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While studying quantum mechanics, I have been calculating problems and examples of quantum mechanical models for a particle by assuming that its underlying Hilbert space is $L^2 (\mathbb{R}^3, \textrm{d}x)$, for example.

Why do we choose Hilbert space to be $L^2$? I understand that it is the only Hilbert space if we choose from spaces $L^p$, but I have trouble understanding the specific choice. I also know that every separable Hilbert space is unitarily isomorphic to each other. But why don't we then use different "version" of Hilbert space on $\mathbb{R}^3$? I guess it probably makes calculations simpler and more "intuitive". But are there examples where another Hilbert space is used to describe particle in, say, three spatial dimensions?

I would appreciate both intuitive answers and some proofs/sources of proofs that are related to this subject.

Daniels Krimans
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The question is basically what basis for the (unique separable) Hilbert space do you use. The $L^2(\mathbb{R}^3,\mathrm dx)$ corresponds to working in eigenbasis of position operator. You can work in any other basis which is convenient for your problem. Say, if you are doing a problem where Hamiltonian has purely discrete spectrum (like harmonic oscillator), you might want to use the energy eigenbasis, and then the Hilbert space "looks" more like square summable sequences $\ell^2$.

Peter Kravchuk
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Short answer: Because of the Schrödinger equation and the invariance of $|\psi|^2$.

Long answer: $L^2(\mathbb{R}^3)$ is the natural choice of Hilbert space if you start off with the one-particle Schrödinger equation $$i \partial_t \psi(x,t) = \left( - \tfrac{\hbar^2}{2m}\triangle + V(x) \right) \psi(x,t). $$ with $x \in \mathbb{R}^3$. Now wonder what properties $\psi$ will have. If you know something about the initial values (e.g. they could be smooth), you might be able to infer something about special properties of $\psi$. But one thing that is always true is the continuity equation $$ \frac{\partial |\psi|^2(t,x)}{\partial_t} = \mathrm{div} \ j.$$ It implies that, as long as no current flows into the boundary of your system, $\int_{\mathbb{R}^3} |\psi|^2$ is constant in time. And if you want it to describe a probability, you better have $$ \int_{\mathbb{R}^3} dx |\psi|^2(x,t) = 1, \quad \forall t \in \mathbb{R}. $$ This directly implies that $\psi$ is an element of $L^2(\mathbb{R}^3)$. In this way of telling the story, which is also the historic one, you see that we were not just looking for some Hilbert space, but the very concrete dynamics of the Schrödinger equation implies that all relevant wave functions are in $L^2$.

Luke
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