Does the Trotter Product Formula apply to the Simple Harmonic Oscillator? We can write the Hamiltonian for the simple harmonic oscillator as $\hat{H} = \hat{T} + \hat{V}$ where $\hat{T} = \dfrac{-\hbar^2}{2 m} \dfrac{\partial^2}{\partial x^2}$ and $\hat{V} = \dfrac{1}{2} m \omega^2 X^2$.
The Trotter product formula (see Thm 20.1 of Hall's Quantum Theory for Mathematicians) states that for $\hat{T}$ and $\hat{V}$ self-adjoint operators, and $\hat{H}$ densely defined and essentially self-adjoint on $\mathrm{Dom}(\hat{T}) \cap \mathrm{Dom}(\hat{V})$, then $\exp{i t \hat{H}} = \lim_{N \to \infty} \Big(\exp{i t \dfrac{\hat{T}}{N}} \exp{i t \dfrac{\hat{V}}{N}}\Big)$.
My question is: do $\hat{T}$ and $\hat{V}$ for the SHO satisfy these conditions?
The reason I ask is that the SHO is such a standard object, but the theorem Hall references (Thm 9.38 of Hall's Quantum Theory for Mathematicians) to discuss which unbounded potentials satisfy the conditions required by the Trotter product formula does NOT work for the SHO, but at the same time the theorem does not state that it is a necessary condition.
I think my difficulty seeing what is going on comes from being very much a beginner in terms of functional analysis, as it is not clear to me what the appropriate domain for $\hat{V}$ in the case of the harmonic oscillator should be.
I would be very grateful for anyone who could point me in the direction of understanding why or why not the Hamiltonian for the SHO satisfies the conditions for the Trotter product formula.
 A: The theorem is in the Google Books excerpt.  
$T$, $V$ and $T+V$  restricted to the span of the Hermite functions are e.s.a. (by Nelson's Analytic Vector Theorem). So the unrestricted $T+V$ must be an  e.s.a. extension of the restriction, since the s.a. extension is the unique maximal symmetric extension.
A: The natural domain of $\hat{V}$ consists of all $f\in L^2$ for which $x^2f  \in L^2$. $\hat{V}$ is self-adjoint on this domain. The natural domain of $\hat{T}$ consists of all twice absolutely continuous functions $f\in L^2$ for which $f''\in L^2$; it is automatic that $f'\in L^2$.
The operator $Hf=-\frac{\hbar}{2m}\frac{d^2f}{dx^2}+\frac{1}{2}m\omega^2x^2f$ is essentially self-adjoint on $\mathcal{D}(\hat{V})\cap\mathcal{D}(\hat{T})$. This has to do with having no non-zero boundary functionals concentrated at $\pm\infty$. So, in fact, $H$ is essentially selfadjoint on $C_{c}^{\infty}(\mathbb{R})$. If this weren't true, the current Quantum Mechanics would have questionable meaning because it would result in non-trivial boundary conditions at $\pm\infty$ with no way to set such conditions in a physically meaningful way.
