Domain Schrödinger equation

Question

If I have a self-adjoint (TIME INDEPENDENT) operator $H: D(H) \subset L^2([0,x]) \rightarrow L^2([0,x])$ with and I want to define the appropriate domain for

$$ (Hf)(x,t) = i \partial_t f(x,t),$$ would it be $f \in (D(H) \oplus C^1[0,T])$?

Answer 1

Normally you would be working on a fixed Hilber space $X$, and $$ H : \mathcal{D}(H)\subseteq X\rightarrow X. $$ Then the solution would be a continuous function $\psi : [0,T]\rightarrow X$ with $\psi(0)=\psi_{0}$. If $\psi_{0}\in\mathcal{D}(H)$, then the solution would be in $C^{1}([0,T],X)$. You were using $x$ in two different ways, and that was a little confusing. If $X=L^{2}[a,b]$, then the solution would be in $C^{1}([0,T],L^{2}[a,b])$. You could interpret the solution $\psi(t)$ as $\Psi(t,x)=(\psi(t))(x)$ because each $\psi(t)$ is a function on $L^{2}[a,b]$, but such a solution should not be automatically be considered classically and jointly smooth in $t,x$. And the initial values are in an $L^{2}[a,b]$ sense as well, meaning that $$\lim_{t\downarrow 0}\|\psi(t)-\psi_{0}\|_{L^{2}[a,b]}=0. $$ Differentiability in $t$ holds for all $t \ge 0$ in the same sense for $\psi_{0}\in\mathcal{D}(H)$: $$ \lim_{h\downarrow 0}\left\|\frac{1}{h}(\psi(t+h)-\psi(t))-\psi'(t)\right\|=0. $$ (The limit also holds for $h\uparrow 0$ if $t > 0$.) In this sense, $i\frac{d}{dx}\psi(t) = H\psi(t)$ for the case of a time-independent Hamiltonian. So the space $C^{1}([0,T],X)$ is the correct space for the time-independent Hamiltonian, provided the initial vector $\psi_{0}$ is in $\mathcal{D}(H)$, which is a prerequisite for any wave function.