AppendicesNotation Memo

The main notations related to the complex structure of the space are reminded in Table 1. Table 2 summarizes the representation of conjugation operations within the Casida and real/imaginary representations. The notations of the most important operators of the paper are gathered in Table 3.

Table 1 Collection of the main notations related to the complexification of spaceTable 2 Summary of the different representations of the operators in YTable 3 Notations related to the main operators of the paperSecond-order expansion of the energy Proof of Proposition 1

The energy is

$$\begin \mathcal (\Psi )&=\sum _^N\left( \frac\int _^3} |\nabla \psi _i(r)|^2 + \int _^3} V_}(r)|\psi _i(r)|^2 dr \right) \\ &\quad + \frac \int _^3} \frac(r) \rho _(r')}drdr' + \int _^3} e_}(\rho _(r)) dr \end$$

We treat these terms in order. The first term is clearly smooth from \(H^\) to \(\mathbb R\). Since \(H^\) is an algebra, the mapping \(\psi _ \mapsto |\psi _|^\) is smooth from \(H^\) to \(H^\), and so using that \(V_\textrm = V_2 + V_\) with \(V_2 \in L^2, V_\infty \in L^\infty \),

$$\begin \left| \int _} (V_ + V_) |\psi _|^ \right|&\le \Vert V_\Vert _}\Vert |\psi _|^\Vert _} + \Vert V_\Vert _} \Vert \psi _\Vert _}^\\&\le (\Vert V_\Vert _} + \Vert V_\Vert _}) \Vert \psi _\Vert _}^, \end$$

the second term is also smooth from \(H^\) to \(\mathbb R\). From Lemma 4, the map from \(\Psi \) to the operator \(L_\) of multiplication by \(\rho _ *\frac\) is smooth from \(H^2\) to \(\mathcal (H^2, H^2)\), and therefore the third term

$$\begin \frac \int _^3} \frac(r) \rho _(r')}drdr' = \frac \sum _^\langle \psi _| L_ \psi _ \rangle \end$$

is smooth from \(H^\) to \(\mathbb R\).

For all \(\Psi \in (H^)^\), \(\rho _\) is in \(L^1 \cap L^\infty \) and the mapping \(\Psi \mapsto \rho _\) is smooth from \(H^\) to \(L^1 \cap L^\infty \). Since \(e_}(0)=0\), \(\int _^3 } e_}(\rho (r))dr\) is well-defined for all \(\rho \in L^1 \cap L^\infty \). As \(e_}\) is \(\mathcal ^2\) by Assumption 2, it is easy to check that \(\rho \mapsto \int _^3 } e_}(\rho (r))dr\) is also \(\mathcal ^2\) on \(L^1 \cap L^\infty \), hence \(\Psi \mapsto \int _^3 } e_}(\rho _(r))dr\) is \(\mathcal ^2\) from \(H^\) to \(\mathbb R\) (Fig. 2).\(\square \)

Proof of Proposition 2

Let \(U^ \in (H^)^ \cap }(1-P_)\). For all \(\varepsilon > 0\), let

$$\begin \Psi (\varepsilon ) = }(\Psi ^ + \varepsilon U^), \end$$

where for any \(\Psi \in (H^)^N\), \(}(\Psi )\) is defined by \( }(\Psi ) = \Psi (\Psi ^\Psi )^\) and for \(\Psi \in (H^2)^N\), \(\Psi ^\Psi = (\langle \Psi _i, \Psi _j \rangle )_\). We have the expansion

$$\begin \Psi (\varepsilon ) = \Psi ^ + \varepsilon U^ + \varepsilon ^ U^ + O(\varepsilon ^) \end$$

in \(H^\).

Since \(\Psi (\varepsilon )^\Psi (\varepsilon )=1\) for all \(\varepsilon \), by identification, we get

$$\begin (U^)^ U^ + (U^)^ U^ = - (U^)^ U^. \end$$

(24)

We then have in \(H^2\)

$$\begin \mathcal (\Psi (\varepsilon ))&= \mathcal (\Psi ^0)+ 2}\langle H_0\Psi ^0 | \varepsilon U^1 +\varepsilon ^2 U^2 \rangle + \langle \varepsilon U^1|H_0|\varepsilon U^1\rangle \\&\quad + }\langle \varepsilon U^1| \mathcal _(\varepsilon U^1)\rangle +o(\varepsilon ^2)\\&= \mathcal (\Psi ^0)\!+\! \varepsilon ^2 \Big (2} \langle H_0 \Psi ^0 | U^2\rangle \!+\! \langle U^1|H_0| U^1\rangle + }\langle U^1 | \mathcal _ ( U^1) \rangle \Big )\\&\quad + o (\varepsilon ^2)\\ 2 }\langle H_0 \Psi ^0 | U^2\rangle&= 2}\left( \sum _^N \lambda _i \langle \psi ^0_i|u_i^2\rangle \right) \\&= -\sum _^N \lambda _i \langle u_i^1| u_i^1\rangle \text \text \end$$

so that

$$\begin \mathcal (\Psi (\varepsilon ))&= \mathcal (\Psi ^0) + \varepsilon ^2 \langle U^1|\mathcal _}( U^1)\rangle +o(\varepsilon ^2). \end$$

On the other hand, we can solve the orthogonal Procustes problem

$$\begin \min _}(N)} \Vert \Psi - \Psi ^ R\Vert ^&= 2 N - 2\max _}(N)} } \langle \Psi , \Psi ^ R \rangle \\&= 2 N - 2\max _}(N)} } \ }((\Psi )^ \Psi ^ ,R)\\&=2N - 2 \max _}(N)} } \langle \Sigma ,R'\rangle \\&= 2N - 2 }(\Sigma ) \end$$

where \(\Sigma \) is the diagonal matrix of the singular values of \((\Psi ^)^ \Psi \). We have

$$\begin (\Psi ^)^ \Psi&= 1 + \varepsilon ^ (U^)^ U^ + O(\varepsilon ^)\\ ((\Psi ^)^ \Psi )^ ((\Psi ^)^ \Psi )&= 1 + \varepsilon ^ \left( (U^)^ U^ + (U^)^ U^\right) + O(\varepsilon ^)\\&= 1 + 2\varepsilon ^ (U^)^U^ + O(\varepsilon ^)\\ }(\Sigma )&= }\left( \sqrt)^ \Psi )^ ((\Psi ^)^ \Psi )} \right) \\&= N + \varepsilon ^ } \left( (U^)^ U^ \right) + O(\varepsilon ^) \end$$

so that

$$\begin \min _}(N)} \Vert \Psi - \Psi ^ R\Vert ^&=2\varepsilon ^ \Vert U^\Vert ^ + O(\varepsilon ^) \end$$

and the result follows. \(\square \)

Control of the Coulomb Terms

We state a useful technical lemma, which ensures stability in \(H^2\) in several occasions through the article.

Lemma 4

Let f, g, h three functions of \(H^2\). Then, the function \((fg*\tfrac)h\) is in \(H^2\), and there exists a constant c such that

$$\begin \Vert \left( fg*\tfrac\right) h\Vert _ \le c \Vert f\Vert _ \Vert g\Vert _ \Vert h\Vert _. \end$$

The proof relies on the Hardy–Littlewood–Sobolev (HLS) inequality in dimension 3:

$$\begin&\forall p, q>0;~ 1< p,q<\infty , ~s.t.~ 1 +\frac = \frac + \frac,\\&\quad \exists C_^>0, ~ \forall a \in L^q: ~ a*\tfrac \in L^p \text \Vert a* \frac\Vert _ \le C_^ \Vert a\Vert _ . \end$$

Lemma 5

Let a and b two complex-valued functions such that a is in \(L^\cap L^}\) and b is in \(L^\), with t in (2, 6). Then, \((a*\tfrac)b\) is \(L^2\), and \(\Vert (a*\tfrac)b\Vert _ \le c (\Vert a\Vert _+ \Vert a\Vert _}})\Vert b\Vert _}\), where c is an unimportant constant.

Fig. 2

Choices of p, q, t to apply Lemma 5

Proof of Lemma 5

Let p such that \(\frac+\frac= \frac\), with t in (2, 6). Then, p is in \((3, +\infty )\). Thus, there exists q in \((1, \frac)\) such that p and q are an admissible pair for the HLS inequality. Since a is in \(L^1 \cap L^}\), it is in \(L^\) as well and we can write:

$$\begin \Vert a* \frac\Vert _} \le C_^ \Vert a\Vert _} \end$$

By the Hölder inequality, \((a* \frac)b\) is \(L^2\) and:

$$\begin \Vert \left( a* \frac\right) b\Vert _}&\le C_^ \Vert a\Vert _} \Vert b\Vert _}\\ \Vert \left( a* \frac\right) b\Vert _}&\le C_^ (\Vert a\Vert _}+\Vert a\Vert _}})\Vert b\Vert _} \end$$

\(\square \)

Proof of Lemma 4

We prove that \(\left( fg*\tfrac\right) h\) is \(L^2\) and then that its second derivative is \(L^2\) as well. The proof is the application of Lemma 5.

fg is in \(H^2\), it thus belongs to \(L^\infty \cap L^2\). It is \(L^1\) as well by the Hölder inequality, since both f and g are \(L^2\). Therefore, fg is \(L^1 \cap L^}\). h is \(L^2 \cap L^\infty \), in particular it is \(L^\) for any t in (2, 6), and we can apply Lemma 5.

We now show that the second derivative of the function belongs to \(L^2\) . It only contains terms of the following form:

\( \left( f \nabla ^2 g * \tfrac\right) h\). f and \(\nabla ^2\,g\) are both \(L^2\), thus their product is \(L^1\). Since f is bounded and \(\nabla ^2g\) is \(L^2\), \(f\nabla ^2\,g\) is \(L^\) as well and Lemma 5 can be applied.

\( \left( \nabla f \nabla g * \tfrac\right) h\). \(\nabla f\) and \(\nabla g\) are both \(L^2\), thus their product is \(L^1\). They are also both in \(H^1\), which injects itself continuously (by Sobolev injection) in \(L^6\) in dimension 3. Their product is thus in \(L^3\).

\( \left( f \nabla g* \tfrac\right) \nabla h\). \(f\nabla g\) is \(L^1 \cap L^2\). \(\nabla h\) belongs to \(H^1\), which injects itself continuously in \(L^6\) by the Sobolev inequality. It also belongs to \(L^2\).

\( \left( fg * \tfrac \right) \nabla ^2 h\). In this case, the HLS inequality cannot be directly applied, since h is only \(L^2\). We would need \(\left( fg*\tfrac\right) \) to be \(L^\infty \) for the Hölder inequality, which is not an admissible p exponent. However, boundedness can easily be obtained like this:

$$\begin&\forall r \in \mathbb ^3, |\left( fg*\tfrac\right) (r)|\\&\quad \le \int _^3} \frac dr' \\&\quad = \int __} \frac dr' + \int _^3 \backslash \mathcal _}\frac dr' \\&\quad \le \int __} \fracdr' \sup _^3} |(fg)(r')|+ \int _^3}|(fg)(r')|dr' \\&\quad \le K\Vert f\Vert _ \Vert g\Vert _ + \Vert f\Vert _ \Vert g\Vert _ \quad \text \quad K= \int __} \fracdr' . \end$$

\(\square \)

Technical Lemmas on \(S_0\) and \(K_0\) Lemma 6

The following assertions are true:

1.

\(S_0: (H^2)^ \rightarrow (H^2)^\) is bounded;

2.

\(K_0: (H^1)^\rightarrow (L^2)^\) is bounded;

3.

\(K_0: (H^2)^\rightarrow (H^2)^\) is bounded;

4.

\(K_0: (H^2)^ \rightarrow (L^2)^\) is compact;

5.

For \(\alpha >0\) small enough, \(K_0: (H_\alpha ^2)^ \rightarrow (L_^2)^\) is bounded.

Proof 1.

\(S_0: (H^2)^ \rightarrow (H^2)^\) is bounded This follows from Assumption 3 and the fact that \(H^2\) is an algebra.

2.

\(K_0: (H^1)^\rightarrow (L^2)^\) is bounded Again by Assumption 3, for \(U \in (H^1)^\), \(S_0(U) \in (H^1)^\), so by assumption on \(v_\textrm\), \(v'_\textrm(\rho _0) S_0(U) \Psi _0 \in (H^1)^\). For \(S_0(U) * \frac\), we use a Hardy inequality to deduce that \(\Vert S_0(U) * \frac\Vert _ \le \Vert U\Vert _\). This proves the assertion.

3.

\(K_0: (H^2)^\rightarrow (H^2)^\) is bounded By assumption on \(v_\textrm\), and since \(H^2\) is an algebra, for \(U \in (H^2)^\), \(v'_\textrm(\rho _0) S_0(U) \Psi _0 \in (H^2)^\). By Lemma 4, we have that \(S_0(U)*\frac \Psi _0 \in (H^2)^\) for \(U \in (H^2)^\). This finishes the proof of this item.

4.

\(K_0: (H^2)^ \rightarrow (L^2)^\) is compact For the proof of the compactness of \(K_0\), we have for \(U \in (H^2)^\)

$$ K_0(U) = \left( S_0(U) * \frac + v'_(\rho ) S_0(U) \right) \Psi _0. $$

\(V \mapsto v'_(\rho ) S_0((-\Delta +1)^)V) \Psi _0\) is an operator of the form \(f(x) g(-\textrm\nabla )\) with \(f,g \in L^2\), hence it is Hilbert–Schmidt [23, Theorem 4.1] thus compact. By Lemma 4, for \(V \in (L^2)^\), \(S_0((-\Delta + 1)^V) * \frac\Psi _0\) is in \((H^2)^\). Using that \(\Psi _0 \in H^2_\alpha \), we have exponential decay of \(\Psi _0\) and its derivative, so by the Rellich–Kondrachov compactness embedding, \(V \mapsto S_0((-\Delta + 1)^V) * \frac\Psi _0\) is compact on \((L^2)^\).

5.

\(K_0: (H_\alpha ^2)^ \rightarrow (L_^2)^\) is bounded This follows from the assumption on the exponential decay of the eigenfunction.\(\square \)