# 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? – user2723984 Nov 18 '20 at 12:49

## 1 Answer

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 $P\mathcal{H}$? – Durd3nT Nov 18 '20 at 16:54
• Also, what do you mean by the subscript $(\partial \Psi / \partial x)_{x=p}$ when $x \in \mathbb{R}^n$ and $p \in \mathcal{M}$? – Durd3nT Nov 18 '20 at 19:07
• The mapping of paths from projective Hilbert space to the unit sphere is definitely possible for unitary time evolutions. – Lorenz Mayer Nov 25 '20 at 10:02
• Thanks a lot for your edits! – Durd3nT Nov 25 '20 at 16:53