem. – Berrick Caleb Fillmore Nov 28 '14 at 6:47
First of all, if you focus on proper functions, instead of elements of $L^2$, the domain is much more tough than your candidate ($L$ extended to our domain is again simply essentially self adjoint but not self-adjoint). The self-adjointness domain contains functions which are nowhere differentiable. A trivial example: If you consider the simpler operator: $$P':= -i \frac{d}{dx} : C_0^\infty(\mathbb R) \to L^2 (\mathbb R)\:,$$ its unique self-adjoint extension is defined on the space of $L^2$ functions which admits weak derivative which, in turn, is $L^2$ (there is another equivalent characterization for $n=1$ but it is not relevant here). The function $e^{-x^2}$ satisfies these requirements, but so does the function $$\psi(x) = e^{-x^2} \quad \mbox{if x is irrational}\quad \psi(x) = x \quad \mbox{otherwise}$$
Coming back to your operator, if you remember that essentially self adjoint is also equivalent to say that $L^\dagger$ (which coincides with the closure $\overline{L}$ of $L$) is self-adjont, you realize that the self-adjoiness domain is simply $D(L^\dagger)$. So $$D(\overline{L}) = \left\{\psi \in L^2(\mathbb R) \:\left|\: \exists \phi_\psi \in L^2(\mathbb R), \: \int L(f)\psi dx \right.= \int f\phi_\psi dx \:, f \in C_0^\infty(\mathbb R)\right\}$$ You see that, if just focusing on the $C^1$ functions, the domain includes all $C^1$ functions which vanish at infinity faster than $1/x$, for instance. This set is much larger than your doamain. But it is not even necessary that these functions are differentiable.