Long-Range Charge Order in the Extended Holstein–Hubbard Model

Abstract

This study investigated the extended Holstein–Hubbard model at half-filling as a model for describing the interplay of electron–electron and electron–phonon couplings. When the electron–phonon and nearest-neighbor electron–electron interactions are strong, we prove the existence of long-range charge order in three or more dimensions at a sufficiently low temperature. As a result, we rigorously justify the phase competition between the antiferromagnetism and charge orders.

Appendices

Appendix 1: Proof of (2.5)

We will show that $\langle q_x\rangle _{\beta , \Lambda }=0$. The hole-particle transformation is a unitary operator u such that

$$\begin{aligned} u c_{x\uparrow } u^{-1} = c_{x\uparrow },\ \ \ u c_{x\downarrow } u^{-1} =(-1)^{\Vert x\Vert }c_{x\downarrow }^*. \end{aligned}$$

(4.1)

We set $\mathbb {H}=u H_{\Lambda } u^{-1}$. Since $uq_xu^{-1}=s_x:=n_{x\uparrow }-n_{x\downarrow }$, we obtain $\mathbb {H}=\mathbb {H}_0+\mathbb {W}$, where

$$\begin{aligned} \mathbb {H}_0&=T+K,\end{aligned}$$

(4.2)

$$\begin{aligned} \mathbb {W}&= U\sum _{x\in \Lambda } s_x^2+V\sum _{\langle x, y\rangle }s_xs_y+g\sum _{x\in \Lambda } s_x(b_x+b_x^*). \end{aligned}$$

(4.3)

Here, T and K are defined by (3.1) and (3.4), respectively. Thus, we have

$$\begin{aligned} \langle q_x\rangle =\langle \!\langle s_x\rangle \!\rangle , \end{aligned}$$

(4.4)

where $\langle \!\langle \cdot \rangle \!\rangle$ is the thermal expectation associated with $\mathbb {H}$.

Let D be a unitary operator such that

$$\begin{aligned} Dc_{x\uparrow } D^{-1} =c_{x\downarrow },\ \ D c_{x\downarrow } D^{-1}=c_{x\uparrow }, \ \ Db_x D^{-1}=-b_x. \end{aligned}$$

(4.5)

Since $D\mathbb {H} D^{-1}=\mathbb {H}$ and $Ds_x D^{-1} =-s_x$, we have $\langle q_x\rangle =\langle \!\langle s_x\rangle \!\rangle =0$. This concludes the proof of (2.5). $\square$

Appendix 2: The Dyson–Lieb–Simon Inequality

Let $\mathfrak {X}_L$ and $\mathfrak {X}_R$ be complex Hilbert spaces and let $\vartheta$ be an antiunitary transformation from $\mathfrak {X}_L$ onto $\mathfrak {X}_R$. Let $A, B, C_j, D_j, j=1,\dots , n$ be linear operators in $\mathfrak {X}_L$. Suppose that A and B are self-adjoint and bounded from below and that $C_j$ and $D_j$ are bounded. We will study the following Hamiltonian:

$$\begin{aligned} H(A, B, \mathbf {C}, \mathbf {D})&=H_0-V,\end{aligned}$$

(4.6)

$$\begin{aligned} H_0&=A\otimes \mathbbm {1}+\mathbbm {1}\otimes \vartheta B \vartheta ^{-1},\end{aligned}$$

(4.7)

$$\begin{aligned} V&= \sum _{j=1}^n \lambda _j\left( C_j\otimes \vartheta D_j\vartheta ^{-1}+C_j^*\otimes \vartheta D_j^* \vartheta ^{-1}\right) . \end{aligned}$$

(4.8)

$H(A, B, \mathbf {C}, \mathbf {D})$ is a self-adjoint operator bounded from below acting in $\mathfrak {X}_L\otimes \mathfrak {X}_R$.

Theorem 3.23

Assume that $e^{-\beta A}$ and $e^{-\beta B}$ are trace class operators for all $\beta >0$ and that $\lambda _j\ge 0$ for all $j\in \{1, \dots , n\}$. Let $Z_{\beta }(A, B, \mathbf {C}, \mathbf {D}) =\mathrm {Tr} \big [ e^{-\beta H(A, B, \mathbf {C}, \mathbf {D})} \big ],\ \beta >0$. We then have

$$\begin{aligned} Z_{\beta }(A, B, \mathbf {C}, \mathbf {D})^2 \le Z_{\beta }(A, A, \mathbf {C}, \mathbf {C})Z_{\beta }(B, B, \mathbf {D}, \mathbf {D}). \end{aligned}$$

(4.9)

Remark 3.24

• (i) In [4], all matrix elements of $A, B, C_j, D_j$ are assumed to be real. However, as noted in [22, 25], this assumption is unnecessary. This point is essential for the present paper.

• (ii) Suppose that $\dim \mathfrak {X}_L<\infty$. Therefore, we set $\mathfrak {X}_L=\mathfrak {X}_R=\mathbb {C}^n$. Let $\vartheta$ be the standard conjugation: $\vartheta \psi =\{\overline{\psi _j}\}_{j=1}^n$ for each $\psi \in \mathfrak {X}_L$. Hence, $\vartheta B\vartheta ^{-1}$ represents the complex conjugation of the matrix elements of B. Now assume the following: (a) $C_j=D_j$ for all j; (b) $C_j$ is self-adjoint for all j ($C_j^*=C_j$); (c) $C_j$ is real for all j ($\vartheta C_j \vartheta ^{-1}=C_j$). In this case, we obtain a finite temperature version of [22, Lemma 14]. $\diamondsuit$

Proof

While this theorem is proven in [25], we present the proof here for reader’ convenience. It suffices to show the assertion when $\dim \mathfrak {X}_L<\infty$.

The following property is fundamental:

$$\begin{aligned} \mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R}\Big [ A\otimes \vartheta B\vartheta ^{-1} \Big ]= \mathrm {Tr}_{\mathfrak {X}_L}[A] \big ( \mathrm {Tr}_{\mathfrak {X}_L}[B] \big )^*. \end{aligned}$$

(4.10)

Especially, we have $\mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R}\Big [ A\otimes \vartheta A\vartheta ^{-1} \Big ]= \big |\mathrm {Tr}_{\mathfrak {X}_L}[A]\big |^2\ge 0$. The reason for this is as follows. Since $\mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R}\Big [ A\otimes \vartheta B\vartheta ^{-1} \Big ]= \mathrm {Tr}_{\mathfrak {X}_L}[A] \mathrm {Tr}_{\mathfrak {X}_R}[\vartheta B \vartheta ^{-1}]$, it suffices to show that $\mathrm {Tr}_{\mathfrak {X}_R}[\vartheta B \vartheta ^{-1}] =(\mathrm {Tr}_{\mathfrak {X}_L}[ B ])^*$. Let $\{e_n\}_{n=1}^{\infty }$ be a complete orthonormal system (CONS) of $\mathfrak {X}_R$. Remarking that $\{\vartheta ^{-1} e_n\}_{n=1}^{\infty }$ is a CONS of $\mathfrak {X}_L$ and since $\langle \vartheta ^{-1} \phi |\vartheta ^{-1} \psi \rangle =\langle \psi |\phi \rangle$, we see that

$$\begin{aligned} \mathrm {Tr}_{\mathfrak {X}_R}[\vartheta B \vartheta ^{-1}]&=\sum _{n=1}^N \langle e_n|\vartheta B\vartheta ^{-1}e_n\rangle =\sum _{n=1}^N \langle \vartheta ^{-1}\vartheta B\vartheta ^{-1}e_n|\vartheta ^{-1}e_n\rangle \nonumber \\&=\sum _{n=1}^N \langle B\vartheta ^{-1} e_n|\vartheta ^{-1} e_n\rangle =(\mathrm {Tr}_{\mathfrak {X}_L}[B])^*. \end{aligned}$$

(4.11)

As a first step, we will prove the assertion by assuming that $C_j$ and $D_j$ are self-adjoint. For simplicity, assume that $\lambda _j=1/2$. By the Duhamel formula,

$$\begin{aligned} \mathrm {e}^{-\beta H(A, B; \mathbf {C}, \mathbf {D})}&=\sum _{N\ge 0}\mathscr {D}_{N, \beta }(A, B; \mathbf {C}, \mathbf {D}),\end{aligned}$$

(4.12)

$$\begin{aligned} \mathscr {D}_{N, \beta }(A, B; \mathbf {C}, \mathbf {D})&=\int _{S_N(\beta )}\ e^{-t_1 H_0}V e^{-t_2 H_0}\ldots e^{-t_N H_0}V e^{-(\beta -\sum _{j=1}^N t_j) H_0}, \end{aligned}$$

(4.13)

where $\int _{S_N(\beta )}=\int _0^{\beta }\mathrm {d}t_1\int _0^{\beta -t_1}\mathrm {d}t_2\ldots \int _0^{\beta -\sum _{j=1}^{N-1}t_j} \mathrm {d}t_N$. Observe that

$$\begin{aligned}&\mathscr {D}_{N, \beta }(A, B; \mathbf {C}, \mathbf {D}) \nonumber \\ =&\sum _{k_1, \dots , k_N\ge 1} \int _{S_N(\beta )}\ \Big [\mathscr {L}_{A; \mathbf {C}}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big ) \Big ] \otimes \vartheta \Big [\mathscr {L}_{B; \mathbf {D}}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big ) \Big ] \vartheta ^{-1}, \end{aligned}$$

(4.14)

where $\mathbf {k}_{(N)}=(k_1,\dots , k_N)\in \mathbb {N}^N,$ $\mathbf {t}_{(N)}=(t_1, \dots , t_N)\in \mathbb {R}^N_+$ and

$$\begin{aligned} \mathscr {L}_{X; \mathbf {Y}}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big )=e^{-t_1 X}Y_{k_1}e^{-t_2 X}\ldots e^{-t_N X}Y_{k_N} e^{-(\beta -\sum _{j=1}^N t_j)X} \end{aligned}$$

(4.15)

with $\mathbf {Y}=\{Y_j\}_j$. By this fact and (4.11), we observe that

$$\begin{aligned}&\mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R}\Big [ \mathscr {D}_{N, \beta }(A, B; \mathbf {C}, \mathbf {D}) \Big ]\nonumber \\ =&\sum _{k_1, \dots , k_N\ge 1}\int _{S_N(\beta )} \Big \{\mathrm {Tr}_{\mathfrak {X}_L}\Big [\mathscr {L}_{A; \mathbf {C}}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)} \big ) \Big ] \Big \} \times \Big \{\mathrm {Tr}_{\mathfrak {X}_L}\Big [\mathscr {L}_{B; \mathbf {D}}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big ) \Big ] \Big \}^*. \end{aligned}$$

(4.16)

Let us introduce an inner product by

$$\begin{aligned} \langle F| G\rangle _{N, \beta }=\sum _{k_1, \dots , k_N\ge 1}\int _{S_N(\beta )}\, F\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big )G\big ( \mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big )^*. \end{aligned}$$

(4.17)

In terms of this inner product, we have

$$\begin{aligned}&\mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R}\Big [ \mathscr {D}_{N, \beta }(A, B; \mathbf {C}, \mathbf {D}) \Big ]=\Big \langle F_{A;\mathbf {C}}^{(N)}\Big | F_{B;\mathbf {D}}^{(N)}\Big \rangle _{N, \beta }, \end{aligned}$$

(4.18)

where

$$\begin{aligned} F_{X;\mathbf {Y}}^{(N)}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big )=\mathrm {Tr}_{\mathfrak {X}_L}\Big [ \mathscr {L}_{X; \mathbf {Y}} \big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big )\Big ]. \end{aligned}$$

(4.19)

By the Schwartz inequality, we have

$$\begin{aligned} \bigg | \mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R}\Big [ e^{-\beta H(A, B, \mathbf {C}, \mathbf {D})} \Big ] \bigg |^2&= \Bigg | \sum _{N\ge 0} \Big \langle F_{A;\mathbf {C}}^{(N)}\Big | F_{B;\mathbf {D}}^{(N)} \Big \rangle _{N, \beta }\Bigg |^2 \nonumber \\&\le \Bigg ( \sum _{N\ge 0}\big \Vert F_{A;\mathbf {C}}^{(N)}\big \Vert ^2_{N, \beta } \Bigg ) \Bigg ( \sum _{N\ge 0}\big \Vert F_{B; \mathbf {D}}^{(N)}\big \Vert ^2_{N, \beta } \Bigg ), \end{aligned}$$

(4.20)

where $\Vert W \Vert ^2_{N, \beta }:=\langle W| W \rangle _{N, \beta }$. Finally, we remark that

$$\begin{aligned} \sum _{N \ge 0}\big \Vert F_{A; \mathbf {C}}^{(N)}\big \Vert ^2_{N, \beta }&=\sum _{N\ge 0} \sum _{k_1, \dots , k_N\ge 1}\int _{S_N(\beta )} \bigg |\mathrm {Tr}_{\mathfrak {X}}\Big [ \mathscr {L}_{A; \mathbf {C}}\big (\mathbf {k}_{(N)}; \mathbf {t}_{(N)}\big ) \Big ] \bigg |^2\nonumber \\&=\sum _{N\ge 0}\mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R} \Big [ \mathscr {D}_{N, \beta }(A, A; \mathbf {C}, \mathbf {C}) \Big ] =\mathrm {Tr}_{\mathfrak {X}_L\otimes \mathfrak {X}_R} \Big [ e^{-\beta H(A, A; \mathbf {C}, \mathbf {C})} \Big ]. \end{aligned}$$

(4.21)

Combining (4.20) and (4.21), we obtain the assertion for the case where $C_j$ and $D_j$ are self-adjoint.

We note that for general $C_j$ and $D_j$, these operators can be written as

$$\begin{aligned} C_j=\mathfrak {R}C_j+i \mathfrak {I}C_j,\ \ D_j=\mathfrak {R}D_j+i \mathfrak {I}D_j, \end{aligned}$$

(4.22)

where $\mathfrak {R}C_j, \mathfrak {R}D_j, \mathfrak {I}C_j$ and $\mathfrak {I}D_j$ are self-adjoint. Since

$$\begin{aligned} C_j\otimes \vartheta D_j\vartheta ^{-1}+C_j^*\otimes \vartheta D_j^* \vartheta ^{-1} = 2\left( \mathfrak {R}C_j\otimes \vartheta \mathfrak {R}D_j\vartheta ^{-1}+\mathfrak {I}C_j\otimes \vartheta \mathfrak {I}D_j \vartheta ^{-1}\right) , \end{aligned}$$

(4.23)

we can reduce the problem to the case where $C_j$ and $D_j$ are self-adjoint. $\square$

Appendix 3: A Useful Lemma

Lemma 3.25

Let B and C be self-adjoint operators. Suppose that $e^{-C}$ is a trace class operator and suppose that B is bounded. We have

$$\begin{aligned} \ln \mathrm {Tr}\big [e^{-(B+C)}\big ] \le \langle -B\rangle +\ln \mathrm {Tr}\big [ e^{-C} \big ], \end{aligned}$$

(4.24)

where $\langle X \rangle =\mathrm {Tr}\big [ X e^{-(B+C)} \big ]\Big /\mathrm {Tr}\big [ e^{-(B+C)} \big ]$.

Proof

Let X and Y be self-adjoint. We know that $F(\lambda )=\ln \mathrm {Tr}[e^{\lambda X+(1-\lambda ) Y}]$ is convex, e.g., as in [32]. Thus, we have $F(1) \ge F'(0)+F(0)$, which implies

$$\begin{aligned} \ln \mathrm {Tr}[e^X]\ge \langle X-Y\rangle _Y+\ln \mathrm {Tr}[e^Y], \end{aligned}$$

(4.25)

where $\langle L \rangle _Y=\mathrm {Tr}[L\, e^Y]\big /\mathrm {Tr}[e^Y]$. Substituting $X=-C$ and $Y=-B-C$, we obtain the desired result. $\square$

Appendix 4: Proof of Proposition 3.10

Let

$$\begin{aligned} \mathscr {S}_n^{(0)}= \big \{ (X_1,\dots , X_n)\, \big |\, X_j\in \Lambda \times \{\uparrow , \downarrow \}, \, j=1, \dots , n\ \text{ and }\ X_i\ne X_j,\, \text{ if }\ i\ne j \big \}. \end{aligned}$$

(4.26)

Let $\mathfrak {S}_n$ be the permutation group on set $\{1, \dots , n\}$. Let $(X_1, \dots , X_n),\, (Y_1, \dots , Y_n)\in \mathscr {S}_n^{(0)}$. If there exists a $\pi \in \mathfrak {S}_n$ such that $(X_{\pi (1)}, \dots , X_{\pi (n)})=(Y_1, \dots , Y_n)$, then we write $(X_1, \dots , X_n) \sim (Y_1, \dots , Y_n)$. The binary relation “$\sim$” on $\mathscr {S}_n^{(0)}$ is an equivalence relation. We denote the quotient set $\mathscr {S}_n^{(0)} \backslash \sim$ by $\mathscr {S}_n$ and for the simplicity of notation, we still denote the equivalence class $[(X_1, \dots , X_n)]$ by $(X_1, \dots , X_n)$.

Set $\displaystyle \mathscr {S}=\bigcup _{n=0}^{2|\Lambda |} \mathscr {S}_n$, where $\mathscr {S}_{n=0}=\{\emptyset \}$. Let

$$\begin{aligned} \mathscr {S}_L&= \big \{ (X_1,\dots , X_n)\in \mathscr {S}\, \big |\, X_j\in \Lambda _L\times \{\uparrow , \downarrow \}, \, j=1, \dots , n,\ n\in \{0\} \cup \mathbb {N}\big \},\end{aligned}$$

(4.27)

$$\begin{aligned} \mathscr {S}_R&= \big \{ (X_1,\dots , X_n)\in \mathscr {S}\, \big |\, X_j\in \Lambda _L\times \{\uparrow , \downarrow \}, \, j=1, \dots , n,\ n\in \{0\}\cup \mathbb {N}\big \}. \end{aligned}$$

(4.28)

Here, if $n=0$, then we understand that $(X_1, \dots , X_n)=\emptyset$. For each $X=(x, \sigma )\in \Lambda \times \{\uparrow , \downarrow \}$, we set $c_X:=c_{x\sigma }$ and $a_X:=a_{x\sigma }$. For each $\mathbf {X}=(X_1, \dots , X_n)\in \mathscr {S}$, we define

$$\begin{aligned} e(\mathbf {X})=c_{X_1}^*\dots c_{X_n}^* \Omega _{\mathrm {f}},\ \ \ f(\mathbf {X})=a_{X_1}^*\ldots a_{X_n}^*\Omega _{\mathrm {f}}, \end{aligned}$$

(4.29)

and $e(\emptyset )=\Omega _{\mathrm {f}}$, $f(\emptyset )=\Omega _{\mathrm {f}}$. The definition (4.29) is independent of the choice of the representative up to the sign factor, and trivially, $\{e(\mathbf {X})\, |\, \mathbf {X}\in \mathscr {S}_R\}$ is a CONS of $\mathfrak {F}_R$. We note that $\{a_X\, |X\in \Lambda \times \{\uparrow , \downarrow \}\}$ satisfies the CARs:

$$\begin{aligned} \{a_X, a_Y^*\}=\delta _{XY},\ \ \ \{a_X, a_Y\}=0. \end{aligned}$$

(4.30)

Moreover, it holds that $a_X\Omega _{\mathrm {f}}=0$ for all $X\in \Lambda \times \{\uparrow , \downarrow \}$. Thus, $\{f(\mathbf {X})\, |\, \mathbf {X}\in \mathscr {S}_L\}$ is a CONS of $\mathfrak {F}_L$.

For each $X=(x, \sigma )\in \Lambda _R\times \{\uparrow , \downarrow \}$, we set $r(X):=(r(x), \sigma )\in \Lambda _L\times \{\uparrow , \downarrow \}$, where r in the right-hand side is defined by (3.42). For each $\mathbf {X}=(X_1, \dots , X_n)\in \mathscr {S}_R$, we further extend the map r as follows:

$$\begin{aligned} r(\mathbf {X}):=(r(X_1), \dots , r(X_n))\in \mathscr {S}_L. \end{aligned}$$

(4.31)

Thus, $\{f(r(\mathbf {X}))\, |\, \mathbf {X}\in \mathscr {S}_R\}$ is a CONS of $\mathfrak {F}_L$. For each $\Psi \in \mathfrak {F}_L$, we have the following expression:

$$\begin{aligned} \Psi =\sum _{\mathbf {X}\in \mathscr {S}_R} \Psi (r(\mathbf {X})) f(r(\mathbf {X})),\ \ \ \ \Psi (r(\mathbf {X}))=\langle f(r(\mathbf {X}))|\Psi \rangle . \end{aligned}$$

(4.32)

Using the expression (4.32), we define an antilinear map $\xi$ from $\mathfrak {F}_L$ onto $\mathfrak {F}_R$ by

$$\begin{aligned} \xi \Psi =\sum _{\mathbf {X}\in \mathscr {S}_R} \overline{\Psi (r(\mathbf {X}))} e(\mathbf {X}) \end{aligned}$$

(4.33)

and $\xi \Omega _{\mathrm {f}}^L=\Omega _{\mathrm {f}}^R$. $\xi ^{-1}$ is given by

$$\begin{aligned} \xi ^{-1} \Phi =\sum _{\mathbf {X}\in \mathscr {S}_R} \overline{\Phi (\mathbf {X})} f(r(\mathbf {X})) \end{aligned}$$

(4.34)

for each $\Phi =\sum _{X\in \mathscr {S}_R} \Phi (\mathbf {X}) e(\mathbf {X})\in \mathfrak {F}_R$. It is not difficult to check that $\langle \xi \Psi _1|\xi \Psi _2\rangle =\overline{\langle \Psi _1|\Psi _2\rangle }$ for all $\Psi _1, \Psi _2\in \mathfrak {F}_L$. Hence, $\xi$ is an antiunitary transformation.

Lemma 3.26

For all $X\in \Lambda _R\times \{\uparrow , \downarrow \}$, it holds that $\xi a_{r(X)}\xi ^{-1}=c_X$.

Proof

For each $\Phi =\sum _{\mathbf {Y}\in \mathscr {S}_R} \Phi (\mathbf {Y})e(\mathbf {Y})\in \mathfrak {F}_R$, we have, by (4.33) and (4.34),

$$\begin{aligned} \xi a_{r(X)}^* \xi ^{-1} \Phi&=\xi a_{r(X)}^* \sum _{\mathbf {Y}\in \mathscr {S}_R} \overline{\Phi (\mathbf {Y})}f(r(\mathbf {Y})) =\xi \sum _{\mathbf {Y}\in \mathscr {S}_R,\ X\notin \mathbf {Y}} \overline{\Phi (\mathbf {Y})}f(r(X, \mathbf {Y}))\nonumber \\&=\sum _{\mathbf {Y}\in \mathscr {S}_R,\ X\notin \mathbf {Y}} \Phi (\mathbf {Y})e(X, \mathbf {Y}) =c_X^* \Phi . \end{aligned}$$

(4.35)

Hence, $\xi a_{r(X)}^* \xi ^{-1}=c_X^*$. $\square$

Recall that $\mathfrak {H}_L=\mathfrak {F}_L\otimes L^2(\mathcal {Q}_L, d\mu _{\Lambda _L})$. We use the following identification:

$$\begin{aligned} \mathfrak {H}_L=\int _{\mathcal {Q}_L}^{\oplus } \mathfrak {F}_L d\mu _{\Lambda _L}({\varvec{\phi }}). \end{aligned}$$

(4.36)

Thus, each vector $\Psi \in \mathfrak {H}_L$ is a $\mathfrak {F}_L$-valued measurable map on $\mathcal {Q}_L$, i.e., ${\varvec{\phi }}\mapsto \Psi ({\varvec{\phi }})$. Now, we define an antiunitary transformation $\vartheta$ from $\mathfrak {H}_L$ onto $\mathfrak {H}_R$ by

$$\begin{aligned} (\vartheta \Psi )({\varvec{\phi }})= (\xi \Psi )(r^{-1}({\varvec{\phi }}))\ \ \text{ a.e. } {\varvec{\phi }}\in \mathcal {Q}_R\text{, } \Psi \in \mathfrak {H}_L, \end{aligned}$$

(4.37)

where, for each ${\varvec{\phi }}=\{\phi _x\}_{x\in \Lambda _R} \in \mathcal {Q}_R$, we define $r^{-1}({\varvec{\phi }})\in \mathcal {Q}_L$ by $\big (r^{-1}({\varvec{\phi }})\big )_x =\phi _{r^{-1}(x)},\, x\in \Lambda _L$.

Remark 3.27

For each measurable function $F({\varvec{\phi }})\ ({\varvec{\phi }}\in \mathcal {Q}_L)$ on $\mathcal {Q}_L$, $F(r^{-1}({\varvec{\phi }}))\ ({\varvec{\phi }}\in \mathcal {Q}_R)$ can be regarded as a function on $\mathcal {Q}_R$. Let $\Psi \in \mathfrak {H}_L$. By (4.32), we have the following expression:

$$\begin{aligned} \Psi ({\varvec{\phi }})= \sum _{\mathbf {X}\in \mathscr {S}_R} \Psi _{r(\mathbf {X})}({\varvec{\phi }}) f(r(\mathbf {X})) \ \ \text{ a.e. } {\varvec{\phi }}\in \mathcal {Q}_L, \end{aligned}$$

(4.38)

where $\Psi _{r(\mathbf {X})}({\varvec{\phi }})=\langle f(r(\mathbf {X}))|\Psi ({\varvec{\phi }})\rangle$. Using this, we have

$$\begin{aligned} (\vartheta \Psi )({\varvec{\phi }})=\sum _{\mathbf {X}\in \mathscr {S}_R} \overline{\Psi _{r(\mathbf {X})}(r^{-1}({\varvec{\phi }}))} e(\mathbf {X})\ \ \text{ a.e. } {\varvec{\phi }}\in \mathcal {Q}_R. \ \ \ \ \diamondsuit \end{aligned}$$

(4.39)

Proposition 3.28

$\vartheta$ satisfies all properties in (3.43) and (3.44).

Proof

By Lemma 3.26, it is easy to check that $\vartheta a_{r(X)}\vartheta ^{-1}=c_X$.

Note that the action of the multiplication operator $\phi _x$ is as follows: For each $\Psi \in \mathfrak {H}_L$ and $x\in \Lambda _L$,

$$\begin{aligned} (\phi _x \Psi )({\varvec{\phi }})= \phi _x \Psi ({\varvec{\phi }}) \ \ \text{ a.e. } {\varvec{\phi }}\in \mathcal {Q}_L. \end{aligned}$$

(4.40)

Thus, we have, for each $\Psi \in \mathfrak {H}_L$ and $x\in \Lambda _R$,

$$\begin{aligned} (\vartheta \phi _{r(x)}\Psi )({\varvec{\phi }})=\phi _x \xi \Psi (r^{-1}({\varvec{\phi }})) =(\phi _x\vartheta \Psi )({\varvec{\phi }})\ \ \text{ a.e. } {\varvec{\phi }}\in \mathcal {Q}_R, \end{aligned}$$

(4.41)

which implies $\vartheta \phi _{r(x)} \vartheta ^{-1}=\phi _x$.

Next, we will prove that $\vartheta \pi _{r(x)} \vartheta ^{-1}=-\pi _x$. Since $\pi _x=-i \frac{\partial }{\partial \phi _x}$, we have, for each $\Psi \in \mathfrak {H}_L$ and $x\in \Lambda _L$,

$$\begin{aligned} (\pi _x \Psi )({\varvec{\phi }})= (-i)\frac{\partial \Psi }{\partial \phi _x}({\varvec{\phi }}) \ \ \text{ a.e. } {\varvec{\phi }}\in \mathcal {Q}_L. \end{aligned}$$

(4.42)

Hence, we have, for each $\Psi \in \mathfrak {H}_L$ and $x\in \Lambda _L$,

$$\begin{aligned} (\vartheta \pi _{x} \Psi )({\varvec{\phi }})=(+i) \overline{\frac{\partial \Psi }{\partial \phi _x}} (r^{-1}({\varvec{\phi }})) =(+i) \frac{\partial (\vartheta \Psi )}{\partial \phi _{r^{-1}(x)}} ({\varvec{\phi }}) =-(\pi _{r^{-1}(x)} \vartheta \Psi )({\varvec{\phi }}) \end{aligned}$$

(4.43)

for a.e. ${\varvec{\phi }}\in \mathcal {Q}_R$. Here, we used the fact that $\xi$ is antilinear. Thus, we conclude that $\vartheta \pi _{x} \vartheta ^{-1}=-\pi _{r^{-1}(x)}$ for each $x\in \Lambda _L$, which implies $\vartheta \pi _{r(x)} \vartheta ^{-1}=-\pi _{x}$ for each $x\in \Lambda _R$. $\square$