2021-06-29

Is this estimate well-known?

Let $H$ be a separable Hilbert space and $\|\cdot\|$ represents the norm induced by the inner product $\langle\cdot,\cdot\rangle$ of $H$ , and let $A$ be a self-adjoint operator on $H$ bounded from below and $D(A)$ denotes the domain of $A$ .
Suppose $\lambda_0\in\mathbb{R}$ is an eigenvalue with multiplicity 1 attaining the value $\inf_{\|x\|=1}\langle Ax,x\rangle$ and $e_0\in H$ is the corresponding eigenvector with $\|e_0\|=1$ .

Proposition. If $\lambda_0$ is an isolated point of the spectrum $\sigma(A)$ and $|\langle x, e_0\rangle - 1/2|\leq 1/2$ , $x\in D(A)$ with $\|x\|=1$ , then
\begin{align*}
\|x-e_0\|^2\leq 2\cdot \frac{\langle Ax,x\rangle - \lambda_0}{\lambda_1-\lambda_0},
\end{align*}
where $\lambda_1=\inf \{\lambda\in\sigma(A)\mid \lambda>\lambda_0\}$ .

Proof. Let $A=\int_{\mathbb{R}}\lambda dE(\lambda)$ be the spectral decomposition of $A$ . Then, for $x\in D(A)$ with $\|x\|=1$ ,
\begin{align*}
\langle Ax,x\rangle - \lambda_0
&= \int \lambda d\langle E(\lambda)x,x\rangle - \lambda_0 \\
&= \lambda_0(|\langle x,e_0\rangle|^2-1) + \int_{\lambda>\lambda_0}\lambda d\langle E(\lambda)x,x\rangle \\
&\geq \lambda_0(|\langle x,e_0\rangle|^2-1) + \lambda_1 \int_{\lambda>\lambda_0}d\langle E(\lambda)x,x\rangle \\
&= \lambda_0(|\langle x,e_0\rangle|^2-1) + \lambda_1(1-|\langle x,e_0\rangle|^2) \\
&= (\lambda_1-\lambda_0)(1-|\langle x,e_0\rangle|^2).
\end{align*}
Since $|\langle x, e_0\rangle - 1/2|\leq 1/2$ implies $|\langle x,e_0\rangle|^2\leq \textrm{Re}\,\langle x,e_0\rangle$ , we obtain
\begin{align*}
\|x-e_0\|^2
&=2(1-\textrm{Re}\,\langle x,e_0\rangle) \\
&\leq 2(1-|\langle x,e_0\rangle|^2) \\
&\leq 2\cdot \frac{\langle Ax,x\rangle - \lambda_0}{\lambda_1-\lambda_0}.
\end{align*}
It completes the proof. $\Box$

2020-06-11

Vespa mandarinia japonica (Japanese giant hornet) - オオスズメバチ

りかえいご

りかのおべんきょうでオオスズメバチを2ひきかい始めました。
どちらも女王で、とてもかわいいです。
たいせつにしたいとおもいます。

I started to keep 2 Japanese giant hornets (Vespa mandarinia japonica), called "Osuzumebachi" in Japanese.
They are queens. So cute!
I will take good care of them.

f:id:megu0210:20200611231436j:plain:w400

f:id:megu0210:20200611231801j:plain:w400

2020-06-05

Is the conjugate of a dense and continuous embedding a dense and continuous one as well?

えいごさんすう

I make a note about the title because it took a hard time finding a proof.
This topic relates to an abstract Wiener space in the Malliavin calculus.

Proposition. Let $B$ be a real separable Banach space and $H$ be a real separable Hilbert space embedded densely and continuously to $B$ . Then, $B^*$ is densely and continuously embedded to $H^*$ .

Proof. Let $\iota$ be a continuous linear injection from $H$ to $B$ such that $\iota(H)$ is dense in $B$ . In order to see that $\iota^*(B^*)$ is dense in $H^*$ , it suffices to show that for $h\in H^{**}$ , $h|_{\iota^*(B^*)}=0$ implies $h=0$ , where $\iota^*$ represents the conjugate of $\iota$ .
In what follows, $\langle\cdot,\cdot\rangle_{H}$ represents the inner product of a (general) Hilbert space $H$ and the natural bilinear form between a (general) Banach space $B$ and its conjugate $B^*$ is denoted by ${}_B\langle \cdot,\cdot \rangle_{B^*}$ . For simplicity, the conjugate space $H^*$ is identified with $H$ by the Riesz representation theorem, and corresponding elements are expressed as the same symbols.
Suppose $h\in H^{**}$ and ${}_{H^*}\langle\iota^*\ell,h\rangle_{H^{**}}=0$ for any $\ell\in B^*$ . Since ${}_{H^*}\langle\iota^*\ell,h\rangle_{H^{**}}=\langle h, \iota^*\ell \rangle_{H^*}={}_{B}\langle \iota h, \ell \rangle_{B^*}$ , the assumption ${}_{H^*}\langle\iota^*\ell,h\rangle_{H^{**}}=0$ implies
\begin{align*}
& {}_{B}\langle \iota h, \ell \rangle_{B^*} = 0. \tag{1}
\end{align*}
Now we set the continuous linear functional $\tilde{\ell}_h$ defined on $\iota(H)$ as follows:
\begin{align*}
\iota(H)\ni\iota g \mapsto \langle g, h \rangle_{H^*}\in\mathbb{R}.
\end{align*}
Since $\iota(H)$ is dense in $B$ , for any $x\in B$ , there is a sequence $(g_n)_n$ in $H$ such that $\iota g_n \to x$ as $n\to \infty$ in $B$ . Thus, we get $\ell_h\in B^*$ , by defining, for $x\in B$ , ${}_B\langle x,\ell_h\rangle_{B^*}:=\lim_{n\to\infty}\tilde{\ell}_h(\iota g_n)$ . We note that the definition of $\ell_h$ is independent of the manner to choose the sequence $(g_n)_n$ . Then, putting $\ell=\ell_h$ in (1) yields $h=0$ . $\Box$

2020-05-12

Solving Linear Equations via VQE (Variational Quantum Eigensolver)

じょうほうさんすうえいご

In this article, we introduce how to solve simultaneous linear equations with NISQ, which is the abbreviation for Noisy Intermediate Scaled Quantum, from Xu et al. [1].

We consider the following linear equation:
\begin{align*}
Ax=b, \tag{1}
\end{align*}
where an invertible matrix $A\in \mathbb{C}^{n\times n}$ , a vector $b\in\mathbb{C}^n$ , and an unknown variable $x\in\mathbb{C}^n$ .

For calculation on quantum computers, let $n$ be a power of 2 and assume $\|b\|=1$ , where $\|\cdot\|$ represents the norm of $\mathbb{C}^n$ induced by the usual inner product $\langle\cdot,\cdot\rangle$ of $\mathbb{C}^n$ .
We note that $\langle\cdot,\cdot\rangle$ is antilinear with respect to the right side in our notation.

We now put
\begin{align*}
H = A^*(I - b\hat{\otimes}b)A,
\end{align*}
where $\hat{\otimes}$ represents the Schatten form; that is, for $x,y\in\mathbb{C}^n$ ,
\begin{align*}
x\hat{\otimes}y\colon \mathbb{C}^n\ni z\mapsto \langle z,y\rangle x \in \mathbb{C}^n,
\end{align*}
which corresponds to the "ket-bra" $|x\rangle\langle y|$ in the quantum mechanics/information.

Proposition. $H$ defined as above is Hermitian and positive semidefinite, and has a minimal eigenvalue 0. Furthermore, the solution to equation (1) is an eigenvector corresponding to eigenvalue 0 of $H$ .

Proof. It is obvious that $H$ is Hermitian. We see the remaining assertion since $I-b\hat{\otimes}b$ is a projection to the orthogonal complement of the subspace spanned by $b$ . $\Box$

Let $e\in \mathbb{C}^n$ be an initial quantum state and $U(\theta)\in\mathbb{C}^{n\times n}$ , $\theta\in\mathbb{R}^d$ be a parametrized unitary operator called "ansatz" and denote $u(\theta)=U(\theta)e$ .
Then, by minimizing
\begin{align*}
L(\theta):= \langle Hu(\theta),u(\theta)\rangle
\end{align*}
with respect to $\theta$ , we get an approximated solution to (1) on a quantum computer.

References

[1] X. Xu et al., Variational algorithms for linear algebra, arXiv:1909.03898, 2019.

2020-04-27

Decomposition of matrices with the Pauli matrices

じょうほうさんすうえいご

りょうしじょうほうで出てきたので整理するよ～

In this article, we introduce that any square matrix can be decomposed as the linear combination of the Pauli matrices.

Denote by $M_n(\mathbb{C})$ the set of all square matrices of order $n\in\mathbb{N}$ whose components value on $\mathbb{C}$ .
Furthermore, we write the 3 Pauli matrices as
\begin{align*}
X =
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}, \quad
Y =
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}, \quad
Z =
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}.
\end{align*}
The following assertion holds obviously:
Proposition. $M_2(\mathbb{C})$ is a Hilbert space under the inner product; for $A,B\in M_2(\mathbb{C})$ ,
\begin{align*}
\langle A, B \rangle_{M_2(\mathbb{C})} = \frac{1}{2} \textrm{Tr}(B^*A) = \frac{1}{2}\sum_{j=1}^2 \langle Ae_j, Be_j\rangle_{\mathbb{C}^2},
\end{align*}
where $\{e_1,e_2\}$ is an orthonormal basis of $\mathbb{C}^2$ . Moreover, $\{I,X,Y,Z\}$ forms an orthonormal basis of $M_2(\mathbb{C})$ .

Consequently, we get the following:
Theorem. Let $n\in\mathbb{N}$ . Then, $\mathcal{B}=\{\bigotimes_{j=1}^n\sigma_j \mid \sigma_j=I,X,Y,Z,~j=1,2,\dots,n\}$ forms an orthonormal basis of the tensor product Hilbert space $\bigotimes^n M_2(\mathbb{C}) \simeq M_{2^n}(\mathbb{C})$ . The members $\bigotimes_{j=1}^n\sigma_j \in \mathcal{B}$ are called $n$ -qubit Pauli matrices.

Thus, for any $A\in\bigotimes^n M_2(\mathbb{C})$ , we get the decomposition $A=\sum_{\sigma\in\mathcal{B}}\langle A, \sigma\rangle\sigma$ , where the inner product represents the one of $\bigotimes^n M_2(\mathbb{C})$ .

2019-05-19

多次元正規分布の条件付き分布

さんすう

ガウス過程に出てくる計算で使う式をメモしとくよ！

$X$ を $n$ 次元正規分布 $N(\mu_n, V_n)$ に従う確率変数とし、 $Y$ を $\mathbb{R}^m$ -値確率変数とする。また、確率変数 $(X,Y)$ は $n+m$ 次元正規分布 $N(\mu, V)$ に従うとする。ただし、
\begin{align*}
\mu =
\begin{pmatrix}
\mu_n \\
\mu_m
\end{pmatrix}, \quad
V =
\begin{pmatrix}
V_n & V_{nm} \\
{}^tV_{nm} & V_m
\end{pmatrix}
\end{align*}
である。このとき、 $X=x$ に関する $Y$ の条件付き分布は
\begin{multline*}
P(Y\in dy|X=x) \\[.5em]
= (2\pi)^{-\frac{m}{2}} (\det \tilde{V}_m)^{-\frac{1}{2}} \exp \left(-\frac{(y-\tilde{\mu}_m)\cdot \tilde{V}_m^{-1}(y-\tilde{\mu}_m)}{2} \right) dy, \\
\mbox{$P_{X}$-a.e.$\,x$}
\end{multline*}
である。ただし、 $\tilde{\mu}_m:=\mu_m+{}^tV_{nm}V_n^{-1}(x-\mu_n)$ , $\tilde{V}_m:=V_m-{}^tV_{nm}V_n^{-1}V_{nm}$ である。

証明. $\det V = \det V_n \cdot \det(V_m - {}^tV_{nm}V_n^{-1}V_{nm})$ と
\begin{align*}
V^{-1} =
\begin{pmatrix}
V_n^{-1} + V_n^{-1}V_{nm}\tilde{V}_m^{-1}{}^tV_{nm}V_n^{-1} & -V_n^{-1}V_{nm}\tilde{V}_m^{-1} \\
-\tilde{V}_m^{-1}{}^tV_{nm}V_n^{-1} & \tilde{V}_m^{-1}
\end{pmatrix}
\end{align*}
を用いる。 $A\in \mathcal{B}(\mathbb{R}^m)$ , $B\in \mathcal{B}(\mathbb{R}^n)$ に対し、
\begin{align*}
& E[\boldsymbol{1}_A(Y) \boldsymbol{1}_B(X)] \\
&= \int dx~\boldsymbol{1}_B(x) (2\pi)^{-\frac{n}{2}} (\det V_n)^{-\frac{1}{2}} \exp\left( -\frac{(x-\mu_n)\cdot V_n^{-1}(x-\mu_n)}{2} \right) \\
& \qquad \times \int dy ~\boldsymbol{1}_A(y)(2\pi)^{-\frac{m}{2}} (\det \tilde{V}_m)^{-\frac{1}{2}} \exp\left( -\frac{(y-\tilde{\mu}_m)\cdot \tilde{V}_m^{-1}(y-\tilde{\mu}_m)}{2} \right) \\
&= \int_B P_X(dx) \int_A dy ~(2\pi)^{-\frac{m}{2}} (\det \tilde{V}_m)^{-\frac{1}{2}} \exp\left( -\frac{(y-\tilde{\mu}_m)\cdot \tilde{V}_m^{-1}(y-\tilde{\mu}_m)}{2} \right)
\end{align*}
となる。□

2019-04-22

マルコフ性のためのおべんきょう

さんすうえいご

In this article, some equalities are introduced for showing the Markov property of stochastic processes. After then, we demonstrate the Markov property of a geometric Brownian motion.

In what follows, $(\Omega,\mathcal{F},P)$ represents a probability space and the expectation under $P$ is denoted by $E[\cdot]$ .
Proposition. Let $(\Omega_1,\mathcal{B}_1)$ and $(\Omega_2,\mathcal{B}_2)$ be measurable spaces, $f\colon \Omega_1\times\Omega_2\to \mathbb{R}$ be a bounded measurable function, and $\mathcal{G}$ be a sub $\sigma$ -field of $\mathcal{F}$ . If an $\Omega_1$ -valued r.v. $X$ is $\mathcal{G}$ -measurable and an $\Omega_2$ -valued r.v. $Y$ is independent of $\mathcal{G}$ , we have the following:
(1) $E[f(X,Y)|\mathcal{G}]=E[f(X,Y)|X]=E[f(x,Y)]\big|_{x=X}$ , a.s.
(2) $E[f(X,Y)|X=x]=E[f(x,Y)]$ , $P_X$ -a.s.

Proof. Recalling the construction of integration in the general measure theory, we may assume that $f$ is the form $\boldsymbol{1}_\Gamma$ for $\Gamma\in\mathcal{B_1}\otimes \mathcal{B_2}$ . Furthermore, the Dynkin's $\pi\text{-}\lambda$ theorem allows $f(x,y)$ to be the form $\boldsymbol{1}_{A}(x)\boldsymbol{1}_B(y)$ for $A\in\mathcal{B_1}$ , $B\in\mathcal{B_2}$ .
(1) Since $X$ is $\mathcal{G}$ -measurable and $Y$ is independent of $\mathcal{G}$ , we see
\begin{align*}
E[\boldsymbol{1}_A(X)\boldsymbol{1}_B(Y)|\mathcal{G}]
&= \boldsymbol{1}_A(X)E[\boldsymbol{1}_B(Y)] \\
&= E[\boldsymbol{1}_A(x)\boldsymbol{1}_B(Y)]\big|_{x=X}, \quad \text{a.s.}
\end{align*}
As for $E[f(X,Y)|X]$ , it is followed by the same argument since $X$ is $\sigma(X)$ -measurable and $Y$ is independent of $\sigma(X)$ .

(2) Suppose $C\in\mathcal{B}_1$ . Then, by the independence between $X$ and $Y$ , we see that
\begin{align*}
E[\boldsymbol{1}_A(X)\boldsymbol{1}_B(Y);X\in C]
&= E[\boldsymbol{1}_B(Y)]E[\boldsymbol{1}_A(X);X\in C] \\
&= \int_{C}P_X(dx) \boldsymbol{1}_A(x) E[\boldsymbol{1}_B(Y)].
\end{align*}
Thus, we obtain
\begin{align*}
E[\boldsymbol{1}_A(X)\boldsymbol{1}_B(Y)|X=x]
&= E[\boldsymbol{1}_A(x)\boldsymbol{1}_B(Y)], \quad \text{$P_X$-a.s.}
\end{align*}
We now complete the proof. □

Remark 1. We note that how to apply the Dynkin's $\pi\text{-}\lambda$ theorem in the proposition.
It is easy to see that
\begin{align*}
\mathcal{P} := \{A\times B~;~ A\in\mathcal{B}_1, B\in\mathcal{B}_2\}
\end{align*}
is a $\pi$ -system in $\Omega_1\times \Omega_2$ ; that is, $\Omega_1\times \Omega_2\in\mathcal{P}$ and it holds $A \cap B \in \mathcal{P}$ for any $A,B\in\mathcal{P}$ .
Denote by $\mathcal{L}$ the collection of $A\in\mathcal{B_1}\otimes\mathcal{B}_2$ satisfying
\begin{align*}
E[\boldsymbol{1}_A(X,Y)|\mathcal{G}]
&= E[\boldsymbol{1}_A(x,Y)]\big|_{x=X}.
\end{align*}
Then, $\mathcal{L}$ is a $\lambda$ -system; that is, all of the followings are fulfilled:

$\Omega_1\times\Omega_2\in\mathcal{L}$ .
$B\setminus A\in\mathcal{L}$ for any $A\subset B$ , $A,B\in\mathcal{L}$ .
$\bigcup_{n=1}^\infty A_n\in\mathcal{L}$ for any $A_1\subset A_2\subset\cdots$ , $A_1,A_2,\dots\in\mathcal{L}$ .

Then one can apply the Dynkin's theorem and get $\sigma(\mathcal{P})=\mathcal{B}_1\otimes\mathcal{B}_2=\mathcal{L}$ . The remainings are also shown in the same way as above.
Remark 2. The proposition above holds even if the assumption of $f$ is relaxed to that $f(X,Y)$ is integrable.
Remark 3. It can be seen directly that $\mathcal{L}$ is a $\sigma$ -field.

We now show the Markov property of a geometric Browinan motion.
Example. Let $X^x_t = xe^{\alpha t + \beta B_t}$ , where $(B_t)_{t\geq 0}$ is a Brownian motion starting at the origin, and let $f\colon\mathbb{R}\to\mathbb{R}$ be a bounded function. And $(\mathcal{F}_t)_{t\geq 0}$ denotes the Brownian filtration. By (1) of the proposition above, we get, for $s < t$ ,
\begin{align*}
E[f(X^x_t)|\mathcal{F}_s]
&= E[f(xe^{\alpha s + \beta B_s}e^{\alpha (t-s) + \beta (B_t-B_s)})|\mathcal{F}_s] \\
&= E[f(ye^{\alpha (t-s) + \beta B_{t-s}})]\big|_{y=X^x_s} \\
&= E[f(X^y_{t-s})]\big|_{y=X^x_s},
\end{align*}
where, in the second equality, we use the fact that $B_t-B_s=B_{t-s}$ holds in distribution.