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

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 $$-a, and $$\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, 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?

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.

• Thanks for the reply! The only thing at seems odd to me is that we are using $e^x$ and $e^{-x}$ in a space in which they are not elements (although I understand in this case the overall wave function remains square-integrable). Does that not cause a problem, or is my understanding incorrect? Commented Feb 2, 2021 at 15:34
• We are not using $e^x$ for all $x$, we are using $e^x$ for $x<-a$, which is an element of $L^2(\mathbb{R})$ (and the same goes for $e^{-x}$). Commented Feb 2, 2021 at 16:02
• A technical comment: the boundary conditions are not an inherent property of the wavefunction, they arise because we split the domain when solving the Schrödinger equation. The wavefunction has to be continuously differentiable everywhere, it's just that at most points we don't have to check that explicitly. Commented Feb 2, 2021 at 16:26
• Can you clarify a little further why it is acceptable to use $e^{x}$ for $x<-a$? It was my understanding that $e^{x}$ was not an element of the space $L^{2}(-\infty,\infty)$ at all, however you are saying it is an element, but only of part of the space? Commented Feb 2, 2021 at 17:41
• Excellent, thank you for the help! Commented Feb 2, 2021 at 19:37