What is the specific Hilbert space of the finite square well?

Question

For a free particle, the QM Hilbert space is $L^2(-\infty,\infty)$ which is the space of all square integrable functions. As a result, this space does not contain functions of the form $e^{\alpha x}$ or $e^{-\alpha x}$ as they are not square integrable, per the bounds. For an infinite square well (of width $a$), the space is $L^2(-a,a)$ and as a result functions of the form $e^{\alpha x}$ and $e^{-\alpha x}$ are perfectly acceptable as they are square integrable between $(-a,a)$ (not to imply these are physically realizable states, but mathematically there is nothing wrong with them). The finite square well (also of width $a$) has known solutions of the form $\psi_{1}=Ae^{\alpha x}$ for $x<-a$, $\psi_{2}=B\sin(kx)+C\cos(kx)$ for $-a<x<a$, and $\psi_{3}=De^{-\alpha x}$ for $x>a$. My confusion then is that the finite square well appears to be 3 separate Hilbert spaces which are piece-wise defined, something along the lines of $L^{2}(-\infty,a)$ for $x<-a$, $L^{2}(-a,a)$ for $-a<x<a$, and $L^{2}(a,\infty)$ for $x>a$. Is this correct? I have never seen this discussed or indicated anywhere and am just looking for a clarification. If this is correct, is there a term for connecting spaces like this, as it doesnt appear to be a tensor sum or tensor product of the spaces?

Answer 1

The Hilbert space is simply $L^2(\mathbb{R})$, there is no piece-wise definition: it is the wave function that is piece-wise defined, but the three pieces $\psi_1$, $\psi_2$ and $\psi_3$ are subject to the boundary conditions in $x=\pm a$ so that the global wave function is continuous and continuously differentiable $\forall x \in \mathbb{R}$. In checking if $\psi\in L^2(\mathbb{R})$, you would square-integrate $\psi_1$ from $-\infty$ to $-a$, $\psi_2$ from $-a$ to $a$, and $\psi_3$ from $a$ to $\infty$, and doing so each of the three integrals yields a finite result.