Some Properties of the Potential-to-Ground State Map in Quantum Mechanics

Abstract

We analyze the map from potentials to the ground state in static many-body quantum mechanics. We first prove that the space of binding potentials is path-connected. Then we show that the map is locally weak–strong continuous and that its differential is compact. In particular, this implies the ill-posedness of the Kohn–Sham inverse problem.

Appendices

Appendix A: Basic Inequalities on Potentials

We recall here in LemmaA.1 several well-known facts about potentials.

Lemma A.1

Take $v,w \in (L^p+L^{\infty })(\mathbb {R}^d)$.

• (i) Taking p as in (1), we have

  

  (37)

• (ii) Let $\mathcal {C}\subset \mathbb {C}$, be a contour in the complex plane which is such that ${{\,\mathrm{dist}\,}}\left( z,\sigma (H_N(v)) \right) \geqslant \eta > 0$ uniformly in $z \in \mathcal {C}$. Let p be as in (1), then the operators

  $$\begin{aligned} \left( -\Delta +1 \right) ^{-\frac{1}{2}} \big ( H_N-z \big ) \left( -\Delta +1 \right) ^{-\frac{1}{2}}, \,\,\,\,\,\,\,\,\,\left( -\Delta +1 \right) ^{\frac{1}{2}} \big ( H_N-z \big )^{-1}\left( -\Delta +1 \right) ^{\frac{1}{2}} \end{aligned}$$

  are uniformly bounded in $z \in \mathcal {C}$.

• (iii) Let $v \in \mathcal {V}_N^{(0)}$, and p as in (1). For $u \in L^p+L^{\infty }$ such that $\left| \! \left| u \right| \! \right| _{L^p+L^{\infty }}$ is small enough, we have $v+u \in \mathcal {V}_N^{(0)}$.

Proof

(i) $\bullet$  If p is as in (1), we have

where we used that $\left| \! \left| (-\Delta _i)^{\frac{1}{2}} (-\Delta _{\mathbb {R}^{dN}} +1)^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} =1$.

$\bullet$  Let us write $v = v_p + v_{\infty }$. We have

$$\begin{aligned}&\left| \! \left| \left( -\Delta +1 \right) ^{-\frac{1}{2}}v\left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} \nonumber \\&\quad \leqslant \left| \! \left| \left( -\Delta +1 \right) ^{-\frac{1}{2}}v_p\left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} \nonumber \\&\quad \quad \quad \quad \quad + \left| \! \left| \left( -\Delta +1 \right) ^{-\frac{1}{2}}v_{\infty }\left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}\nonumber \\&\quad \leqslant \left| \! \left| \left( -\Delta +1 \right) ^{-\frac{1}{2}}v_p\left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} + \left| \! \left| v_{\infty } \right| \! \right| _{L^{\infty }} \nonumber \\&\quad \leqslant \left| \! \left| \sqrt{\left| v_p\right| } \left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}^2 + \left| \! \left| v_{\infty } \right| \! \right| _{L^{\infty }} . \end{aligned}$$

(38)

In the last inequality, we used that

$$\begin{aligned} \left( -\Delta +1 \right) ^{-\frac{1}{2}} v_p \left( -\Delta +1 \right) ^{-\frac{1}{2}} = \left( -\Delta +1 \right) ^{-\frac{1}{2}} \sqrt{\left| v_p\right| } {{\,\mathrm{sgn}\,}}(v_p) \sqrt{\left| v_p\right| }\left( -\Delta +1 \right) ^{-\frac{1}{2}}, \end{aligned}$$

where ${{\,\mathrm{sgn}\,}}(v)$ is equal to 1 if $v > 0$, $-1$ if $v <0$ and 0 if $v =0$, it satisfies $\left| \! \left| {{\,\mathrm{sgn}\,}}(v) \right| \! \right| _{L^2 \rightarrow L^2} \leqslant 1$, hence

$$\begin{aligned} \left| \! \left| \left( -\Delta +1 \right) ^{-\frac{1}{2}} v_p \left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} \leqslant \left| \! \left| \sqrt{\left| v_p\right| } \left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}^2. \end{aligned}$$

As for the first term in (38), for $d \geqslant 3$, with $p = d/2$ we have

$$\begin{aligned} \left| \! \left| \sqrt{\left| v_p\right| } \left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}&\leqslant \left| \! \left| \sqrt{\left| v_p\right| } (-\Delta )^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} \leqslant c_{d} \left| \! \left| \sqrt{\left| v_p\right| } \right| \! \right| _{L^{2p}} \\&= c_{d} \sqrt{ \left| \! \left| v_p \right| \! \right| _{L^{p}} }, \end{aligned}$$

where we used the Hardy–Littlewood–Sobolev inequality [40, Theorem 4.3] in the last inequality. For $d \in \left\{ 1,2 \right\}$, we can use the Kato–Seiler–Simon inequality [53, Theorem 4.1] to get

$$\begin{aligned} \left| \! \left| \sqrt{\left| v_p\right| } \left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}&\leqslant \left| \! \left| \sqrt{\left| v_p\right| } \left( -\Delta +1 \right) ^{-\frac{1}{2}} \right| \! \right| _{\mathfrak {S}_{2p}} \\&\leqslant (2\pi )^{-d/(2p)} \left| \! \left| \sqrt{\left| v_p\right| } \right| \! \right| _{L^{2p}} \left| \! \left| \big ( \left| x\right| ^2+1 \big )^{-\frac{1}{2}} \right| \! \right| _{L^{2p}} \\&\leqslant c_{d,p} \sqrt{ \left| \! \left| v_p \right| \! \right| _{L^{p}} }. \end{aligned}$$

(ii) Take $c \geqslant 0$ and let us define . We remark that

$$\begin{aligned} H + c = (-\Delta +c)^{\frac{1}{2}} \left( 1+ (-\Delta +c)^{-\frac{1}{2}} A (-\Delta +c)^{-\frac{1}{2}} \right) (-\Delta +c)^{\frac{1}{2}}, \end{aligned}$$

hence we only need to show that $\left| \! \left| (-\Delta +c)^{-\frac{1}{2}} A (-\Delta +c)^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} <1$. For instance we will show that

$$\begin{aligned} \left| \! \left| (-\Delta _{\mathbb {R}^d} + c)^{-\frac{1}{2}} v (-\Delta _{\mathbb {R}^d} + c)^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2} \leqslant \left| \! \left| \sqrt{\left| v\right| } (-\Delta _{\mathbb {R}^d} + c)^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}^2 \end{aligned}$$

is as small as we want. For any $\epsilon > 0$, there exists $c_{\epsilon } \geqslant 0$ such that $\left| v\right| \leqslant \epsilon (-\Delta ) + c_{\epsilon }$ in the sense of forms, hence for all $u \in \mathcal {C}^{\infty }$, we have

$$\begin{aligned} \left| \! \left| \sqrt{\left| v\right| } (-\Delta + c)^{-\frac{1}{2}} u \right| \! \right| _{L^2}^2&\leqslant \epsilon \left| \! \left| (-\Delta )^{\frac{1}{2}} (-\Delta + c)^{-\frac{1}{2}} u \right| \! \right| _{L^2}^2 + c_{\epsilon } \left| \! \left| (-\Delta + c)^{-\frac{1}{2}} u \right| \! \right| _{L^2}^2 \\&\leqslant \left( \epsilon + \frac{c_{\epsilon }}{c} \right) \left| \! \left| u \right| \! \right| _{L^2}^2. \end{aligned}$$

We can first choose $\epsilon$ small and then choose c large so that the quantity $\left| \! \left| \sqrt{\left| v\right| } (-\Delta + c)^{-\frac{1}{2}} \right| \! \right| _{L^2 \rightarrow L^2}$ is arbitrarily small.

(iii) The statement follows from the resolvent formula

and Cauchy’s formula

see for instance [36, 50]. $\quad \square$

Appendix B: Weak–Strong Continuity and Compactness

We recall here relations between weak–strong continuity and compactness. Following [23, Definition 7.6], we say that a map is compact if it maps bounded sets into relatively compact sets. The link between ill-posedness of a problem and its linearization can be involved, see for instance [52] and [7, Appendix]. We start by considering standard results, and adapt them to the case when the image space is an embedded submanifold.

Lemma B.1

Let X and Y be Banach spaces, $U \subset X$ an open set, a closed embedded submanifold of Y, and a map $f : U \rightarrow M$.

• (i) If f is compact, continuous and differentiable on U, then $\mathrm{d}_x f$ is compact for any $x \in U$.

• (ii) If $U = X$ is the dual of a Banach space, and if f is weak–strong continuous, then f is compact.

• (iii) If f is compact and M is infinite-dimensional, then f(X) is a countable union of compact sets, and f(X) has empty interior.

• (iv) If f is compact and X is infinite-dimensional, then $f^{-1}$ is discontinuous.

Proof

The only difference in the proof, compared to the case $M=Y$, is (i).

• (i) In the case $M = Y$, this is proved in [23]. We apply it to $\iota _{M \rightarrow Y} \circ f : U \rightarrow Y$ and get that $\mathrm{d}_x \left( \iota _{M \rightarrow Y} \circ f \right) \left( X \cap \left\{ \left| \! \left| \cdot \right| \! \right| _{L^2 \rightarrow L^2} \leqslant 1 \right\} \right) = \iota _{\mathrm{T}_{f(x)} M \rightarrow Y} \circ \left( \mathrm{d}_x f \right) \left( X \cap \left\{ \left| \! \left| \cdot \right| \! \right| _{L^2 \rightarrow L^2} \leqslant 1 \right\} \right)$ is compact. A map is proper if preimages of relatively compact open sets are relatively compact open sets [56, Definition 16.26]. One can prove that for a Banach space F and a closed subset $E \subset F$, the inclusion map is proper. Since $\iota _{\mathrm{T}_{f(x)} M \rightarrow Y}$ is proper, then $\left( \mathrm{d}_x f \right) \left( X \cap \left\{ \left| \! \left| \cdot \right| \! \right| _{L^2 \rightarrow L^2} \leqslant 1 \right\} \right)$ is relatively compact. We remark that we only used that is an embedded submanifold of Y, we did not use the closed condition.

• (ii) Let $G \subset B_0(r) \subset X$ be a bounded set and $x_n \in G$ a sequence. By Banach–Alaoglu’s theorem, $x_n \rightharpoonup x$ for some $x \in B_0(r)$ and up to a subsequence. By weak–strong continuity of f, $f(x_n) \rightarrow f(x)$ strongly.

• (iii) We define the sets $X_r :=X \cap \left\{ x \in X \, \bigr \vert \, \left| \! \left| x \right| \! \right| _{X} \leqslant r \right\}$, for $r \geqslant 0$. Since f is compact, then the $\overline{f(X_r)}$’s are compact and thus have empty interiors by Riesz’s theorem [4, Theorem 6.5], which applies in our case because M is locally a normed vector space. We have

  $$\begin{aligned} f(X) = \cup _{r\in \mathbb {N}} f(X_r) \subset \cup _{r\in \mathbb {N}} \overline{f(X_r)}. \end{aligned}$$

  Finally, by Baire’s theorem [4, Theorem 2.1] f(X) has empty interior. We recall that a closed subset of a compact space is compact.

• (iv) Let $B \subset X$ be a ball, f(B) is relatively compact. Assuming that $f^{-1}$ is continuous, $f^{-1}\left( f(B) \right) \supset B$, is also relatively compact, and hence B as well. But this contradicts [4, Theorem 6.5]. The inverse $f^{-1}$ is thus discontinuous.

$\square$

Here is a summary of the relations between compactness and weak–strong continuity for a map and its differential.

We also remark that $\mathrm{d}_x f$ weak–strong continuous for any $x \in U$ does not imply that f is weak–strong continuous, a simple counterexample is $L^2(\mathbb {R}^n) \ni x \mapsto \left| \! \left| x \right| \! \right| _{L^2}^2$, and this is also the case for $v \mapsto \Psi (v)$.