# 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.

1. 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:

1. 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?

2. 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 $$x^{-1}$$?

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

• hey, could you swap the link to the paper for the link to the abstract, so that those without APS access can still see what it is? Nov 18, 2020 at 12:49

1. 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.
2. 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, $$\frac{d\Phi_t}{dt}|_{t=0} = \frac{\partial \Psi(x)}{\partial x_i}|_{x = x(0)} \frac{d x^i(t)}{dt}|_{t=0} \ .$$
3. 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 $$\mathcal{M}$$.
• Thank you for this nice reply! One question; I learned to define a tangent vector as $v_{\Phi, p} : C^\infty \rightarrow \mathbb{R}$ where $v_{\Phi, p}(f) = (f \circ \Phi)^\prime(0)$ (with $\Phi(t)$ the curve, as in your reply). How does this relate to your notation of a tangent vector $d \Phi / dt |_{t=0}$, which seems to not take any function as argument? Oh and also; Can I always write $S \mathcal{H}$, the unit sphere in $\mathcal{H}$, to be the space of all rays? Is this equivalent to the projective Hilbert space $P\mathcal{H}$? Nov 18, 2020 at 16:54
• Also, what do you mean by the subscript $(\partial \Psi / \partial x)_{x=p}$ when $x \in \mathbb{R}^n$ and $p \in \mathcal{M}$? Nov 18, 2020 at 19:07