Where the time-dependent wavefunction $\Psi(\vec{x},t)$ lies? Supose $\vec{x}=(x,y,z)\in \mathbb{R}^3$. The state of a physical system is described by the function $\Psi(\vec{x},t)$, where it must satisfy
$$\int_{\mathbb{R}^3} d^3\vec{x}\;\vert\Psi(\vec{x},t)\vert^2 = 1,$$
but, where this $\Psi$ lies? Most textbooks say that the state of a physical system lies (when the Schrödinger equation is time-independent) in $L^2(\mathbb{R}^3, d^3\vec{x})$ but this is not precise, since $\Psi(\vec{x},t)=\Psi(x,y,z,t).$ 
Even when the solution to the Schrödinger equation is separable, and, therefore is something like
$$
\Psi(\vec{x},t)=\psi(\vec{x})e^{-i\omega t}
$$
$\Psi$ does not lie in $L^2(\mathbb{R}^3)$, although, $\psi(\vec{x})$ does.
So, my question is what is the mathematical space that $\Psi(\vec{x},t)$ lies? My guess is that it lies in some space like $L^2(\mathbb{R}^3)\times C(\mathbb{R})$ but I'm not sure. 
 A: The space of states of usual quantum mechanics is indeed the space of time-independent functions $L^2(\mathbb{R}^3,\mathrm{d}^3x)$. $\psi(\vec x,t)$ is not a single state, it is the quantum version of a trajectory - it gives the full time evolution of a state, i.e. for every $t$, it gives a state. Therefore, $\psi(\vec x,t)$ should be thought of as a (smooth) map
$$ \mathbb{R}\to L^2(\mathbb{R}^3), t\mapsto \psi(\dot{},t)$$
where $\psi(\dot{},t)$ (for fixed $t$!) now is an element of $L^2(\mathbb{R}^3)$.
A: Quantum mechanics is studied often in either the Schrödinger picture or in the Heisenberg picture. In the Heisenberg picture (which most people like to use more often), state vectors are time-independent and all time dependence lies in the operators. In the Heisenberg picture, the state of a system is given at a fixed time slice and therefore the wave-function $\psi(\vec{x})$ is an element of $L^2({\mathbb R}^3)$. The operators are now functions of $t$, with ${\hat A}_H(t) = e^{i {\hat H} t} {\hat A}_S e^{-i {\hat H} t}$. The expectation value of any operator is then
$$
A_\psi(t) = \frac{\displaystyle \int \mathrm d^3 x \,\,\psi^*(\vec{x}) {\hat A}_H(t) \psi(\vec{x})  }{ \displaystyle \int \mathrm d^3 x \,\,\psi^*(\vec{x}) \psi(\vec{x})  } 
$$
In the Schrödinger picture, while the state is time-dependent. We label each state by a time-independent wave-function corresponding to the state of the system at a particular fixed time $t = t_0$. Of course, unitarity implies that there is a unique invertible map from this state to the state at any other time. Thus, even in the Schroedinger picture, though not directly obvious, the space of states is taken to be time-independent. 
