Stability of Ferromagnetism in Many-Electron Systems

Abstract

We construct a model-independent framework describing stabilities of ferromagnetism in strongly correlated electron systems; Our description relies on the operator theoretic correlation inequalities. Within the new framework, we reinterpret the Marshall–Lieb–Mattis theorem and Lieb’s theorem; in addition, from the new perspective, we prove that Lieb’s theorem still holds true even if the electron–phonon and electron–photon interactions are taken into account. We also examine the Nagaoka–Thouless theorem and its stabilities. These examples verify the effectiveness of our new viewpoint.

Appendices

Construction of Self-dual Cones

In this section, we define some self-dual cones which are important in the proofs of theorems in Sect. 4.

1.1 A Canonical Cone in ${\mathscr {L}}^2(\mathfrak {H})$

Let $\mathfrak {H}$ be a complex Hilbert space. The set of all Hilbert–Schmidt class operators on $\mathfrak {H}$ is denoted by ${\mathscr {L}}^2(\mathfrak {H})$, i.e., ${\mathscr {L}}^2(\mathfrak {H})=\{ \xi \in {\mathscr {B}}(\mathfrak {H})\, |\, \mathrm {Tr}[\xi ^* \xi ]<\infty \}$. Henceforth, we regard ${\mathscr {L}}^2(\mathfrak {H})$ as a Hilbert space equipped with the inner product $\langle \xi | \eta \rangle _{{\mathscr {L}}^2}=\mathrm {Tr}[\xi ^* \eta ],\, \xi , \eta \in {\mathscr {L}}^2(\mathfrak {H})$.

For each $A\in {\mathscr {B}}(\mathfrak {H})$, the left multiplication operator is defined by

$$\begin{aligned} {\mathcal {L}}(A)\xi =A\xi ,\ \ \xi \in {\mathscr {L}}^2(\mathfrak {H}). \end{aligned}$$

(A.1)

Similarly, the right multiplication operator is defined by

$$\begin{aligned} {\mathcal {R}}(A)\xi =\xi A, \ \ \xi \in {\mathscr {L}}^2(\mathfrak {H}). \end{aligned}$$

(A.2)

Note that ${\mathcal {L}}(A)$ and ${\mathcal {R}}(A)$ belong to ${\mathscr {B}}({\mathscr {L}}^2(\mathfrak {H}))$. It is not difficult to check that

$$\begin{aligned} {\mathcal {L}}(A){\mathcal {L}}(B)={\mathcal {L}}(AB),\ \ {\mathcal {R}}(A){\mathcal {R}}(B)={\mathcal {R}}(BA),\ \ A, B\in {\mathscr {B}}(\mathfrak {H}). \end{aligned}$$

(A.3)

Let $\vartheta$ be an antiunitary operator on $\mathfrak {H}$.¹⁰ Let $\Phi _{\vartheta }$ be an isometric isomorphism from ${\mathscr {L}}^2(\mathfrak {H})$ onto $\mathfrak {H}\otimes \mathfrak {H}$ defined by $\Phi _{\vartheta }(|x\rangle \langle y|)=x\otimes \vartheta y\ \ \ \forall x,y\in \mathfrak {H}.$ Then,

$$\begin{aligned} {\mathcal {L}}(A) =\Phi _{\vartheta }^{-1} A\otimes 1\Phi _{\vartheta } ,\ \ \ {\mathcal {R}}(\vartheta A^*\vartheta )=\Phi _{\vartheta }^{-1} 1 \otimes A\Phi _{\vartheta } \end{aligned}$$

(A.4)

for each $A\in {\mathscr {B}}(\mathfrak {H})$. We write these facts simply as

$$\begin{aligned} \mathfrak {H}\otimes \mathfrak {H}={\mathscr {L}}^2(\mathfrak {H}),\ \ A\otimes 1 ={\mathcal {L}}(A),\ \ 1 \otimes A={\mathcal {R}}(\vartheta A^*\vartheta ), \end{aligned}$$

(A.5)

if no confusion arises. If A is self-adjoint, then ${\mathcal {L}}(A)$ and ${\mathcal {R}}(A)$ are self-adjoint.

Recall that a bounded linear operator $\xi$ on $\mathfrak {H}$ is said to be positive if $\langle x| \xi x\rangle _{\mathfrak {H}} \ge 0$ for all $x\in \mathfrak {H}$. We write this as $\xi \ge 0$.

Definition A.1

A canonical cone in ${\mathscr {L}}^2(\mathfrak {H})$ is given by

$$\begin{aligned} {\mathscr {L}}^2(\mathfrak {H})_+= \Big \{\xi \in {\mathscr {L}}^2(\mathfrak {H})\, \Big |\,\xi \text{ is } \text{ self-adjoint } \text{ and } \xi \ge 0 \text{ as } \text{ an } \text{ operator } \text{ on } \mathfrak {H}\Big \}. \end{aligned}$$

(A.6)

$\diamondsuit$

Theorem A.2

${\mathscr {L}}^2(\mathfrak {H})_+$ is a self-dual cone in ${\mathscr {L}}^2(\mathfrak {H})$.

Proof

Apply [36, Proposition 2.5] and Theorem H.1. $\square$

Proposition A.3

Let $A\in {\mathscr {B}}(\mathfrak {H})$. We have ${\mathcal {L}}(A^*){\mathcal {R}}(A)\unrhd 0$ w.r.t. ${\mathscr {L}}^2(\mathfrak {H})_+$.

Proof

For each $\xi \in {\mathscr {L}}^2(\mathfrak {H})_+$, we have ${\mathcal {L}}(A^*){\mathcal {R}}(A)\xi =A^*\xi A \ge 0.$$\square$

Remark A.4

• In [10], Gross studied a theory of noncommutative integration in the fermionic Fock space; his framework is related to this subsection.

• Proposition A.3 is closely connected with reflection positivity, see, e.g., [6, 7].

$\diamondsuit$

1.2 The Hole-Particle Transformations

Before we proceed, we introduce an important unitary operator W as follows: The hole-particle transformation is a unitary operator W on ${\mathfrak {E}}$ satisfying the following (i) and (ii):

• (i) For all $x\in \Lambda$, $W c_{x\uparrow } W^* =c_{x\uparrow }$ and $Wc_{x \downarrow }W^*=\gamma (x) c_{x\downarrow }^*,$ where $\gamma (x)=1$ if $x\in A$, $\gamma (x)=-1$ if $x\in B$.

• (ii) $\displaystyle W \Omega =\prod _{x\in \Lambda }^{\#} c_{x\downarrow }^*\Omega ,$ where $\Omega$ is the fermionic Fock vacuum in ${\mathfrak {E}}$, and $\prod _{x\in \Lambda }^{\#}$ indicates the product taken over all sites in $\Lambda$ with an arbitrarily fixed order.

The hole-particle transformations on ${\mathfrak {E}} \otimes \mathfrak {F}_{\mathrm {ph}}$ and ${\mathfrak {E}}\otimes \mathfrak {F}_{\mathrm {rad}}$ are defined by $W\otimes 1$.

Notation A.5

Let X be a linear operator on ${\mathfrak {Z}}$. Suppose that X commutes with $N_{\mathrm {el}}$ and $S_{\mathrm {tot}}^{(3)}$. We will use the following notations:

• ${\tilde{X}}=WXW^{-1}$;

• ${\tilde{X}}_{n}=WX_{n}W^{-1}$;

• ${\tilde{X}}_{n}[M]=W X_{n}[M]W^{-1}={\tilde{X}}\upharpoonright W {\mathfrak {Z}}_{n}[M]$. $\diamondsuit$

1.3 Some Useful Expressions of $W{\mathfrak {E}}_{n=|\Lambda |}[M]$ and $W{\mathfrak {H}}_{Q}$

Let ${\mathfrak {E}}({\mathfrak {X}})$ be the fermionic Fock space over ${\mathfrak {X}}$. Remark the factorization property: ${\mathfrak {E}}({\mathfrak {X}}_1\oplus {\mathfrak {X}}_2)={\mathfrak {E}}({\mathfrak {X}}_1)\otimes {\mathfrak {E}}( {\mathfrak {X}}_2)$. By this fact, we have

$$\begin{aligned} {\mathfrak {E}}={\mathfrak {E}}(\ell ^2(\Lambda ) \oplus \ell ^2(\Lambda )) ={\mathfrak {E}}(\ell ^2(\Lambda ))\otimes {\mathfrak {E}}(\ell ^2(\Lambda )). \end{aligned}$$

(A.7)

In this expression, the N-electron subspace becomes

$$\begin{aligned} {\mathfrak {E}}_N=\bigoplus _{N_1+N_2=N} {\mathcal {E}}^{N_1} \otimes {\mathcal {E}}^{N_2}, \end{aligned}$$

(A.8)

where ${\mathcal {E}}^{K}=\bigwedge ^{K} \ell ^2(\Lambda )$. Moreover, the M-subspace can be written as

$$\begin{aligned} {\mathfrak {E}}_N[M]={\mathcal {E}}^{\frac{N}{2}+M} \otimes {\mathcal {E}}^{\frac{N}{2}-M}. \end{aligned}$$

(A.9)

Because ${\tilde{N}}_{\mathrm {el}}=WN_{\mathrm {el}}W^*=2S^{(3)}_{\mathrm {tot}}+|\Lambda |$ and ${\tilde{S}}_{\mathrm {tot}}^{(3)}= WS_{\mathrm {tot}}^{(3)} W^*=\frac{N_{\mathrm {el}}}{2}-\frac{|\Lambda |}{2},$ we obtain, by (A.9), that

$$\begin{aligned} W {\mathfrak {E}}_{n=|\Lambda |} [M]={\mathfrak {E}}_{n=2M+|\Lambda |}[M=0]={\mathcal {E}}^{M^{\dagger }} \otimes {\mathcal {E}}^{M^{\dagger }}, \end{aligned}$$

(A.10)

where $M^{\dagger }=M+|\Lambda |/2$. This expression will play an important role in the present paper. Using this, we have

$$\begin{aligned} W{\mathfrak {H}}_{Q}[M]={\tilde{Q}} {\mathcal {E}}^{M^{\dagger }} \otimes {\mathcal {E}}^{M^{\dagger }}. \end{aligned}$$

(A.11)

The following formula will be useful:

$$\begin{aligned} {\tilde{Q}}=WQW^* =\prod _{x\in \Lambda } (n_x-1)^2. \end{aligned}$$

(A.12)

We derive a convenient expression of $W{\mathfrak {H}}_{Q}[M]$ for later use. To this end, we set ${\mathcal {S}}=\{0, 1\}^{\Lambda }$. For each ${\varvec{m} } =({\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }) \in {\mathcal {S}} \times {\mathcal {S}}$ with ${\varvec{m}}_{\sigma }=\{m_{x\sigma }\}_{x\in \Lambda }\, (\sigma =\uparrow , \downarrow )$, we define

$$\begin{aligned} |{\varvec{m}}\rangle = \prod ^{\#}_{x\in \Lambda } \Big ( c_{x\uparrow }^* \Big )^{m_{x\uparrow }} \Big ( c_{x\downarrow }^* \Big )^{m_{x\downarrow }} \Omega . \end{aligned}$$

(A.13)

Clearly, $\{ |{\varvec{m}}\rangle \, |\, {\varvec{m}} \in {\mathcal {S}} \times {\mathcal {S}}\}$ is a complete orthonormal system (CONS) of ${\mathfrak {E}}$. Using $|{\varvec{m}}\rangle$, we can represent $W{\mathfrak {H}}_{Q}[M]$ as

$$\begin{aligned} W {\mathfrak {H}}_Q[M]= \mathrm {Lin} \Big \{ | {\varvec{m}} \rangle \ \Big |\ {\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow }, \ \ |{\varvec{m}}_{\uparrow }|= |{\varvec{m}} _{\downarrow }|=M^{\dagger }\Big \}, \end{aligned}$$

(A.14)

where $|{\varvec{m}_{\sigma }}|=\sum _{x\in \Lambda } m_{x\sigma }$.

1.4 A Self-dual Cone in $W{\mathfrak {E}}_{n=|\Lambda |}[M]$

Corresponding to the identification (A.7), we have

$$\begin{aligned} c_{x\uparrow }={\mathsf {c}}_x\otimes 1,\ \ \ c_{x\uparrow }=(-1)^{{\mathsf {N}}}\otimes {\mathsf {c}}_x, \end{aligned}$$

(A.15)

where ${\mathsf {c}}_x$ is the annihilation operator on ${\mathfrak {E}}(\ell ^2(\Lambda ))$, and ${\mathsf {N}}=\sum _{x\in \Lambda } {\mathsf {n}}_x$ with ${\mathsf {n}}_x={\mathsf {c}}_x^* {\mathsf {c}}_x$.

Let us define a natural self-dual cone in $W{\mathfrak {E}}_{n=|\Lambda |}[M]$. By the identifications (A.5) and (A.10), we have

$$\begin{aligned} W{\mathfrak {E}}_{n=|\Lambda |} [M]={\mathscr {L}}^2({\mathcal {E}}^{M^{\dagger }}) . \end{aligned}$$

(A.16)

Here, the antiunitary $\vartheta$ for this expression is defined as

$$\begin{aligned} \vartheta {\mathsf {c}}_x\vartheta ={\mathsf {c}}_x,\ \ \ \ \vartheta \Omega =\Omega , \end{aligned}$$

(A.17)

where $\Omega$ is the Fock vacuum in ${\mathfrak {E}}(\ell ^2(\Lambda ))$. (As for $\vartheta$, see Sect. A.1.) Now, we define a natural self-dual cone ${\mathfrak {P}}_{\mathrm {H}}[M]$ in ${\mathscr {L}}^2({\mathcal {E}}^{M^{\dagger }})$ by

$$\begin{aligned} {\mathfrak {P}}_{\mathrm {H}} [M]={\mathscr {L}}^2({\mathcal {E}}^{M^{\dagger }})_+. \end{aligned}$$

(A.18)

1.5 A Self-dual Cone in $W{\mathfrak {E}}_{n=|\Lambda |} [M]\otimes \mathfrak {F}_{\mathrm {ph}}$

We set $p_x=i \sqrt{\frac{\omega }{2}}(b_x^*-b_x)$ and $q_x=\frac{1}{\sqrt{2\omega }}(b_x^*+b_x).$ Both operators are essentially self-adjoint. We denote their closures by the same symbols. Remark the following identification: $\mathfrak {F}_{\mathrm {ph}}=L^2({\mathcal {Q}}, d{\varvec{q}})=L^2({\mathcal {Q}}),$ where ${\mathcal {Q}}=\mathbb {R}^{|\Lambda |},\ d{\varvec{q}} =\prod _{x\in \Lambda } dq_x$ is the $|\Lambda |$-dimensional Lebesgue measure on ${\mathcal {Q}}$, and $L^2({\mathcal {Q}})$ is the Hilbert space of the square integrable functions on ${\mathcal {Q}}$. Under this identification, $q_x$ and $p_x$ can be viewed as multiplication and partial differential operators, respectively. This expression of $p_x$ and $q_x$ in $L^2({\mathcal {Q}})$ is called the Schrödinger representation. The Hilbert space $W{\mathfrak {E}}_{n=|\Lambda |}[M] \otimes \mathfrak {F}_{\mathrm {ph}}$ can be identified with $W{\mathfrak {E}}_{n=|\Lambda |}[M] \otimes L^2({\mathcal {Q}})$ in this representation.

A natural self-dual cone in $W{\mathfrak {E}}_{n=|\Lambda |}[M] \otimes \mathfrak {F}_{\mathrm {ph}}$ is defined by

$$\begin{aligned} {\mathfrak {P}}_{\mathrm {HH}}[M]=\int _{{\mathcal {Q}}}^{\oplus } {\mathfrak {P}}_{\mathrm {H}}[M] d{\varvec{q}}, \end{aligned}$$

(A.19)

where the RHS of (A.19) is the direct integral of ${\mathfrak {P}}_{\mathrm {H}}[M]$, see Appendix I for details.

1.5.1 A Self-dual Cone in $W{\mathfrak {H}}_{Q}[M]$

Using the expression (A.14), we define a self-dual cone in $W {\mathfrak {H}}_Q[M]$ by

$$\begin{aligned} {\mathfrak {P}}_{\mathrm {Heis}}[M] = \mathrm {Coni} \Big \{ |{\varvec{m}}\rangle \ \Big |\ {\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow },\ |{\varvec{m}}_{\uparrow }|=|{\varvec{m}}_{\downarrow }| =M^{\dagger }\Big \}. \end{aligned}$$

(A.20)

The following property will be useful:

Proposition A.6

Under the identification (A.16), we have ${\mathfrak {P}}_{\mathrm {Heis}}[M]={\tilde{Q}} {\mathfrak {P}}_{\mathrm {H}}[M],$ where ${\tilde{Q}}$ is given by (A.12).

Proof

Let $\varphi \in W{\mathfrak {E}}_{n=|\Lambda |}[M]$. $\varphi$ can be expressed as $\varphi =\sum _{{\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }\in {\mathcal {S}}(M^{\dagger })} \varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }}|{\varvec{m}}\rangle ,$ where ${\mathcal {S}}(M^{\dagger })=\{ {\varvec{m}\in {\mathcal {S}}\, |\, |{\varvec{m}}|=M^{\dagger }} \}$. We say that $\{ \varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }} \}$ is positive semidefinite, if $\sum _{{\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }\in {\mathcal {S}}(M^{\dagger })} \varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }} z^*_{{\varvec{m}}_{\uparrow }} z_{{\varvec{m}}_{\downarrow }} \ge 0$ for all ${\varvec{z}}=\{ z_{{\varvec{m}}} \}_{{\varvec{m}} \in {\mathcal {S}}(M^{\dagger })} \in \mathbb {C}^{|{\mathcal {S}}(M^{\dagger })|}$. Remark that $\varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow } }\ge 0$ if ${\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow }$ in this case.

We note that $\varphi \in \mathfrak {P}_{\mathrm {H}}[M]$ if and only if $\{\varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }} \}$ is positive semidefinite. To see this, we just remark that, by the definition of $\Phi _{\vartheta }$ in Sect. A.1, $\varphi$ can be written as $\varphi =\sum _{{\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }\in {\mathcal {S}}(M^{\dagger })} \varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }}|{\varvec{m}}_{\uparrow }\rangle \langle {\varvec{m}}_{\downarrow }|.$ Here, we used the fact that $|\vartheta {\varvec{m}}_{\downarrow }\rangle =|{\varvec{m}}_{\downarrow }\rangle$ by (A.17).

For each $x\in \Lambda$, we observe that $(n_x-1)^2|{\varvec{m}} \rangle \ne 0$, if and only if $m_{x\uparrow }=m_{x\downarrow }$. Hence, ${\tilde{Q}}|{\varvec{m}}\rangle \ne 0$, if and only if ${\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow }$. Therefore, for each $\varphi \in {\mathfrak {P}}_{\mathrm {H}}[M]$, we obtain

$$\begin{aligned} {\tilde{Q}} \varphi =\sum _{{\varvec{m}}_{\uparrow }= {\varvec{m}}_{\downarrow }\in {\mathcal {S}}(M^{\dagger })} \varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }}|{\varvec{m}}\rangle . \end{aligned}$$

(A.21)

Because $\varphi _{ {\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }} \ge 0$ provided that ${\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow }$, the RHS of (A.21) belongs to ${\mathfrak {P}}_{\mathrm {Heis}}[M]$. $\square$

Proof of Theorems 4.3 and 4.5

In this appendix, we will prove Theorems 4.3 and 4.5. Because several notations are defined in Appendix A, the reader is suggested to study Appendix A first.

1.1 Basic Properties of $H_{\mathrm {Heis}}$

For each $x, y\in \Lambda$ with $x\ne y$ and $\sigma \in \{\uparrow , \downarrow \}$, set $A_{xy\sigma } =c_{x\sigma }c_{y\sigma }^*.$ One can express $H_{\mathrm {Heis}}$ as

$$\begin{aligned} H_{\mathrm {Heis}}&=\sum _{x, y\in \Lambda }\frac{J_{xy}}{2} \bigg \{-(A_{xy \uparrow }A_{xy \downarrow }^*+A_{xy\uparrow }^* A_{xy \downarrow }) +\frac{1}{2}(n_{x\uparrow }-n_{x\downarrow })(n_{y\uparrow }-n_{y\downarrow })\bigg \}. \end{aligned}$$

(B.1)

By the hole-particle transformation W, we have ${\tilde{H}}_{\mathrm {Heis}}=WH_{\mathrm {Heis}} W^{-1}=-T+V,$ where

$$\begin{aligned} T=\sum _{x, y\in \Lambda }\frac{J_{xy}}{2} (A_{xy \uparrow }^*A_{xy \downarrow }^*+A_{xy\uparrow } A_{xy \downarrow }),\ \ \ V= \sum _{x, y\in \Lambda } \frac{J_{xy}}{4} (n_x-1)(n_y-1). \end{aligned}$$

(B.2)

For each $x, y\in \Lambda$, we set $T_{xy}=\frac{J_{xy}}{2} A_{xy\uparrow } A_{xy \downarrow }.$ Trivially, we have $T=\sum _{x,y\in \Lambda }(T_{xy}+T_{xy}^*)$.

Lemma B.1

Let $x, y\in \Lambda$ with $x\ne y$. For all ${\varvec{m}} =({\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }) \in {\mathcal {S}} \times {\mathcal {S}}$ with ${\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow }$, we have the following:

• (i) $\displaystyle T_{xy}|{\varvec{m}}\rangle = \frac{J_{xy}}{2} \big |{\varvec{m}_{\uparrow }-{\varvec{\delta }}_{x}+{\varvec{\delta }}_{y}}, {\varvec{m}_{\downarrow }-{\varvec{\delta }}_{x}+{\varvec{\delta }}_{y}} \big \rangle$, where ${\varvec{\delta }}_a=\{ \delta _{ax}\}_{x\in \Lambda }$.

• (ii) $\displaystyle T_{xy}^* |{\varvec{m}}\rangle = \frac{J_{xy}}{2} \big |{\varvec{m}_{\uparrow }+{\varvec{\delta }}_{x}-{\varvec{\delta }}_{y}}, {\varvec{m}_{\downarrow }+{\varvec{\delta }}_{x}-{\varvec{\delta }}_{y}} \big \rangle$.

In the above, we understand that

$$\begin{aligned} \big |{\varvec{m}_{\uparrow }-{\varvec{\delta }}_{x}+{\varvec{\delta }}_{y}}, {\varvec{m}_{\downarrow }-{\varvec{\delta }}_{x}+{\varvec{\delta }}_{y}} \big \rangle =0 \end{aligned}$$

(B.3)

if $m_{\uparrow x}=0$ or $m_{\uparrow y}=1$ or $m_{\downarrow x}=0$ or $m_{\downarrow y}=1$, and

$$\begin{aligned} \big |{\varvec{m}_{\uparrow }+{\varvec{\delta }}_{x}-{\varvec{\delta }}_{y}}, {\varvec{m}_{\downarrow }+{\varvec{\delta }}_{x}-{\varvec{\delta }}_{y}} \big \rangle =0 \end{aligned}$$

(B.4)

if $m_{\uparrow x}=1$ or $m_{\uparrow y}=0$ or $m_{\downarrow x}=1$ or $m_{\downarrow y}=0$.

Proof

This lemma immediately follows from (A.13) and (B.1). $\square$

Let $\mathfrak {P}_{\mathrm {Heis}}[M]$ be the self-dual cone in $W{\mathfrak {H}}_Q[M]$ defined by (A.20).

Lemma B.2

We have $T_{xy} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$. Therefore, we have $T\unrhd 0$ and $e^{\beta T} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta \ge 0$.

Proof

By Lemma B.1, we immediately get $T_{xy} \unrhd 0$ w.r.t. ${\mathfrak {P}}_{\mathrm {Heis}}[M]$, which implies that $T\unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$. Thus, applying Proposition H.7, we conclude that $e^{\beta T} \unrhd 0$ w.r.t. ${\mathfrak {P}}_{\mathrm {Heis}}[M]$ for all $\beta \ge 0$. $\square$

Lemma B.3

$e^{-\beta V} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta \ge 0$.

Proof

Since $n_x|{\varvec{m}}\rangle =(m_{x\uparrow }+m_{x\downarrow }) |{\varvec{m}}\rangle$, every $|{\varvec{m}}\rangle$ is an eigenvector of V: $V|{\varvec{m}}\rangle =V({\varvec{m}}) |{\varvec{m}}\rangle$, where $V({\varvec{m}})$ is the corresponding eigenvalue. Hence, $e^{-\beta V} |{\varvec{m}}\rangle =e^{-\beta V({\varvec{m}})} |{\varvec{m}}\rangle$, which implies $e^{-\beta V} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$. $\square$

Proposition B.4

$e^{-\beta {\tilde{H}}_{\mathrm {Heis}}[M]} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta \ge 0$.

Proof

By Lemmas B.2, B.3 and Theorem H.9, we arrive at the assertion in the proposition. $\square$

Theorem B.5

Assume (C. 1) and (C. 2). We have $\big ( {\tilde{H}}_{\mathrm {Heis}}[M]+s\big )^{-1} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $s >-E\big ( {\tilde{H}}_{\mathrm {Heis}}[M] \big )$. In particular, $\big ( {\tilde{H}}_{\mathrm {MLM}}[M]+s\big )^{-1} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $s>-E\big ( {\tilde{H}}_{\mathrm {MLM}}[M] \big )$.

Proof

Before we will enter the proof, we remark the following. Let $\psi \in \mathfrak {P}_{\mathrm {Heis}}[M]$. Set $S_M=\{ {\varvec{m}}=({\varvec{m}}_{\uparrow }, {\varvec{m}}_{\downarrow }) \in {\mathcal {S}}\times {\mathcal {S}}\, |\, {\varvec{m}}_{\uparrow }={\varvec{m}}_{\downarrow },\ |{\varvec{m}}_{\uparrow }|=|{\varvec{m}}_{\downarrow }|=M^{\dagger }\}$. We can express $\psi$ as $\psi =\sum _{{\varvec{m}} \in S_M} \psi ({\varvec{m}}) |{\varvec{m}}\rangle$ with $\psi ({\varvec{m}}) \ge 0$ for all ${\varvec{m}}\in S_M$.

We will apply Theorem H.15 with $A=-T$ and $B=V$. To this end, we will check all assumptions in the theorem.

The assumption (b) is satisfied by Lemma B.3. To check (a) and (c), we set $V_n=(1-e^{-n})V, \ n\in \mathbb {N}$. Trivially, $-T+V_n$ converges to ${\tilde{H}}_{\mathrm {Heis}}[M]$, and ${\tilde{H}}_{\mathrm {Heis}}[M]-V_n$ converges to $-T$ in the uniform topology as $n\rightarrow \infty$. Thus, the assumption (a) is fulfilled. To see (c), take $\psi , \psi '\in \mathfrak {P}_{\mathrm {Heis}}[M]$. Suppose that $\langle \psi |\psi '\rangle =0$. We can express these as

$$\begin{aligned} \psi =\sum _{{\varvec{m}} \in S_M} \psi ({\varvec{m}}) |{\varvec{m}}\rangle ,\ \ \ \psi '=\sum _{{\varvec{m}} \in S_M} \psi '({\varvec{m}}) |{\varvec{m}}\rangle \end{aligned}$$

(B.5)

with $\sum _{{\varvec{m}} \in S_M} \psi ({\varvec{m}}) \psi '({\varvec{m}})=0$. Because $\psi ({\varvec{m}})$ and $\psi '({\varvec{m}})$ are nonnegative, we conclude $\psi (\varvec{m}) \psi '(\varvec{m})=0$ for all ${\varvec{m}} \in S_M$. Thus,

$$\begin{aligned} \langle \psi |e^{-\beta V_n} \psi '\rangle =\sum _{{\varvec{m}} \in S_M} \psi (\varvec{m}) \psi '(\varvec{m}) e^{-\beta (1-e^{-n})V({\varvec{m}})}=0, \end{aligned}$$

(B.6)

where $V({\varvec{m}})$ is defined in the proof of Lemma B.3. Therefore, (c) is satisfied.

Choose $\psi , \psi ' \in \mathfrak {P}_{\mathrm {Heis}}[M] \backslash \{0\}$, arbitrarily. Again we express these vectors as (B.5). Since $\psi$ and $\psi '$ are nonzero, there exist ${\varvec{m}}, {\varvec{m}}' \in S_M$ such that $\psi ({\varvec{m}})>0$ and $\psi '({\varvec{m}'})>0$. Because $\psi \ge \psi ({\varvec{m}}) |{\varvec{m}}\rangle$ and $\psi ' \ge \psi '({\varvec{m}}') |{\varvec{m}}'\rangle$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$, we see that

$$\begin{aligned} \langle \psi |e^{\beta T} |\psi '\rangle \ge \psi ({\varvec{m}}) \psi '({\varvec{m}}') \langle {\varvec{m}}|e^{\beta T} |{\varvec{m}}'\rangle \end{aligned}$$

(B.7)

by Lemma B.2.

By Lemma B.2 again, we have

$$\begin{aligned} e^{\beta T} =\sum _{n =0}^{\infty } \frac{\beta ^{n }}{n!} T^{n} \unrhd \frac{\beta ^{\ell }}{\ell !} T^{\ell } \end{aligned}$$

(B.8)

w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta \ge 0$ and $\ell \in \mathbb {N}_0$. On the other hand, using the assumption (C. 1) and Lemma B.2, there are sequencces $\{x_1, x_2, \dots , x_{\ell }; y_1, y_2, \dots , y_{\ell } \}$ and $\{\#_1, \dots , \#_{\ell }\}\in \{{\pm }\}^{\ell }$such that

$$\begin{aligned} \Big \langle {\varvec{m}} \Big | T_{x_1y_1}^{\#_1} T_{x_2y_2}^{\#_2}\cdots T_{x_{\ell } y_{\ell }}^{\#_{\ell }}\Big |{\varvec{m}}'\Big \rangle >0, \end{aligned}$$

(B.9)

where $a^{\#}=a$ if $\#=-$, $a^{\#}=a^*$ if $\#=+$. Since $T \unrhd T_{xy}$ for all $x, y\in \Lambda$, we have

$$\begin{aligned} T^{\ell }\unrhd T_{x_1y_1}^{\#_1} T_{x_2y_2}^{\#_2}\cdots T_{x_{\ell }y_{ \ell }}^{\#_{\ell }} \end{aligned}$$

(B.10)

w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta \ge 0$. Combining (B.8), (B.9) and (B.10), we obtain

$$\begin{aligned} \langle {\varvec{m}}|e^{\beta T} |{\varvec{m}}'\rangle \ge \frac{\beta ^{\ell }}{\ell !} \Big \langle {\varvec{m}} \Big | T_{x_1y_1}^{\#_1} T_{x_2y_2}^{\#_2}\cdots T_{x_{\ell } y_{\ell }}^{\#_{\ell }}\Big |{\varvec{m}}'\Big \rangle >0. \end{aligned}$$

(B.11)

By this and (B.7), we conclude that $e^{\beta T} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta >0$. Therefore, we conclude that, by Theorem H.15, $e^{-\beta {\tilde{H}}_{\mathrm {Heis}}[M]} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta >0$. Because $H_{\mathrm {MLM}}$ is a special case of the Heisenberg model, we also get $e^{-\beta {\tilde{H}}_{\mathrm {MLM}}[M]} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {Heis}}[M]$ for all $\beta >0$. By applying Proposition H.6, we obtain the desired result in Theorem B.5. $\square$

1.2 Proof of Theorem 4.3

We begin with the following proposition [23, 26]:

Proposition B.6

The ground state of $H_{\mathrm {MLM}}$ has total spin $S=\frac{1}{2}\big ||A|-|B|\big |$ and is unique apart from the trivial $(2S+1)$-degeneracy.

Proof

By Theorems H.10 and B.5, the ground state of $H_{\mathrm {MLM}}[M]$ is unique for all $M\in \mathrm {spec}\Big (S_{\mathrm {tot}, |\Lambda |}^{(3)}\Big )$. Let us work in the $M=0$ subspace. Since $2H_{\mathrm {MLM}}=({\varvec{S}}_A+ {\varvec{S}}_B)^2-{\varvec{S}}_A^2-{\varvec{S}}_B^2$, we see that the ground state of $H_{\mathrm {MLM}}[M=0]$ has total spin $S=\frac{1}{2}\big ||A|-|B|\big |$. We denote by $\varphi _0$ the ground state of $H_{\mathrm {MLM}}[0]$. Let $S_{\mathrm {tot}}^{(\pm )}=S_{\mathrm {tot}}^{(1)}\pm i S_{\mathrm {tot}}^{(2)}.$ We set $\varphi _M=(S_{\mathrm {tot}}^{(+)})^M \varphi _0$ for $M>0$ and $\varphi _M=(S_{\mathrm {tot}}^{(-)})^{|M|} \varphi _0$ for $M<0$. Because $H_{\mathrm {MLM}}$ commutes with $S_{\mathrm {tot}}^{(\pm )}$, we know that $\varphi _M$ is the unique ground state of $H_{\mathrm {MLM}}[M]$ which has total spin $S=\frac{1}{2}\big ||A|-|B|\big |$. Thus, the ground state of $H_{\mathrm {MLM}}$ is unique apart from the trivial $(2S+1)$-degeneracy. $\square$

Completion of Proof of Theorem 4.3

By applying Theorem 3.5 and Proposition B.6, one obtains Theorem 4.3. $\square$

1.3 Proof of Theorem 4.5

By Theorem B.5, we can check all conditions in Definition 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}}, \ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\\&\quad = \Big ({\tilde{H}}_{\mathrm {Heis}}[M],\, {\tilde{H}}_{\mathrm {MLM}}[M];\, W{\mathfrak {H}}_{Q}[M],\ W{\mathfrak {H}}_{Q}[M];\\&\quad 1; \, {\mathfrak {P}}_{\mathrm {Heis}}[M],\, {\mathfrak {P}}_{\mathrm {Heis}}[M]; {\tilde{S}}_{\mathrm {tot}, |\Lambda |}^2[M]; 1\Big ). \end{aligned}$$

Hence, ${\tilde{H}}_{\mathrm {Heis}}[M] \leadsto {\tilde{H}}_{\mathrm {MLM}}[M]$ for all $M\in \mathrm {spec}\Big (S_{\mathrm {tot}, |\Lambda |}^{(3)}\Big )$.

By interchanging the roles of $H_{\mathrm {Heis}}$ and $H_{\mathrm {MLM}}$ in the above, we have ${\tilde{H}}_{\mathrm {MLM}}[M] \leadsto {\tilde{H}}_{\mathrm {Heis}}[M]$ for all $M\in \mathrm {spec}\Big (S_{\mathrm {tot}, |\Lambda |}^{(3)}\Big )$. Thus, ${\tilde{H}}_{\mathrm {MLM}}[M] \equiv {\tilde{H}}_{\mathrm {Heis}}[M]$. By applying Proposition 2.10 with $V=W$, we conclude that $H_{\mathrm {MLM}}[M] \equiv H_{\mathrm {Heis}}[M]$. Especially, $H_{\mathrm {Heis}}$ belongs to the Marshall–Lieb–Mattis stability class. $\square$

Proof of Theorem 4.7

Because several notations are defined in Appendix A, the reader is suggested to study Appendix A first.

1.1 Basic Materials

1.1.1 Properties of ${\tilde{Q}}$

Let $\mathfrak {P}_{\mathrm {H}}[M]$ be the self-dual cone in $W{\mathfrak {E}}_{n=|\Lambda |}[M]$ defined by (A.18). Recall that ${\tilde{Q}}$ is defined by ${\tilde{Q}}=WQW^{-1}$, see Notation A.5 for more details.

Proposition C.1

We have ${\tilde{Q}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ for all $M\in \mathrm {spec}\Big (S_{\mathrm {tot}, |\Lambda |}^{(3)}\Big )$.

Proof

Recall the formula (A.12). Set $q_x=(1-n_x)^2$. Since

$$\begin{aligned} q_x=1-n_x+2n_{x\uparrow } n_{x\downarrow } =1-{\mathcal {L}}({{\mathsf {n}}}_x)-{\mathcal {R}}({\mathsf {n}}_x)+2{\mathcal {L}}({\mathsf {n}}_x) {\mathcal {R}}({\mathsf {n}}_x), \end{aligned}$$

(C.1)

we have, by Theorem H.12,

$$\begin{aligned} e^{tq_x} \unrhd 0\ \ \ \text{ w.r.t. } \mathfrak {P}_{\mathrm {H}}[M]\hbox { for all } t\ge 0. \end{aligned}$$

(C.2)

Because $q_x$ is an orthogonal projection, we have

$$\begin{aligned} e^{tq_x} =1+q_x(e^t-1)=q_x^{\perp } +q_x e^t. \end{aligned}$$

(C.3)

By (C.2) and (C.3), we obtain $e^{-t}q_x^{\perp } +q_x \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$. Taking $t\rightarrow \infty$, we arrive at $q_x\unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$, which implies that $\displaystyle {\tilde{Q}} =\prod _{x\in \Lambda } q_x \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$. $\square$

1.1.2 Properties of ${\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]$

First, we remark that, by the hole-particle transformation, the Coulomb interaction term becomes

$$\begin{aligned} W \sum _{x,y\in \Lambda } \frac{U_{xy}}{2} (n_x-1)(n_y-1) W^{-1} = \sum _{x,y\in \Lambda } \frac{U_{xy}}{2} (n_{x\uparrow }-n_{x\downarrow })(n_{y\uparrow }-n_{y\downarrow }). \end{aligned}$$

(C.4)

Using the identification (A.16), we can express ${\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]$ as ${\tilde{H}}_{\mathrm {H}, |\Lambda |}[M] ={\mathbb {T}}-{\mathbb {U}},$ where

$$\begin{aligned} {\mathbb {T}}={\mathcal {L}}(T)+{\mathcal {R}}(T),\ \ \ T=\sum _{x, y\in \Lambda } \bigg ( t_{xy} {\mathsf {c}}_{x}^* {\mathsf {c}}_y+\frac{U_{xy}}{2} {\mathsf {n}}_x{\mathsf {n}}_y \bigg ) \end{aligned}$$

(C.5)

and ${\mathbb {U}}=\sum _{x, y\in \Lambda } U_{xy} {\mathcal {L}}({\mathsf {n}}_x) {\mathcal {R}}({\mathsf {n}}_y).$ Here, recall that ${\mathsf {c}}_x$ and ${\mathsf {n}}_x$ are defined in Sect. A.4. Also, recall that ${\mathcal {L}}(\cdot )$ and ${\mathcal {R}}(\cdot )$ are defined in Sect. A.1.

Lemma C.2

We have ${\mathbb {U}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$.

Proof

Since ${\mathbf {U}}=\{U_{xy}\}_{x, y}$ is a positive definite matrix by (A. 3), all eigenvalues $\{\lambda _x\}_{x\in \Lambda }$ of ${\mathbf {U}}$ are strictly positive, and there exists an orthogonal matrix ${\mathbf {P}}$ such that ${\mathbf {U}}={\mathbf {P}} {\mathbf {D}} {\mathbf {P}}^{\mathrm {T}}$, where ${\mathbf {D}}=\mathrm {diag}(\lambda _x)$. We set ${\mathbf {n}}=\{{\mathsf {n}}_x\}_{x\in \Lambda }$ and set $\hat{{\mathbf {n}}}={\mathbf {P}}^{\mathrm {T}} {\mathbf {n}}$. Denoting $\hat{{\mathbf {n}}}=\{\hat{{\mathsf {n}}}_x\}_{x\in \Lambda }$, we have

$$\begin{aligned} {\mathbb {U}} =\big \langle {\mathcal {L}}({\mathbf {n}})|{\mathbf {U}} {\mathcal {R}}({\mathbf {n}}) \big \rangle =\big \langle {\mathcal {L}}(\hat{{\mathbf {n}}})|{\mathbf {D}} {\mathcal {R}}(\hat{{\mathbf {n}}}) \big \rangle =\sum _{x\in \Lambda } \lambda _x {\mathcal {L}}(\hat{{\mathsf {n}}}_x){\mathcal {R}}(\hat{{\mathsf {n}}}_x)\unrhd 0. \end{aligned}$$

(C.6)

This completes the proof. $\square$

Proposition C.3

We have $\big ( {\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]+s\big )^{-1} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ for all $s>-E\big ( {\tilde{H}}_{\mathrm {H}, |\Lambda |}[M] \big )$.

Proof

We remark that ${\mathbb {U}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ by Lemma C.2, and

$$\begin{aligned} e^{-\beta {\mathbb {T}}} ={\mathcal {L}} (e^{-\beta T}) {\mathcal {R}} (e^{-\beta T}) \unrhd 0 \ \ \text{ w.r.t. } \mathfrak {P}_{\mathrm {H}}[M]. \end{aligned}$$

(C.7)

Therefore, by Theorem H.8, we have $e^{-\beta {\tilde{H}}_{\mathrm {H}, |\Lambda |}[M] } \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ for all $\beta >0$. By applying Proposition H.5, we conclude the assertion in Proposition C.3. $\square$

In [33, Theorem 4.6], we proved the following stronger result:

Theorem C.4

We have $\big ({\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]+s\big )^{-1} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ for all $s >-E\big ({\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]\big )$.

1.2 Completion of Proof of Theorem 4.7

We will check all conditions in Definition 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}},\ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\\&\quad = \Big ({\tilde{H}}_{\mathrm {H}, |\Lambda |}[M], \, {\tilde{H}}_{\mathrm {Heis}}[M];\, W{\mathfrak {E}}_{n=|\Lambda |}[M],\ W{\mathfrak {H}}_Q[M];\, {\tilde{Q}};\\&\qquad \quad {\mathfrak {P}}_{\mathrm {H}}[M],\, {\mathfrak {P}}_{\mathrm {Heis}}[M];\ {\tilde{S}}_{\mathrm {tot}, |\Lambda |}^2[M]; 1\Big ). \end{aligned}$$

By Proposition C.1, (i) of Definition 2.9 is satisfied. (ii) of Definition 2.9 follows from Proposition A.6. (iii) and (iv) of Definition 2.9 follow from Theorem C.4 and Theorem B.5, respectively. Taking Proposition 2.10 (with $V=W$) into consideration, we obtain Theorem 4.7. $\square$

Proof of Theorem 4.10

1.1 Basic Materials

1.1.1 Properties of $P_{\mathrm {ph}}$

Let $P_{\mathrm {ph}}=1\otimes |\Omega _{\mathrm {ph}} \rangle \langle \Omega _{\mathrm {ph}}|$. Let $\mathfrak {P}_{\mathrm {HH}}[M]$ be the self-dual cone in $W{\mathfrak {E}}_{n=|\Lambda |}[M]\otimes \mathfrak {F}_{\mathrm {ph}}$ defined by (A.19).

Proposition D.1

We have $P_{\mathrm {ph}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$ for all $M\in \mathrm {spec}\Big (S_{\mathrm {tot}, |\Lambda |}^{(3)}\Big )$.

Proof

Remark that, in the Schrödinger representation, the Fock vacuum $\Omega _{\mathrm {ph}}$ can be expressed as $\Omega _{\mathrm {ph}}({\varvec{q}})=\pi ^{-|\Lambda |/4}\exp (-{\varvec{q}}^2/2)$, which is strictly positive for all ${\varvec{q}}$. Hence, for each $\varphi \in L^2({\mathcal {Q}})_+$, we have $\langle \Omega _{\mathrm {ph}}|\varphi \rangle \ge 0$, which implies that $|\Omega _{\mathrm {ph}}\rangle \langle \Omega _{\mathrm {ph}}| \unrhd 0$ w.r.t. $L^2({\mathcal {Q}})_+$. Thus, by Corollary I.4, we conclude the assertion in Proposition D.1. $\square$

Proposition D.2

Under the identification by $\tau$ in Sect. 4.6, we have $P_{\mathrm {ph}} {\mathfrak {P}}_{\mathrm {HH}}[M]={\mathfrak {P}}_{\mathrm {H}}[M]$ for all $M\in \mathrm {spec}\Big (S_{\mathrm {tot}, |\Lambda |}^{(3)}\Big )$.

Proof

It is easily checked that $P_{\mathrm {ph}} {\mathfrak {P}}_{\mathrm {HH}}[M] ={\mathfrak {P}}_{\mathrm {H}}[M]\otimes \Omega _{\mathrm {ph}}$. Hence, by the identification in Sect. 4.6, we obtain the desired result. $\square$

1.1.2 Properties of ${\tilde{H}}_{\mathrm {HH}, |\Lambda |} [M]$

We work in the Schrödinger representation introduced in Sect. A.5. Let

$$\begin{aligned} L=-i \sqrt{2} \omega ^{-3/2}\sum _{x, y\in \Lambda } g_{xy} (n_{x\uparrow }-n_{x\downarrow }) p_y. \end{aligned}$$

(D.1)

L is essentially anti-self-adjoint.¹¹ We denote its closure by the same symbol. A unitary operator $e^L$ is called the Lang-Firsov transformation [21]. We remark the following properties:

$$\begin{aligned} e^L c_{x\uparrow } e^{-L}&=\exp \Bigg \{+ i\sqrt{2} \omega ^{-3/2} \sum _{y\in \Lambda } g_{xy} p_y \Bigg \} c_{x\uparrow }, \end{aligned}$$

(D.2)

$$\begin{aligned} e^L c_{x\downarrow } e^{-L}&=\exp \Bigg \{- i\sqrt{2} \omega ^{-3/2} \sum _{y\in \Lambda } g_{xy} p_y \Bigg \} c_{x\downarrow }, \end{aligned}$$

(D.3)

$$\begin{aligned} e^L b_x e^{-L}&= b_x-\omega ^{-1} \sum _{y\in \Lambda }g_{xy} (n_{y\uparrow }-n_{y\downarrow }). \end{aligned}$$

(D.4)

The following facts will be used:

$$\begin{aligned} e^{-i \frac{\pi }{2}N_{\mathrm {ph}}} q_x e^{i \frac{\pi }{2}N_{\mathrm {ph}}} =\omega ^{-1}p_x,\ \ \ e^{-i \frac{\pi }{2}N_{\mathrm {ph}}} p_x e^{i \frac{\pi }{2}N_{\mathrm {ph}}} =\omega q_x, \end{aligned}$$

(D.5)

where $N_{\mathrm {ph}}=\sum _{x\in \Lambda } b_x^*b_x=\frac{1}{2}\sum _{x\in \Lambda } (p_x^2+ \omega ^2q_x^2-1)$.

Lemma D.3

Set ${\mathscr {U}}= e^{-i\frac{\pi }{2} N_{\mathrm {ph}}} e^L$. We have

$$\begin{aligned} {\mathscr {U}}{\tilde{H}}_{\mathrm {HH}, |\Lambda |} {\mathscr {U}}^* =T_{+g, \uparrow }+T_{-g, \downarrow }+{\tilde{U}}_{\mathrm {eff}} +E_{\mathrm {ph}} +\mathrm {Const.}, \end{aligned}$$

(D.6)

where

$$\begin{aligned} T_{\pm g, \sigma }&= \sum _{x, y\in \Lambda } t_{xy} \exp \big \{ \pm i \Phi _{xy}\big \} c_{x\sigma }^*c_{y\sigma }, \end{aligned}$$

(D.7)

$$\begin{aligned} \Phi _{xy}&= \sqrt{2} \omega ^{-1/2} \sum _{z\in \Lambda } (g_{xz} -g_{yz}) q_z, \end{aligned}$$

(D.8)

$$\begin{aligned} E_{\mathrm {ph}}&= \frac{1}{2} \sum _{x\in \Lambda } (p_x^2+\omega ^2q_x^2), \end{aligned}$$

(D.9)

$$\begin{aligned} {\tilde{U}}_{\mathrm {eff}}&= \frac{1}{2} \sum _{x, y\in \Lambda } U_{\mathrm {eff}, xy} (n_{x\uparrow }-n_{x\downarrow })(n_{y\uparrow }-n_{y\downarrow }). \end{aligned}$$

(D.10)

Proof

By the hole-particle transformation, we have

$$\begin{aligned} {\tilde{H}}_{\mathrm {HH}, |\Lambda |} ={\tilde{H}}_{\mathrm {H}, |\Lambda |}+\sum _{x, y\in \Lambda } g_{xy} (n_{x\uparrow }-n_{x\downarrow }) \sqrt{2\omega } q_y +E_{\mathrm {ph}}+\frac{ |\Lambda | \omega }{2}. \end{aligned}$$

(D.11)

By applying (D.2)–(D.5), we obtain the desired expression in Lemma D.3. $\square$

In what follows, we ignore the constant term in (D.6). By Lemma D.3 and taking the identification (A.16) into consideration, we arrive at the following:

Corollary D.4

Let ${\mathbb {H}}_{\mathrm {HH}}[M]={\mathscr {U}} {\tilde{H}}_{\mathrm {HH}, |\Lambda |}[M] {\mathscr {U}}^*$. We have ${\mathbb {H}}_{\mathrm {HH}}[M]={\mathbb {T}}_{\mathrm {HH}}-{\mathbb {U}}_{\mathrm {eff}} +E_{\mathrm {ph}},$ where

$$\begin{aligned} {\mathbb {T}}_{\mathrm {HH}}&= \int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {L}}\big ({\mathbf {T}}_{+g}({\varvec{q}}) \big )d{\varvec{q}} +\int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {R}}\big ({\mathbf {T}}_{+g}({\varvec{q}}) \big )d{\varvec{q}}, \end{aligned}$$

(D.12)

$$\begin{aligned} {\mathbf {T}}_{+g}({\varvec{q}})&=\sum _{x, y\in \Lambda } t_{xy} {\mathsf {c}}^*_x {\mathsf {c}}_y\exp \big \{ +i\Phi _{xy} ({\varvec{q}}) \big \}+\frac{1}{2} \sum _{x, y\in \Lambda } U_{\mathrm {eff}, xy} {\mathsf {n}}_x {\mathsf {n}}_y, \end{aligned}$$

(D.13)

$$\begin{aligned} \Phi _{xy} ({\varvec{q}})&=\sqrt{2} \omega ^{-1/2} \sum _{z\in \Lambda } (g_{xz} -g_{yz}) q_z,\ \ {\varvec{q}}=\{q_x\}_{x\in \Lambda } \in {\mathcal {Q}} \end{aligned}$$

(D.14)

and

$$\begin{aligned} {\mathbb {U}}_{\mathrm {eff}} =\sum _{x, y\in \Lambda } U_{\mathrm {eff}, xy} {\mathcal {L}}({\mathsf {n}}_x) {\mathcal {R}}({\mathsf {n}}_y). \end{aligned}$$

(D.15)

Proof

By the identifications (A.15) and (A.16), we have

$$\begin{aligned} c_{x\uparrow }^* c_{y\uparrow }&={\mathsf {c}}_x^*{\mathsf {c}}_{y} \otimes 1={\mathcal {L}}( {\mathsf {c}}_x^*{\mathsf {c}}_{y}), \end{aligned}$$

(D.16)

$$\begin{aligned} c_{x\downarrow }^* c_{y\downarrow }&=1\otimes {\mathsf {c}}_x^*{\mathsf {c}}_{y} ={\mathcal {R}}\big ( \vartheta ( {\mathsf {c}}_x^*{\mathsf {c}}_{y})^*\vartheta \big ) ={\mathcal {R}}( {\mathsf {c}}_y^*{\mathsf {c}}_{x}). \end{aligned}$$

(D.17)

Here, we used the fact $\vartheta {\mathsf {c}}_x\vartheta ={\mathsf {c}}_x$.

By (D.16), we have

$$\begin{aligned} T_{+g, \uparrow }&= \sum _{x, y\in \Lambda } t_{xy} \int ^{\oplus }_{{\mathcal {Q}}} \exp \big \{ +i \Phi _{xy}({\varvec{q}}) \big \} {\mathcal {L}}({\mathsf {c}}_{x}^*{\mathsf {c}}_y) d{\varvec{q}}\nonumber \\&=\int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {L}}\Bigg ( \sum _{x, y\in \Lambda } t_{xy} \exp \big \{ +i \Phi _{xy}({\varvec{q}}) \big \} {\mathsf {c}}_{x}^*{\mathsf {c}}_y \Bigg ) \, d{\varvec{q}}, \end{aligned}$$

(D.18)

where we used the linearity of ${\mathcal {L}}(\cdot ):\ {\mathcal {L}}(aX+bY)=a {\mathcal {L}}(X)+b{\mathcal {L}}(Y)$.

On the other hand, we have, by (D.17),

$$\begin{aligned} T_{-g, \downarrow }&=\sum _{x, y\in \Lambda } t_{xy} \int ^{\oplus }_{{\mathcal {Q}}} \exp \big \{ -i \Phi _{xy}({\varvec{q}}) \big \} {\mathcal {R}}({\mathsf {c}}_{y}^*{\mathsf {c}}_x) d{\varvec{q}}\nonumber \\&=\int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {R}}\Bigg ( \sum _{x, y\in \Lambda } t_{xy} \exp \big \{ -i \Phi _{xy}({\varvec{q}}) \big \} {\mathsf {c}}_{y}^*{\mathsf {c}}_x \Bigg ) \, d{\varvec{q}} \end{aligned}$$

(D.19)

$$\begin{aligned}&=\int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {R}}\Bigg ( \sum _{x, y\in \Lambda } t_{yx} \exp \big \{ +i \Phi _{yx}({\varvec{q}}) \big \} {\mathsf {c}}_{y}^*{\mathsf {c}}_x \Bigg ) \, d{\varvec{q}}\nonumber \\&=\int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {R}}\Bigg ( \sum _{x, y\in \Lambda } t_{xy} \exp \big \{ +i \Phi _{xy}({\varvec{q}}) \big \} {\mathsf {c}}_{x}^*{\mathsf {c}}_y \Bigg ) \, d{\varvec{q}}. \end{aligned}$$

(D.20)

Here, we used the following properties: $t_{xy}=t_{yx}$ and $\Phi _{yx}({\varvec{q}})=-\Phi _{xy}({\varvec{q}})$. (Note that the last equality in (D.20) comes from the relabeling.) Using these observations, we can prove (D.12). $\square$

Lemma D.5

We have the following:

• (i) $e^{-\beta {\mathbb {T}}_{\mathrm {HH}}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$ for all $\beta \ge 0$.

• (ii) ${\mathbb {U}}_{\mathrm {eff}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$.

• (iii) $e^{-\beta E_{\mathrm {ph}}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$ for all $\beta \ge 0$.

Proof

(i) By (i) of Lemma I.6, we see that

$$\begin{aligned} e^{-\beta {\mathbb {T}}_{\mathrm {HH}}} =\int ^{\oplus }_{{\mathcal {Q}}} {\mathcal {L}}\Big (e^{-\beta {\mathbb {T}}_{\mathrm {HH}}({\varvec{q}})}\Big ) {\mathcal {R}}\Big (e^{-\beta {\mathbb {T}}_{\mathrm {HH}}({\varvec{q}})}\Big ) d{\varvec{q}} \unrhd 0\ \ \ \text{ w.r.t. } \mathfrak {P}_{\mathrm {HH}}[M]. \end{aligned}$$

(D.21)

Since ${\mathbb {U}}_{\mathrm {eff}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ by Lemma C.2, we obtain (ii) by (ii) of Lemma I.6. Finally, because $e^{-\beta E_{\mathrm {rad}} }\unrhd 0$ w.r.t. $L^2({\mathcal {Q}})_+$ for all $\beta \ge 0$ by Example 6, (iii) follows from Proposition I.5. $\square$

Proposition D.6

We have $\big ( {\mathbb {H}}_{\mathrm {HH}}[M]+s\big )^{-1} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$ for all $s \ge -E\big ( {\mathbb {H}}_{\mathrm {HH}}[M] \big )$.

Proof

By Lemma D.5 and Theorem H.8 with $A={\mathbb {T}}_{\mathrm {HH}}+E_{\mathrm {rad}}$ and $B={\mathbb {U}}_{\mathrm {eff}}$, we obtain that $e^{-\beta {\mathbb {H}}_{\mathrm {HH}}[M] } \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$ for all $\beta \ge 0$. Hence, by using Proposition H.5, we conclude the desired result in Proposition D.6. $\square$

Remark that, though subjects are different, there are some similarities between the idea here and that of [25].

In [36], the following much stronger property was proven:

Theorem D.7

We have ${\mathscr {U}} \big ( {\tilde{H}}_{\mathrm {HH}, |\Lambda |}[M] +s\big )^{-1}{\mathscr {U}}^* = \big ({\mathbb {H}}_{\mathrm {HH}}[M]+s\big )^{-1} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {HH}}[M]$ for all $s>-E\big ( {\tilde{H}}_{\mathrm {HH}, |\Lambda |}[M] \big )$.

1.2 Completion of Proof of Theorem 4.10

We will check all conditions in Definition 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}},\ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\nonumber \\&\quad = \Big ({\tilde{H}}_{\mathrm {HH}, |\Lambda |}[M], \, {\tilde{H}}_{\mathrm {H}, |\Lambda |}[M];\, W{\mathfrak {E}}_{n=|\Lambda |}[M]\otimes L^2({\mathcal {Q}}),\ W{\mathfrak {E}}_{n=|\Lambda |}[M];\nonumber \\&\qquad P_{\mathrm {ph}};\, {\mathfrak {P}}_{\mathrm {HH}}[M],\, {\mathfrak {P}}_{\mathrm {H}}[M];\ {\tilde{S}}_{\mathrm {tot}, |\Lambda |}^2[M]; {\mathscr {U}}\Big ). \end{aligned}$$

By Proposition D.1, (i) of Definition 2.9 is satisfied. By Proposition D.2, (ii) of Definition 2.9 is fulfilled. (iii) and (iv) of Definition 2.9 follow from Theorems D.7 and C.4, respectively. By Proposition 2.10 with $V=W$, one obtains Theorem 4.10$\square$

Proof of Theorem 4.14

1.1 Preliminaries

A path integral approach to the quantized radiation fields was initiated by Feynman [4]. There is still a lack of mathematical justification of it, but its euclidean version (i.e., the imaginary time path integral of the radiation fields) has been studied rigorously, see, e.g., [11, 24, 34, 49]. In these works, field operators are expressed as linear operators acting on a certain $L^2$-space. In this subsection, we give such an expression of the radiation fields. Our construction is based on Feynman’s original work [4], and has the benefit of usability.

1.1.1 Second Quantization

Let ${\mathfrak {X}}$ be a complex Hilbert space. The bosonic Fock space over ${\mathfrak {X}}$ is given by $\mathfrak {F}({\mathfrak {X}})=\bigoplus _{n=0}^{\infty } \otimes ^n_{\mathrm {s}} {\mathfrak {X}}$. Let A be a positive self-adjoint operator on ${\mathfrak {X}}$. The second quantization of A is defined by

$$\begin{aligned} d\Gamma (A)=0\oplus \bigoplus _{n=1}^{\infty } \Bigg [\sum _{j=1}^n 1\otimes \cdots \otimes \underbrace{ A }_{j^{\mathrm {th}}}\otimes \cdots \otimes 1\Bigg ]. \end{aligned}$$

(E.1)

We denote by a(f) the annihilation operator in $\mathfrak {F}({\mathfrak {X}})$ with test vector f [43, Sect. X. 7]. By definition, a(f) is densely defined, closed, and antilinear in f.

We first recall the following factorization property of the bosonic Fock space:

$$\begin{aligned} \mathfrak {F}({\mathfrak {X}}_1\oplus {\mathfrak {X}}_2 ) =\mathfrak {F}({\mathfrak {X}}_1) \otimes \mathfrak {F}({\mathfrak {X}}_2). \end{aligned}$$

(E.2)

Corresponding to (E.2), we have

$$\begin{aligned} d\Gamma (A\oplus B)&=d\Gamma (A) \otimes 1+1\otimes d\Gamma (B), \end{aligned}$$

(E.3)

$$\begin{aligned} a(f\oplus g)&=a(f)\otimes 1+1\otimes a(g). \end{aligned}$$

(E.4)

Let $E_{\mathrm {rad}}$ be the closure of $\sum _{\lambda =1,2}\sum _{k\in V^*} \omega (k)a(k, \lambda )^*a(k, \lambda )$. We denote by $[\omega ]$ a multiplication operator $\omega \oplus \omega$ on $\ell ^2(V^*) \oplus \ell ^2(V^*)$. Remark that $E_{\mathrm {rad}}$ is positive, self-adjoint, and

$$\begin{aligned} E_{\mathrm {rad}}=d\Gamma ([\omega ]), \end{aligned}$$

(E.5)

where $d\Gamma (A)$ is the second quantization of A in $\mathfrak {F}_{\mathrm {rad}}$.

1.1.2 The Ultraviolet Cutoff Decomposition

Let $V_{\le \kappa }^* =\{k\in V^*\, |\, |k| \le \kappa \}$, where $\kappa$ is the ultraviolet cutoff introduced in Sect. 1.3.3. Since $\ell ^2(V^*)=\ell ^2(V_{\le \kappa }^*)\oplus \ell ^2((V_{\le \kappa }^*)^c)$ with $(V_{\le \kappa }^*)^c$, the complement of $V_{\le \kappa }^*$, we have the identification $\ell ^2(V^*)\oplus \ell ^2(V^*)= \Big (\ell ^2(V_{\le \kappa }^*) \oplus \ell ^2(V_{\le \kappa }^*)\Big ) \oplus \Big ( \ell ^2((V_{\le \kappa }^*)^c) \oplus \ell ^2((V_{\le \kappa }^*)^c) \Big )$, which implies

$$\begin{aligned} \mathfrak {F}=\mathfrak {F}_{ \le \kappa } \otimes \mathfrak {F}_{>\kappa }, \end{aligned}$$

(E.6)

by (E.2), where $\mathfrak {F}_{ \le \kappa }$ and $\mathfrak {F}_{>\kappa }$ are the bosonic Fock spaces over $\ell ^2(V_{\le \kappa }^*) \oplus \ell ^2(V_{\le \kappa }^*)$ and $\ell ^2((V_{\le \kappa }^*)^c) \oplus \ell ^2((V_{\le \kappa }^*)^c)$, respectively.

Let $d\Gamma _{\le \kappa }(A)$ and $d\Gamma _{>\kappa }(B)$ be the second quantized operators in $\mathfrak {F}_{\le \kappa }$ and $\mathfrak {F}_{>\kappa }$, respectively. We have

$$\begin{aligned} E_{\mathrm {rad}}=d\Gamma _{\le \kappa }([\omega ]) \otimes 1+1\otimes d\Gamma _{>\kappa }([\omega ]) \end{aligned}$$

(E.7)

by (E.3).

We introduce two closed subspaces of $\ell ^2(V^*_{\le \kappa })$ as follows:

$$\begin{aligned} \ell ^2_{ {\varvec{\varepsilon }}_1 }(V^*_{\le \kappa })= \overline{\bigcup _{j=1, 2, 3} \mathrm {ran}(\varepsilon _{1j}) }, \ \ \ \ell ^2_{ {\varvec{\varepsilon }}_2 }(V^*_{\le \kappa })= \overline{ \bigcup _{j=1,2,3} \mathrm {ran}(\varepsilon _{2j}) }, \end{aligned}$$

(E.8)

where ${\varvec{\varepsilon }}_{\lambda }=(\varepsilon _{\lambda 1}, \varepsilon _{\lambda 2}, \varepsilon _{\lambda 3})$ are the polarization vectors given by (1.16). Here, we identify the functions $\varepsilon _{1j}$ and $\varepsilon _{2j}$ with the corresponding multiplication operators acting on $\ell ^2(V_{\le \kappa }^*)$.

Lemma E.1

Let ${\mathcal {H}}_{\varepsilon }=\ell ^2_{ {\varvec{\varepsilon }}_1 }(V^*_{\le \kappa })\oplus \ell ^2_{ {\varvec{\varepsilon }}_2 }(V^*_{\le \kappa })$. Let ${\mathcal {I}}=\{k=(k_1, k_2, k_3)\in V^*_{\le \kappa }\, |\, k_1=k_2=0\}$. We have the following identifications:

$$\begin{aligned} {\mathcal {H}}_{\varepsilon }&\cong \ell ^2(V^*_{ \kappa }) \oplus \ell ^2(V^*_{ \kappa }), \ \ V_{\kappa }^*:=V_{\le \kappa }^*\backslash {\mathcal {I}}, \end{aligned}$$

(E.9)

$$\begin{aligned} {\mathcal {H}}_{\varepsilon }^{\perp }&\cong \ell ^2({\mathcal {I}}) \oplus \ell ^2({\mathcal {I}}). \end{aligned}$$

(E.10)

Proof

Since

$$\begin{aligned} \Bigg (\overline{\bigcup _{j=1, 2, 3} \mathrm {ran}(\varepsilon _{1j}) }\Bigg )^{\perp } =\bigcap _{j=1,2,3}\Big (\mathrm {ran}(\varepsilon _{1j})\Big )^{\perp } =\bigcap _{j=1,2,3} \ker (\varepsilon _{1j}), \end{aligned}$$

(E.11)

we have, by (1.16)

$$\begin{aligned} {\mathcal {H}}_{\varepsilon }^{\perp }&= \{ f\in \ell ^2(V^*_{\le \kappa }) \oplus \ell ^2(V^*_{\le \kappa })\, |\, f(k)=0\ \ \text{ for } \forall k\in {\mathcal {I}}^c \} \cong \ell ^2({\mathcal {I}}) \oplus \ell ^2({\mathcal {I}}). \end{aligned}$$

(E.12)

Thus, we obtain

$$\begin{aligned} {\mathcal {H}}_{\varepsilon }&=\{ f\in \ell ^2(V^*_{\le \kappa }) \oplus \ell ^2(V^*_{\le \kappa })\, |\, f(k)=0 \ \ \text{ for } \forall k\in {\mathcal {I}} \}\cong \ell ^2(V_{\kappa }^*) \oplus \ell ^2(V_{\kappa }^*). \end{aligned}$$

(E.13)

This completes the proof. $\square$

By (E.2) and Lemma E.1, we obtain

$$\begin{aligned} \mathfrak {F}_{\le \kappa }=\mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}})\otimes \mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp }). \end{aligned}$$

(E.14)

Lemma E.2

Corresponding to the decomposition (E.14), we have

$$\begin{aligned} d\Gamma _{\le \kappa }([\omega ])=d\Gamma _{{\varvec{\varepsilon }}}([\omega ])\otimes 1+1\otimes d\Gamma _{\perp }([\omega ]) , \end{aligned}$$

(E.15)

where $d\Gamma _{{\varvec{\varepsilon }}}([\omega ])$ is the second quantization of $[\omega ]\upharpoonright {\mathcal {H}}_{{\varvec{\varepsilon }}}$ in $\mathfrak {F}( {\mathcal {H}}_{{\varvec{\varepsilon }}})$ and $d\Gamma _{\perp }([\omega ])$ is the second quantization of $[\omega ] \upharpoonright {\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp }$ in $\mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp }).$

Proof

Remark that $[\omega ]=\big ([\omega ] \upharpoonright {\mathcal {H}}_{{\varvec{\varepsilon }}}\big ) \oplus \big ([\omega ] \upharpoonright {\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp }\big )$. Using (E.3), we obtain

$$\begin{aligned} d\Gamma _{\le \kappa }([\omega ])=d\Gamma _{\le \kappa }\Big (\big ([\omega ]\upharpoonright {\mathcal {H}}_{\varepsilon } \big ) \oplus \big ([\omega ] \upharpoonright {\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp } \big )\Big ) = \text{ RHS } \text{ of } (\mathrm{E.15}). \end{aligned}$$

(E.16)

This completes the proof. $\square$

Let

$$\begin{aligned} \ell ^2_{{\mathrm {even}}}(V^*_{\kappa })&=\{f\in \ell ^2(V^*_{\kappa })\, |\, f(-k)=f(k)\}, \end{aligned}$$

(E.17)

$$\begin{aligned} \ell ^2_{{\mathrm {odd}}}(V^*_{\kappa })&=\{f\in \ell ^2(V^*_{\kappa })\, |\, f(-k)=-f(k)\}. \end{aligned}$$

(E.18)

We set

$$\begin{aligned} {\mathfrak {h}}_1={\mathfrak {h}}_3=\ell ^2_{\mathrm {even}}(V^*_{\kappa }),\ \ \ \ {\mathfrak {h}}_2={\mathfrak {h}}_4=\ell ^2_{\mathrm {odd}}(V^*_{\kappa }). \end{aligned}$$

(E.19)

Since $\ell ^2_{{\varvec{\varepsilon }}_1}(V^*_{\kappa })={\mathfrak {h}}_1\oplus {\mathfrak {h}}_2$ and $\ell ^2_{{\varvec{\varepsilon }}_2}(V^*_{ \kappa })={\mathfrak {h}}_3\oplus {\mathfrak {h}}_4$, we obtain

$$\begin{aligned} {\mathcal {H}}_{{\varvec{\varepsilon }}}={\mathfrak {h}}_1\oplus {\mathfrak {h}}_2\oplus {\mathfrak {h}}_3\oplus {\mathfrak {h}}_4. \end{aligned}$$

(E.20)

Using the decomposition (E.20), the annihilation operator on $\mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}})$ can be expressed as $a(f_1, f_2, f_3, f_4)$. In what follows, we use the following notations:

$$\begin{aligned} a(f, 0, 0, 0)&=a_1(f), \ a(0, f, 0, 0)=a_2(f), \end{aligned}$$

(E.21)

$$\begin{aligned} a(0, 0, f, 0)&=a_3(f),\ a(0,0,0, f)=a_4(f). \end{aligned}$$

(E.22)

Thus, $a(f_1, f_2, f_3, f_4)=\sum _{r=1}^4 a_r(f_r)$.

By (E.2) and (E.20), we have the following:

$$\begin{aligned} \mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}})=\bigotimes _{r=1}^4\mathfrak {F}({\mathfrak {h}}_r). \end{aligned}$$

(E.23)

We will often use the following identifications corresponding to (E.23):

$$\begin{aligned} a_1(f)=a_1(f)\otimes 1\otimes 1\otimes 1,\ \ a_2(f)=1\otimes a_2(f)\otimes 1\otimes 1,\ \ \mathrm {etc.} \end{aligned}$$

(E.24)

For notational convenience, we express $a_r(f)$ as

$$\begin{aligned} a_r(f)=\sum _{k\in V_{ \kappa }^*} f(k)^* a_r(k),\ \ f\in {\mathfrak {h}}_r. \end{aligned}$$

(E.25)

The following lemma will be useful:

Lemma E.3

• (i) Let $d\Gamma _r(\omega )$ be the second quantization of $\omega$ in $\mathfrak {F}({\mathfrak {h}}_r)$. We have

  $$\begin{aligned} d\Gamma _{{\varvec{\varepsilon }}}([\omega ])=&d\Gamma _1(\omega )\otimes 1 \otimes 1 \otimes 1 + 1 \otimes d\Gamma _2(\omega )\otimes 1 \otimes 1\nonumber \\&+ 1 \otimes 1 \otimes d\Gamma _3(\omega )\otimes 1 + 1\otimes 1 \otimes 1 \otimes d\Gamma _4(\omega ). \end{aligned}$$

  (E.26)

• (ii) We have

  $$\begin{aligned} A(x)=&A_1(x)\otimes 1 \otimes 1 \otimes 1+1 \otimes A_2(x)\otimes 1 \otimes 1 \nonumber \\&+1 \otimes 1 \otimes A_3(x)\otimes 1 +1 \otimes 1 \otimes 1 \otimes A_4(x), \end{aligned}$$

  (E.27)

  where

  $$\begin{aligned} A_1(x)&=\sum _{k\in V_{\kappa }^*}{\varvec{\varepsilon }}_1(k) \chi _{\kappa }(k) \cos (k\cdot x)\phi _1(k),\ A_2(x)=\sum _{k\in V_{\kappa }^*}{\varvec{\varepsilon }}_1(k) \chi _{\kappa }(k) \sin (k\cdot x)\pi _2(k),\nonumber \\ A_3(x)&=\sum _{k\in V_{\kappa }^*}{\varvec{\varepsilon }}_2(k) \chi _{\kappa }(k) \cos (k\cdot x)\phi _3(k),\ A_4(x)=\sum _{k\in V_{\kappa }^*}{\varvec{\varepsilon }}_2(k) \chi _{\kappa }(k) \sin (k\cdot x)\pi _4(k),\nonumber \\ \pi _r(k)&=\frac{i}{\sqrt{2\omega (k)}}(a_r(k)-a_r(k)^*),\ \ \phi _r(k)=\frac{1}{\sqrt{2\omega (k)}}(a_r(k)+a_r(k)^*). \end{aligned}$$

  (E.28)

1.1.3 The Feynman–Schrödinger Representation

We set $N=\# V_{ \kappa }^*$, the cardinality of $V_{\kappa }^*$. Let us consider a Hilbert space $L^2(\mathbb {R}^N)$. For each $\varphi \in L^2(\mathbb {R}^N)$, we define a multiplication operator q(k) by

$$\begin{aligned} \big ( q(k) \varphi \big )({{\varvec{q}}})=q(k) \varphi ({{\varvec{q}}}),\ \ \ {{\varvec{q}}}=\{q(k)\}_{k\in V_{ \kappa }^*} \in \mathbb {R}^N. \end{aligned}$$

(E.29)

We also define p(k) by $p(k)=-i \partial /\partial q(k)$.

Now, let us switch to a larger Hilbert space $L^2(\mathbb {R}^N) \otimes L^2(\mathbb {R}^N) = L^2(\mathbb {R}^{2N})$. We set

$$\begin{aligned} q_1(k)=q(k) \otimes 1, \ \ q_2(k)=1\otimes q(k). \end{aligned}$$

(E.30)

Similarly, we define $p_1(k)$ and $p_2(k)$. It is easy to see that $[q_r(k), p_{r'}(k')] =i \delta _{rr'} \delta _{kk'}$. The annihilation operator is defined by $b_r(k)=\sqrt{\frac{\omega (k)}{2}} q_r(k)+\frac{1}{\sqrt{2\omega (k)}} ip_r(k)$. Remark that $[b_r(k), b_{r'}(k')^*]=\delta _{rr'} \delta _{kk'}$ holds. Let

$$\begin{aligned} \Phi _0(\{{{\varvec{q}}}_r\}_{r=1}^2) =\pi ^{-N/2} \exp \Bigg \{ -\sum _{r=1}^2 \sum _{k\in V_{\kappa }^*} q_r(k)^2 \Bigg \},\ \ {{\varvec{q}}}_r=\{q_r(k)\}_{k\in V_{\kappa }^*}\in \mathbb {R}^N. \end{aligned}$$

(E.31)

Then, we have $b_r(k) \Phi _0=0$.

We define linear operators $b_1, b_2, b_3$ and $b_4$ by

$$\begin{aligned} b_{r}(g)=\sum _{k\in V_{\kappa }^*} g(k)^* {\tilde{b}}_r(k), \ \ r=1, 2, 3, 4 \end{aligned}$$

(E.32)

for each $g\in {\mathfrak {h}}_r$, where ${\tilde{b}}_1(k)={\tilde{b}}_2(k)=b_1(k)$ and ${\tilde{b}}_2(k)={\tilde{b}}_4(k)=b_2(k)$. It is easily verified that $[b_r(g), b_{r'}(g')^*]=\delta _{rr'} \langle g|g'\rangle$.

The following lemma is easily checked:

Lemma E.4

We have

$$\begin{aligned} L^2(\mathbb {R}^{2N})= \overline{\mathrm {Lin}}\Big \{ b_{r_1}(g_1)^* \cdots b_{r_n}(g_n)^* \Phi _0, \Phi _0\, \Big |\, g_j\in {\mathfrak {h}}_{r_j},\, r_j=1, \dots , 4,\, j=1, \dots , n\in \mathbb {N}\Big \}, \end{aligned}$$

(E.33)

where $\overline{\mathrm {Lin}}(S)$ represents the closure of $\mathrm {Lin}(S)$.

Next, we define a unitary operator U from $\bigotimes _{r=1}^4\mathfrak {F}({\mathfrak {h}}_r)$ onto $L^2(\mathbb {R}^{2N})$ by

$$\begin{aligned} U \Omega _{\mathrm {rad}}&=\Phi _0, \end{aligned}$$

(E.34)

$$\begin{aligned} Ua_{r_1}(g_1)^*\cdots a_{r_n}(g_n)^* \Omega _{\mathrm {rad}}&=b_{r_1}(g_1)^*\cdots b_{r_n}(g_n)^* \Phi _0. \end{aligned}$$

(E.35)

Let $N_r=d\Gamma _r(1), r=1,\dots , 4$ . Let ${\mathbf {U}}$ be a unitary operator given by

$$\begin{aligned} {\mathbf {U}}=U\exp \{-i \pi (N_2+N_4)/2\}. \end{aligned}$$

(E.36)

For each real-valued$g\in {\mathfrak {h}}_r\, (r=1,2,3,4)$, we set

$$\begin{aligned} \phi _r(g)=\sum _{k\in V_{\kappa }^*} g(k) \phi _r(k), \ \ \pi _r(g)=\sum _{k\in V_{\kappa }^*} g(k) \pi _r(k), \end{aligned}$$

(E.37)

where $\phi _r(k)$ and $\pi _r(k)$ are defined by (E.28). Similarly, we set $q_{\mu }(g)=\sum _{k\in V_{ \kappa }^*} g(k) q_{\mu }(k)$ for each $\mu =1,2$. By the definition of ${\mathbf {U}}$, we have the following:

• ${\mathbf {U}} \phi _1(g){\mathbf {U}}^{-1} =q_1(g)$ for each $g\in {\mathfrak {h}}_1$, real-valued;

• ${\mathbf {U}} \pi _2(g) {\mathbf {U}}^{-1} =q_1(g)$ for each $g\in {\mathfrak {h}}_2$, real-valued;

• ${\mathbf {U}} \phi _3(g){\mathbf {U}}^{-1} =q_2(g)$ for each $g\in {\mathfrak {h}}_3$, real-valued;

• ${\mathbf {U}} \pi _4(g) {\mathbf {U}}^{-1} =q_2(g)$ for each $g\in {\mathfrak {h}}_4$, real-valued.

The point here is that $\phi _1(g),\, \pi _2(g), \, \phi _3(g)$ and $\pi _4(g)$ can be identified with the multiplication operators$q_1(g)$ and $q_2(g)$.

Proposition E.5

We have the following:

• (i)
  $$\begin{aligned} {\mathbf {U}}d\Gamma _{{\varvec{\varepsilon }}}([\omega ]) {\mathbf {U}}^{-1} =\sum _{\mu =1,2} \sum _{k\in V_{ \kappa }^*} \frac{1}{2}\Big \{ p_{\mu }(k)^2+\omega (k)^2q_{\mu }(k)^2 \Big \}-\sum _{k\in V_{\kappa }^*} \omega (k). \end{aligned}$$

  (E.38)
• (ii)
  $$\begin{aligned} {\mathbf {U}} A(x){\mathbf {U}}^{-1}=&\sum _{k\in V^*_{\kappa }} \chi _{\kappa }(k)\Big \{ {\varvec{\varepsilon }}_1(k) \Big (\cos (k\cdot x)+\sin (k\cdot x)\Big ) q_{1}(k)\nonumber \\&+ {\varvec{\varepsilon }}_2(k) \Big (\cos (k\cdot x)+\sin (k\cdot x)\Big ) q_2(k) \Big \}. \end{aligned}$$

  (E.39)

Proof

(i) is easy to check. (ii) immediately follows from (ii) of Lemma E.3 and properties just above Proposition E.5. $\square$

1.2 Natural Self-dual Cones

By [48, Theorem I.11], we can identify $\mathfrak {F}_{>\kappa }$ with $L^2({\mathcal {M}}, d\mu )$, where $d\mu$ is some Gaussian probability measure. Also, by (E.10), we can identify $\mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp })$ as $\mathfrak {F}({{\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp }})=\mathfrak {F}(\mathbb {C}^{2|{\mathcal {I}}|})=L^2(\mathbb {R}^{2|{\mathcal {I}}|}; d{\varvec{\eta }})=L^2(\mathbb {R}^{2|{\mathcal {I}}|})$. Combining these with the Feynman-Schrödinger representation in Sect. E.1.3, we obtain

$$\begin{aligned} \mathfrak {F}_{\mathrm {rad}} =\mathfrak {F}_{\le \kappa } \otimes \mathfrak {F}_{>\kappa } =\mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}})\otimes \mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}}^{\perp }) \otimes \mathfrak {F}_{>\kappa } =L^2(\mathbb {R}^{2N}) \otimes L^2(\mathbb {R}^{2|{\mathcal {I}}|}) \otimes L^2({\mathcal {M}}, d\mu ). \end{aligned}$$

(E.40)

Thus, denoting ${\mathscr {R}}=\mathbb {R}^{2N}\times \mathbb {R}^{2|{\mathcal {I}}|} \times {\mathcal {M}}$ and $d\nu = d{{\varvec{q}}} d{\varvec{\eta }} d\mu$, we have $\mathfrak {F}_{\mathrm {rad}}=L^2({\mathscr {R}}, d\nu )$. Moreover, ${\mathbf {U}} E_{\mathrm {rad}}{\mathbf {U}}^{-1}= L_0+L_1,$ where

$$\begin{aligned} L_0&=\sum _{\mu =1,2} \sum _{k\in V_{ \kappa }^*} \frac{1}{2}\Big \{ p_{\mu }(k)^2+\omega (k)^2q_{\mu }(k)^2 \Big \}-\sum _{k\in V_{\kappa }^*} \omega (k), \end{aligned}$$

(E.41)

$$\begin{aligned} L_1&=\sum _{\mu =1,2} \sum _{k\in {\mathcal {I}}} \frac{1}{2}\bigg \{ -\frac{\partial ^2}{\partial \eta _{\mu }^2(k)}+\omega (k)^2\eta _{\mu }(k)^2 \bigg \}-\sum _{k\in {\mathcal {I}}} \omega (k) +d\Gamma _{>\kappa }([\omega ]). \end{aligned}$$

(E.42)

In this representation, we have ${\mathfrak {E}}_{n=|\Lambda |}[M]\otimes \mathfrak {F}_{\mathrm {rad}} ={\mathfrak {E}}_{n=|\Lambda |}[M]\otimes L^2({\mathscr {R}}, d\nu )$ and ${\mathbf {U}} H_{\mathrm {rad}}{\mathbf {U}}^{-1} =K_{\mathrm {rad}}+L_1,$ where

$$\begin{aligned} K_{\mathrm {rad}}=\sum _{x, y\in \Lambda }\sum _{\sigma =\uparrow , \downarrow } t_{xy} \exp \bigg \{ i \int _{C_{xy}} dr\cdot {\mathscr {A}}(r) \bigg \} c_{x\sigma }^*c_{y\sigma } +\sum _{x, y\in \Lambda } \frac{U_{xy}}{2}(n_x-1)(n_y-1)+L_0. \end{aligned}$$

(E.43)

Here, ${\mathscr {A}}(x)=({\mathscr {A}}_1(x), {\mathscr {A}}_2(x), {\mathscr {A}}_3(x))$ is the triplet of the multiplication operator defined by the RHS of (E.39).

Let $\mathfrak {P}_{\mathrm {H}}[M]$ be the self-dual cone in $W{\mathfrak {E}}_{n=|\Lambda |}[M]$ defined by (A.18). The following self-dual cones are important in the remainder of this section:

• $\displaystyle {\mathfrak {P}}_{\mathrm {rad}}[M]=\int ^{\oplus }_{{\mathscr {R}}}{\mathfrak {P}}_{\mathrm {H}}[M] d\nu$ in ${\mathfrak {E}}_{n=|\Lambda |}[M] \otimes \mathfrak {F}_{\mathrm {rad}}$;

• $\displaystyle {\mathfrak {P}}_{{\varvec{\varepsilon }}}[M]=\int ^{\oplus }_{{\mathscr {Q}}} {\mathfrak {P}}_{\mathrm {H}}[M] d{\varvec{q}}$ in ${\mathfrak {E}}_{n=|\Lambda |}[M] \otimes \mathfrak {F}({\mathcal {H}}_{{\varvec{\varepsilon }}})$, where ${\mathscr {Q}}=\mathbb {R}^{2N}$.

Proposition E.6

Let $P_{\mathrm {rad}}=1\otimes |\Omega _{\mathrm {rad}}\rangle \langle \Omega _{\mathrm {rad}}|$. We have the following:

• (i) $P_{\mathrm {rad}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {rad}}[M]$;

• (ii) Under the identification in Sect. 4.7, we have $P_{\mathrm {rad}} \mathfrak {P}_{\mathrm {rad}}[M]=\mathfrak {P}_{\mathrm {H}}[M]$.

Proof

Proofs of (i) and (ii) are similar to those of Propositions D.1 and D.2. $\square$

We will study the hole-particle transformed Hamiltonians ${\tilde{H}}_{\mathrm {rad}, |\Lambda |}$ and ${\tilde{K}}_{\mathrm {rad}, |\Lambda |}$.

Proposition E.7

If $\big ( {\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M]+s\big )^{-1} \rhd 0$ w.r.t. ${\mathfrak {P}}_{{\varvec{\varepsilon }}}[M]$ for all $s>- E\big ( {\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M] \big )$, then ${\mathbf {U}}\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M]+s\big )^{-1} {\mathbf {U}}^{-1}\rhd 0$ w.r.t. ${\mathfrak {P}}_{\mathrm {rad}}[M]$ for all $s> -E\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M] \big )$.

Proof

Using (E.40), we rewrite ${\mathfrak {E}}_{n=|\Lambda |}[M]\otimes \mathfrak {F}_{\mathrm {rad}}$ as

$$\begin{aligned} {\mathfrak {E}}_{n=|\Lambda |}[M]\otimes \mathfrak {F}_{\mathrm {rad}} ={\mathfrak {X}} \otimes L^2(\mathbb {R}^{2|{\mathcal {I}}|} \times {\mathcal {M}}, d{\varvec{\eta }}d\mu ), \end{aligned}$$

(E.44)

where ${\mathfrak {X}}={\mathfrak {E}}_{n=|\Lambda |}[M]\otimes L^2({\mathscr {Q}})$. We set

$$\begin{aligned} L_1^{(1)}&=\sum _{\mu =1,2} \sum _{k\in {\mathcal {I}}} \frac{1}{2}\bigg \{ -\frac{\partial ^2}{\partial \eta _{\mu }^2(k)}+\omega (k)^2\eta _{\mu }(k)^2 \bigg \}-\sum _{k\in {\mathcal {I}}} \omega (k), \end{aligned}$$

(E.45)

$$\begin{aligned} L_1^{(2)}&= d\Gamma _{>\kappa }([\omega ]). \end{aligned}$$

(E.46)

By Example 6 and [48, Theorem I.16], we have $e^{-\beta L_1^{(1)}} \rhd 0$ w.r.t. $L^2(\mathbb {R}^{2|{\mathcal {I}}|}, d{\varvec{\eta }})_+$ and $e^{-\beta L_1^{(2)}} \rhd 0$ w.r.t. $L^2({\mathcal {M}}, d\mu )_+$ for all $\beta >0$. Thus, the ground states of $L_1^{(1)}$ and $L_1^{(2)}$ are unique by Theorem H.10. Let $\Omega ^{(1)}$ (resp. $\Omega ^{(2)}$) be the unique ground state of $L_1^{(1)}$ (resp. $L_1^{(2)}$). Trivially, $\Omega _1=\Omega ^{(1)} \otimes \Omega ^{(2)}$ is the unique ground state of $L_1$. By Corollary I.8, we know that $\Omega _1>0$ w.r.t. $\int _{{\mathcal {M}}}^{\oplus } L^2(\mathbb {R}^{2|{\mathcal {I}}|}, d{\varvec{\eta }})_+d\mu =L^2(\mathbb {R}^{2|{\mathcal {I}}|\times {\mathcal {M}}}, d{\varvec{\eta }}d\mu )_+$.

By the assumption and Theorem H.10, the ground state of ${\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M]$ is unique. Let $\psi$ be the unique ground state of ${\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M]$. We also know that $\psi >0$ w.r.t. $\mathfrak {P}_{{\varvec{\varepsilon }}}[M]$. Because ${\mathbf {U}}{\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M] {\mathbf {U}}^{-1} ={\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M]\otimes 1+1\otimes L_1$, $\psi \otimes \Omega _1$ is the unique ground state of ${\mathbf {U}}{\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M] {\mathbf {U}}^{-1}$. Because $\mathfrak {P}_{\mathrm {rad}}[M]= \int ^{\oplus }_{ \mathbb {R}^{2|{\mathcal {I}}|} \times {\mathcal {M}}} \mathfrak {P}_{{\varvec{\varepsilon }}}[M] d\mu d{\varvec{\eta }}$, we have $\psi \otimes \Omega _1>0$ w.r.t. $\mathfrak {P}_{\mathrm {rad}}[M]$ by Corollary I.8. By applying Theorem H.10 again, we conclude that ${\mathbf {U}}\big ({\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M]+s\big )^{-1} {\mathbf {U}}^{-1}\rhd 0$ w.r.t. ${\mathfrak {P}}_{\mathrm {rad}}[M]$ for all $s> -E\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M] \big )$. $\square$

We wish to prove ${\mathbf {U}}\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M]+s\big )^{-1} {\mathbf {U}}^{-1} \rhd 0$ w.r.t. ${\mathfrak {P}}_{\mathrm {rad}}[M]$ for all $s>-E\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M] \big )$. By Proposition E.7, it suffices to show that $\big ( {\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M] +s\big )^{-1} \rhd 0$ w.r.t. ${\mathfrak {P}}_{{\varvec{\varepsilon }}}[M]$ for all $s>-E\big ( {\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M] \big )$. To this end, we need some preparations.

Lemma E.8

We have ${\tilde{K}}_{\mathrm {rad}, |\Lambda |}[M] =S_{+, \uparrow }+S_{-, \downarrow }+{\tilde{U}}+L_0,$ where

$$\begin{aligned} S_{\pm , \sigma }&=\sum _{x, y\in \Lambda } t_{xy} \exp \Big \{\pm i\Phi _{x, y}^{{\mathscr {A}}}\Big \} c_{x\sigma }^*c_{y\sigma }, \end{aligned}$$

(E.47)

$$\begin{aligned} \Phi _{xy}^{{\mathscr {A}}}&=\int _{C_{xy}} dr\cdot {\mathscr {A}}(r), \end{aligned}$$

(E.48)

$$\begin{aligned} {\tilde{U}}&= \sum _{x, y\in \Lambda }\frac{U_{xy}}{2}(n_{x\uparrow }-n_{x\downarrow })(n_{y\uparrow }-n_{y\downarrow }). \end{aligned}$$

(E.49)

Proof

Observe that

$$\begin{aligned}&W \sum _{x, y\in \Lambda } t_{xy} \exp \big \{ +i \Phi _{xy}^{{\mathscr {A}}} \big \}c_{x\downarrow }^*c_{y\downarrow } W^{-1}\nonumber \\&\quad =\sum _{x, y\in \Lambda } t_{xy} \exp \big \{ +i \Phi _{xy}^{{\mathscr {A}}} \big \} \gamma (x)\gamma (y) c_{x\downarrow }c_{y\downarrow }^*. \end{aligned}$$

(E.50)

Since $\Lambda$ is bipartite in terms of $\{t_{xy}\}$, we have $\gamma (x)\gamma (y)t_{xy}=-t_{xy}$. Thus, because $\Phi _{xy}^{{\mathscr {A}}}=-\Phi _{yx}^{{\mathscr {A}}}$, we obtain

$$\begin{aligned} \text{ RHS } \text{ of } (\mathrm{E.50}) =&\sum _{x, y\in \Lambda }t_{xy} \exp \big \{ +i \Phi _{xy}^{{\mathscr {A}}} \big \} c_{y\downarrow }^*c_{x\downarrow }\nonumber \\ =&\sum _{x, y\in \Lambda }t_{xy} \exp \big \{ -i \Phi _{yx}^{{\mathscr {A}}} \big \} c_{y\downarrow }^*c_{x\downarrow }\nonumber \\ =&S_{-, \downarrow }. \end{aligned}$$

(E.51)

Similarly, we see that $W \sum _{x, y\in \Lambda } t_{xy} \exp \big \{ +i \Phi _{xy}^{{\mathscr {A}}} \big \}c_{x\uparrow }^*c_{y\uparrow } W^{-1}=S_{+, \uparrow }$. $\square$

Corollary E.9

Let ${\mathbb {K}}[M] = {\tilde{K}}_{\mathrm {rad}, |\Lambda |} [M]$. We have ${\mathbb {K}}[M]={\mathbb {S}}-{\mathbb {U}}+L,$ where

$$\begin{aligned} {\mathbb {S}}&=\int ^{\oplus }_{{\mathscr {Q}}} {\mathcal {L}}({\mathbf {S}}_+({\varvec{q}})) d{\varvec{q}} +\int ^{\oplus }_{{\mathscr {Q}}} {\mathcal {R}}({\mathbf {S}}_+({\varvec{q}}))d{\varvec{q}}, \end{aligned}$$

(E.52)

$$\begin{aligned} {\mathbf {S}}_+({\varvec{q}})&= \sum _{x, y\in \Lambda }t_{xy} \exp \Big \{ +i \Phi _{xy}^{{\mathscr {A}}}({\varvec{q}}) \Big \} {\mathsf {c}}_x^*{\mathsf {c}}_y +\sum _{x, y\in \Lambda } \frac{U_{xy}}{2} {\mathsf {n}}_x{\mathsf {n}}_y, \end{aligned}$$

(E.53)

$$\begin{aligned} \Phi _{xy}^{{\mathscr {A}}}(\varvec{q})&=\int _{C_{xy}} dr \cdot {\mathscr {A}}(r)[{\varvec{q}}], \end{aligned}$$

(E.54)

$$\begin{aligned} {\mathscr {A}}(r)[{{\varvec{q}}}]&=\sum _{k\in V^*_{\kappa }} \chi _{\kappa }(k)\Big \{ {\varvec{\varepsilon }}_1(k) \Big (\cos (k\cdot x)+\sin (k\cdot x)\Big ) q_{1}(k)\nonumber \\&\quad + {\varvec{\varepsilon }}_2(k) \Big (\cos (k\cdot x)+\sin (k\cdot x)\Big ) q_2(k) \Big \},\ \ {\varvec{q}}=\{q_1(k), q_2(k)\}_{k\in V_{\kappa }^*} \in {\mathscr {Q}}, \end{aligned}$$

(E.55)

$$\begin{aligned} {\mathbb {U}}&= \sum _{x, y\in \Lambda } U_{xy} {\mathcal {L}}({\mathsf {n}}_x) {\mathcal {R}}({\mathsf {n}}_y). \end{aligned}$$

(E.56)

Proof

The proof is similar to that of Corollary D.4. So we only explain a point to which attention should be paid. We observe that, by (E.47),

$$\begin{aligned} S_{-, \downarrow }&=\sum _{x, y\in \Lambda } t_{xy} \int ^{\oplus }_{{\mathscr {Q}}} \exp \big \{ -i \Phi _{xy}^{{\mathscr {A}}}({\varvec{q}}) \big \} {\mathcal {R}}({\mathsf {c}}_y^* {\mathsf {c}}_x) d{\varvec{q}}\nonumber \\&=\sum _{x, y\in \Lambda } t_{yx} \int ^{\oplus }_{{\mathscr {Q}}} \exp \big \{ +i \Phi _{yx}^{{\mathscr {A}}}({\varvec{q}}) \big \} {\mathcal {R}}({\mathsf {c}}_y^* {\mathsf {c}}_x) d{\varvec{q}}\nonumber \\&=\sum _{x, y\in \Lambda } t_{xy} \int ^{\oplus }_{{\mathscr {Q}}} \exp \big \{ +i \Phi _{xy}^{{\mathscr {A}}}({\varvec{q}}) \big \} {\mathcal {R}}({\mathsf {c}}_x^* {\mathsf {c}}_y) d{\varvec{q}}, \end{aligned}$$

(E.57)

where we used the fact $\Phi _{xy}^{{\mathscr {A}}}({\varvec{q}})=-\Phi _{yx}^{{\mathscr {A}}}({\varvec{q}})$, and the last equality in (E.57) comes from relabeling the indecies. Using this, we obtain the second term in (E.52). $\square$

Lemma E.10

We have the following:

• (i) $e^{-\beta {\mathbb {S}}} \unrhd 0$ w.r.t. $\mathfrak {P}_{{\varvec{\varepsilon }}}[M]$ for all $\beta \ge 0$.

• (ii) ${\mathbb {U}}\unrhd 0$ w.r.t. $\mathfrak {P}_{{\varvec{\varepsilon }}}[M]$.

• (iii) $e^{-\beta E_{\mathrm {rad}}} \unrhd 0$ w.r.t. $\mathfrak {P}_{{\varvec{\varepsilon }}}[M]$ for all $\beta \ge 0$.

Proof

(i) By (i) of Lemma I.6 and (E.52), we see that

$$\begin{aligned} e^{-\beta {\mathbb {S}}} =\int ^{\oplus }_{{\mathscr {Q}}} {\mathcal {L}}\Big (e^{-\beta {\mathbf {S}}_+({\varvec{q}})}\Big ) {\mathcal {R}}\Big (e^{-\beta {\mathbf {S}}_+({\varvec{q}})}\Big ) d{\varvec{q}} \unrhd 0\ \ \ \text{ w.r.t. } \mathfrak {P}_{{\varvec{\varepsilon }}}[M]. \end{aligned}$$

(E.58)

Since ${\mathbb {U}} \unrhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ by Lemma C.2, and $e^{-\beta E_{\mathrm {rad}}} \unrhd 0$ w.r.t. $L^2({\mathscr {Q}})_+$ by Example 6, (ii) and (iii) immediately follow from (ii) of Lemma I.6 and Proposition I.5, respectively. $\square$

Proposition E.11

We have $\big ({\mathbb {K}}[M]+s\big )^{-1}\unrhd 0$ w.r.t. $\mathfrak {P}_{{\varvec{\varepsilon }}}[M]$ for all $s>-E \big ( {\mathbb {K}}[M] \big )$.

Proof

By Proposition H.5, Theorem H.8 and Lemma E.10, we obtain the desired result in Proposition E.11. $\square$

Applying the method developed in [36], we can prove the following:

Theorem E.12

We have $\big ({\mathbb {K}}[M]+s\big )^{-1} \rhd 0$ w.r.t. $\mathfrak {P}_{{\varvec{\varepsilon }}}[M]$ for all $s>E \big ( {\mathbb {K}}[M] \big )$.

Proof

Since the proof is similar to [36, Theorem 4.1], we omit it. $\square$

By Proposition E.7 and Theorem E.12, we finally obtain the following:

Corollary E.13

${\mathbf {U}}\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M]+s\big )^{-1} {\mathbf {U}}^{-1}\rhd 0$ w.r.t. ${\mathfrak {P}}_{\mathrm {rad}}[M]$ for all $s> E\big ( {\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M] \big )$.

1.3 Completion of Proof of Theorem 4.10

We will check all conditions in Definition 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}},\ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\nonumber \\&\quad = \Big ({\tilde{H}}_{\mathrm {rad}, |\Lambda |}[M], \, {\tilde{H}}_{\mathrm {H}, |\Lambda |}[M];\, W{\mathfrak {E}}_{n=|\Lambda |}[M]\otimes L^2({\mathscr {R}}, d\nu ),\ W{\mathfrak {E}}_{n=|\Lambda |}[M];\nonumber \\&\qquad P_{\mathrm {rad}};\, {\mathfrak {P}}_{\mathrm {rad}}[M],\, {\mathfrak {P}}_{\mathrm {H}}[M];\ {\tilde{S}}_{\mathrm {tot}, |\Lambda |}^2[M]; {\mathbf {U}}\Big ). \end{aligned}$$

By Proposition E.6, (i) and (ii) of Definition 2.9 are satisfied. (iii) and (iv) of Definition 2.9 follow from Theorem C.4 and Corollary E.13, respectively. By applying Proposition 2.10 with $V=W$, we obtain Theorem 4.10. $\square$

Proof of Thereom 6.1

(i) Let us consider the on-site Coulomb interaction $\{U\delta _{xy}\}$. Clearly, $\{U\delta _{xy}\}$ is positive definite provided that $U>0$. Thus, by Theorem C.4, it holds that $\big ( {\tilde{H}}^{(U)}_{\mathrm {H}, |\Lambda | }[M]+s\big )^{-1} \rhd 0$ w.r.t. $\mathfrak {P}_{\mathrm {H}}[M]$ for all $s>-E\big ( {\tilde{H}}^{(U)}_{\mathrm {H}, |\Lambda | }[M] \big )$. Therefore, we can check all conditions in Definitions 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}},\ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\nonumber \\&\quad = \Big ({\tilde{H}}_{\mathrm {H}, |\Lambda |}[M], \, {\tilde{H}}_{\mathrm {H}, |\Lambda |}^{(U)}[M];\, W{\mathfrak {E}}_{n=|\Lambda |}[M],\ W{\mathfrak {E}}_{n=|\Lambda |}[M] ;\\&\qquad 1;\, {\mathfrak {P}}_{\mathrm {H}}[M],\, {\mathfrak {P}}_{\mathrm {H}}[M];\ {\tilde{S}}_{\mathrm {tot}, |\Lambda |}^2[M]; 1\Big ), \end{aligned}$$

see Appendices B and C for definitions of the above operators. Thus, we have ${\tilde{H}}_{\mathrm {H}, |\Lambda |}[M] \leadsto {\tilde{H}}_{\mathrm {H}, |\Lambda |}^{(U)}[M]$. By interchanging the roles of ${\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]$ and ${\tilde{H}}_{\mathrm {H}, |\Lambda |}^{(U)}[M]$, we obtain ${\tilde{H}}_{\mathrm {H}, |\Lambda |}^{(U)}[M] \leadsto {\tilde{H}}_{\mathrm {H}, |\Lambda |}[M]$.

In a similar way, we can prove (ii) and (iii). $\square$

Proof of Theorems 5.2, 5.4 and 5.6

1.1 Self-dual Cones

In this subsection, we construct some self-dual cones to prove the stabilities of the Nagaoka–Thouless theorem in Sect. 5. In the rest of this section, we continue to assume (B. 1), (B. 2) and (B. 3).

1.1.1 Self-dual Cone in ${\mathfrak {H}}_{\mathrm {NT}}[M]$

For each $(x, {\varvec{\sigma }})\in {\mathcal {C}}_M$, we set ${\varvec{\sigma }}'=\{\sigma _z' \}_{z\in \Lambda } \in \{\uparrow , \downarrow \}^{\Lambda }$ by

$$\begin{aligned} \sigma _z'= {\left\{ \begin{array}{ll} \uparrow &{} \text{ if } z=x\\ \sigma _z &{} \text{ otherwise }, \end{array}\right. } \end{aligned}$$

(G.1)

where ${\mathcal {C}}_M$ is defined in Sect. 1.5.1. With this notation, we define a complete orthonormal system (CONS) $\{|x, {\varvec{\sigma }} \rangle \, |\, (x, {\varvec{\sigma }}) \in {\mathcal {C}}_M\} \subseteq {\mathfrak {H}}_{\mathrm {NT}}[M]$ by $\displaystyle |x, {\varvec{\sigma }}\rangle =c_{x\uparrow } \prod _{z\in \Lambda }' c_{z\sigma _z'}^*\Omega ,$ where $\displaystyle \prod _{z\in \Lambda }'$ indicates the ordered product according to an arbitrarily fixed order in $\Lambda$. Remark that this CONS was introduced by Tasaki [52, 53].

Definition G.1

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, a canonical self-dual cone in ${\mathfrak {H}}_{\mathrm {NT}}[M]$ is defined by ${\mathfrak {Q}}_{\mathrm {H}}[M]=\mathrm {Coni} \{ |x, {\varvec{\sigma }}\rangle \, \, |\, (x, {\varvec{\sigma }}) \in {\mathcal {C}}_M \}.$ Recall that this type of self-dual cone is defined in Example 2 of Sect. 2. $\diamondsuit$

1.1.2 Self-dual Cone in ${\mathfrak {H}}_{\mathrm {NT}}[M] \otimes \mathfrak {F}_{\mathrm {ph}}$

We switch to the Schrödinger representation already discussed in Sect. A.5: $\mathfrak {F}_{\mathrm {ph}}=L^2({\mathcal {Q}})$. Under the identification ${\mathfrak {H}}_{\mathrm {NT}}[M] \otimes \mathfrak {F}_{\mathrm {ph}}= {\mathfrak {H}}_{\mathrm {NT}}[M] \otimes L^2({\mathcal {Q}})$, we define a self-dual cone ${\mathfrak {Q}}_{\mathrm {HH}}[M]$ by

$$\begin{aligned} {\mathfrak {Q}}_{\mathrm {HH}}[M] =\int ^{\oplus }_{{\mathcal {Q}}} {\mathfrak {Q}}_{\mathrm {H}}[M] d{\varvec{q}}. \end{aligned}$$

(G.2)

Remark that the right hand side of (G.2) is a direct integral of ${\mathfrak {Q}}_{\mathrm {H}}[M]$, see Appendix I for details.

1.1.3 Self-dual Cone in ${\mathfrak {H}}_{\mathrm {NT}}[M] \otimes \mathfrak {F}_{\mathrm {rad}}$

We work in the Feynman–Schrödinger representation introduced in Sect. E.1.3: $\mathfrak {F}_{\mathrm {rad}}=L^2({\mathscr {R}}, d\nu )$. Under the identification ${\mathfrak {H}}_{\mathrm {NT}}[M] \otimes \mathfrak {F}_{\mathrm {rad}}= {\mathfrak {H}}_{\mathrm {NT}}[M] \otimes L^2({\mathscr {R}}, d\nu )$, we define a self-dual cone ${\mathfrak {Q}}_{\mathrm {rad}}[M]$ by

$$\begin{aligned} {\mathfrak {Q}}_{\mathrm {rad}}[M] =\int ^{\oplus }_{{\mathscr {R}}} {\mathfrak {Q}}_{\mathrm {H}}[M] d\nu . \end{aligned}$$

(G.3)

1.2 The Nagaoka–Thouless Theorem

Here, we will give a brief review of the Nagaoka–Thouless theorem. Remark that a strategy below is mathematically equivalent to Tasaki’s work [52].

In [37], Miyao proved the following:

Theorem G.2

$e^{-\beta H^{\infty }_{\mathrm {H}}[M]} \rhd 0$ w.r.t. ${\mathfrak {Q}}_{\mathrm {H}}[M]$ for all $\beta >0$.

As a corollary of Theorem G.2, we get the Nagaoka–Thouless theorem:

Corollary G.3

(Theorem 1.11) The ground state of $H_{\mathrm {H}}^{\infty }$ has total spin $S=\frac{1}{2}(|\Lambda |-1)$ and is unique apart from the trivial $(2S+1)$-degeneracy.

Proof

For each $x\in \Lambda$, let us introduce a spin configuration by

$$\begin{aligned} (\Uparrow _x)_y={\left\{ \begin{array}{ll} \uparrow &{} y\ne x\\ 0 &{} y=x. \end{array}\right. } \end{aligned}$$

(G.4)

Then $\{ |x, \Uparrow _x\rangle \, |\, x\in \Lambda \}$ is a CONS of ${\mathfrak {H}}_{\mathrm {NT}}[M=\frac{1}{2}(|\Lambda |-1)]$. Clearly, each $|x, \Uparrow _x\rangle$ has total spin $S=\frac{1}{2}(|\Lambda |-1)$.

By Theorems H.6, H.10 and G.2, the ground state of $H^{\infty }_{\mathrm {H}}[M]$ with $M=\frac{1}{2}(|\Lambda |-1)$ is unique and strictly positive w.r.t. ${\mathfrak {Q}}_{\mathrm {H}}[M=\frac{1}{2}(|\Lambda |-1)].$ The ground state of $H^{\infty }_{\mathrm {H}}[\tfrac{1}{2}(|\Lambda |-1)]$ is denoted by $\psi$. Since ${\mathfrak {Q}}_{\mathrm {H}} [\frac{1}{2}(|\Lambda |-1)]= \mathrm {Coni}\{ |x, \Uparrow _x\rangle \, |\, x\in \Lambda \}$, we have $\psi =\sum _{x\in \Lambda } \psi _x |x, \Uparrow _x\rangle$ with $\psi _x>0$. Thus, $\psi$ has total spin $S=\frac{1}{2}(|\Lambda |-1)$.

We set ${\mathcal {S}}^{(\pm )}={\mathcal {S}}^{(1)}\pm i {\mathcal {S}}^{(2)}$ as usual. Let $\psi _{\ell }=(S^{(-)})^{\ell } \psi ,\ \ell =0, 1, \dots , |\Lambda |-2$. Then $\psi _{\ell }$ is the unique ground state of $H^{\infty }_{\mathrm {H}}[\frac{1}{2}(|\Lambda |-\ell -1)]$ for each $\ell =0, 1,\dots , |\Lambda |-2$, and has total spin $S=\frac{1}{2}(|\Lambda |-1)$.

Completion of Proof of Theorem 5.2

By Theorem 3.5 and Corollary G.3, one obtains Theorem 5.2. $\square$

1.3 Proof of Theorem 5.4

Corollaries 5.5 and 5.7 were already proved in [37]. Here, we prove these results in the context of the stability class introduced in Sect. 2.

Proposition G.4

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, we have $P_{\mathrm {ph}} \unrhd 0$ w.r.t. ${\mathfrak {Q}}_{\mathrm {HH}}[M]$.

Proof

In the Schrödinger representation, we have $\Omega _{\mathrm {ph}}({\varvec{q}}) =\pi ^{-|\Lambda |/4} \exp (-{\varvec{q}}^2/2)$. Thus, $\Omega _{\mathrm {ph}} >0$ w.r.t. $L^2({\mathcal {Q}})_+$, which implies that $|\Omega _{\mathrm {ph}} \rangle \langle \Omega _{\mathrm {ph}}| \unrhd 0$ w.r.t. $L^2({\mathcal {Q}})_+$. Hence, by Proposition I.5, we obtain the desired assertion in the lemma. $\square$

Proposition G.5

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, we have $P_{\mathrm {ph}} {\mathfrak {Q}}_{\mathrm {HH}}[M]= \mathfrak { Q}_{\mathrm {H}}[M]$.

Proof

It is not hard to check that $P_{\mathrm {ph}}{\mathfrak {Q}}_{\mathrm {HH}}[M]={\mathfrak {Q}}_{\mathrm {H}}[M]\otimes \Omega _{\mathrm {ph}}$. Thus, using the identification ${\mathfrak {H}}_{\mathrm {NT}}[M]= {\mathfrak {H}}_{\mathrm {NT}}[M] \otimes \Omega _{\mathrm {ph}}$ mentioned in Sect. 5.3, we have $P_{\mathrm {ph}}{\mathfrak {Q}}_{\mathrm {HH}}[M]={\mathfrak {Q}}_{\mathrm {H}}[M]$. $\square$

Recall the definition of the Lang-Firsov transformation $e^L$ in Sect. D.1.2. In [37], we proved the following:

Theorem G.6

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, we have $e^L \exp \Big \{-\beta H_{\mathrm {HH}}^{\infty }[M]\Big \} e^{-L} \rhd 0$ w.r.t. ${\mathfrak {Q}}_{\mathrm {HH}}[M]$ for all $\beta >0$.

Proof

See [37, Theorem 5.10]. $\square$

Completion of Proof of Theorem 5.2

We will check all conditions in Definition 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}},\ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\nonumber \\&\quad =\Big ( H_{\mathrm {HH}}^{\infty }[M], H_{\mathrm {H}}^{\infty }[M];\ {\mathfrak {H}}_{\mathrm {NT}}[M] \otimes L^2({\mathcal {Q}}), {\mathfrak {H}}_{\mathrm {NT}}[M];\ \nonumber \\&\qquad P_{\mathrm {ph}};\ {\mathfrak {Q}}_{\mathrm {HH}}[M], {\mathfrak {Q}}_{\mathrm {H}}[M];\ {\mathcal {S}}^2;\ e^L \Big ). \end{aligned}$$

(G.5)

By Propositions G.4 and G.5, we can confirm (i) and (ii) of Definition 2.9. By Theorems G.2, G.6 and H.6, (iii) and (iv) of Definition 2.9 are satisfied. Hence, by Proposition 2.10, we obtain Theorem 5.2. $\square$

1.4 Proof of Theorem 5.6

Because the idea of the proof is similar to that of Theorem 5.4, we provide a sketch only.

Let $\Omega _{\mathrm {rad}}$ be the Fock vacuum in $\mathfrak {F}_{\mathrm {rad}}$. We set $P_{\mathrm {rad}}=1\otimes |\Omega _{\mathrm {rad}}\rangle \langle \Omega _{\mathrm {rad}}|$.

In a similar way as in Sect. G.3, we can prove the following two propositions.

Proposition G.7

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, we have $P_{\mathrm {rad}} \unrhd 0$ w.r.t. ${\mathfrak {Q}}_{\mathrm {rad}}[M]$.

Proposition G.8

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, we have $P_{\mathrm {rad}}{\mathfrak {Q}}_{\mathrm {rad}}[M]={\mathfrak {Q}}_{\mathrm {H}}[M]$.

In [37], we proved the following:

Theorem G.9

For each $M\in \mathrm {spec}\big ( {\mathcal {S}}^{(3)} \big )$, we have ${\mathsf {F}} \exp \Big \{-\beta H_{\mathrm {rad}}^{\infty }[M]\Big \} {\mathsf {F}}^{-1}\rhd 0$ w.r.t. ${\mathfrak {Q}}_{\mathrm {rad}}[M]$ for all $\beta >0$, where ${\mathsf {F}}=e^{i\pi N_{\mathrm {rad}}/2} {\mathbf {U}}$ with $N_{\mathrm {rad}}=d\Gamma (1)$. Here, ${\mathbf {U}}$ is defined by (E.36).

Completion of Proof of Theorem 5.4

We will check all conditions in Definition 2.9 with

$$\begin{aligned}&(H, \ H_*;\ {\mathfrak {H}},\ {\mathfrak {H}}_*;\ P;\ {\mathfrak {P}}, \ {\mathfrak {P}}_*; O; U)\nonumber \\&\quad =\Big ( H_{\mathrm {rad}}^{\infty }[M], H_{\mathrm {H}}^{\infty }[M];\ {\mathfrak {H}}_{\mathrm {NT}}[M] \otimes L^2({\mathscr {R}}, d\nu ), {\mathfrak {H}}_{\mathrm {NT}}[M];\ \nonumber \\&\qquad P_{\mathrm {rad}};\ \mathfrak {Q_{\mathrm {rad}}}[M], {\mathfrak {Q}}_{\mathrm {H}}[M]; \ {\mathcal {S}}^2;\ {\mathsf {F}} \Big ). \end{aligned}$$

(G.6)

By Propositions G.7 and G.8, we can confirm (i) and (ii) of Definition 2.9. (iv) of Definition 2.9 is satisfied by Theorems G.9 and H.6. By Theorems G.2 and H.6, we know that (iii) of Definition 2.9 is fulfilled. Therefore, by Proposition 2.10, we obtain Theorem 5.4. $\square$

Basic Properties of Operator Theoretic Correlation Inequalities

We begin with the following theorem.

Theorem H.1

Let $\mathfrak {P}$ be a convex cone in $\mathfrak {H}$. $\mathfrak {P}$ is self-dual if and only if:

• (i) $\langle \xi | \eta \rangle \ge 0$ for all $\xi , \eta \in \mathfrak {P}$.

• (ii) Let $\mathfrak {H}_{\mathbb {R}}$ be a real closed subspace of $\mathfrak {H}$ generated by $\mathfrak {P}$ . Then for all $\xi \in \mathfrak {H}_{\mathbb {R}}$, there exist $\xi _+, \xi _-\in \mathfrak {P}$ such that $\xi =\xi _+-\xi _-$ and $\langle \xi _+| \xi _-\rangle =0$.

• (iii) $\mathfrak {H}=\mathfrak {H}_{\mathbb {R}}+i \mathfrak {H}_{{\mathbb {R}}}= \{\xi + i \eta \, |\, \xi , \eta \in \mathfrak {H}_{\mathbb {R}}\}$.

Proof

For the reader’s convenience, we provide a sketch of the proof.

Assume that $\mathfrak {P}$ is self-dual. Then, by [1] or [2, Proof of Proposition 2.5.28], we easily check that the conditions (i)–(iii) are fulfilled.

Conversely, suppose that $\mathfrak {P}$ satisfies (i)–(iii). We see that $\mathfrak {P}\subseteq \mathfrak {P}^{\dagger }$ by (i). We will show the inverse. Let $\xi \in \mathfrak {P}^{\dagger }$. By (ii) and (iii), we can write $\xi$ as $\xi =(\xi _{R, +}-\xi _{R, -})+i(\xi _{I, +}-\xi _{I, -})$ with $\xi _{R, \pm }, \xi _{I, \pm } \in \mathfrak {P}$, $\langle \xi _{R, +}|\xi _{R, -}\rangle =0$ and $\langle \xi _{I, +}|\xi _{I, -}\rangle =0$. Assume that $\xi _{I, +}\ne 0$. Then, $\langle \xi |\xi _{I, +}\rangle$ is a complex number, which contradicts with the fact that $\langle \xi |\eta \rangle \ge 0$ for all $\eta \in \mathfrak {P}$. Thus, $\xi _{I, +}$ must be 0. Similarly, we have $\xi _{I, -}=0$. Next, assume that $\xi _{R, -}\ne 0$. Because $\xi _{R, -}\in \mathfrak {P}$, we have $0\le \langle \xi |\xi _{R, -}\rangle =-\Vert \xi _{R, -}\Vert ^2<0,$ which is a contradiction. Hence, we conclude that $\xi =\xi _{R, +}\in \mathfrak {P}$. $\square$

Corollary H.2

Let $\mathfrak {P}$ be a self-dual cone in $\mathfrak {H}$. For each $\xi \in \mathfrak {H}$, we have the following decomposition:

$$\begin{aligned} \xi =(\xi _1-\xi _2)+i(\xi _3-\xi _4), \end{aligned}$$

(H.1)

where $\xi _1, \xi _2, \xi _3$ and $\xi _4$ satisfy $\xi _1, \xi _2, \xi _3, \xi _4\in \mathfrak {P}$, $\langle \xi _1|\xi _2\rangle =0$ and $\langle \xi _3|\xi _4\rangle =0$.

Definition H.3

Suppose that $A\mathfrak {H}_{\mathbb {R}}\subseteq \mathfrak {H}_{\mathbb {R}}$ and $B\mathfrak {H}_{\mathbb {R}} \subseteq \mathfrak {H}_{\mathbb {R}}$. If $(A-B) \mathfrak {P}\subseteq \mathfrak {P}$, then we write this as $A \unrhd B$ w.r.t. $\mathfrak {P}$. $\diamondsuit$

The following proposition is fundamental in the present paper

Proposition H.4

Let $A, B, C, D\in {\mathscr {B}}(\mathfrak {H})$ and let $a, b\in \mathbb {R}$.

• (i) If $A\unrhd 0, B\unrhd 0$ w.r.t. $\mathfrak {P}$ and $a, b\ge 0$, then $aA +bB \unrhd 0$ w.r.t. $\mathfrak {P}$.

• (ii) If $A \unrhd B \unrhd 0$ and $C\unrhd D \unrhd 0$ w.r.t. $\mathfrak {P}$, then $AC\unrhd BD \unrhd 0$ w.r.t. $\mathfrak {P}$.

• (iii) If $A \unrhd 0$ w.r.t. $\mathfrak {P}$, then $A^*\unrhd 0$ w.r.t. $\mathfrak {P}$.

Proof

See, e.g., [39, Lemmas 2.6 and 2.7]. $\square$

Proposition H.5

Let A be a positive self-adjoint operator. The following statements are mutually equivalent:

• (i) $e^{-tA} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $t\ge 0$.

• (ii) $(A+s)^{-1} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $s>-E(A)$.

Proof

The proposition immediately follows from the following formulas:

$$\begin{aligned} (A+s)^{-1} =\int _0^{\infty } e^{-t(A+s)} dt,\ \ \ e^{-tA}=\mathrm {s}\text{- }\displaystyle \lim _{n\rightarrow \infty }\Big (1+\frac{s}{n}A\Big )^{-n}, \end{aligned}$$

(H.2)

where $\mathrm {s}\text{- }\displaystyle \lim _{n\rightarrow \infty }$ indicates the strong limit. $\square$

Proposition H.6

Let A be a positive self-adjoint operator. The following statements are mutually equivalent:

• (i) The semigroup $e^{-tA}$ is ergodic, that is, for every $\xi , \eta \in \mathfrak {P}\backslash \{0\}$, there exists a $t_0\ge 0$ such that $\langle \xi |e^{-t_0A} \eta \rangle >0$. Note that $t_0$ could depend on $\xi$ and $\eta$.

• (ii) $(A+s)^{-1} \rhd 0$ w.r.t. $\mathfrak {P}$ for all $s>-E(A)$.

In particular, if $e^{-t A} \rhd 0$ w.r.t. $\mathfrak {P}$ for all $t>0$, then $(A+s)^{-1} \rhd 0$ w.r.t. $\mathfrak {P}$ for all $s>-E(A)$.

Proof

Use the elementary formula: $(A+s)^{-1} =\int _0^{\infty } e^{-t(A+s)} dt$. $\square$

Proposition H.7

Assume that $A \unrhd 0$ w.r.t. $\mathfrak {P}$. Then $\mathrm {e}^{\beta A} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

See, e.g., [35, Proposition A. 3]. $\square$

Theorem H.8

Let A be a self-adjoint positive operator on $\mathfrak {H}$ and $B\in {\mathscr {B}}(\mathfrak {H})$. Suppose that

• (i) $e^{-\beta A} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$;

• (ii) $B\unrhd 0$ w.r.t. $\mathfrak {P}$.

Then we have $e^{-\beta (A-B)}\unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

See, e.g., [35, Proposition A. 5]. $\square$

Proposition H.9

Assume that $\mathrm {e}^{\beta A} \unrhd 0$ and $\mathrm {e}^{\beta B} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$. Then $\mathrm {e}^{\beta (A+B)} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

See, e.g., [35, Proposition A. 4]. $\square$

The following theorem plays an important role in the present study.

Theorem H.10

(Perron–Frobenius–Faris) Let A be a self-adjoint operator, bounded from below. Let $E(A)=\inf \mathrm {spec}(A)$, where $\mathrm {spec}(A)$ is spectrum of A. Suppose that $0\unlhd e^{-tA}$ w.r.t. $\mathfrak {P}$ for all $t\ge 0$, and that E(A) is an eigenvalue. Let $P_A$ be the orthogonal projection onto the closed subspace spanned by eigenvectors associated with E(A). Then the following statements are equivalent:

• (i) $\dim \mathrm {ran}P_A=1$ and $P_A\rhd 0$ w.r.t. $\mathfrak {P}$.

• (ii) $(A+s)^{-1}\rhd 0$ w.r.t. $\mathfrak {P}$ for all $s>-E(A)$.

Proof

See, e.g., [3, 32, 44]. $\square$

Remark H.11

By (i), there exists a unique $\xi \in \mathfrak {H}$ such that $\xi >0$ w.r.t. $\mathfrak {P}$ and $P_A=|\xi \rangle \langle \xi |$. Of course, $\xi$ satisfies $A\xi =E(A)\xi$. Hence, (i) implies that the lowest eigenvalue of A is nondegenerate, and the corresponding eigenvector is strictly positive. $\diamondsuit$

Theorem H.12

Let us consider the case where $\mathfrak {P}={\mathscr {L}}^2({\mathfrak {H}})_+$ given in Definition A.1. Let $A, C_j\in {\mathscr {B}}({\mathfrak {H}})$. Suppose that A is self-adjoint. We set $H={\mathcal {L}}(A)+{\mathcal {R}}(A)-\sum _{j=1}^n {\mathcal {L}}(C_j) {\mathcal {R}}(C_j^*).$ Then $e^{-\beta H} \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$.

Proof

We set $H_0={\mathcal {L}}(A)+{\mathcal {R}}(A)$ and $V=\sum _{j=1}^n {\mathcal {L}}(C_j) {\mathcal {R}}(C_j^*).$ Trivially, $H=H_0-V$. By Proposition A.3, we have $e^{-\beta H_0}={\mathcal {L}}(e^{-\beta A}) {\mathcal {R}}(e^{-\beta A}) \unrhd 0$ w.r.t. $\mathfrak {P}$ for all $\beta \ge 0$. Because $V\unrhd 0$ w.r.t. $\mathfrak {P}$, we can apply Theorem H.8. $\square$

Remark H.13

Jaffe and Pedrocchi give an alternative proof within an algebraic setting [16]. $\diamondsuit$

Proposition H.14

Let $A\in {\mathscr {B}}({\mathfrak {H}})$. The following statements are equivalent:

• (i) $A\rhd 0$ w.r.t. $\mathfrak {P}$.

• (ii) $A^*\rhd 0$ w.r.t. $\mathfrak {P}$.

Proof

(i) $\Rightarrow$ (ii): Let $\xi , \eta \in \mathfrak {P}\backslash \{0\}$. Because $A\xi >0$ w.r.t. $\mathfrak {P}$, we have $\langle \xi |A^*\eta \rangle =\langle A\xi |\eta \rangle >0.$ Since $\xi$ is arbitrary, we have $A^*\eta >0$ w.r.t. $\mathfrak {P}$, which implies that $A^*\rhd 0$ w.r.t. $\mathfrak {P}$. Simiraly, we can prove that (ii) $\Rightarrow$ (i). $\square$

Theorem H.15

Let A and B be self-adjoint operators, bounded from below. Assume the following conditions:

• (a) There exists a sequence of bounded self-adjoint operator $C_n$ such that $A+C_n$ converges to B in the strong resolvent sense and $B-C_n$ converges to A in the strong resolvent sense as $n\rightarrow \infty$;

• (b) $e^{-tC_n}\unrhd 0$ w.r.t. $\mathfrak {P}$ for all $t\in \mathbb {R}$ and $n\in \mathbb {N}$;

• (c) For all $\xi , \eta \in \mathfrak {P}$ such that $\langle \xi |\eta \rangle =0$, it holds that $\langle \xi |e^{-tC_n}\eta \rangle =0$ for all $n\in \mathbb {N}$ and $t\ge 0$.

The following (i) and (ii) are mutually equivalent:

• (i) $(A+s)^{-1} \rhd 0$ w.r.t. $\mathfrak {P}$ for all $s>-E(A)$;

• (ii) $(B+s)^{-1} \rhd 0$ w.r.t. $\mathfrak {P}$ for all $s>-E(B)$.

Proof

See, e.g., [3, Theorem 3] and [37, Theorem A.1]. $\square$

Direct Integrals of Self-dual Cones

Let ${\mathfrak {H}}$ be a complex Hilbert space. Let $(M, {\mathfrak {M}}, \mu )$ be a $\sigma$-finite measure space. The Hilbert space of $L^2(M, d\mu ; {\mathfrak {H}})$ of square integrable ${\mathfrak {H}}$-valued functions [44, Sect. XIII.16] is called a constant fiber direct integral, and is written as $\displaystyle \int ^{\oplus }_M{\mathfrak {H}} d\mu .$ The inner product on $\displaystyle \int ^{\oplus }_M{\mathfrak {H}} d\mu$ is given by $\displaystyle \langle \Phi |\Psi \rangle =\int _M\langle \Phi (m)|\Psi (m)\rangle _{{\mathfrak {H}} }d\mu ,$ where $\langle \cdot |\cdot \rangle _{{\mathfrak {H}}}$ is the inner product on ${\mathfrak {H}}$. Remark that $L^2(M, d\mu ; {\mathfrak {H}})$ can be naturally identified with ${\mathfrak {H}} \otimes L^2(M, d\mu )$:

$$\begin{aligned} {\mathfrak {H}}\otimes L^2(M, d\mu )=\int _M^{\oplus } {\mathfrak {H}} d\mu . \end{aligned}$$

(I.1)

$L^{\infty }(M, d\mu ; {\mathscr {B}}(\mathfrak {H}))$ denotes the space of measurable functions from M to ${\mathscr {B}}(\mathfrak {H})$ with

$$\begin{aligned} \Vert A\Vert _{\infty }=\mathrm {ess.sup} \Vert A(m)\Vert _{{\mathscr {B}}(\mathfrak {H})}<\infty . \end{aligned}$$

(I.2)

A bounded operator A on $\int ^{\oplus }_M{\mathfrak {H}} d\mu$ is said to be decomposed by the direct integral decomposition, if and only if there is a function $A(\cdot )\in L^{\infty }(M, d\mu ; {\mathscr {B}}({\mathfrak {H}}))$ such that

$$\begin{aligned} (A\Psi )(m)=A(m)\Psi (m) \end{aligned}$$

(I.3)

for all $\Psi \in \int ^{\oplus }_M{\mathfrak {H}} d\mu$. In this case, we call Adecomposable and write

$$\begin{aligned} A=\int ^{\oplus }_MA(m)d\mu . \end{aligned}$$

(I.4)

Example 18

Let $B\in {\mathscr {B}}({\mathfrak {H}})$. Under the identification (I.1), we have

$$\begin{aligned} B\otimes 1=\int ^{\oplus }_MBd\mu . \end{aligned}$$

(I.5)

$\diamondsuit$

Lemma I.1

If $A(\cdot ) \in L^{\infty }(M, d\mu ; {\mathscr {B}}({\mathfrak {H}}))$, then there is a unique decomposable operator $A \in {\mathscr {B}}(\int ^{\oplus }_M{\mathfrak {H}} d\mu )$ such that (I.3) holds.

Proof

See [44, Theorem XIII. 83]. $\square$

Let $\mathfrak {P}$ be a self-dual cone in ${\mathfrak {H}}$. We set

$$\begin{aligned} \int ^{\oplus }_M \mathfrak {P}d\mu =\bigg \{ \Psi \in \int ^{\oplus }_M {\mathfrak {H}} d\mu \, \Big |\, \Psi (m) \ge 0\hbox { w.r.t. }\mathfrak {P}\hbox { for }\mu \hbox {-a.e.} \bigg \}. \end{aligned}$$

(I.6)

It is not hard to check that $\int ^{\oplus }_M\mathfrak {P}d\mu$ is a self-dual cone in $\int ^{\oplus }_M{\mathfrak {H}} d\mu$. We call $\int ^{\oplus }_M\mathfrak {P}d\mu$ a direct integral of$\mathfrak {P}$.

The following lemma is useful:

Lemma I.2

Let $\Psi \in \int ^{\oplus }_M{\mathfrak {H}}d\mu$. The following statements are equivalent:

• (i) $\Psi \ge 0$ w.r.t. $\int ^{\oplus }_M \mathfrak {P}d\mu$.

• (ii) $\langle \xi \otimes f|\Psi \rangle \ge 0$ for all $\xi \in \mathfrak {P},\ f\in L^2(M, d\mu )_+$.

Proof

To show (i) $\Rightarrow$ (ii) is easy. Let us show the inverse. We set $G_{\xi }(m):=\langle \xi |\Psi (m)\rangle$. By (ii), we have $\langle \xi \otimes f|\Psi \rangle =\int _MG_{\xi }(m) f(m) d\mu \ge 0$ for each $f\in L^2(M, d\mu )_+.$ Since f is arbitrary, we conclude that $G_{\xi }(m) \ge 0$. Because $\xi$ is arbitrary, we finally arrive at $\Psi (m) \ge 0$ w.r.t. $\mathfrak {P}$. $\square$

Proposition I.3

Let $A=\int ^{\oplus }_MA(m) d\mu$ be a decomposable operator on $\int ^{\oplus }_M{\mathfrak {H}} d\mu$. If $A(m) \unrhd 0$ w.r.t. $\mathfrak {P}$ for $\mu$-a.e., then $A\unrhd 0$ w.r.t. $\int ^{\oplus }_M\mathfrak {P}d\mu$.

Proof

For each $\Psi \in \int ^{\oplus } _M\mathfrak {P}d\mu$, we have $(A\Psi )(m)=A(m) \Psi (m)\ge 0$ w.r.t. $\mathfrak {P}$ for $\mu$-a.e.. Hence, $A\Psi \ge 0$ w.r.t. $\int ^{\oplus }_M\mathfrak {P}d\mu$. $\square$

Corollary I.4

Let $B\in {\mathscr {B}}({\mathfrak {H}})$. Under the identification (I.1), if $B\unrhd 0$ w.r.t. $\mathfrak {P}$, then $B\otimes 1 \unrhd 0$ w.r.t. $\int ^{\oplus }_M\mathfrak {P}d\mu$.

Proof

Use (I.5). $\square$

Proposition I.5

Let C be a bounded linear operator on $L^2(M, d\mu )$. Under the identification (I.1), if $C\unrhd 0$ w.r.t. $L^2(M, d\mu )_+$, then $1\otimes C \unrhd 0$ w.r.t. $\int ^{\oplus }_M\mathfrak {P}d\mu$.

Proof

Let $\Psi \in \int ^{\oplus }_M\mathfrak {P}d\mu$. For each $\xi \in \mathfrak {P}$ and $f\in L^2(M, d\mu )_+$, we have

$$\begin{aligned} \langle 1\otimes C\Psi |\xi \otimes f\rangle =\langle \Psi |\xi \otimes C^*f\rangle . \end{aligned}$$

(I.7)

Because $C^*\unrhd 0$ w.r.t. $L^2(M, d\mu )_+$, $C^*f\ge 0$ w.r.t. $L^2(M, d\mu )_+$, which implies that $\xi \otimes C^*f \ge 0$. Thus, the right hand side of (I.7) is positive. By Lemma I.2, we conclude that $1\otimes C \Psi \ge 0$ w.r.t. $\int ^{\oplus }_M \mathfrak {P}d\mu$. $\square$

The following lemma is useful in the present study:

Lemma I.6

Let ${\mathcal {Z}}=\mathbb {R}^n$. Let $\mathfrak {P}={\mathscr {L}}^2({\mathfrak {X}})_+$ be a natural self-dual cone in ${\mathfrak {X}}\otimes {\mathfrak {X}}={\mathscr {L}}^2({\mathfrak {X}})$, see Section A.1 for notations.

• (i) Let $B: {\mathcal {Z}}\rightarrow {\mathscr {B}}({\mathfrak {X}});\ {\varvec{q}}\mapsto B({\varvec{q}})$ be continuous with $\sup _{\varvec{q}} \Vert B({\varvec{q}})\Vert <\infty$. We have

  $$\begin{aligned} \int _{{\mathcal {Z}}}^{\oplus } {\mathcal {L}}(B({\varvec{q}})^*){\mathcal {R}}(B({\varvec{q}}))d{\varvec{q}} \unrhd 0 \ \ \ \text{ w.r.t. } \displaystyle \int _{{\mathcal {Z}}} ^{\oplus }\mathfrak {P}d{\varvec{q}}. \end{aligned}$$

  (I.8)

• (ii) Let $C\in {\mathscr {B}}({\mathfrak {X}})$. We have ${\mathcal {L}}(C^*) {\mathcal {R}}(C) \otimes 1 \unrhd 0$ w.r.t. $\int ^{\oplus }_{{\mathcal {Z}}} \mathfrak {P}d{\varvec{q}}$.

Proof

Set $A({\varvec{q}}):= {\mathcal {L}}(B({\varvec{q}})^*){\mathcal {R}}(B({\varvec{q}})) \in {\mathscr {B}}({\mathscr {L}}^2({\mathfrak {X}}))$. We have $\mathrm {ess.sup}_{\varvec{q}} \Vert A({\varvec{q}})\Vert \le ( \sup _{\varvec{q}} \Vert B({\varvec{q}})\Vert )^2<\infty .$ Thus, we can define a bounded linear operator $\int _{{\mathcal {Z}}}^{\oplus } {\mathcal {L}}(B({\varvec{q}})^*){\mathcal {R}}(B({\varvec{q}}))d{\varvec{q}}$ on $\int ^{\oplus }_{{\mathcal {Z}}} {\mathscr {L}}^2({\mathfrak {X}}) d{\varvec{q}}$ by Lemma I.1. Since $A({\varvec{q}}) \unrhd 0$ w.r.t. $\mathfrak {P}$ by Proposition A.3, we can apply Proposition I.3 and obtain (i).

(ii) immediately follows from Propositions A.3 and Corollary I.4. $\square$

Lemma I.7

Let $\Psi \in \int _M {\mathfrak {H}} d\mu$. The following statements are equivalent:

• (i) $\Psi >0$ w.r.t. $\int _M^{\oplus } \mathfrak {P}d\mu$.

• (ii) $\langle \xi \otimes f|\Psi \rangle >0$ for all $\xi \in \mathfrak {P}\backslash \{0\}$ and $f\in L^2(M, d\mu )_+ \backslash \{0\}$.

Proof

To prove that (i) $\Rightarrow$ (ii) is easy. Let us prove the converse. We set $G_{\xi }(m) =\langle \xi |\Psi (m)\rangle _{{\mathfrak {H}}}$. We have $\langle \xi \otimes f|\Psi \rangle =\int _MG_{\xi }(m) f(m) d\mu >0$ for every $f\in L^2(M, d\mu )_+\backslash \{0\}.$ Thus, we have $G_{\xi }(m)>0$ for $\mu$-a.e.. Because $\xi$ is arbitrary, we conclude that $\Psi (m)>0$ w.r.t. $\mathfrak {P}$ for $\mu$-a.e.. $\square$

Corollary I.8

Let $\xi \in \mathfrak {P}$ and let $f\in L^2(M, d\mu )_+$. If $\xi >0$ w.r.t. $\mathfrak {P}$ and $f>0$ w.r.t. $L^2(M, d\mu )_+$, then it holds that $\xi \otimes f>0$ w.r.t. $\int _M^{\oplus } \mathfrak {P}d\mu$.