A squeezing invariant measurement to test displacement of quantum Gaussian states

TL;DR

This paper introduces a quantum measurement method invariant under squeezing operations to test displacement parameters of Gaussian states, comparing its effectiveness with classical heterodyne-based tests.

Contribution

It develops a squeezing-invariant quantum measurement for hypothesis testing of displacement parameters in Gaussian states, providing a new approach for quantum state analysis.

Findings

Calculated the type II error probability for pure states.

Demonstrated the invariance property of the measurement under squeezing.

Compared the quantum test with classical heterodyne-based tests.

Abstract

We consider a hypothesis testing problem for displacement parameters of n independent copies of an m-mode squeezed quantum Gaussian state whose mixture parameter is known. Given n>1, we construct a quantum measurement as a test using an observable which is invariant by n-fold tensor product of any m-mode squeezing operator. For a pure state case, we calculate the type II error probability of this test. We compare this test with a Hotelling's T-squared test which is based on heterodyne measurements.

Equations

$$H_0:\hat{\rho }\in {\mathcal{S}}_0\text{mbox}versus{H}_1:\hat{\rho }\in {\mathcal{S}}_1.$$

$${\beta }_{\hat{\rho }}[T_1]\le {\beta }_{\hat{\rho }}[T_2]({}^{\forall }\hat{\rho }\in {\mathcal{S}}_1)\text{mbox}and{\beta }_{\hat{\rho }}[T_1]<{\beta }_{\hat{\rho }}[T_2]({}^{\exists }\hat{\rho }\in {\mathcal{S}}_1)$$

$$\underset{\xi \in \Xi }{\sup }{\beta }_{{\hat{\rho }}_{\theta ,\xi }}[T_1]\le \underset{\xi \in \Xi }{\sup }{\beta }_{{\hat{\rho }}_{\theta ,\xi }}[T_2]({}^{\forall }\theta \in {\Theta }_1)$$

$$\underset{\xi \in \Xi }{\sup }{\beta }_{{\hat{\rho }}_{\theta ,\xi }}[T_1]<\underset{\xi \in \Xi }{\sup }{\beta }_{{\hat{\rho }}_{\theta ,\xi }}[T_2]({}^{\exists }\theta \in {\Theta }_1).$$

$${\beta }_{{\hat{\rho }}_{\theta ,\xi }}[T]=\mathrm{Tr}[{\hat{\rho }}_{\theta ,{\xi }_0}{\hat{\Pi }}_0]({}^{\forall }\theta \in {\Theta }_1,{}^{\forall }\xi \in \Xi ).$$

$${\mathrm{Sqz}}^m=\left\{\left(\begin{array}{cc}A& S\\ \bar{S}& \bar{A}\end{array}\right)|A\in {\mathrm{Ant}}_{\mathbb{C}}^m,S\in {\mathrm{Sym}}_{\mathbb{C}}^m\right\}.$$

$$|\theta ⟩(x)=\frac{e^{-|\theta {|}^2/2}}{{\pi }^{1/4}}{e}^{-x^2/2+\sqrt{2}\theta x-{\theta }^2/2}.$$

$${\hat{\rho }}_{\theta ,N}=\{\begin{array}{ll}|\theta ⟩⟨\theta |& \text{mbox}ifN=0,\\ \frac{1}{\pi N}{\iint }_{{\mathbb{R}}^2}{e}^{-|r+is-\theta {|}^2/N}|r+is⟩⟨r+is|drds& \text{mbox}ifN>0,\end{array}$$

$$\hat{a}|\theta ⟩=\theta |\theta ⟩.$$

$$\hat{s}_{\eta}=\sum_{i=1}^{m}\sum_{j=1}^{m}\Big{(}A_{i,j}\hat{a}_{i}^{*}\hat{a}_{j}+\frac{1}{2}S_{i,j}\hat{a}_{i}^{*}\hat{a}_{j}^{*}-\frac{1}{2}\bar{S}_{i,j}\hat{a}_{i}\hat{a}_{j}\Big{)}.$$

$$H_0:\vec{\theta }={\vec{0}}_m\text{mbox}versus{H}_1:\vec{\theta }\ne {\vec{0}}_m,$$

$${\hat{a}}_{i,j}={\hat{I}}^{\otimes (jm-m)}\otimes {\hat{a}}_i\otimes {\hat{I}}^{\otimes (mn-jm)}={\hat{I}}^{\otimes (i-1+jm-m)}\otimes \hat{a}\otimes {\hat{I}}^{\otimes (m-i+mn-jm)}.$$

$${\hat{d}}_{j,k}=\sqrt{-1}\sum _{i=1}^m({\hat{a}}_{i,j}^{\ast }{\hat{a}}_{i,j}-{\hat{a}}_{i,k}^{\ast }{\hat{a}}_{i,k}).$$

$${\mathcal{K}}_t=\{f\in {\mathcal{H}}^{\otimes mn}\mid f^{\ast }{\hat{T}}_{\mathrm{SI}}f\le t\Vert f{\Vert }^2\}.$$

$$\{\begin{array}{l}1-\alpha =(1-w)\mathrm{Tr}[{\hat{\rho }}_{{\vec{0}}_m,N}^{\otimes n}{\hat{K}}_s]+w\mathrm{Tr}[{\hat{\rho }}_{{\vec{0}}_m,N}^{\otimes n}{\hat{K}}_t],\\ s<t\text{mbox}and0<w\le 1.\end{array}$$

$$\mathcal{F}\ni M\mapsto \frac{1}{\pi }{\iint }_M|x+iy⟩⟨x+iy|dxdy,$$

$${\vec{\mu }}_{\vec{\theta }}=\left(\begin{array}{c}\mathrm{Re}(\vec{\theta })\\ \mathrm{Im}(\vec{\theta })\end{array}\right)\text{mbox}and{G}_{\eta }=\exp \left(\begin{array}{cc}\mathrm{Re}(A)+\mathrm{Re}(S)& -\mathrm{Im}(A)+\mathrm{Im}(S)\\ \mathrm{Im}(A)+\mathrm{Im}(S)& \mathrm{Re}(A)-\mathrm{Re}(S)\end{array}\right),$$

$${\vec{\mu }}_{\vec{\theta },\eta }=G_{\eta }{\vec{\mu }}_{\vec{\theta }}\text{mbox}and{\Sigma }_{\eta ,N}=\frac{2N+1}{4}{G}_{\eta }({}^t{G}_{\eta })+\frac{1}{4}{I}_{2m},$$

$$p_{\lambda }(f)=\sum _{k=0}^{\infty }\frac{e^{-\lambda /2}(\lambda /2{)}^k/k!}{B\left(k+\mu /2,\nu /2\right)}{\left(\frac{\mu f}{\mu f+\nu }\right)}^{k+\mu /2}{\left(\frac{\nu }{\mu f+\nu }\right)}^{\nu /2}\frac{1}{f},$$

$${\hat{\Pi }}_0=\frac{1}{{\pi }^{mn}}{\stackrel{2mn}{\overbrace{\int \cdots \int }}}_{F_{\mathrm{HH}}\le c}|Z⟩⟨Z|\prod _{i=1}^m\prod _{j=1}^{n}d{x}_{i,j}d{y}_{i,j},$$

$$\gamma (r)=\frac{1}{N+1-N{e}^{ir}}\text{mbox}and{\psi }_s(r)=\exp [\gamma (r)(e^{ir}-1)s^2]$$

$${\phi }_Y(r)=[\gamma (r)\gamma (-r){]}^m{\psi }_{\Vert \vec{\theta }\Vert }(r){\psi }_{\Vert \vec{\theta }\Vert }(-r).$$

$${\beta }_{{\hat{\rho }}_{\vec{\theta },\eta ,0}^{\otimes n}}[T_{\alpha }^{\mathrm{SI}}]=(1-\alpha )\frac{e^{-n\Vert \vec{\theta }{\Vert }^2}}{B\left(\frac{n-1}{2},\frac{1}{2}\right)}{\int }_0^{\pi }{e}^{n\Vert \vec{\theta }{\Vert }^2\cos \phi }(\sin \phi {)}^{n-2}d\phi ,$$

$$[X,L]=O_{\mu ,\nu },$$

$${\mathrm{tr}}_{\mu }[XC]=\sum _{i=1}^{\mu }\sum _{j=1}^{\mu }{X}_{i,j}{C}_{j,i}={\mathrm{tr}}_{\mu }[CX].$$

$${\vec{a}}_j=\left(\begin{array}{c}{\hat{a}}_{1,j}\\ ⋮\\ {\hat{a}}_{m,j}\end{array}\right)\text{mbox}and\text{underaccent}\stackrel{~}{}{\vec{a}}_j=\left(\begin{array}{c}{\vec{a}}_j\\ {}^t{\vec{a}}_j^{\ast }\end{array}\right),$$

$$K_{\mu }=\left(\begin{array}{cc}{I}_{\mu }& O_{\mu }\\ O_{\mu }& -I_{\mu }\end{array}\right)\in {\mathrm{Mat}}_{\mathbb{R}}^{2m}.$$

$$\underaccent{\tilde}{\hat u}_{\eta}=\frac{1}{2}\mathrm{tr}_{n}[\underaccent{\tilde}{a}_{m,n}^{*}K_{m}\eta\underaccent{\tilde}{a}_{m,n}]-n\underaccent{\tilde}{c}_{\eta}\hat{I}^{\otimes mn}.$$

$${\hat{u}}_A={\mathrm{tr}}_n[{\tilde{a}}_{m,n}^{\ast }A{\tilde{a}}_{m,n}].$$

$${\hat{v}}_A={\mathrm{tr}}_m[{}^t{\tilde{a}}_{m,n}^{\ast }A({}^t{\tilde{a}}_{m,n})].$$

$$J_{j,k}=\left(\begin{array}{ccccc}{O}_{j-1}& & & & \\ & 0& & -1& \\ & & O_{k-j-1}& & \\ & 1& & 0& \\ & & & & O_{n-k}\end{array}\right).$$

Full text

1 Introduction

In quantum hypothesis testing [1], it is important to test whether the displacement parameter of several independent copies of a quantum Gaussian state [2] is zero or not. Kumagai and Hayashi [3] have studied this problem in a situation that the mixture parameter is unknown and that the state is not squeezed. They constructed a Positive Operator Valued Measurement (POVM) as a test which is invariant by some unitary actions. Using the invariance, they proved that their test is optimal in a minimax criterion. On the other hand, if the states are squeezed by an unknown squeezing action, then, even if the mixture parameter is known, we have not obtained an optimal test. For example, we will need such a test when we should check a displacement channel by setting a squeezed probe which is generated by a squeezing device which is not completely controlled.

One of reasonable measurements for this problem might be the Hotelling’s $T$-squared test [4] using data from heterodyne measurements [5]. We call this test the Heterodyne-Hotelling (HH) test. Let $\hat{\rho}_{\vec{\theta},N}$ be an $m$-mode non-squeezed quantum Gaussian state parameterized by a displacement vector $\vec{\theta}\in\mathbb{C}^{m}$ and by a mixture value $N\geq 0$. Let $\hat{S}_{\eta}$ be an $m$-mode squeezing operator parameterized by a matrix $\eta$. The $m$-mode squeezed quantum Gaussian state is defined by $\hat{\rho}_{\vec{\theta},\eta,N}=\hat{S}_{\eta}\hat{\rho}_{\vec{\theta},N}\hat{S}_{\eta}^{*}$. If there are $n$ independent copies of $\hat{\rho}_{\vec{\theta},\eta,N}$, and if $mn$ independent heterodyne measurements are applied to $\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}$, then we obtain $n$ independent random vectors obeying a common $2m$-dimensional normal distribution. If $n\geq 2m+1$, then the sample covariance matrix $\bar{\Sigma}_{n}$ is invertible by probability one, and the Hotelling’s $T$-squared statistic $T^{2}$ can be defined using $\bar{\Sigma}_{n}^{-1}$. Let $\mu=2m$, and let $\nu=n-\mu$. If $\vec{\theta}$ is the zero vector $\vec{0}_{m}\in\mathbb{C}^{m}$, then $F_{\mathrm{HH}}=(n-1)^{-1}(\nu/\mu)T^{2}$ obeys $F_{\mu,\nu}$, the central $F$ distribution with $\mu$ and $\nu$ degrees of freedom. (See [6].) Choose a constant $\alpha$ as a level (of significance). (See [7].) Define a critical point $c$ as a solution to the equation $\Pr\{F_{\mathrm{HH}}>c\mid\vec{\theta}=\vec{0}_{m}\}=\alpha$. The HH test $T^{\mathrm{HH}}_{\alpha}$ of level $\alpha$ is a decision rule by which $\vec{\theta}=\vec{0}_{m}$ is accepted if $F_{\mathrm{HH}}\leq c$ is observed. Even if $N$ is unknown, $T^{\mathrm{HH}}_{\alpha}$ can be defined. However, if $n\leq 2m$, then $T^{\mathrm{HH}}_{\alpha}$ can not be defined for the sake of non-invertibility of $\bar{\Sigma}_{n}$. Moreover, even if $N$ is zero, $T^{\mathrm{HH}}_{0}$ is a trivial test in a sense that $c$ is infinity. Furthermore, in a minimax criterion, $T^{\mathrm{HH}}_{\alpha}$ is not optimal.

In this paper, assuming $N$ is known and $\eta$ is unknown, we propose a new test. Since the new test is invariant by the action of $\hat{S}_{\eta}^{\otimes n}$, it is said to be Squeezing Invariant (SI). The SI test of level $\alpha$ is denoted by $T^{\mathrm{SI}}_{\alpha}$. If $N$ is unknown, then $T^{\mathrm{SI}}_{\alpha}$ is not defined. However, for the following three reasons, $T^{\mathrm{SI}}_{\alpha}$ is superior to $T^{\mathrm{HH}}_{\alpha}$. First, $n\geq 2$ is enough to define $T^{\mathrm{SI}}_{\alpha}$. Second, if $N=0$, then $T^{\mathrm{SI}}_{0}$ is not a trivial test. Third, if $N=0$, then, in a minimax criterion, $T^{\mathrm{SI}}_{\alpha}$ dominates $T^{\mathrm{HH}}_{\alpha}$.

In Sec. 2, we will setup the problem by defining words and symbols. In Sec. 3, we will construct $T^{\mathrm{HH}}_{\alpha}$ and $T^{\mathrm{SI}}_{\alpha}$. In Sec. 4, we will give six theorems and a numerical comparison. In Sec. 5, we will give proofs of the theorems.

2 Setups

We define words and symbols.

2.1 What is quantum hypothesis testing?

Let $\mathcal{H}$ be a Hilbert space. For $f\in\mathcal{H}$, let $f^{*}$ be the dual vector, and let $\|f\|=\sqrt{f^{*}f}$ be the norm. Let $\mathcal{L}(\mathcal{H})$ be the set of linear operators on $\mathcal{H}$. Let $\hat{I}_{\mathcal{H}}\in\mathcal{L}(\mathcal{H})$ be the identity. Let $X^{*}$ be the adjoint of $X\in\mathcal{L}(\mathcal{H})$. Let $\mathcal{U}(\mathcal{H})\subset\mathcal{L}(\mathcal{H})$ be the set of unitary operators. If $X\in\mathcal{L}(\mathcal{H})$ is positive, then we write $X\geq 0$. Let $\mathrm{Tr}[X]$ be the trace of $X\in\mathcal{L}(\mathcal{H})$. The set of density operators is given by $\mathcal{S}(\mathcal{H})=\{\hat{\rho}\in\mathcal{L}(\mathcal{H})\mid\hat{\rho}\geq 0,\ \mathrm{Tr}[\hat{\rho}]=1\}$.

Consider a quantum system described by $\mathcal{H}$ whose state $\hat{\rho}\in\mathcal{S}(\mathcal{H})$ is unknown. Let $\mathcal{S}_{0}$ and $\mathcal{S}_{1}$ be subsets of $\mathcal{S}(\mathcal{H})$, where the intersection $\mathcal{S}_{0}\cap\mathcal{S}_{1}$ is empty. Assume that either $\hat{\rho}\in\mathcal{S}_{0}$ or $\hat{\rho}\in\mathcal{S}_{1}$ is true. When we need to accept the null hypothesis $H_{0}:\hat{\rho}\in\mathcal{S}_{0}$ or the alternative hypothesis $H_{1}:\hat{\rho}\in\mathcal{S}_{1}$, it is said that we test

[TABLE]

Let $\hat{\Pi}_{0}$ and $\hat{\Pi}_{1}$ be positive operators satisfying $\hat{\Pi}_{0}+\hat{\Pi}_{1}=\hat{I}_{\mathcal{H}}$. A test for (1) is a two-valued POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$ by which $H_{k}$ is accepted if $\hat{\Pi}_{k}$ is observed.

Let $T$ be a test with POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$. There are two types of mistakes caused by $T$. The type I error is acceptance of $H_{1}$ while $H_{0}$ is true. The type II error is acceptance of $H_{0}$ while $H_{1}$ is true. Hence, the type I error probability is defined by $\alpha_{\hat{\rho}}[T]=\mathrm{Tr}[\hat{\rho}\hat{\Pi}_{1}]$ as a function of $\hat{\rho}\in\mathcal{S}_{0}$, and the type II error probability is defined by $\beta_{\hat{\rho}}[T]=\mathrm{Tr}[\hat{\rho}\hat{\Pi}_{0}]$ as a function of $\hat{\rho}\in\mathcal{S}_{1}$. A level (of significance) is an upper bound for $\alpha_{\hat{\rho}}[T]$. If $\sup_{\hat{\rho}\in\mathcal{S}_{0}}\alpha_{\hat{\rho}}[T]\leq\alpha$ holds, then $T$ is called a test of level $\alpha$. If $\alpha$ is small, and if $H_{1}$ is accepted, then one may be confident that $H_{1}$ is really true because the risk of type I error is negligible. This method was proposed by [1] as a generalization of the classical statistical hypothesis testing theory, which is described in [7].

For any $\alpha$ with $0\leq\alpha\leq 1$, there exists a test of level $\alpha$. Let $T^{\mathrm{triv}}_{\alpha}$ be a test defined by $\hat{\Pi}_{0}=(1-\alpha)\hat{I}_{\mathcal{H}}$. For any $\hat{\rho}\in\mathcal{S}_{0}$, it holds that $\alpha_{\hat{\rho}}[T^{\mathrm{triv}}_{\alpha}]=\alpha$. Hence, $T^{\mathrm{triv}}_{\alpha}$ is a test of level $\alpha$, and is called a trivial test of level $\alpha$.

If $T_{1}$ and $T_{2}$ are tests of a common level $\alpha$, then they are compared by the type II error probabilities. If

[TABLE]

hold, then we conclude that $T_{1}$ dominates $T_{2}$. If a test $T$ dominates $T^{\mathrm{triv}}_{\alpha}$, then there exists $\hat{\rho}\in\mathcal{S}_{1}$ such that $\beta_{\hat{\rho}}[T]<1-\alpha$.

In many cases, however, the condition (2) is so strict that we can not complete the comparison. Hence, we use a minimax criterion instead of (2). Consider a case where $\hat{\rho}\in\mathcal{S}_{0}\cup\mathcal{S}_{1}$ is parameterized by $\theta\in\Theta$ and $\xi\in\Xi$ as $\hat{\rho}_{\theta,\xi}$. If there exists $\Theta_{1}\subset\Theta$ such that $\mathcal{S}_{1}=\{\hat{\rho}_{\theta,\xi}\mid\theta\in\Theta_{1},\ \xi\in\Xi\}$, then we are not interested in the true value of $\xi$. In such a case, $\theta$ is called the parameter of interest, and $\xi$ is called the parameter of nuisance. (In this sense, the displacement is the parameter of interest, and the squeezing is the parameter of nuisance.) The condition (2) is modified as

[TABLE]

If $T_{1}$ and $T_{2}$ of a common level $\alpha$ satisfy the condition (3), then we conclude that $T_{1}$ domintates $T_{2}$ in the minimax criterion. If a test $T$ of level $\alpha$ satisfies $\sup_{\xi\in\Xi}\beta_{\hat{\rho}_{\theta,\xi}}[T]=1-\alpha$ $({{}^{\forall}}\theta\in\Theta_{1})$, then, in the minimax criterion, $T$ is no better than $T^{\mathrm{triv}}_{\alpha}$.

Kumagai and Hayashi [3] studied a theory of the minimax criterion in quantum hypothesis testing, and they showed that an optimality in the minimax criterion is concerned with unitary invariance. Let $\theta\in\Theta$ be the parameter of interest, and let $\xi\in\Xi$ be the parameter of nuisance. Let $U:\Xi\to\mathcal{U}(\mathcal{H})$ be a map given as $\xi\mapsto U_{\xi}$. Assume that there exists $\xi_{0}\in\Xi$ such that $\hat{\rho}_{\theta,\xi}=U_{\xi}\hat{\rho}_{\theta,\xi_{0}}U_{\xi}^{*}$ holds for any $\theta\in\Theta$ and for any $\xi\in\Xi$. A test $T$ whose POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$ satisfies $U_{\xi}^{*}\hat{\Pi}_{0}U_{\xi}=\hat{\Pi}_{0}$ $({{}^{\forall}}\xi\in\Xi)$ is said to be invariant by $U$. If $T$ is invariant by $U$, then it holds that

[TABLE]

Kumagai and Hayashi [3] proved that a test is invariant by such $U$ if the test is optimal in the minimax criterion.

We do not prove optimality of $T^{\mathrm{SI}}_{\alpha}$. However, we prove that $T^{\mathrm{SI}}_{\alpha}$ is SI, and that, if the mixture $N$ is zero, then $T^{\mathrm{SI}}_{\alpha}$ dominates $T^{\mathrm{triv}}_{\alpha}$ in the minimax criterion. Moreover, we prove that $T^{\mathrm{HH}}_{\alpha}$ does not satisfy (4), which is a necessary condition for $T$ to be SI. Furthermore, we prove that $T^{\mathrm{HH}}_{\alpha}$ is no better than $T^{\mathrm{triv}}_{\alpha}$ in the minimax criterion. Hence, $T^{\mathrm{SI}}_{\alpha}$ dominates $T^{\mathrm{HH}}_{\alpha}$ in the minimax criterion.

2.2 Notations of sets of matrices

To parameterize multi-mode squeezing operators, we use several sets of matrices. Let $\mathbb{N}$ be the set of positive integers.. For any $m,n\in\mathbb{N}$, let $\mathrm{Mat}_{\mathbb{K}}^{m,n}$ be the set of $m$-by-$n$ matrices whose entries belong to $\mathbb{K}$, which will be $\mathbb{C}$ or $\mathbb{R}$. For $X\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, the transpose is denoted by ${{}^{t}}X\in\mathrm{Mat}_{\mathbb{C}}^{n,m}$, the entry-wise complex conjugate is denoted by $\bar{X}\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, and the adjoint ${{}^{t}}\bar{X}$ is denoted by $X^{*}\in\mathrm{Mat}_{\mathbb{C}}^{n,m}$. Let $\mathrm{Mat}_{\mathbb{K}}^{m}$ be $\mathrm{Mat}_{\mathbb{K}}^{m,m}$. The set of anti-hermitian matrices is defined by $\mathrm{Ant}_{\mathbb{K}}^{m}=\{A\in\mathrm{Mat}_{\mathbb{K}}^{m}\mid A=-A^{*}\}$. The set of symmetric matrices is defined by $\mathrm{Sym}_{\mathbb{K}}^{m}=\{S\in\mathrm{Mat}_{\mathbb{K}}^{m}\mid S={{}^{t}}S\}$. Define $\mathrm{Sqz}^{m}\subset\mathrm{Mat}_{\mathbb{C}}^{2m}$ by

[TABLE]

For $\eta\in\mathrm{Sqz}^{m}$, the upper-left submatrix $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$ is called the anti-hermitian part of $\eta$, and the upper-right submatrix $S\in\mathrm{Sym}_{\mathbb{C}}^{m}$ is called the symmetric part of $\eta$.

2.3 Multi-mode squeezed quantum Gaussian states

Let $\mathcal{H}$ be $L^{2}(\mathbb{R})$, the set of $\mathbb{C}$-valued square-integrable functions of a real coordinate variable $x\in\mathbb{R}$. The inner product of $f,g\in\mathcal{H}$ is defined by $f^{*}g=\int_{\mathbb{R}}\overline{f(x)}g(x)dx$, where $\bar{z}$ is the conjugate of $z\in\mathbb{C}$. A single-mode electromagnetic field is described by $\mathcal{H}$. (See [8] and [9].) For $\theta\in\mathbb{C}$, the coherent vector $|\theta\rangle\in\mathcal{H}$ is defined by

[TABLE]

Let $\langle\theta|$ be $|\theta\rangle^{*}$. For $\theta\in\mathbb{C}$, and for $N\geq 0$, the single-mode non-squeezed quantum Gaussian state $\hat{\rho}_{\theta,N}\in\mathcal{S}(\mathcal{H})$ is defined by

[TABLE]

where $i=\sqrt{-1}$. (See [2].) The state is pure if $N=0$.

For $m\in\mathbb{N}$, an $m$-mode system is described by $\mathcal{H}^{\otimes m}$. For $\vec{\theta}={{}^{t}}(\theta_{1},\theta_{2},...,\allowbreak\theta_{m})\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, and for $N\geq 0$, the $m$-mode non-squeezed quantum Gaussian state $\hat{\rho}_{\vec{\theta},N}\in\mathcal{S}(\mathcal{H}^{\otimes m})$ is defined by $\hat{\rho}_{\vec{\theta},N}=\hat{\rho}_{\theta_{1},N}\otimes\hat{\rho}_{\theta_{2},N}\otimes\cdots\otimes\hat{\rho}_{\theta_{m},N}$.

Define $\hat{q}\in\mathcal{L}(\mathcal{H})$ by $\hat{q}f(x)=xf(x)$, and define $\hat{p}\in\mathcal{L}(\mathcal{H})$ by $\hat{p}f(x)=-idf(x)/dx$, where $i=\sqrt{-1}$. They satisfy $\hat{q}=\hat{q}^{*}$, $\hat{p}=\hat{p}^{*}$ and $[\hat{q},\hat{p}]=\hat{q}\hat{p}-\hat{p}\hat{q}=i\hat{I}$, where $\hat{I}\in\mathcal{L}(\mathcal{H})$ is the identity. The annihilation operator $\hat{a}\in\mathcal{L}(\mathcal{H})$ is defined by $\hat{a}=(\hat{q}+i\hat{p})/\sqrt{2}$. It holds that $[\hat{a},\hat{a}^{*}]=\hat{I}$, and that

[TABLE]

For $i\in\{1,2,...,m\}$, the $i$-th annihilation operator $\hat{a}_{i}\in\mathcal{L}(\mathcal{H}^{\otimes m})$ is defined by $\hat{a}_{i}=\hat{I}^{\otimes(i-1)}\otimes\hat{a}\otimes\hat{I}^{\otimes(m-i)}$. For $\eta\in\mathrm{Sqz}^{m}$, let $A_{i,j}$ and $S_{i,j}$ be the $(i,j)$-th entries of the anti-hermitian part and of the symmetric part, respectively, and let

[TABLE]

An $m$-mode squeezing operator $\hat{S}_{\eta}\in\mathcal{U}(\mathcal{H}^{\otimes m})$ is defined by $\hat{S}_{\eta}=\exp(\hat{s}_{\eta})$. The $m$-mode squeezed quantum Gaussian state $\hat{\rho}_{\vec{\theta},\eta,N}\in\mathcal{S}(\mathcal{H}^{\otimes m})$ is defined by $\hat{\rho}_{\vec{\theta},\eta,N}=\hat{S}_{\eta}\hat{\rho}_{\vec{\theta},N}\hat{S}_{\eta}^{*}$.

2.4 Our hypothesis testing problem

Suppose that a quantum state of the form $\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}\in\mathcal{S}(\mathcal{H}^{\otimes mn})$ is given. We call $m\in\mathbb{N}$ the mode size, $n\in\mathbb{N}$ the sample size, $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$ the displacement parameter, $\eta\in\mathrm{Sqz}^{m}$ the squeezing parameter and $N\geq 0$ the mixture parameter. We assume that $\vec{\theta}$ and $\eta$ are unknown, and that $N$ is known. Our problem is to test

[TABLE]

where $\vec{0}_{m}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$ is the zero vector.

The squeezing parameter $\eta\in\mathrm{Sqz}^{m}$ is a nuisance parameter because $(\ref{eq_hypo})$ does not depend on $\eta$. Hence, the minimax criterion (3) is specified by $\Theta_{1}=\{\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}\mid\vec{\theta}\neq\vec{0}_{m}\}$ and by $\Xi=\mathrm{Sqz}^{m}$. For a subspace $\mathcal{K}\subset\mathcal{H}^{\otimes mn}$, and for $\hat{L}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$, let $\hat{L}\mathcal{K}\subset\mathcal{H}^{\otimes mn}$ be $\{\hat{L}f\mid f\in\mathcal{K}\}$. If $\hat{S}_{\eta}^{\otimes n}\mathcal{K}=\mathcal{K}$ holds for any $\eta\in\mathrm{Sqz}^{m}$, then $\mathcal{K}$ is said to be Squeezing Invariant (SI). If $\hat{L}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ satisfies $\hat{L}=(\hat{S}_{\eta}^{\otimes n})^{*}\hat{L}\hat{S}_{\eta}^{\otimes n}$ $({{}^{\forall}}\eta\in\mathrm{Sqz}^{m})$, then $\hat{L}$ is said to be SI. A test with POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$ is said to be SI if $\hat{\Pi}_{0}$ is SI.

2.5 Definition of $a_{i,j}$

For $i\in\{1,2,...,m\}$ and for $j\in\{1,2,...,n\}$, the $(i,j)$-th annihilation operator $\hat{a}_{i,j}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ is defined by

[TABLE]

3 Constructions of $T^{\mathrm{SI}}_{\alpha}$ and $T^{\mathrm{HH}}_{\alpha}$

3.1 Construction of $T^{\mathrm{SI}}_{\alpha}$

Assume that $n\geq 2$. We first construct an observable $\hat{T}_{\mathrm{SI}}\in\mathcal{L}(\mathcal{H}^{mn})$, which is positive and SI. For $j,k\in\{1,2,...,n\}$, let $\hat{v}_{j,k}=\sum_{i=1}^{m}(\hat{a}_{i,k}^{*}\hat{a}_{i,j}-\hat{a}_{i,j}^{*}\hat{a}_{i,k})$. We will show, in Theorem 1, that $\hat{v}_{j,k}$ is SI. Moreover, by Lemma 14, $\hat{v}_{j,k}$ is unitarily equivalent to

[TABLE]

For $k\in\{1,2,...,n-1\}$, let $\hat{r}_{k}=\arctan(\sqrt{k})\hat{v}_{k,k+1}$, and let $\hat{R}_{k}=\exp(\hat{r}_{k})\in\mathcal{U}(\mathcal{H}^{\otimes mn})$. Let $\hat{R}=\hat{R}_{n-1}\hat{R}_{n-2}\cdots\hat{R}_{1}\in\mathcal{U}(\mathcal{H}^{\otimes mn})$. Let $\hat{T}_{\mathrm{SI}}=\sum_{k=1}^{n-1}\hat{R}^{*}\hat{v}_{k,n}\hat{v}_{k,n}^{*}\hat{R}$. Because of $\hat{v}_{k,n}\hat{v}_{k,n}^{*}\geq 0$, we have $\hat{T}_{\mathrm{SI}}\geq 0$. Moreover, since $\hat{v}_{j,k}$ is SI, $\hat{T}_{\mathrm{SI}}$ is SI.

Next, we construct the SI test of level $\alpha\in[0,1]$. For $t\in\mathbb{R}$, define a Hilbert subspace $\mathcal{K}_{t}\subset\mathcal{H}^{\otimes mn}$ by

[TABLE]

Since $\hat{T}_{\mathrm{SI}}$ is SI, $\mathcal{K}_{t}$ is SI. Since $\hat{T}_{\mathrm{SI}}\geq 0$ holds, $s<0$ implies $\mathcal{K}_{s}=\{0\}$. Let $\hat{K}_{t}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ be the projection on $\mathcal{K}_{t}$. Since $\mathcal{K}_{t}$ is SI, $\hat{K}_{t}$ is SI. Hence, for any $\eta\in\mathrm{Sqz}^{m}$, it holds that $\mathrm{Tr}[\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}\hat{K}_{t}]=\mathrm{Tr}[\hat{\rho}_{\vec{\theta},N}^{\otimes n}\hat{K}_{t}]$.

Assume that $N$ is known. For $\alpha\in[0,1]$, let $s,t,w\in\mathbb{R}$ be solutions to

[TABLE]

Let $\hat{\Pi}_{0}=(1-w)\hat{K}_{s}+w\hat{K}_{t}$, and let $\hat{\Pi}_{1}=\hat{I}^{\otimes mn}-\hat{\Pi}_{0}$. The SI test $T^{\mathrm{SI}}_{\alpha}$ is defined by the POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$.

3.2 Construction of $T^{\mathrm{HH}}_{\alpha}$

Let $\mathcal{F}$ be the set of Borel subsets of $\mathbb{R}^{2}$. A single-mode heterodyne measurement is a POVM defined by

[TABLE]

where $i=\sqrt{-1}$. For $\eta\in\mathrm{Sqz}^{m}$, let $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$ be the anti-hermitian part, and let $S\in\mathrm{Sym}_{\mathbb{C}}^{m}$ be the symmetric part. Define $\vec{\mu}_{\vec{\theta}}\in\mathrm{Mat}_{\mathbb{R}}^{2m,1}$ and $G_{\eta}\in\mathrm{Mat}_{\mathbb{R}}^{2m}$ by

[TABLE]

respectively. Define $\vec{\mu}_{\vec{\theta},\eta}\in\mathrm{Mat}_{\mathbb{R}}^{2m,1}$ and $\Sigma_{\eta,N}\in\mathrm{Mat}_{\mathbb{R}}^{2m}$ by

[TABLE]

respectively. By Lemma 30, applying $mn$ independent single-mode heterodyne measurements to $\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}$, we obtain $n$ independent $2m$-dimensional random vectors $\vec{X}_{1},\vec{X}_{2},...,\vec{X}_{n}$ according to a common $2m$-dimensional normal distribution whose mean vector is $\vec{\mu}_{\vec{\theta},\eta}$ and whose covariance matrix is $\Sigma_{\eta,N}$; say $\mathit{N}_{2m}(\vec{\mu}_{\vec{\theta},\eta},\Sigma_{\eta,N})$. Let $\bar{X}_{n}$ be the sample mean vector $n^{-1}\sum_{j=1}^{n}\vec{X}_{j}$, and let $\bar{\Sigma}_{n}$ be the sample covariance matrix $(n-1)^{-1}\sum_{j=1}^{n}(\vec{X}_{j}-\bar{X}_{n})({{}^{t}}\vec{X}_{j}-{{}^{t}}\bar{X}_{n})$.

Assume that $n\geq 2m+1$. Then, $\bar{\Sigma}_{n}$ has the inverse $\bar{\Sigma}_{n}^{-1}$ by probability one. The Hotelling’s $T$-squared statistic is defined by $T^{2}=n({{}^{t}}\bar{X}_{n})\bar{\Sigma}_{n}^{-1}\bar{X}_{n}$. Let $\mu=2m$, $\nu=n-2m$, $F_{\mathrm{HH}}=(n-1)^{-1}(\nu/\mu)T^{2}$ and $\lambda=n({{}^{t}}\vec{\mu}_{\vec{\theta},\eta})\Sigma_{\eta,N}^{-1}\vec{\mu}_{\vec{\theta},\eta}$. Then, $F_{\mathrm{HH}}$ obeys $F_{\mu,\nu;\lambda}$, the non-central $F$ distribution with $\mu$ and $\nu$ degrees of freedom and with non-centrality $\lambda$. The probability density function of $F_{\mu,\nu;\lambda}$ is

[TABLE]

where $B(x,y)$ is the beta function. (See [6].) If $\vec{\theta}=\vec{0}_{m}$, then $\lambda=0$ and thus $F_{\mathrm{HH}}$ obeys $F_{\mu,\nu}=F_{\mu,\nu;0}$ the central $F$ distribution. Define the critical point $c$ as a solution to the equation $\int_{c}^{\infty}p_{0}(f)df=\alpha$.

The HH test $T^{\mathrm{HH}}_{\alpha}$ is a test by which $H_{0}$ is accepted if $F_{\mathrm{HH}}\leq c$ is observed.

For $\vec{z}={{}^{t}}(z_{1},z_{2},...,z_{m})\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, define \big{|}\vec{z}\big{\rangle}\in\mathcal{H}^{\otimes m} by \big{|}\vec{z}\big{\rangle}=|z_{1}\rangle\otimes|z_{2}\rangle\otimes\cdots\otimes|z_{m}\rangle. For $Z=(\vec{z}_{1},\vec{z}_{2},...,\vec{z}_{n})\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, define $|Z\rangle\in\mathcal{H}^{\otimes mn}$ by |Z\rangle=|\vec{z}_{1}\rangle\otimes\big{|}\vec{z}_{2}\big{\rangle}\otimes\cdots\otimes\big{|}\vec{z}_{n}\big{\rangle}. Let $\langle Z|$ be $|Z\rangle^{*}$. The POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$ for $T^{\mathrm{HH}}_{\alpha}$ is given by

[TABLE]

where $x_{i,j}$ and $y_{i,j}$ are $(i,j)$-th entries of $\mathrm{Re}(Z)$ and $\mathrm{Im}(Z)$, respectively.

4 Theorems and a numerical comparison

Let $m$ be the mode size, $n$ be the sample size, $\vec{\theta}$ be the displacement parameter, $\eta$ be the squeezing parameter, $N$ be the mixture parameter, and $\alpha$ be the level. The first theorem implies that $\hat{v}_{j,k}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$, $\hat{R}=\hat{R}_{n-1}\hat{R}_{n-2}\cdots\hat{R}_{1}\in\mathcal{U}(\mathcal{H}^{\otimes mn})$, $\hat{T}_{\mathrm{SI}}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$, $\mathcal{K}_{t}\subset\mathcal{H}^{\otimes mn}$, $\hat{K}_{t}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ and $T^{\mathrm{SI}}_{\alpha}$ are all SI.

Theorem 1**.**

For any $\eta\in\mathrm{Sqz}^{m}$, and for any $j,k\in\{1,2,...,n\}$, it holds that $(\hat{S}_{\eta}^{\otimes n})^{*}\hat{v}_{j,k}\hat{S}_{\eta}^{\otimes n}=\hat{v}_{j,k}$.

Proof.

See Sec. 5.2. ∎

A vector $f\in\mathcal{H}^{\otimes mn}$ is regarded as a function $f:\mathrm{Mat}_{\mathbb{R}}^{m,n}\to\mathbb{C}$, where the $(i,j)$-th entry $x_{i,j}$ of the matrix variable $X\in\mathrm{Mat}_{\mathbb{R}}^{m,n}$ is specified by $\hat{a}_{i,j}f=(x_{i,j}f+\partial f/\partial x_{i,j})/\sqrt{2}$. If $f(X)=f(Xe^{A})$ holds for any $A\in\mathrm{Ant}_{\mathbb{R}}^{n}$, then $f(X)$ is said to be Mode-wisely Rotationally Invariant (MwRI). If $m=1$ and if $n\geq 2$, then an MwRI function is a radial function of $n$ variables.

Theorem 2**.**

The Hilbert subspace $\mathcal{K}_{0}$ is the set of square-integrable MwRI functions.

Proof.

See Sec. 5.4. ∎

Consider the case of $n=2$. Then, $\hat{T}_{\mathrm{SI}}=\hat{v}_{1,2}\hat{v}_{1,2}^{*}$ holds because $\hat{R}^{*}\hat{v}_{1,2}\hat{R}=\hat{v}_{1,2}$ holds. Let $X$ be a random variable given by observing $\hat{T}_{\mathrm{SI}}$. Let $\hat{T}=-i\hat{v}_{1,2}$. where $i=\sqrt{-1}$. It holds that $\hat{T}^{*}=\hat{T}$ and that $\hat{T}_{\mathrm{SI}}=\hat{T}^{2}$. Let $Y$ be a random variable given by observing $\hat{T}$. For any state, the probability distribution of $X$ is equal to that of $Y^{2}$.

For a random variable $Z$, the probability distribution can be identified by the characteristic function $\varphi_{Z}(r)=E[e^{irZ}]$, where $E[W]$ is the expected value of a random variable $W$. Let $\mathit{NB}_{m}(p)$ be the negative binomial distribution whose probability function is $f(x)=\begin{pmatrix}m+x-1\\ x\end{pmatrix}(1-p)^{m}p^{x}$. If $Z$ obeys $\mathit{NB}_{m}(p)$, then it holds that $\varphi_{Z}(r)=(1-p)^{m}(1-pe^{ir})^{-m}$. Let $\mathit{Poi}(\lambda)$ be the Poisson distribution whose probability function is $f(x)=e^{-\lambda}\lambda^{x}/(x!)$. If $W$ obeys $\mathit{Poi}(\lambda)$, then it holds that $\varphi_{W}(r)=e^{\lambda(e^{ir}-1)}$. Define $\gamma(r)$ and $\psi_{s}(r)$ by

[TABLE]

Theorem 3**.**

(i) If $n=2$, then it holds that

[TABLE]

(ii) Assume that $F$, $G$, $P_{k}$ and $Q_{k}$ $(k\in\mathbb{N})$ are mutually independent random variables, where $F$ and $G$ obey $\mathit{NB}_{m}(N/(N+1))$, and where $P_{k}$ and $Q_{k}$ obey $\mathit{Poi}\left(\frac{\|\theta\|^{2}N^{k-1}}{(N+1)^{k+1}}\right)$. If $n=2$, then the probability distribution of $Y$ is that of $F-G+\sum_{k=1}^{\infty}(kP_{k}-kQ_{k})$.

Proof.

See Sec. 5.7. ∎

The following theorem shows that $T^{\mathrm{SI}}_{\alpha}$ dominates $T^{\mathrm{triv}}_{\alpha}$ if $N=0$.

Theorem 4**.**

If $N=0$, then, for any $m\geq 1$ and for any $n\geq 2$, the type II error probability of $T^{\mathrm{SI}}_{\alpha}$ is

[TABLE]

where $B(x,y)$ is the beta function.

Proof.

See Sec. 5.8. ∎

For a test to be SI, it is necessary that the type II error probability does not depend on $\eta\in\mathrm{Sqz}^{m}$. In the following theorem, (i) implies that $T^{\mathrm{HH}}_{\alpha}$ is not SI. Moreover, (ii),implies that $T^{\mathrm{HH}}_{\alpha}$ is no better than $T^{\mathrm{triv}}_{\alpha}$ in the minimax criterion.

Theorem 5**.**

*(i) If $\vec{\theta}\neq\vec{0}_{m}$, then $\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]$ depends on $\eta\in\mathrm{Sqz}^{m}$.

(ii) It holds that $\sup_{\eta\in\mathrm{Sqz}^{m}}\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]=1-\alpha$.*

Proof.

See Sec. 5.9. ∎

For the case of $m=1$, $n=3$ and $N=0$, the type II error probabilities of $T^{\mathrm{SI}}_{\alpha}$ and $T^{\mathrm{HH}}_{\alpha}$ are plotted in Figure. The figure says that $\beta_{\hat{\rho}_{\theta,\eta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]<\beta_{\hat{\rho}_{\theta,\eta,0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]$ if $\theta\fallingdotseq 0$, and that $\beta_{\hat{\rho}_{\theta,\eta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]>\beta_{\hat{\rho}_{\theta,\eta,0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]$ if $\theta\gg 0$. If $\eta=O_{2}$ We can evaluate $\beta_{\hat{\rho}_{\theta,O_{2},0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]$ with $\beta_{\hat{\rho}_{\theta,O_{2},0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]$ as follows.

Theorem 6**.**

*Consider the case of $m=1$, $n=3$, $\eta=O_{2}$, $N=0$ and $0<\alpha<1$.

(i) There exists $s>0$ such that, if $0<|\theta|<s$, then $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]<\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]$ holds.

(ii) There exists $t>0$ such that, if $\theta>t$, then $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]>\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]$ holds.*

Proof.

See Sec. 5.10. ∎

5 Proofs of the theorems

5.1 Matrix notations of operators

Let $m$ be the mode size, and let $n$ be the sample size. For any $\mu,\nu\in\mathbb{N}$, let $\mathrm{Mat}_{m,n}^{\mu,\nu}$ be the set of $\mu$-by-$\nu$ matrices whose entries belong to $\mathcal{L}(\mathcal{H}^{\otimes mn})$. Identifying $1$ with $\hat{I}^{\otimes mn}$, we regard that $\mathrm{Mat}_{\mathbb{C}}^{\mu,\nu}$ is a vector subspace of $\mathrm{Mat}_{m,n}^{\mu,\nu}$. Let $\mathrm{Mat}_{m,n}^{\mu}$ be $\mathrm{Mat}_{m,n}^{\mu,\mu}$. We will use $L\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ which can be written by a linear combination of entries of $X\in\mathrm{Mat}_{m,n}^{\mu,\nu}$.

Define a linear map $\mathrm{tr}_{\mu}:\mathrm{Mat}_{m,n}^{\mu}\to\mathcal{L}(\mathcal{H}^{\otimes mn})$ by $\mathrm{tr}_{\mu}[X]=\sum_{i=1}^{\mu}X_{i,i}$, where $X_{i,j}$ is the $(i,j)$-th entry of $X$.

For $X\in\mathrm{Mat}_{m,n}^{\mu,\nu}$ and for $L\in\mathcal{L}(\mathcal{H}^{\otimes mn})$, define $XL,LX,[X,L]\in\mathrm{Mat}_{m,n}^{\mu,\nu}$ by $(XL)_{i,j}=X_{i,j}L$, $(LX)_{i,j}=LX_{i,j}$ and $[X,L]=XL-LX$, respectively. If $X\in\mathrm{Mat}_{\mathbb{C}}^{\mu,\nu}$ and $L\in\mathcal{L}(\mathcal{H}^{\otimes mn})$, then it holds that

[TABLE]

where $O_{\mu,\nu}\in\mathrm{Mat}_{\mathbb{C}}^{\mu,\nu}$ is the zero matrix. For $X\in\mathrm{Mat}_{m,n}^{\lambda,\mu}$ and for $Y\in\mathrm{Mat}_{m,n}^{\mu,\nu}$, define the product $XY\in\mathrm{Mat}_{m,n}^{\lambda,\nu}$ by $(XY)_{i,j}=\sum_{k=1}^{\mu}X_{i,k}Y_{k,j}$. For $X\in\mathrm{Mat}_{m,n}^{\mu,\nu}$, define $X^{*}\in\mathrm{Mat}_{m,n}^{\nu,\mu}$ by $(X^{*})_{i,j}=(X_{j,i})^{*}$, and define ${{}^{t}}X\in\mathrm{Mat}_{m,n}^{\nu,\mu}$ by $({{}^{t}}X)_{i,j}=X_{j,i}$. The operations $X\mapsto X^{*}$ and $X\mapsto{{}^{t}}X$ are commutative as $({{}^{t}}X)^{*}={{}^{t}}(X^{*})$, and it holds that $({{}^{t}}X^{*})_{i,j}=X_{i,j}^{*}$. For $X\in\mathrm{Mat}_{m,n}^{\lambda,\mu}$, for $Y\in\mathrm{Mat}_{m,n}^{\mu,\nu}$ and for $C\in\mathrm{Mat}_{\mathbb{C}}^{\mu,\nu}$, it holds that $(XY)^{*}=Y^{*}X^{*}$ and that ${{}^{t}}(XC)=({{}^{t}}C)({{}^{t}}X)$. Moreover, for $X\in\mathrm{Mat}_{m,n}^{\mu}$ and for $C\in\mathrm{Mat}_{\mathbb{C}}^{\mu}$, it holds that

[TABLE]

For $j\in\{1,2,...,n\}$, define $\vec{a}_{j}\in\mathrm{Mat}_{m,n}^{m,1}$ and \underaccent{\tilde}{\vec a}_{j}\in\mathrm{Mat}_{m,n}^{2m,1} by

[TABLE]

respectively. Define $\tilde{a}_{m,n}\in\mathrm{Mat}_{m,n}^{m,n}$ by $\tilde{a}_{m,n}=(\vec{a}_{1},\vec{a}_{2},...,\vec{a}_{n})$. Define \underaccent{\tilde}{a}_{m,n}\in\mathrm{Mat}_{m,n}^{2m,n} by \underaccent{\tilde}{a}_{m,n}=(\underaccent{\tilde}{\vec a}_{1},\underaccent{\tilde}{\vec a}_{2},...,\underaccent{\tilde}{\vec a}_{n}). For $\mu\in\mathbb{N}$, let $I_{\mu}\in\mathrm{Mat}_{\mathbb{R}}^{\mu}$ be the identity matrix, let $O_{\mu}\in\mathrm{Mat}_{\mathbb{R}}^{\mu}$ be the zero matrix, and let

[TABLE]

If $n=1$, then, it holds that \hat{s}_{\eta}=2^{-1}\underaccent{\tilde}{\vec a}_{1}^{*}K_{m}\eta\underaccent{\tilde}{\vec a}_{1}-\underaccent{\tilde}{c}_{\eta}\hat{I}^{\otimes m}, where \underaccent{\tilde}{c}_{\eta}=4^{-1}\mathrm{tr}_{2m}[K_{m}\eta]\in\sqrt{-1}\mathbb{R}. For any $m,n\in\mathbb{N}$, and for any $\eta\in\mathrm{Sqz}^{m}$, define \underaccent{\tilde}{\hat u}_{\eta}\in\mathcal{L}(\mathcal{H}^{\otimes mn}) by

[TABLE]

Then, it holds that \exp(\underaccent{\tilde}{\hat u}_{\eta})=\hat{S}_{\eta}^{\otimes n}. For any $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$, define $\hat{u}_{A}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ by

[TABLE]

For $\eta\in\mathrm{Sqz}^{m}$, if the anti-hermitian part is $A$, and if the symmetric part is $O_{m}$, then it holds that \hat{u}_{A}=\underaccent{\tilde}{\hat u}_{\eta}.

For $A\in\mathrm{Ant}_{\mathbb{C}}^{n}$, define $\hat{v}_{A}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ by

[TABLE]

For $j,k\in\{1,2,...,n\}$ with $j<k$, define $J_{j,k}\in\mathrm{Ant}_{\mathbb{R}}^{n}$ by

[TABLE]

If $j=k$, let $J_{j,k}=O_{n}$. If $j>k$, let $J_{j,k}=-J_{k,j}$. It holds that $\hat{v}_{J_{j,k}}=\hat{v}_{j,k}$.

By Eq. (5), for any $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$ and for any $B\in\mathrm{Ant}_{\mathbb{C}}^{n}$, it holds that

[TABLE]

5.2 Proof of Theorem 1 ($\hat{v}_{j,k}$ is SI)

We first consider how $\hat{s}_{\eta}$ is represented in the case of $n=1$. To simplify the notation, \underaccent{\tilde}{\vec a}_{1}\in\mathrm{Mat}_{m,1}^{2m,1} is denoted by \underaccent{\tilde}{\vec a}={{}^{t}}(\hat{a}_{1},\hat{a}_{2},...,\hat{a}_{m},\hat{a}_{1}^{*},\hat{a}_{2}^{*},...,\hat{a}_{m}^{*}).

Lemma 1**.**

If $n=1$, then, for any $m\in\mathbb{N}$, and for any $\eta\in\mathrm{Sqz}^{m}$, it holds that [\underaccent{\tilde}{\vec a},\underaccent{\tilde}{\hat s}_{\eta}]=\eta\underaccent{\tilde}{\vec a}.

Proof.

Let $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$ be the anti-hermitian part of $\eta$, and let $S\in\mathrm{Sym}_{\mathbb{C}}^{m}$ be the symmetric part of $\eta$. For $i\in\{1,2,...,m\}$, let

[TABLE]

Then, for any $i\leq m$, it holds that

[TABLE]

where $\eta_{i}\in\mathrm{Mat}_{\mathbb{C}}^{1,2m}$ is the $i$-th row vector of $\eta$. Similarly, let

[TABLE]

This definition is equivalent to

[TABLE]

For any $i\leq m$, it holds that

[TABLE]

By Eqs. (16) and (17), we obtain [\underaccent{\tilde}{\vec a},\hat{s}_{\eta}]=\eta\underaccent{\tilde}{\vec a}. ∎

Then, for any $n\geq 1$, we obtain a representation of \underaccent{\tilde}{\hat u}_{\eta} on \underaccent{\tilde}{a}_{m,n}.

Lemma 2**.**

For any $m,n\in\mathbb{N}$, and for any $\eta\in\mathrm{Sqz}^{m}$, it holds that [\underaccent{\tilde}{a}_{m,n},\underaccent{\tilde}{\hat u}_{\eta}]=\eta\underaccent{\tilde}{a}_{m,n}.

Proof.

For any $i\in\{1,2,...,m\}$, and for any $j,k\in\{1,2,...,n\}$ with $j\neq k$, it holds that [\hat{a}_{i,j},\underaccent{\tilde}{\vec a}_{k}]=[\hat{a}_{i,j}^{*},\underaccent{\tilde}{\vec a}_{k}]=O_{2m,1}. Hence, by Lemma 1, we obtain [\underaccent{\tilde}{a}_{m,n},\underaccent{\tilde}{\hat u}_{\eta}]=\eta\underaccent{\tilde}{a}_{m,n}. ∎

Lemma 3**.**

For any $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$, it holds that $[\tilde{a}_{m,n},\hat{u}_{A}]=A\tilde{a}_{m,n}$ and that $[{{}^{t}}\tilde{a}_{m,n}^{*},\hat{u}_{A}]=\bar{A}({{}^{t}}\tilde{a}_{m,n}^{*})$.

Proof.

Applying Lemma 2 to the case where the symmetric part of $\eta$ is $O_{m}$, we obtain the statement. ∎

Lemma 4**.**

For any $A\in\mathrm{Ant}_{\mathbb{C}}^{n}$, it holds that $[\tilde{a}_{m,n},\hat{v}_{A}]=-\tilde{a}_{m,n}\bar{A}$ and that $[{{}^{t}}\tilde{a}_{m,n}^{*},\hat{v}_{A}]=-{{}^{t}}\tilde{a}_{m,n}^{*}A$.

Proof.

By Lemma 3, it holds that $[{{}^{t}}\tilde{a}_{m,n},\hat{v}_{A}]=A({{}^{t}}\tilde{a}_{m,n})$ and that $[\tilde{a}_{m,n}^{*},\hat{v}_{A}]=\bar{A}\tilde{a}_{m,n}^{*}$. By transposing, we obtain the statement. ∎

Lemma 5**.**

For any $m,n\in\mathbb{N}$, and for any $B\in\mathrm{Ant}_{\mathbb{R}}^{n}$, it holds that [\underaccent{\tilde}{a}_{m,n},\hat{v}_{B}]=-\underaccent{\tilde}{a}_{m,n}B.

Proof.

By Lemma 4, and by $\bar{B}=B$, we have [\underaccent{\tilde}{a}_{m,n},\hat{v}_{B}]=-\underaccent{\tilde}{a}_{m,n}B. ∎

Lemma 6**.**

For any $\eta\in\mathrm{Sqz}^{m}$ and for any $B\in\mathrm{Ant}_{\mathbb{R}}^{n}$, it holds that [\underaccent{\tilde}{\hat u}_{\eta},\hat{v}_{B}]=0.

Proof.

By Eq. (12), it holds that $[K_{m}\eta,\hat{v}_{B}]=O_{2m}$. Hence, we have

[TABLE]

By Lemma 5, we have

[TABLE]

By Eq. (13), the right-hand side is zero. ∎

Proof of Theorem 1.

Using $J_{j,k}\in\mathrm{Ant}_{\mathbb{R}}^{n}$ of (14), it holds that $\hat{v}_{j,k}=\hat{v}_{J_{j,k}}$. Hence, by Lemma 6, it holds that [\underaccent{\tilde}{\hat u}_{\eta},\hat{v}_{j,k}]=0. As \hat{S}_{\eta}^{\otimes n}=\exp(\underaccent{\tilde}{\hat u}_{\eta}), we obtain $(\hat{S}_{\eta}^{\otimes n})^{*}\hat{v}_{j,k}\hat{S}_{\eta}^{\otimes n}=\hat{v}_{j,k}$. ∎

5.3 Other properties of $\hat{u}_{A}$ and $\hat{v}_{B}$

We will use the following lemma in Secs. 5.4, 5.7 and 5.8.

Lemma 7**.**

For any $m,n\in\mathbb{N}$, and for any $A,B\in\mathrm{Ant}_{\mathbb{C}}^{n}$, it holds that $[\hat{v}_{A},\hat{v}_{B}]=\hat{v}_{[A,B]}$.

Proof.

By Eq. (12), it holds that $[A,\hat{v}_{B}]=O_{n}$. Hence, we have

[TABLE]

Moreover, by Lemma 4, we have

[TABLE]

Hence, we have $[\hat{v}_{A},\hat{v}_{B}]=\hat{v}_{[A,B]}$. ∎

We will use the following lemma in Sec. 5.5.

Lemma 8**.**

For any $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$ and for any $B\in\mathrm{Ant}_{\mathbb{C}}^{n}$, it holds that $[\hat{u}_{A},\hat{v}_{B}]=0$.

Proof.

By Eq. (12), it holds that $[A,\hat{v}_{B}]=O_{m}$. Hence, we have

[TABLE]

By Lemma 4, we have

[TABLE]

By Eq. (13), the right-hand side is zero. ∎

5.4 Proof of Theorem 2 (MwRI functions)

We consider how $\hat{v}_{j,k}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ is represented by $f(X)\in\mathcal{H}^{\otimes mn}$ where $X\in\mathrm{Mat}_{\mathbb{R}}^{m,n}$.

Lemma 9**.**

For any $r\in\mathbb{R}$, it holds that $e^{r\hat{v}_{j,k}}f(X)=f(Xe^{rJ_{j,k}})$.

Proof.

Let $x_{i,j}$ be the $(i,j)$-th entry of $X$. It holds that $\hat{v}_{j,k}=\sum_{i=1}^{m}(\hat{a}_{i,k}^{*}\hat{a}_{i,j}-\hat{a}_{i,j}^{*}\hat{a}_{i,k})=\sum_{i=1}^{m}(x_{i,k}\partial/\partial x_{i,j}-x_{i,j}\partial/\partial x_{i,k})$. Let $(s_{i,j,k},\varphi_{i,j,k})$ be a polar coordinate of $(x_{i,j},x_{i,k})$ defined by $x_{i,j}=s_{i,j,k}\cos\varphi_{i,j,k}$ and $x_{i,k}=s_{i,j,k}\sin\varphi_{i,j,k}$. Then, it holds that

[TABLE]

and so that $\hat{v}_{j,k}=-\sum_{i=1}^{m}\partial/\partial\varphi_{i,j,k}$. As $e^{-t\partial/\partial\varphi}=\sum_{k=0}^{\infty}(-t\partial/\partial\varphi)^{k}/k!$ holds, it holds that $e^{-t\partial/\partial\varphi}\varphi^{n}=\sum_{k=0}^{n}\begin{pmatrix}n\\ k\end{pmatrix}\varphi^{n-k}(-t)^{k}=(\varphi-t)^{n}$. Hence, we obtain \exp(r\hat{v}_{j,k})f(X)=f({{}^{t}}[\exp(-rJ_{j,k})({{}^{t}}X)])=f\big{(}X\exp(rJ_{j,k})\big{)}. ∎

For $\hat{L}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$, let $\mathrm{Null}(\hat{L})\subset\mathcal{H}^{\otimes mn}$ be the nullspace $\{f\in\mathcal{H}^{\otimes mn}\mid\hat{L}f=0\}$. The following lemma shows that $\bigcap_{k=1}^{n-1}\mathrm{Null}(\hat{v}_{k,n})$ is the set of square-integrable MwRI functions.

Lemma 10**.**

It holds that $\bigcap_{k=1}^{n-1}\mathrm{Null}(\hat{v}_{k,n})=\bigcap_{A\in\mathrm{Ant}_{\mathbb{R}}^{n}}\mathrm{Null}(\hat{v}_{A})$.

Proof.

By Lemma 7, it holds that $[\hat{v}_{i,j},\hat{v}_{j,k}]=\hat{v}_{k,i}$ for any $i,j,k\in\{1,2,...,n\}$. Hence, it holds that

[TABLE]

Hence, we have $\bigcap_{k=1}^{n-1}\mathrm{Null}(\hat{v}_{k,n})=\bigcap_{i=1}^{n}\allowbreak\bigcap_{j=1}^{n}\mathrm{Null}(\hat{v}_{i,j})$. Since $J_{j,k}\in\mathrm{Ant}_{\mathbb{R}}^{n}$ holds, $\bigcap_{i=1}^{n}\bigcap_{j=1}^{n}\mathrm{Null}(\hat{v}_{i,j})\supset\bigcap_{A\in\mathrm{Ant}_{\mathbb{R}}^{n}}\mathrm{Null}(\hat{v}_{A})$ holds. Since any $\hat{v}_{A}$ is a linear combination of $\hat{v}_{j,k}$, we have $\bigcap_{i=1}^{n}\bigcap_{j=1}^{n}\mathrm{Null}(\hat{v}_{i,j})\subset\bigcap_{A\in\mathrm{Ant}_{\mathbb{R}}^{n}}\mathrm{Null}(\hat{v}_{A})$. Hence, we obtain the statement. ∎

Proof of Theorem 2.

Let $\hat{T}=\sum_{k=1}^{n-1}\hat{v}_{k,n}\hat{v}_{k,n}^{*}$. Then, it holds that $\mathrm{Null}(\hat{T})=\bigcap_{k=1}^{n-1}\mathrm{Null}(\hat{v}_{k,n}\hat{v}_{k,n}^{*})$. Since $\mathrm{Null}(\hat{v}_{k,n}\hat{v}_{k,n}^{*})=\mathrm{Null}(\hat{v}_{k,n}^{*}\hat{v}_{k,n})=\mathrm{Null}(\hat{v}_{k,n})$ holds, we have $\bigcap_{k=1}^{n-1}\mathrm{Null}(\hat{v}_{k,n}\hat{v}_{k,n}^{*})=\bigcap_{k=1}^{n-1}\mathrm{Null}(\hat{v}_{k,n})$. By Lemma 10, $\mathrm{Null}(\hat{T})$ is the set of square-integrable MwRI functions. Since $\hat{T}_{\mathrm{SI}}=\hat{R}^{*}\hat{T}\hat{R}$ holds, we have $\mathcal{K}_{0}=\mathrm{Null}(\hat{T}_{\mathrm{SI}})=\hat{R}^{*}\mathrm{Null}(\hat{T})$. By $\hat{R}=\hat{R}_{n-1}\hat{R}_{n-2}\cdots\hat{R}_{1}$, and $R_{k}=\exp(\arctan(\sqrt{k})\hat{v}_{k,k+1})$, we have $\hat{R}^{*}\mathrm{Null}(\hat{T})=\mathrm{Null}(\hat{T})$. ∎

5.5 Actions of $\hat{u}_{A}$ and $\hat{v}_{B}$ on $\vec{\theta}$

For $\theta\in\mathbb{C}$, the displacement operator $\hat{D}_{\theta}\in\mathcal{U}(\mathcal{H})$ is defined by $\hat{D}_{\theta}=\exp(\hat{d}_{\theta})$, where $\hat{d}_{\theta}=\theta\hat{a}^{*}-\bar{\theta}\hat{a}$. It holds that

[TABLE]

because of $[\hat{a},\hat{d}_{\theta}]=\theta\hat{I}$, and $[\hat{a}^{*},\hat{d}_{\theta}]=\bar{\theta}\hat{I}$, respectively. Using Taylor expansion, for $r\in\mathbb{R}$, we have $\hat{D}_{r}f(x)=\exp(-ir\sqrt{2}\hat{p})f(x)=f(x-\sqrt{2}r)$. By Baker-Hausdorff formula [8], it holds that $\hat{D}_{r+is}=e^{-irs}e^{i\sqrt{2}s\hat{q}}e^{-i\sqrt{2}r\hat{p}}$ for $r,s\in\mathbb{R}$. Hence, we have

[TABLE]

For $\vec{z}={{}^{t}}(z_{1},z_{2},...,z_{m})\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, let \big{|}\vec{z}\big{\rangle}=|z_{1}\rangle\otimes|z_{2}\rangle\otimes\cdots\otimes|z_{m}\rangle\in\mathcal{H}^{\otimes m}. For $Z=(\vec{Z}_{1},\vec{Z}_{2},...,\vec{Z}_{n})\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, let $|Z\rangle=|\vec{Z}_{1}\rangle\otimes|\vec{Z}_{2}\rangle\otimes\cdots\otimes|\vec{Z}_{n}\rangle\in\mathcal{H}^{\otimes mn}$. For $W\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, let

[TABLE]

and define a displacement operator $\hat{D}_{W}\in\mathcal{U}(\mathcal{H}^{\otimes mn})$ by $\hat{D}_{W}=\exp(\hat{d}_{W})$. By Eq. (19), it holds that \hat{D}_{W}\big{|}O_{m,n}\big{\rangle}=|W\rangle, where $O_{m,n}\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$ is the zero matrix. For $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$, let $\hat{U}_{A}=\exp(\hat{u}_{A})\in\mathcal{U}(\mathcal{H}^{\otimes mn})$. For $B\in\mathrm{Ant}_{\mathbb{C}}^{n}$, let $\hat{V}_{B}=\exp(\hat{v}_{B})\in\mathcal{U}(\mathcal{H}^{\otimes mn})$.

Lemma 11**.**

For any $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$, for any $B\in\mathrm{Ant}_{\mathbb{C}}^{n}$ and for any $Z\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, it holds that

[TABLE]

Proof.

By Lemma 8, it holds that $\hat{U}_{A}\hat{V}_{B}=\hat{V}_{B}\hat{U}_{A}$. Hence, we just need to prove \hat{U}_{A}\hat{V}_{B}|Z\rangle=\big{|}e^{A}Ze^{-\bar{B}}\big{\rangle}. By Lemmas 3 and 4, it holds that $\hat{U}_{A}^{*}\hat{V}_{B}^{*}\hat{d}_{W}\hat{V}_{B}\hat{U}_{A}=\hat{d}_{e^{-A}We^{\bar{B}}}$. By Eq (15), it holds that \hat{U}_{A}\hat{V}_{B}\big{|}O_{m,n}\big{\rangle}=\big{|}O_{m,n}\big{\rangle}. Hence,

[TABLE]

is obtained. ∎

For $Z=(\vec{Z}_{1},\vec{Z}_{2},...,\vec{Z}_{n})\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$, and for $N\geq 0$, let $\hat{\rho}_{Z,N}\in\mathcal{S}(\mathcal{H}^{\otimes mn})$ be $\hat{\rho}_{\vec{Z}_{1},N}\otimes\hat{\rho}_{\vec{Z}_{2},N}\otimes\cdots\otimes\hat{\rho}_{\vec{Z}_{n},N}$.

Lemma 12**.**

For any $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$, for any $B\in\mathrm{Ant}_{\mathbb{C}}^{n}$, for any $Z\in\mathrm{Mat}_{\mathbb{C}}^{m,n}$ and for any $N\geq 0$, it holds that

[TABLE]

Proof.

Let $\hat{\sigma}=\hat{U}_{A}\hat{V}_{B}\hat{\rho}_{Z,N}\hat{V}_{B}^{*}\hat{U}_{A}^{*}$. In the case of $N=0$, by Lemma 11, we obtain $\hat{\sigma}=\hat{\rho}_{e^{A}Ze^{-\bar{B}},0}$. Consider the case of $N>0$. By Lemma 8, it holds that $\hat{\sigma}=\hat{V}_{B}\hat{U}_{A}\hat{\rho}_{Z,N}\hat{U}_{A}^{*}\hat{V}_{B}^{*}$. By Lemma 11, it holds that

[TABLE]

where $\|W-Z\|^{2}=\mathrm{tr}_{m}[(W-Z)(W^{*}-Z^{*})]$, and where $u_{i,j}$ and $v_{i,j}$ are the $(i,j)$-th entries of $\mathrm{Re}(W)$ and $\mathrm{Im}(W)$, respectively. Replace $\mathrm{Re}(e^{A}We^{-\bar{B}})$ by $R\in\mathrm{Mat}_{\mathbb{R}}^{m,n}$, and replace $\mathrm{Im}(e^{A}We^{-\bar{B}})$ by $S\in\mathrm{Mat}_{\mathbb{R}}^{m,n}$. Let $\vec{u}_{j},\vec{v}_{j},\vec{r}_{j}$ and $\vec{s}_{j}\in\mathrm{Mat}_{\mathbb{R}}^{m,1}$ be $j$-th column vectors of $\mathrm{Re}(W)$, $\mathrm{Im}(W)$, $R$ and $S$, respectively. Define $\vec{u},\vec{v},\vec{r}$ and $\vec{s}\in\mathrm{Mat}_{\mathbb{R}}^{mn,1}$ by

[TABLE]

respectively. Let $r_{i,j}$ and $s_{i,j}$ be the $(i,j)$-th entries of $R$ and $S$, respectively. Then, it holds that $\vec{r}+\sqrt{-1}\vec{s}=C(\vec{u}+\sqrt{-1}\vec{v})$, where $C\in\mathrm{Mat}_{\mathbb{C}}^{mn}$ is a unitary matrix given by the Kronecker product of $e^{B}$ and $e^{A}$ as $C=e^{B}\otimes e^{A}$. Let $D=B\otimes I_{m}+I_{n}\otimes A\in\mathrm{Ant}_{\mathbb{C}}^{mn}$. Then, it holds that $C=e^{D}$. Define $\vec{w},\vec{t}\in\mathrm{Mat}_{\mathbb{R}}^{2mn,1}$ and $E\in\mathrm{Ant}_{\mathbb{R}}^{2mn}$ by

[TABLE]

respectively. Then, it holds that $\vec{t}=e^{E}\vec{w}$ and that $\det[e^{E}]=\exp(\mathrm{tr}_{2mn}[E])=1$. Hence, the Jacobian of the replacement of $e^{E}\vec{w}$ by $\vec{t}$ is one. Hence, we have

[TABLE]

where $T=e^{A}We^{-\bar{B}}$. Since

[TABLE]

holds, we obtain $\sigma=\hat{\rho}_{e^{A}Ze^{-\bar{B}},N}$. ∎

Lemma 13**.**

For any $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, it holds that $\hat{R}\hat{\rho}_{\vec{\theta},N}^{\otimes n}\hat{R}^{*}=\hat{\rho}_{\vec{0}_{m},N}^{\otimes(n-1)}\otimes\hat{\rho}_{\sqrt{n}\vec{\theta},N}$.

Proof.

Let $\vec{1}_{n}\in\mathrm{Mat}_{\mathbb{R}}^{n,1}$ be a vector whose entries are all one, and let $\vec{e}_{n}\in\mathrm{Mat}_{\mathbb{R}}^{n,1}$ be a unit vector whose $n$-th entry is one. It holds that $\hat{\rho}_{\vec{\theta},N}^{\otimes n}=\hat{\rho}_{\vec{\theta}({{}^{t}}\vec{1}_{n}),N}$ and that $\hat{\rho}_{\vec{0}_{m},N}^{\otimes(n-1)}\otimes\hat{\rho}_{\sqrt{n}\vec{\theta},N}=\hat{\rho}_{\sqrt{n}\vec{\theta}({{}^{t}}\vec{e}_{n}),N}$. Let $R_{k}=\exp[(\arctan\sqrt{k})J_{k,k+1}]\in\mathrm{Mat}_{\mathbb{R}}^{n}$, and let $R=R_{n-1}R_{n-2}\cdots R_{1}\in\mathrm{Mat}_{\mathbb{R}}^{n}$. By Lemma 12, it holds that $\hat{R}\hat{\rho}_{\vec{\theta},N}^{\otimes n}\hat{R}^{*}=\hat{\rho}_{\vec{\theta}[{{}^{t}}(R\vec{1}_{n})],N}$. If $t\in[0,\pi/2)$, then $\cos t=1/\sqrt{1+\tan^{2}t}$ and $\sin t=\tan t/\sqrt{1+\tan^{2}t}$ hold. Hence, $\cos(\arctan\sqrt{k})=1/\sqrt{k+1}$ and $\sin(\arctan\sqrt{k})=\sqrt{k}/\sqrt{k+1}$ hold. Hence, we have

[TABLE]

The $k$-th and $k+1$-th row vectors of $R_{k}R_{k-1}\cdots R_{1}$ are recursively obtained by calculating the $2$-by-$2$ submatrix as

[TABLE]

Hence, the $n$-th row vector of $R$ is ${{}^{t}}(1/\sqrt{n}\vec{1}_{n})$. Since $R$ is an orthogonal matrix, it holds that $R\vec{1}_{n}=\sqrt{n}\vec{e}_{n}$. ∎

5.6 Some properties related to the Fock vectors

Let $\hat{N}\in\mathcal{L}(\mathcal{H})$ be the number operator $\hat{a}^{*}\hat{a}$. It holds that $[\hat{a},\hat{N}]=\hat{a}$, and so that $e^{-ir\hat{N}}\hat{a}e^{ir\hat{N}}=e^{ir}\hat{a}$ (${{}^{\forall}}r\in\mathbb{R}$), where $i=\sqrt{-1}$. Let $\mathbb{N}_{0}$ be $\mathbb{N}\cup\{0\}$. For $n\in\mathbb{N}_{0}$, the $n$-th Fock vector, or the $n$-th number vector, $f_{n}\in\mathcal{H}$ is defined by $f_{n}=(1/\sqrt{n!})(\hat{a}^{*})^{n}|0\rangle$. It holds that $\hat{a}f_{n}=\sqrt{n}f_{n-1}$, $\hat{a}^{*}f_{n}=\sqrt{n+1}f_{n+1}$ and $\hat{N}f_{n}=nf_{n}$.

By Baker-Hausdorff formula, the displacement operator $\hat{D}_{\theta}\in\mathcal{U}(\mathcal{H})$ satisfies $\hat{D}_{\theta}=e^{-|\theta|^{2}/2}e^{\theta\hat{a}^{*}}e^{-\bar{\theta}\hat{a}}$ for $\theta\in\mathbb{C}$. Hence we have

[TABLE]

Hence, by the power series calculation, we have

[TABLE]

Moreover, $\{f_{n}\}_{n=0}^{\infty}$ is a complete orthonormal basis of $\mathcal{H}$ because $f_{n}^{*}g=0$ $({{}^{\forall}}n\in\mathbb{N}_{0})$ is equivalent to $\int_{\mathbb{R}}e^{i\sqrt{2}sx}g(x)dx=0$ $({{}^{\forall}}s\in\mathbb{R})$. Hence, it holds that $\hat{N}=\sum_{n=0}^{\infty}nf_{n}f_{n}^{*}$. Calculating the Gaussian mixture of $|\theta\rangle\langle\theta|$ using the form of $(\ref{eq_fock_coherent})$, we have

[TABLE]

This equation will be used in Lemma 17.

For $j,k\in\{1,2,...,n\}$ with $j<k$, define $K_{j,k}\in\mathrm{Ant}_{\mathbb{C}}^{n}$ by

[TABLE]

For $k>j$, let $K_{j,k}=-K_{k,j}$. Let $K_{j,j}=O_{n}$. The operator $\hat{d}_{j,k}$ defined in (7) is equal to $\hat{v}_{K_{j,k}}$. In Sec. 5.7, we will use the following lemma.

Lemma 14**.**

111This lemma was suggested by Prof. K. Matsumoto.

For any $j,k\in\{1,2,..,n\}$, let $\hat{U}\in\mathcal{U}(\mathcal{H}^{\otimes mn})$ be $\exp((\pi/4)\hat{v}_{J_{j,k}})$, and let $\hat{V}\in\mathcal{U}(\mathcal{H}^{\otimes mn})$ be $\exp((\pi/4)\hat{v}_{K_{j,k}})$. Then, it holds that $\hat{U}^{*}\hat{V}^{*}\hat{v}_{j,k}\hat{V}\hat{U}=\hat{d}_{j,k}$.

Proof.

Let U=\exp\Big{[}\frac{\pi}{4}\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}\Big{]}, and let V=\exp\Big{[}\frac{\pi}{4}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\Big{]}. As $U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}$ and $V=\begin{pmatrix}e^{i\pi/4}&0\\ 0&e^{-i\pi/4}\end{pmatrix}$, it holds that $U^{*}V^{*}J_{j,k}VU=K_{j,k}$. By Lemma 7, we have $\hat{U}^{*}\hat{V}^{*}\hat{v}_{j,k}\hat{V}\hat{U}=\hat{d}_{j,k}$. ∎

In quantum optical experiments, the unitary operators $\hat{U}$ and $\hat{V}$ of the above lemma can be realized by phase-shifting and beam-splitting, respectively. Moreover, $\hat{T}_{j,k}=-\sqrt{-1}\hat{d}_{j,k}$ is an observable whose POVM can be realized by arithmetic subtraction of data obtained by number measurements. Hence, if $n=2$, then $T^{\mathrm{SI}}_{\alpha}$ can be realized by beam-splitters and photon counters.

5.7 Proof of Theorem 3 (Negative binomial and Poisson distributions)

Let $\hat{N}\in\mathcal{L}(\mathcal{H})$ be $\hat{a}^{*}\hat{a}$. Let $h_{\theta,N}(r)$ be $\mathrm{Tr}[\hat{\rho}_{\theta,N}e^{ir\hat{N}}]$, where $i=\sqrt{-1}$. We write $h_{\theta,N}(r)$ using $\gamma(r)$ and $\psi_{s}(r)$ of (11).

Lemma 15**.**

For any $N\geq 0$, it holds that $h_{0,N}(r)=\gamma(r)$.

Proof.

By Eq. $(\ref{eq_gauss_fock})$, it holds that

[TABLE]

Hence, we have $h_{0,N}(r)=\gamma(r)$. ∎

Lemma 16**.**

If $N=0$, then it holds that $h_{\theta,0}(r)=\psi_{|\theta|}(r)$.

Proof.

By Eq. (20), it holds that

[TABLE]

Since $\gamma(r)$ is one, $h_{\theta,0}(r)$ is $\psi_{|\theta|}(r)$. ∎

Lemma 17**.**

For any $N\geq 0$, it holds that $h_{\theta,N}(r)=\gamma(r)\psi_{|\theta|}(r)$.

Proof.

If $N=0$, then $\gamma(r)$ is one, and by Lemma 16, we have $h_{\theta,0}(r)=\gamma(r)\psi_{|\theta|}(r)$. Hereafter, assume that $N>0$. For $x,y\in\mathbb{R}$, let $z\in\mathbb{C}$ be $x+iy$, and let $f_{\theta,N}(z)$ be $\exp(-|z-\theta|^{2}/N)$. It holds that

[TABLE]

By Lemma 11, we have e^{ir\hat{N}}|z\rangle=\big{|}e^{ir}z\big{\rangle}. By Eq. (21), we have $\langle z|e^{ir}z\rangle=\exp(e^{ir}|z|^{2}-|z|^{2})$. Hence, $\langle z|e^{ir\hat{N}}|z\rangle f_{\theta,N}(z)$ is $\exp(-g_{\theta,N}(z))$, where $g_{\theta,N}(z)=|z-\theta|^{2}/N+(1-e^{ir})|z|^{2}$. Let $c_{N,r}$ be $1/N+1-e^{ir}$. For $w\in\mathbb{C}$, let $\vec{\mu}_{w}\in\mathrm{Mat}_{\mathbb{R}}^{2,1}$ be ${{}^{t}}(\mathrm{Re}(w),\mathrm{Im}(w))$. Then, it holds that

[TABLE]

We can factorize $h_{\theta,N}(r)$ to $\psi_{|\theta|}(r)$ and $h_{0,N}(r)$ as

[TABLE]

By Lemma 15, $h_{0,N}(r)$ is $\gamma(r)$. Hence, we have $h_{\theta,N}(r)=\gamma(r)\psi_{|\theta|}(r)$. ∎

Next, we prove (i) of Theorem 3 for the case of $m=1$.

Lemma 18**.**

If $m=1$, and if $n=2$, then it holds that $\varphi_{Y}(r)=\gamma(r)\gamma(-r)\allowbreak\psi_{|\theta|}(r)\psi_{|\theta|}(-r)$.

Proof.

By Lemma 13, it holds that $\varphi_{Y}(r)=\mathrm{Tr}[\hat{\rho}_{0,N}\otimes\hat{\rho}_{\sqrt{2}\theta,N}\exp(r\hat{v}_{1,2})]$. Let $\hat{U}=\exp((\pi/4)\hat{v}_{J_{1,2}})$, and let $\hat{V}=\exp((\pi/4)\hat{v}_{K_{1,2}})$. By Lemma 12, it holds that $\hat{U}^{*}\hat{V}^{*}\hat{\rho}_{0,N}\otimes\hat{\rho}_{\sqrt{2}\theta,N}\hat{V}\hat{U}=\hat{\rho}_{e^{i\pi/4}\theta,N}^{\otimes 2}$. By Lemma 14, it holds that $\hat{U}^{*}\hat{V}^{*}\hat{v}_{1,2}\hat{V}\hat{U}=\hat{d}_{j,k}$. Hence, we have

[TABLE]

By Lemma 17, we have $\varphi_{Y}(r)=\gamma(r)\gamma(-r)\psi_{|\theta|}(r)\psi_{|\theta|}(-r)$. ∎

For any random variable $Z$, and for any constant $c\in\mathbb{R}$, the characteristic functions of $Z$ and $cZ$ satisfy

[TABLE]

For any mutually independent random variables $Z$ and $W$, the characteristic functions of $Z$, $W$ and $S=Z+W$ satisfy

[TABLE]

Proof of Theorem 3.

(i) Let $\theta_{k}$ be the $k$-th entry of $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$. By Lemma 18, and by Eq. (24), we have

[TABLE]

(ii) Let $S=F-G+\sum_{k=1}^{\infty}(kP_{k}-kQ_{k})$. By Eqs. (23) and (24), it holds that

[TABLE]

Since $F$ and $G$ obey $NB_{m}(N(N+1)^{-1})$, it holds that $\varphi_{F}(r)=\varphi_{G}(r)=\gamma(r)$. Let $\lambda_{k}=\|\vec{\theta}\|^{2}N^{k-1}(N+1)^{-k-1}$. Since $P_{k}$ and $Q_{k}$ obey $\mathit{Poi}(\lambda_{k})$, it holds that $\varphi_{P_{k}}(r)=\varphi_{Q_{k}}(r)=\exp[\lambda_{k}(e^{ir}-1)]$. As

[TABLE]

holds, we have $\varphi_{S}(r)=\gamma(r)\gamma(-r)\psi_{\|\vec{\theta}\|}(r)\psi_{\|\vec{\theta}\|}(-r)$. ∎

5.8 Proof of Theorem 4 (Type II error probability of $T^{\mathrm{SI}}_{\alpha}$ for $N=0$)

For $\mu,\nu\in\mathbb{N}$, let $T^{\mu,\nu}_{X}$ be the tangent space of $\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$ at $X\in\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$. Let $x_{i,j}$ be the coordinate variable for the $(i,j)$-th entry of $X\in\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$. Then, a basis of $T^{\mu,\nu}_{X}$ is $\{\partial/\partial x_{i,j}|_{X}\mid 1\leq i\leq\mu,1\leq j\leq\nu\}$. For $Y\in\mathrm{Mat}_{\mathbb{R}}^{\mu}$ and for $Z\in\mathrm{Mat}_{\mathbb{R}}^{\nu}$, let $f_{Y,Z}:\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}\to\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$ be a map given by $X\mapsto YX({{}^{t}}Z)$. The pushforward $df_{Y,Z}|_{X}:T^{\mu,\nu}_{X}\to T^{\mu,\nu}_{f_{Y,Z}(X)}$ at $X\in\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$ is given by $df_{Y,Z}|_{X}(\partial/\partial x_{i,j}|_{X})=\sum_{m=1}^{\mu}\sum_{n=1}^{\nu}Y_{m,i}Z_{n,j}\partial/\partial x_{m,n}|_{f_{Y,Z}(X)}$, where $M_{k,l}$ is the $(k,l)$-th entry of a matrix $M$. For $X\in\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$ and for $K\in\mathbb{N}$, let $\mathcal{P}=\{P_{1}(X),P_{2}(X),...,P_{K}(X)\}\subset\mathrm{Mat}_{\mathbb{R}}^{\mu}$ and $\mathcal{Q}=\{Q_{1}(X),Q_{2}(X),...,Q_{K}(X)\}\subset\mathrm{Mat}_{\mathbb{R}}^{\nu}$ be sets of $K$ matrices which satisfy $\sum_{k=1}^{K}\mathrm{tr}_{\nu}[{{}^{t}}MP_{k}(X)MQ_{k}(X)]>0$ for any non-zero matrix $M\in\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$. The inner product $(u,v)^{\mathcal{P},\mathcal{Q}}_{X}$ of $u,v\in T^{\mu,\nu}_{X}$ is defined by $(\partial/\partial x_{p,q}|_{X},\partial/\partial x_{r,s}|_{X})^{\mathcal{P},\mathcal{Q}}_{X}=\sum_{k=1}^{K}P_{k,p,r}(X)Q_{k,s,q}(X)$, where $P_{k,i,j}(X)$ and $Q_{k,i,j}(X)$ are the $(i,j)$-th entries of $P_{k}(X)$ and $Q_{k}(X)$, respectively. Let $\|u\|^{\mathcal{P},\mathcal{Q}}_{X}$ be $\sqrt{(u,u)^{\mathcal{P},\mathcal{Q}}_{X}}$. If $\mathcal{P}=\{I_{\mu}\}$ and $\mathcal{Q}=\{I_{\nu}\}$ for any $X\in\mathrm{Mat}_{\mathbb{R}}^{\mu,\nu}$, then the inner product is said to be Euclidean. The pullback of $(\cdot,\cdot)^{\mathcal{P},\mathcal{Q}}_{f_{Y,Z}(X)}$ by $f_{Y,Z}$ is $(\cdot,\cdot)^{\mathcal{P}^{\prime},\mathcal{Q}^{\prime}}_{X}$, where $\mathcal{P}^{\prime}\subset\mathrm{Mat}_{\mathbb{R}}^{\mu}$ and $\mathcal{Q}^{\prime}\subset\mathrm{Mat}_{\mathbb{R}}^{\nu}$ consist of ${{}^{t}}YP_{k}(f_{Y,Z}(X))Y$ and ${{}^{t}}ZQ_{k}(f_{Y,Z}(X))Z$, respectively.

Let $\mathrm{SO}_{\mathbb{R}}^{n}$ be $\{e^{A}\mid A\in\mathrm{Ant}_{\mathbb{R}}^{n}\}$, the set of special orthogonal matrices. For $R\in\mathrm{SO}_{\mathbb{R}}^{n}$, let $T_{R}\subset T^{n,n}_{R}$ be the tangent space of $\mathrm{SO}_{\mathbb{R}}^{n}\subset\mathrm{Mat}_{\mathbb{R}}^{n}$ at $R$. We use $(\cdot,\cdot)^{\mathcal{P},\mathcal{Q}}_{R}$ by restricting the space $T^{n,n}_{R}$ to $T_{R}$. The dimension of $T_{R}$ is $d_{n}=n(n-1)/2$. If $C\subset T_{R}$ is a cuboid framed by edge vectors $v_{1},v_{2},...,v_{d_{n}}\in T_{R}$ which are mutually orthogonal with respect to $(\cdot,\cdot)^{\mathcal{P},\mathcal{Q}}_{R}$, then the volume $\mathrm{vol}^{\mathcal{P},\mathcal{Q}}_{R}[C]$ is $\prod_{k=1}^{d_{n}}\|v_{k}\|^{\mathcal{P},\mathcal{Q}}_{R}$. Let $\mathcal{F}_{\mathrm{SO}_{\mathbb{R}}^{n}}$ be the set of Borel subsets of $\mathrm{SO}_{\mathbb{R}}^{n}$. Let $\mu^{\mathcal{P},\mathcal{Q}}:\mathcal{F}_{\mathrm{SO}_{\mathbb{R}}^{n}}\to[0,\infty)$ be a measure on $\mathrm{SO}_{\mathbb{R}}^{n}$ given by $\mathrm{vol}^{\mathcal{P},\mathcal{Q}}_{R}[\cdot]$. For $L\in\mathrm{SO}_{\mathbb{R}}^{n}$, and for $M\subset\mathrm{SO}_{\mathbb{R}}^{n}$, let $LM\subset\mathrm{SO}_{\mathbb{R}}^{n}$ be $\{Lx\mid x\in M\}$. The following lemma implies that, if the inner product is Euclidean, then $\mu^{\mathcal{P},\mathcal{Q}}$ is a left-invariant Haar measure.

Lemma 19**.**

If $\mathcal{P}=\mathcal{Q}=\{I_{n}\}$, then, for any $L\in\mathrm{SO}_{\mathbb{R}}^{n}$, and for any $M\in\mathcal{F}_{\mathrm{SO}_{\mathbb{R}}^{n}}$ it holds that $\mu^{\mathcal{P},\mathcal{Q}}(LM)=\mu^{\mathcal{P},\mathcal{Q}}(M)$.

Proof.

Let $f_{L}:\mathrm{SO}_{\mathbb{R}}^{n}\to\mathrm{SO}_{\mathbb{R}}^{n}$ be $U\mapsto LU$, and let $df_{L}|_{U}:T_{U}\to T_{LU}$ be the pushforward. For any $U\in\mathrm{SO}_{\mathbb{R}}^{n}$ and for any $u,v\in T_{U}$, we have $(df_{L}(u),df_{L}(v))^{\{I_{n}\},\{I_{n}\}}_{LU}=(u,v)^{\{I_{n}\},\{I_{n}\}}_{U}$. Let $C\subset M$ be any infinitesimal cuboid, and let $R\in C$ be one of the vertices. Then, $C$ is identified by mutually orthogonal edge vectors $v_{1},v_{2},...,v_{d_{n}}\in T_{R}$. It holds that $\mathrm{vol}^{\mathcal{P},\mathcal{Q}}_{LR}[LC]=\mathrm{vol}^{\mathcal{P},\mathcal{Q}}_{R}[C]$. Hence, we have $\mu^{\mathcal{P},\mathcal{Q}}(LM)=\mu^{\mathcal{P},\mathcal{Q}}(M)$. ∎

Let $\mu:\mathcal{F}_{\mathrm{SO}_{\mathbb{R}}^{n}}\to[0,\infty)$ be $\mu^{\mathcal{P},\mathcal{Q}}$ with $\mathcal{P}=\mathcal{Q}=\{I_{n}\}$. For $U\in\mathrm{SO}_{\mathbb{R}}^{n}$, let \underaccent{\tilde}{\hat V}_{U}\in\mathcal{U}(\mathcal{H}^{\otimes mn}) be $\exp(\hat{v}_{\log U})$. By Lemma 7, for any $L,R\in\mathrm{SO}_{\mathbb{R}}^{n}$, it holds that \underaccent{\tilde}{\hat V}_{L}\underaccent{\tilde}{\hat V}_{R}=\underaccent{\tilde}{\hat V}_{LR}. Moreover, by Lemma 9, it holds that \underaccent{\tilde}{\hat V}_{L}\underaccent{\tilde}{\hat V}_{R}f(X)=\underaccent{\tilde}{\hat V}_{L}f(XR)=f(XLR). Hence, $\mathrm{SO}_{\mathbb{R}}^{n}$ acts on the left of $\mathcal{H}^{\otimes mn}$ by \underaccent{\tilde}{\hat V}:\mathrm{SO}_{\mathbb{R}}^{n}\to\mathcal{U}(\mathcal{H}^{\otimes mn}). Let $\hat{W}\in\mathcal{L}(\mathcal{H}^{\otimes mn})$ be \mu(\mathrm{SO}_{\mathbb{R}}^{n})^{-1}\int_{\mathrm{SO}_{\mathbb{R}}^{n}}\underaccent{\tilde}{\hat V}_{U}\mu(dU).

Lemma 20**.**

For any mode size $m\geq 1$, it holds that $\hat{K}_{0}=\hat{W}$.

Proof.

For any $f\in\mathcal{H}^{\otimes mn}$, and for any $L\in\mathrm{SO}_{\mathbb{R}}^{n}$, it holds that \underaccent{\tilde}{\hat V}_{L}\hat{W}f=\mu(\mathrm{SO}_{\mathbb{R}}^{n})^{-1}\int_{\mathrm{SO}_{\mathbb{R}}^{n}}f(XLU)\mu(dU). Let $U^{\prime}$ be $LU$. By Lemma 19, it holds that $\mu(dU)=[\mu(dU)/\mu(dU^{\prime})]\mu(dU^{\prime})=\mu(dU^{\prime})$. Hence, it holds that \underaccent{\tilde}{\hat V}_{L}\hat{W}f=\hat{W}f. By Theorem 2, we have $\hat{W}f\in\mathcal{K}_{0}$. If $g\in\mathcal{K}_{0}$, then it holds that \underaccent{\tilde}{\hat V}_{U}g=g for any $U\in\mathrm{SO}_{\mathbb{R}}^{n}$. Hence, we have $\hat{W}g=\mu(\mathrm{SO}_{\mathbb{R}}^{n})^{-1}\int_{\mathrm{SO}_{\mathbb{R}}^{n}}g\mu(dU)=g$. As a result, $\hat{W}$ is the projection on $\mathcal{K}_{0}$. ∎

Let $\mathcal{S}^{n-1}\subset\mathrm{Mat}_{\mathbb{R}}^{1,n}$ be the set of unit row vectors. For $M\subset\mathcal{S}^{n-1}$, and for $U\in\mathrm{SO}_{\mathbb{R}}^{n}$, let $MU\subset\mathcal{S}^{n-1}$ be $\{\vec{v}U\mid\vec{v}\in M\}$. Let $\mathcal{F}_{\mathcal{S}^{n-1}}$ be the set of Borel subsets of $\mathcal{S}^{n-1}$. If a measure $\lambda:\mathcal{F}_{\mathcal{S}^{n-1}}\to[0,\infty)$ satisfies $\lambda(MU)=\lambda(M)$ (${{}^{\forall}}M\in\mathcal{F}_{\mathcal{S}^{n-1}}$, ${{}^{\forall}}U\in\mathrm{SO}_{\mathbb{R}}^{n}$), then $\lambda$ is said to be rotationally uniform. For $\vec{v}\in\mathcal{S}^{n-1}$, let $T^{\mathcal{S}}_{\vec{v}}\subset T^{1,n}_{\vec{v}}$ be the tangent space of $\mathcal{S}^{n-1}\subset\mathrm{Mat}_{\mathbb{R}}^{1,n}$ at $\vec{v}$. We use the inner product $(\cdot,\cdot)^{\mathcal{P},\mathcal{Q}}_{\vec{v}}$ by restricting the space to $T^{\mathcal{S}}_{\vec{v}}$. Since elements of $\mathcal{P}$ are scalars, we assume, without loss of generality, that $\mathcal{P}=\{I_{1}\}$ and that $\mathcal{Q}=\{Q(\vec{v})\}$, where $Q(\vec{v})\in\mathrm{Sym}_{\mathbb{R}}^{n}$ is a positive matrix which may depend on $\vec{v}$. Let $(\cdot,\cdot)^{\mathcal{Q}}_{\vec{v}}=(\cdot,\cdot)^{\{I_{1}\},\{Q(\vec{v})\}}_{\vec{v}}$, and let $\|\cdot\|^{\mathcal{Q}}_{\vec{v}}=\|\cdot\|^{\{I_{1}\},\{Q(\vec{v})\}}_{\vec{v}}$. The dimension of $T^{\mathcal{S}}_{\vec{v}}$ is $n-1$. If $C\in T^{\mathcal{S}}_{\vec{v}}$ is a cuboid with edge vectors $v_{1},v_{2},...,v_{n-1}\in T^{\mathcal{S}}_{\vec{v}}$ which are mutually orthogonal with respect to $(\cdot,\cdot)^{\mathcal{Q}}_{\vec{v}}$. then the volume $\mathrm{vol}^{\mathcal{Q}}_{\vec{v}}[C]$ is $\prod_{k=1}^{n-1}\|v_{k}\|^{\mathcal{Q}}_{\vec{v}}$. Let $\lambda^{\mathcal{Q}}:\mathcal{F}_{\mathcal{S}^{n-1}}\to[0,\infty)$ be the measure given by $\mathrm{vol}^{\mathcal{Q}}_{\vec{v}}[\cdot]$.

Lemma 21**.**

If $\lambda^{\mathcal{Q}}:\mathcal{F}_{\mathcal{S}^{n-1}}\to[0,\infty)$ is rotationally uniform, then there exists $c>0$ such that, for any $\vec{v}\in\mathcal{S}^{n-1}$, $Q(\vec{v})$ is $cI_{n}$.

Proof.

For $U\in\mathrm{SO}_{\mathbb{R}}^{n}$, define $f_{U}:\mathcal{S}^{n-1}\to\mathcal{S}^{n-1}$ by $\mathcal{S}^{n-1}\ni\vec{v}\mapsto\vec{v}U^{-1}=\vec{v}({{}^{t}}U)$. For any $U\in\mathrm{SO}_{\mathbb{R}}^{n}$, for any $\vec{v}\in\mathcal{S}^{n-1}$, and for any $u,v\in T^{\mathcal{S}}_{\vec{v}}$, it holds that $(df_{U}(u),df_{U}(v))^{\{Q(\vec{v})\}}_{\vec{v}U^{-1}}=(u,v)^{\{{{}^{t}}UQ(\vec{v})U\}}_{\vec{v}}$. The group action $(U,\vec{v})\mapsto f_{U}(\vec{v})$ is transitive, that is, for any $\vec{v}\in\mathcal{S}^{n-1}$, the orbit $\{f_{U}(\vec{v})\mid U\in\mathrm{SO}_{\mathbb{R}}^{n}\}$ is $\mathcal{S}^{n-1}$. For $\vec{v}\in\mathcal{S}^{n-1}$, let $\Sigma_{\vec{v}}\subset\mathrm{SO}_{\mathbb{R}}^{n}$ be $\{U\in\mathrm{SO}_{\mathbb{R}}^{n}\mid f_{U}(\vec{v})=\vec{v}\}$, the set of stabilizers of $\vec{v}$. For any $\vec{v}\in\mathcal{S}^{n-1}$, $\Sigma_{\vec{v}}$ is isomorphic to $\mathrm{SO}_{\mathbb{R}}^{n-1}$. For $\vec{v}\in\mathcal{S}^{n-1}$, define a group representation $\rho_{\vec{v}}:\Sigma_{\vec{v}}\to\mathrm{GL}(T^{\mathcal{S}}_{\vec{v}})$ by $U\mapsto df_{U}|_{\vec{v}}$. Then, $\rho_{\vec{v}}$ is irreducible. Since $\lambda^{\mathcal{Q}}$ is rotationally uniform, it is necessary that, for any $\vec{v}\in\mathcal{S}^{n-1}$, for any $R\in\Sigma_{\vec{v}}$, and for any $u,v\in T^{\mathcal{S}}_{\vec{v}}$, $(\rho_{\vec{v}}(R)u,\rho_{\vec{v}}(R)v)^{\mathcal{Q}}_{\vec{v}}$ is $(u,v)^{\mathcal{Q}}_{\vec{v}}$, and so that, for any $\vec{v}\in\mathcal{S}^{n-1}$, and for any $R\in\Sigma_{\vec{v}}$, ${{}^{t}}RQ(\vec{v})R$ is $Q(\vec{v})$. By Schur’s lemma, there exists a function $\varphi:\mathcal{S}^{n-1}\to\mathbb{R}$ such that $Q(\vec{v})=\varphi(\vec{v})I_{n}$. (See [10].) By the positivity of the inner product, $\varphi(\vec{v})$ is positive for any $\vec{v}\in\mathcal{S}^{n-1}$. By the transitivity of $f_{U}$, $\varphi(\vec{v})$ is constant. ∎

For $\vec{v}\in\mathcal{S}^{n-1}$, define a measure $\lambda_{\vec{v}}:\mathcal{F}_{\mathcal{S}^{n-1}}\to[0,\infty)$ by \lambda_{\vec{v}}(M)=\mu\big{(}\{U\in\mathrm{SO}_{\mathbb{R}}^{n}\mid\vec{v}U^{-1}\allowbreak\in M\}\big{)}.

Lemma 22**.**

For any $\vec{v}\in\mathcal{S}^{n-1}$, $\lambda_{\vec{v}}$ is rotationally uniform.

Proof.

Choose $M\in\mathcal{F}_{\mathcal{S}^{n-1}}$ and $U\in\mathrm{SO}_{\mathbb{R}}^{n}$, arbitrarily. It holds that $\{L\in\mathrm{SO}_{\mathbb{R}}^{n}\mid\vec{v}L^{-1}\in MU\}=U^{-1}\{R\in\mathrm{SO}_{\mathbb{R}}^{n}\mid\vec{v}R^{-1}\in M\}$. By Lemma 19, we have $\lambda_{\vec{v}}(MU)=\mu(U^{-1}\{R\mid\vec{v}R^{-1}\in M\})=\mu(\{R\mid\vec{v}R^{-1}\in M\})=\lambda_{\vec{v}}(M)$. ∎

Let $\lambda_{1}:\mathcal{F}_{\mathcal{S}^{n-1}}\to[0,\infty)$ be the rotationally uniform measure with $\lambda_{1}(\mathcal{S}^{n-1})=1$. Consider the case of $m=1$, and, for $\vec{r}=(r_{1},r_{2},...,r_{n})\in\mathrm{Mat}_{\mathbb{R}}^{1,n}$, let \big{|}\vec{r}\big{\rangle}\in\mathcal{H}^{\otimes n} be $|r_{1}\rangle\otimes|r_{2}\rangle\otimes\cdots\otimes|r_{n}\rangle$.

Lemma 23**.**

If the mode size $m$ is one, then, for any $r\geq 0$, and for any $\vec{u}\in\mathcal{S}^{n-1}$, it holds that \hat{K}_{0}\big{|}r\vec{u}\big{\rangle}=\int_{\mathcal{S}^{n-1}}\big{|}r\vec{v}\big{\rangle}\lambda_{1}(d\vec{v}).

Proof.

By Lemma 20, it holds that \hat{K}_{0}\big{|}r\vec{u}\big{\rangle}=\mu(\mathrm{SO}_{\mathbb{R}}^{n})^{-1}\int_{\mathrm{SO}_{\mathbb{R}}^{n}}\underaccent{\tilde}{\hat V}_{U}\big{|}r\vec{u}\big{\rangle}\mu(dU). By Lemma 11, it holds that \underaccent{\tilde}{\hat V}_{U}\big{|}r\vec{u}\big{\rangle}=\big{|}r\vec{u}U^{-1}\big{\rangle}, and so that \hat{K}_{0}\big{|}r\vec{u}\big{\rangle}=\mu(\mathrm{SO}_{\mathbb{R}}^{n})^{-1}\int_{\mathcal{S}^{n-1}}\big{|}r\vec{v}\big{\rangle}\lambda_{\vec{u}}(d\vec{v}). By Lemma 22, we have \hat{K}_{0}\big{|}r\vec{u}\big{\rangle}=\int_{\mathcal{S}^{n-1}}\big{|}r\vec{v}\big{\rangle}\allowbreak\lambda_{1}(d\vec{v}). ∎

Let $\mathbb{T}\subset\mathbb{R}^{n-2}$ be $[0,\pi)^{n-2}$, and let $\mathcal{T}\subset\mathbb{R}^{n-1}$ be $\mathbb{T}\times[0,2\pi)$. For $k\in\{1,2,...,n-1\}$, let $t_{k}$ be the coordinate variable of the $k$-th entry of $\vec{t}\in\mathcal{T}$. We define unit row vectors $\vec{\nu}_{0},\vec{\nu}_{1},...,\vec{\nu}_{n-1}\in\mathcal{S}^{n-1}$ which are parameterized by $\vec{t}\in\mathcal{T}$, as follows. Let $\vec{\nu}_{0}\in\mathcal{S}^{n-1}$ be $(1,O_{1,n-1})$. For $k\in\{1,2,...,n-1\}$, let $\vec{\nu}_{k}\in\mathcal{S}^{n-1}$ be

[TABLE]

Define $\varphi:\mathcal{T}\to\mathcal{S}^{n-1}$ by $\varphi(\vec{t})=\vec{\nu}_{n-1}$. For $k\in\{1,2,...,n\}$, let $x_{k}$ be the coordinate variable of the $k$-th entry of $\vec{v}\in\mathcal{S}^{n-1}$. The pushforward $d\varphi(\partial/\partial t_{i}|_{\vec{t}})$ is $\sum_{j=1}^{n}(\partial x_{j}/\partial t_{i})\partial/\partial x_{j}|_{\varphi(\vec{t})}$. Let $\vec{J}_{i}\in\mathrm{Mat}_{\mathbb{R}}^{1,n}$ be a row vector whose $j$-th entry is $\partial x_{j}/\partial t_{i}$. Then, as the pullback of the Euclidean inner product for $T^{\mathcal{S}}_{\varphi(\vec{t})}$, the inner product of $\partial/\partial t_{i}|_{\vec{t}}$ and $\partial/\partial t_{j}|_{\vec{t}}$ is given by

[TABLE]

For $k\in\{1,2,...,n-1\}$, let

[TABLE]

and let $D_{k}=\partial R_{k}/\partial t_{k}\in\mathrm{Mat}_{\mathbb{R}}^{n}$. Let $R_{j,k}=R_{j}R_{j+1}\cdots R_{k}$. If $j>k$, let $R_{j,k}$ be $I_{n}$. It holds that $\vec{\nu}_{k}=\vec{\nu}_{0}R_{1,k}$ and that $\vec{J}_{i}=\vec{\nu}_{i-1}D_{i}R_{i+1,n-1}$.

Lemma 24**.**

It holds that $\|d\varphi(\partial/\partial t_{k}|_{\vec{t}})\|^{\{I_{n}\}}_{\vec{\nu}_{n-1}}=c_{k}$, where

[TABLE]

Proof.

Define $E_{k}\in\mathrm{Mat}_{\mathbb{R}}^{n}$ by

[TABLE]

It holds that $D_{k}({{}^{t}}D_{k})=E_{k}$. Since the $k$-th entry of $\vec{\nu}_{k-1}$ is $c_{k}$, it holds that $\vec{\nu}_{k-1}E_{k}({{}^{t}}\vec{\nu}_{k-1})=c_{k}^{2}$. Hence, we have $\vec{J}_{k}({{}^{t}}\vec{J}_{k})=c_{k}^{2}$. Since, for $j<k$, $t_{j}$ belongs to $[0,\pi)$, it holds that $\sin t_{j}\geq 0$, and so that $\sqrt{c_{k}^{2}}=c_{k}$. ∎

Lemma 25**.**

222This lemma was suggested by Prof. F. Tanaka.

If $j\neq k$, then $d\varphi(\partial/\partial t_{j}|_{\vec{t}})$ is orthogonal to $d\varphi(\partial/\partial t_{k}|_{\vec{t}})$ with respect to the Euclidean inner product.

Proof.

Define $A_{k}\in\mathrm{Ant}_{\mathbb{R}}^{n}$ and $F_{k}\in\mathrm{Mat}_{\mathbb{R}}^{n}$ by

[TABLE]

respectively. Then, it hoolds that $R_{k}({{}^{t}}D_{k})=A_{k}$ and that $F_{k}A_{k}F_{k}=O_{n}$. Assume that $j<k$. Let $\vec{\nu}_{j,k}=\vec{\nu}_{j-1}D_{j}R_{j+1,k}$. Then, it holds that $\vec{J}_{j}({{}^{t}}\vec{J}_{k})=\vec{\nu}_{j,k-1}A_{k}({{}^{t}}\vec{\nu}_{k-1})$, and that

[TABLE]

Hence, we have $\vec{J}_{j}({{}^{t}}\vec{J}_{k})=0$. ∎

Let $P\subset T^{\mathcal{S}}_{\vec{\nu}_{n-1}}$ be a parallelepiped framed by $\{d\varphi(\partial/\partial t_{k}|_{\vec{t}})\mid k=1,2,...,n-1\}$. Let

[TABLE]

Lemma 26**.**

*If the inner product of $T^{\mathcal{S}}_{\vec{\nu}_{n-1}}$ is Euclidean, then

(i) the volume of $P$ is $\prod_{k=1}^{n-2}\sin^{n-k-1}t_{k}$,

and

(ii) the area $|\mathcal{S}^{n-1}|$ of $\mathcal{S}^{n-1}$ is $\prod_{k=1}^{n-1}s_{k}$,*

Proof.

(i) By Lemma 25, $P$ is a cuboid with respect to the Euclidean inner product. By Lemmas 24, we have $\mathrm{vol}^{\{I_{n}\}}_{\vec{\nu}_{n-1}}[P]=\prod_{k=1}^{n-2}\sin^{n-k-1}t_{k}$.

(ii) By taking integral of (i), we obtain (ii). ∎

For $r\geq 0$, define $g_{r}\in\mathcal{H}^{\otimes n}$ by

[TABLE]

Lemma 27**.**

If the mode size $m$ is one, then for any $\vec{r}\in\mathrm{Mat}_{\mathbb{R}}^{1,n}$, it holds that \hat{K}_{0}\big{|}\vec{r}\big{\rangle}=g_{\|\vec{r}\|}.

Proof.

By Lemma 23, we have \hat{K}_{0}\big{|}\vec{r}\big{\rangle}=\int_{\mathcal{S}^{n-1}}\big{|}\|\vec{r}\|\vec{v}\big{\rangle}\lambda_{1}(d\vec{v}). By Lemma 21, the integration by $\lambda_{1}$ is calculated by the Euclidean inner product. By Lemma 26, we have \hat{K}_{0}\big{|}\vec{r}\big{\rangle}=g_{\|\vec{r}\|}.

∎

Lemma 28**.**

Assume that $m=1$. For $\vec{r}\in\mathrm{Mat}_{\mathbb{R}}^{1,n}$, it holds that

[TABLE]

Proof.

By Eq. $(\ref{eq_inner_product})$, for any $r\geq 0$, it holds that

[TABLE]

By Lemma 27, it holds that

[TABLE]

Calculating the beta integration $s_{1}=\int_{0}^{\pi}\sin^{n-2}tdt$, we obtain (25). ∎

Proof of Theorem 4.

Since the POVM $\{\hat{\Pi}_{0},\hat{\Pi}_{1}\}$ of $T^{\mathrm{SI}}_{\alpha}$ is given by $\hat{\Pi}_{0}=(1-\alpha)\hat{K}_{0}$, we have $\beta_{\hat{\rho}_{\vec{\theta},0}^{\otimes n}}[T^{\mathrm{SI}}_{\alpha}]=(1-\alpha)\beta_{\hat{\rho}_{\vec{\theta},0}^{\otimes n}}[T^{\mathrm{SI}}_{0}]$. Let $\vec{u}_{m}\in\mathrm{Mat}_{\mathbb{R}}^{m,1}$ be the unit vector whose first entry is one. For $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, let $A\in\mathrm{Ant}_{\mathbb{C}}^{n}$ be a solution to the equation $e^{A}\vec{\theta}=\|\vec{\theta}\|\vec{u}_{m}$. Let $E_{m}\in\mathrm{Mat}_{\mathbb{R}}^{m,n}$ be $(\overbrace{\vec{u}_{m},\vec{u}_{m},...,\vec{u}_{m}}^{n})$. By Lemma 12, it holds that $\hat{U}_{A}\hat{\rho}_{\vec{\theta},0}^{\otimes n}\hat{U}_{A}^{*}=\hat{\rho}_{\|\vec{\theta}\|E_{m},0}$. Let $\hat{K}^{\prime}_{0}\in\mathcal{L}(\mathcal{H}^{\otimes n})$ be $\hat{K}_{0}$ for $m=1$. Then, it holds that $\mathrm{Tr}[\hat{\rho}_{\|\vec{\theta}\|E_{m},0}\hat{K}_{0}]=\mathrm{Tr}[\hat{\rho}_{\|\vec{\theta}\|E_{1},0}\hat{K}^{\prime}_{0}]\mathrm{Tr}[\hat{\rho}_{0E_{1},0}\hat{K}^{\prime}_{0}]^{m-1}$. By Lemma 28, we have the statement. ∎

5.9 Proof of Theorem 5 ($T^{\mathrm{HH}}_{\alpha}$ is not SI)

By Eq. (3.12) of Leonhardt [8], the Fourier transform of the Wigner function of $\hat{\rho}\in\mathcal{S}(\mathcal{H})$ is $F_{\hat{\rho}}(u,v)=\mathrm{Tr}[\hat{\rho}\exp(-iu\hat{q}-iv\hat{p})]$, where $i=\sqrt{-1}$, $\hat{q}=(\hat{a}+\hat{a}^{*})/\sqrt{2}$ and $\hat{p}=-i(\hat{a}-\hat{a}^{*})/\sqrt{2}$. For $k\in\{1,2,...,m\}$, let $\hat{q}_{k}=(\hat{a}_{k}+\hat{a}_{k}^{*})/\sqrt{2}\in\mathcal{L}(\mathcal{H}^{\otimes m})$ and let $\hat{p}_{k}=-i(\hat{a}_{k}-\hat{a}_{k}^{*})/\sqrt{2}\in\mathcal{L}(\mathcal{H}^{\otimes m})$. Let

[TABLE]

Let

[TABLE]

The Fourier transform of the Wigner function of $\hat{\rho}\in\mathcal{S}(\mathcal{H}^{\otimes m})$ is $F_{\hat{\rho}}=\mathrm{Tr}[\hat{\rho}\exp(-i({{}^{t}}\vec{w})\vec{r})]$. Let $F_{\vec{\theta},\eta,N}(\vec{u},\vec{v})=F_{\hat{\rho}_{\vec{\theta},\eta,N}}(\vec{u},\vec{v})$. We use $\vec{\mu}_{\vec{\theta}}$ and $G_{\eta}$ defined in (8).

Lemma 29**.**

It holds that

[TABLE]

Proof.

First, define $L\in\mathrm{Mat}_{\mathbb{C}}^{2m}$ and \underaccent{\tilde}{\vec a}\in\mathrm{Mat}_{m,1}^{2m,1} by

[TABLE]

respectively. Let $\vec{z}=-i(\vec{u}+i\vec{v})/\sqrt{2}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, and let \underaccent{\tilde}{\vec z}={{}^{t}}({{}^{t}}\vec{z},\vec{z}^{*})\in\mathrm{Mat}_{\mathbb{C}}^{2m,1}. Let \hat{D}_{\vec{z}}=\exp(\underaccent{\tilde}{\vec a}^{*}K_{m}\underaccent{\tilde}{\vec z})\in\mathcal{U}(\mathcal{H}^{\otimes m}). Then, we have

[TABLE]

By Eq. (19), we have \hat{D}_{\vec{z}}\big{|}\vec{0}_{m}\big{\rangle}=\big{|}\vec{z}\big{\rangle}. By Eq. (21), we have \big{\langle}\vec{0}_{m}\big{|}\vec{z}\big{\rangle}=\exp(-\|\vec{u}\|^{2}/4-\|\vec{v}\|^{2}/4). Hence, we obtain

[TABLE]

Next, we have $L^{*}e^{\eta}L=\exp(L^{*}\eta L)=G_{\eta}$. Moreover, it holds that F_{\vec{\theta},\eta,0}=\big{\langle}\vec{0}_{m}\big{|}\hat{D}_{\vec{\theta}}^{*}\hat{S}_{\eta}^{*}\exp(-i({{}^{t}}\vec{w})L^{*}\underaccent{\tilde}{\vec a})\hat{S}_{\eta}\hat{D}_{\vec{\theta}}\big{|}\vec{0}_{m}\big{\rangle}. By Lemma 1, it holds that \hat{S}_{\eta}^{*}\underaccent{\tilde}{\vec a}\hat{S}_{\eta}=e^{\eta}\underaccent{\tilde}{\vec a}. By Eq. (18), it holds that \hat{D}_{\vec{\theta}}^{*}\underaccent{\tilde}{\vec a}\hat{D}_{\vec{\theta}}=\underaccent{\tilde}{\vec a}+\sqrt{2}L\vec{\mu}_{\vec{\theta}}\hat{I}^{\otimes m}. Hence, we have

[TABLE]

By (26) and (27), we obtain

[TABLE]

Next, for $\vec{x}={{}^{t}}(x_{1},x_{2},...,x_{m})\in\mathrm{Mat}_{\mathbb{R}}^{m,1}$ and for $\vec{y}={{}^{t}}(y_{1},y_{2},...,y_{m})\in\mathrm{Mat}_{\mathbb{R}}^{m,1}$, let $g_{\vec{\theta},N}(\vec{x},\vec{y})=(\pi N)^{-1}e^{-\|\vec{x}+i\vec{y}-\vec{\theta}\|^{2}/N}$. Then, it holds that

[TABLE]

By (28), the integrated factor of (29) is the Fourier transform of $g_{\vec{\theta},N}(\vec{x},\vec{y})$. Hence, we obtain the statement. ∎

We use $\Sigma_{\eta,N}$ defined in (9).

Lemma 30**.**

It holds that

[TABLE]

Proof.

Let $W_{\hat{\rho}}(x,y)$ be the Wigner function of $\hat{\rho}\in\mathcal{S}(\mathcal{H})$. By the overlap formula, it holds that $\mathrm{Tr}[\hat{\rho}\hat{\sigma}]=2\pi\iint_{\mathbb{R}^{2}}W_{\hat{\rho}}(x,y)W_{\hat{\sigma}}(x,y)dxdy$. (See (3.22) of [8].) Moreover, by the Parseval’s formula, it holds that $\mathrm{Tr}[\hat{\rho}\hat{\sigma}]=(2\pi)^{-1}\iint_{\mathbb{R}^{2}}\overline{F_{\hat{\rho}}(u,v)}F_{\hat{\sigma}}(u,v)dudv$. Let $\vec{\nu}=\vec{\mu}_{\vec{z}}-G_{\eta}\vec{\mu}_{\vec{\theta}}$. By Lemma 29, we have

[TABLE]

Let $\vec{w}^{\prime}=\vec{w}+i\Sigma_{\eta,N}^{-1}\vec{\nu}/\sqrt{2}$. Then, the argument of the exponential function of (30) is equal to $-({{}^{t}}\vec{w}^{\prime})\Sigma_{\eta,N}\vec{w}^{\prime}-({{}^{t}}\vec{\nu})\Sigma_{\eta,N}^{-1}\vec{\nu}/2$. Calculating the Gausian intagration and multiplying $\pi^{-m}$, we obtain the statement. ∎

Let $\kappa={{}^{t}}\vec{\mu}_{\vec{\theta},\eta}\Sigma_{\eta,N}^{-1}\vec{\mu}_{\vec{\theta},\eta}$.

Lemma 31**.**

For any $r>0$, for any $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}\backslash\{\vec{0}_{m}\}$ and for any $N\geq 0$, there exists $\eta\in\mathrm{Sqz}^{m}$ such that

[TABLE]

Proof.

Let $\zeta\in\mathrm{Sqz}^{m}$ be $\log(r)\begin{pmatrix}O_{m}&I_{m}\\ I_{m}&O_{m}\end{pmatrix}$. Then, $G_{\zeta}$ is $\begin{pmatrix}rI_{m}&O_{m}\\ O_{m}&r^{-1}I_{m}\end{pmatrix}$. Let $\vec{u}\in\mathrm{Mat}_{\mathbb{R}}^{m,1}$ be any real column vector. Then, it holds that $\vec{\mu}_{\vec{u}}=\vec{\mu}_{\vec{u},O_{2m}}=\begin{pmatrix}\vec{u}\\ \vec{0}_{m}\end{pmatrix}$, and that

[TABLE]

Let $A\in\mathrm{Ant}_{\mathbb{C}}^{m}$ be a diagonal matrix satisfying $e^{-A}\vec{\theta}\in\mathrm{Mat}_{\mathbb{R}}^{m,1}$. Let $\vec{s}=e^{-A}\vec{\theta}$. Let $U=\begin{pmatrix}e^{A}&O_{m}\\ O_{m}&e^{-A}\end{pmatrix}$. Let $\eta\in\mathrm{Sqz}^{m}$ be $U\zeta U^{*}$. There exists an orthogonal matrix $R\in\mathrm{Mat}_{\mathbb{R}}^{2m}$ such that $G_{\eta}=RG_{\zeta}({{}^{t}}R)$ and ${{}^{t}}R\vec{\mu}_{\vec{\theta}}=\begin{pmatrix}\vec{s}\\ \vec{0}_{m}\end{pmatrix}$. Hence, it holds that

[TABLE]

Hence, we obtain (31). ∎

Let $\lambda$ be $n\kappa$. Let $p_{\lambda}(f)$ be the probability density function of $F_{\mu,\nu;\lambda}$ defined in (10). Let $c$ be the critical point of level $\alpha$, that is, the solution to $\int_{c}^{\infty}p_{0}(f)df=\alpha$.

Lemma 32**.**

For any $m\geq 1$, for any $n\geq 2m+1$, for any $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, for any $\eta\in\mathrm{Sqz}^{m}$, for any $N\geq 0$ and for any $\alpha\in(0,1)$, there exists $\delta>0$ such that $\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]=1-\alpha-(1-\alpha-\delta)\lambda/2+o(\lambda)$ holds as $\lambda$ goes to zero.

Proof.

Let q(f)=\partial p_{\lambda}(f)/\partial\lambda\big{|}_{\lambda=0}. It holds that

[TABLE]

Let $\delta=(\mu+\nu)\int_{0}^{c}(\mu f+\nu)^{-1}fp_{0}(f)df$. Then, it holds that $\int_{0}^{c}q(f)df=(\delta-1+\alpha)/2$. Hence, we have $\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]=1-\alpha+(\delta-1+\alpha)\lambda/2+o(\lambda)$ as $\lambda\to 0$. ∎

Proof of Theorem 5.

(i) By Lemma 31, $\lambda$ depends on $\eta$. By Lemma 32, $\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]=\int_{0}^{c}p_{\lambda}(f)df$ depends on $\eta$.

(ii) By Eq. (10), $\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]=\int_{0}^{c}p_{\lambda}(f)df$ is greater than $e^{-\lambda/2}(1-\alpha)$. By Lemma 31, for any $\varepsilon\in\mathbb{R}$ with $0<\varepsilon<1-\alpha$, there exists $\eta\in\mathrm{Sqz}^{m}$ such that ${{}^{t}}\vec{\mu}_{\vec{\theta},\eta}\Sigma_{\eta,N}^{-1}\vec{\mu}_{\vec{\theta},\eta}<2\log[(1-\alpha)/(1-\alpha-\varepsilon)]$, that is, $\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]>1-\alpha-\varepsilon$. Hence, we have $\sup_{\eta\in\mathrm{Sqz}^{m}}\beta_{\hat{\rho}_{\vec{\theta},\eta,N}^{\otimes n}}[T^{\mathrm{HH}}_{\alpha}]=1-\alpha$. ∎

5.10 Proof of Theorem 6(Comparison of $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]$ with $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]$)

We compare orders of type II error probabilities for $\|\vec{\theta}\|\fallingdotseq 0$ and for $\|\vec{\theta}\|\gg 0$.

Lemma 33**.**

For any $m\geq 1$, for any $n\geq 2$ and for any $\alpha\in[0,1)$, it holds that $\beta_{\hat{\rho}_{\vec{\theta},0}^{\otimes n}}[T^{\mathrm{SI}}_{\alpha}]=1-\alpha-(1-\alpha)n\|\vec{\theta}\|^{2}+o(\|\vec{\theta}\|^{2})$ as $\|\vec{\theta}\|^{2}$ goes to zero.

Proof.

Let $f(r)=e^{-nr}\int_{0}^{\pi}e^{nr\cos\varphi}(\sin\varphi)^{n-2}d\varphi$. It holds that

[TABLE]

and that $f(0)=B((n-1)/2,1/2)$. By Theorem 4, it holds that $\beta_{\hat{\rho}_{\vec{\theta},0}^{\otimes n}}[T^{\mathrm{SI}}_{\alpha}]=(1-\alpha)f(\|\vec{\theta}\|^{2})/B((n-1)/2,1/2)$. Hence, we have $\beta_{\hat{\rho}_{\vec{\theta},0}^{\otimes n}}[T^{\mathrm{SI}}_{\alpha}]-1+\alpha=-(1-\alpha)n\|\vec{\theta}\|^{2}+o(\|\vec{\theta}\|^{2})$ as $\|\vec{\theta}\|^{2}\to 0$. ∎

Lemma 34**.**

If $m=1$, $n=3$ and $N=0$, then, for any $\alpha\in[0,1)$, it holds that $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]=O(|\theta|^{-2})$ as $|\theta|$ goes to infinity.

Proof.

Let $r=3|\theta|^{2}$, let $s=e^{-r}/B(1,1/2)$, and let $t=\int_{0}^{\pi}e^{r\cos\varphi}\sin\varphi d\varphi$. By Theorem 4, it holds that $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]=(1-\alpha)st$. We can calculate $t$ as

[TABLE]

Hence, we have $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]=(1-\alpha)(1-e^{-2r})/[rB(1,1/2)]=O(|\theta|^{-2})$ as $\|\vec{\theta}\|^{2}\to\infty$. ∎

Let $\lambda=n({{}^{t}}\vec{\mu}_{\vec{\theta},\eta})\Sigma_{\eta,N}^{-1}\vec{\mu}_{\vec{\theta},\eta}$. Let $p_{\lambda}(f)$ be the probability density function of $f$ given in (10).

Lemma 35**.**

For any $m\geq 1$, for any $n\geq 2m+1$, for any $\vec{\theta}\in\mathrm{Mat}_{\mathbb{C}}^{m,1}$, for any $\eta\in\mathrm{Sqz}^{m}$, for any $N\geq 0$ and for any $c>0$, there exists $t>0$ such that $\int_{0}^{c}p_{\lambda}(f)df=o(e^{-t\lambda})$ as $\lambda$ goes to infinity.

Proof.

Let $A_{0}=\int_{0}^{c}p_{\lambda}(f)df$. Replace $\mu f/(\mu f+\nu)$ by $x$. The Jacobian of this replacement is $df/dx=(dx/df)^{-1}=x^{-1}(1-x)^{-1}f$. Let $b_{0}=\mu c/(\nu+\mu c)$. Let

[TABLE]

It holds that $A_{0}=\int_{0}^{b_{0}}\sum_{k=0}^{\infty}q_{k}r_{k}(x)dx$. Choose $b_{1}\in(b_{0},1)$ arbitrarily. There exists $k_{0}\in\mathbb{N}$ such that, for any $k\geq k_{0}$ and for any $x\in(0,b_{0})$,

[TABLE]

holds. Let $R_{k}=\int_{0}^{b_{0}}r_{k}(x)dx$. By (32), for any $k\geq k_{0}$, it holds that $R_{k+1}<b_{1}R_{k}$. Let $A_{1}=\sum_{k=0}^{k_{0}-1}q_{k}R_{k}$, and let $A_{2}=A_{0}-A_{1}=\sum_{k=k_{0}}^{\infty}q_{k}R_{k}$. It holds that

[TABLE]

For any $s>0$, it holds that

[TABLE]

Let $t=2^{-1}\min\{(1-b_{1})/2,(1-e^{-s})/2\}$. By (33) and (34), we have $A_{0}=o(e^{-t\lambda})$ as $\lambda\to\infty$. ∎

Proof of Theorem 6.

(i) By Lemmas 33 and 32, there exists $\delta>0$ such that $\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{HH}}_{\alpha}]-\beta_{\hat{\rho}_{\theta,0}^{\otimes 3}}[T^{\mathrm{SI}}_{\alpha}]=3\delta|\theta|^{2}+o(|\theta|^{2})$ holds as $|\theta|\to 0$.

(ii) By Lemmas 34 and 35, we obtain the statement. ∎

Acknowledgements

The author thanks Prof. Keiji Matsumoto of NII and Prof. Fuyuhiko Tanaka of Osaka University for comments and suggestions.