Subspace of Hilbert space as manifold for variational state I have a question on the geometrical description of time-dependent quantum states and variatioanl states. I will outline my problem and ask questions along the way.

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*Assume you have a time-dependent state $\Psi(t) \in \mathcal{H}$ can you say that $\Psi(t)$ is a curve on a manifold parameterized with $t$?

Now we want to approximate this time-dependent state by means of a time-dependent variational state $\Phi(t)$. In this paper they say that this variational state is part of a submanifold $\mathcal{M} \subset \mathcal{H}$ and that there is a smooth parameterization of this state such that $\Phi(t) = \Phi(x(t))$ with $x \in \mathbb{R}^n$. They go on to state that the vectors $\partial \Phi / \partial x_i |_{x=x(t)}$ span a tangent space $T_t M$.
What is the reasoning behind this? I have several problems with this as I am missing connecting pieces in the mathematical notation.
My biggest question:


*Why can I construct a basis of the tangent space from the parameterization $x$? Doesn't $x$ need to be a chart map for this?


*How can I obtain such a parameterization $x(t)$? Can I say that $x: \mathcal{M} \rightarrow \mathbb{R}^n$ is chart map and that the parameterization of $\Phi(t)$ is somehow through the inverse $x^{-1}$?


*If I can say that $\Phi(t)$ is a curve on $\mathcal{M}$, how do I get to the notation of a tangent space?
 A: *

*Fundamentally, quantum states are rays in Hilbert space $\mathcal{H}$, i.e. of the form $r = \mathbb{C} \Psi$ for $\Psi \in \mathcal{H}$. Therefore, a time-dependent state is a path of rays $r(t)$. However, assuming $r(0)$ to not be the zero ray (i.e. there is a non-zero vector in $r(0)$), we can pick a representative $\Psi_0 \in r(0)$ of unit length. As time evolution is by unitaries, we get a function $t\mapsto \Psi_t \in r(t)$, where all the $\Psi_t$ have unit length. Therefore, we can consider $t\mapsto \Psi_t$ a map into the unit sphere $S\mathcal{H}$ inside $\mathcal{H}$. Usually one assumes time-evolution to be strongly continuous, and therefore $t\mapsto \Psi_t$ is continuous. It might happen that this curve is in fact smooth, for example if $\mathcal{H}$ is finite dimensional.


*Suppose we have a map $\Psi : \mathbb{R}^n \mapsto S\mathcal{H}$, i.e. we have a map that assigns to a given value of the parameters $x \in \mathbb{R}^n$ a wavefunction $\Psi(x) \in S\mathcal{H}$. We can now simply define a subset $S\mathcal{H}\supset\mathcal{M} := \Psi(\mathbb{R}^n)$. Note that $\mathcal{M}$ is not, in general, a manifold. This has to be checked for the specific $x\mapsto \Psi(x)$ given. However, if this has been checked, we automatically get a basis of the tangent space $T_p\mathcal{M}$.
There are two equivalent definitions, one in terms of germs of curves, the other in terms of derivations. For the definition in terms of curves, consider curves $(-\epsilon,\epsilon) \ni t \mapsto \Phi_t \in \mathcal{M}$ such that $\Phi_0 = p$. Such maps are called smooth if $\varphi \circ \Phi_t$ is smooth for every chart $\varphi$. Two curves $\Phi^1_t, \Phi^2_t$ are equivalent if the derivatives of $\varphi \circ \Phi^1_t$ and $\varphi \circ \Phi^2_t$ coincide at $t=0$ for every chart. The equivalence class is denoted as $\frac{d \Phi_t}{dt}|_{t=0}$. To see that this is a vector space, i.e. that the sum of two vectors and the multiple of a vector are again vectors, one has to introduce a chart. In $\mathbb{R}^n$, one can to any given inital vector, $v$, find a curve through the origin having $v$ as its tangent vector at the origin. Then this curve can be lifted to the manifold. Furthermore, if we have a tangent vector in this sense (an equivalence class of curves), we can also consider them as derivations on the smooth functions; indeed, let $f\in C^\infty(\mathcal{M})$. Then for a given curve $\Phi_t$ define a derivation $D(f):= \frac{d}{dt}|_{t=0} (f\circ\Phi_t)$.
The definition in terms of curves is useful for our purposes, because by defintion of \mathcal{M}, all curves $(\Phi_t)_t \subset \mathcal{M}$ are of the form $\Phi_t = \Psi(x(t))$, where $t \mapsto x(t)$ is a curve in $\mathbb{R}^n$. Consequently,
$$
\frac{d\Phi_t}{dt}|_{t=0} = \frac{\partial \Psi(x)}{\partial x_i}|_{x = x(0)} \frac{d x^i(t)}{dt}|_{t=0} \ .
$$


*Crucially, for a general $\Phi_t \in S\mathcal{H}$, we will not be able to find a parametrization $\Phi_t = \Psi(x(t))$, simply as the latter exists only in the subset $\mathcal{M}$.
