quantum mechanics - What is a Hilbert Space?

The Hilbert Space is the space where wavefunction live. But how would I describe it in words? Would it be something like:

The infinite dimensional vector space consisting of all functions of position $\psi(\vec x)$ given the conditions that $\psi(\vec x)$ is a smooth, continuous function.

(I am not saying this is right, it is merely an example). Further more how does $\Psi(\vec x, t)$ fit into this? Would it be appropriate to say that at any $\Psi(\vec x, t)$ itself is not in the Hilbert space, but at any given time $t_0$ then the function $\psi_0 (\vec x)\equiv \Psi(\vec X,t_0)$ is a member of the Hilbert Space? And all operators $\hat Q$ have to act on members in the Hilbert Space and therefore cannot have time-derivatives but can have time dependencies.

Answer

Your idea of what $\psi(\vec x,t)$ is supposed to be is essentially correct. Given a space of states $\mathcal{H}$, the "Schrödinger state" is a map

$$\psi : \mathbb{R}\to\mathcal{H}, t\mapsto\lvert\psi(t)\rangle$$ where $\lvert \psi(t)\rangle\in\mathcal{H}$ for every instant $t$. If $\mathcal{H}$ is a space of functions in a variable $\vec x$, then $\lvert \psi(t)\rangle$ is often written $\psi(\vec x,t)$.

The space of wavefunctions in usual quantum mechanics is crucially not the space of smooth functions $C^\infty(\mathbb{R}^3,\mathbb{C})$, but the space of equivalence classes of square integrable functions $L^2(\mathbb{R^3},\mathbb{C})$(link to Wikipedia article on $L^p$-spaces). This space contains the smooth compactly supported functions $C_c^\infty(\mathbb{R}^3,\mathbb{C})$ (non-compactly supported functions are not necessarily square-integrable) and also all smooth square-integrable functions, but is larger.

The reason for this is twofold: For one, there is no reason to demand a generic wavefunction be continuous - the whole physical content of the wavefunction is encapsulated in the probability density

$$\rho(\vec x) = \lvert\psi(x)\rvert^2$$ and this needs to have $\int\rho(\vec x)\mathrm{d}^3x = 1$ to be probability density, hence $\psi$ must be square integrable, but nothing else.

Another reason is that the smooth compactly supported functions do not form a Hilbert space under the scalar product

$$(f,g) = \int \overline{f(x)}g(x)\mathrm{d}^3 x$$ since they are not complete - there are sequences which are Cauchy with respect to the $L^2$-norm induced by this inner product, but which do not converge. The space of square-integrable functions is precisely the completion of the smooth compactly supported functions under this norm, and hence a Hilbert space. In other words, every wavefunction may be arbitrarily accurately be approximated by a smooth compactly supported function, but is itself not guaranteed to be a smooth function.

Among other difficulties commonly overlooked, this means that, strictly speaking, one cannot evaluate wavefunctions at points, since the $L^2$-space elements are only defined up to a zero measure set, and points are of zero measure. This is again meaningful when considering $\rho(\vec x)$: The value of a probability density on a zero measure set is physically meaningless, since only the integration of it over a set of non-zero measure gives a physically relevant probability.