Is this estimate well-known?
Let be a separable Hilbert space and represents the norm induced by the inner product of , and let be a self-adjoint operator on bounded from below and denotes the domain of .
Suppose is an eigenvalue with multiplicity 1 attaining the value and is the corresponding eigenvector with .
Proposition. If is an isolated point of the spectrum and , with , then
\begin{align*}
\|x-e_0\|^2\leq 2\cdot \frac{\langle Ax,x\rangle - \lambda_0}{\lambda_1-\lambda_0},
\end{align*}
where .
Proof. Let be the spectral decomposition of . Then, for with ,
\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 implies , 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.
Vespa mandarinia japonica (Japanese giant hornet) - オオスズメバチ
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.
Is the conjugate of a dense and continuous embedding a dense and continuous one as well?
This topic relates to an abstract Wiener space in the Malliavin calculus.
Proposition. Let be a real separable Banach space and be a real separable Hilbert space embedded densely and continuously to . Then, is densely and continuously embedded to .
Proof. Let be a continuous linear injection from to such that is dense in . In order to see that is dense in , it suffices to show that for , implies , where represents the conjugate of .
In what follows, represents the inner product of a (general) Hilbert space and the natural bilinear form between a (general) Banach space and its conjugate is denoted by . For simplicity, the conjugate space is identified with by the Riesz representation theorem, and corresponding elements are expressed as the same symbols.
Suppose and for any . Since , the assumption implies
\begin{align*}
& {}_{B}\langle \iota h, \ell \rangle_{B^*} = 0. \tag{1}
\end{align*}
Now we set the continuous linear functional defined on as follows:
\begin{align*}
\iota(H)\ni\iota g \mapsto \langle g, h \rangle_{H^*}\in\mathbb{R}.
\end{align*}
Since is dense in , for any , there is a sequence in such that as in . Thus, we get , by defining, for , . We note that the definition of is independent of the manner to choose the sequence . Then, putting in (1) yields .
Solving Linear Equations via VQE (Variational Quantum Eigensolver)
We consider the following linear equation:
\begin{align*}
Ax=b, \tag{1}
\end{align*}
where an invertible matrix , a vector , and an unknown variable .
For calculation on quantum computers, let be a power of 2 and assume , where represents the norm of induced by the usual inner product of .
We note that 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 represents the Schatten form; that is, for ,
\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" in the quantum mechanics/information.
Proposition. 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 .
Proof. It is obvious that is Hermitian. We see the remaining assertion since is a projection to the orthogonal complement of the subspace spanned by .
Let be an initial quantum state and , be a parametrized unitary operator called "ansatz" and denote .
Then, by minimizing
\begin{align*}
L(\theta):= \langle Hu(\theta),u(\theta)\rangle
\end{align*}
with respect to , 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.
Decomposition of matrices with the Pauli matrices
In this article, we introduce that any square matrix can be decomposed as the linear combination of the Pauli matrices.
Denote by the set of all square matrices of order whose components value on .
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. is a Hilbert space under the inner product; for ,
\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 is an orthonormal basis of . Moreover, forms an orthonormal basis of .
Consequently, we get the following:
Theorem. Let . Then, forms an orthonormal basis of the tensor product Hilbert space . The members are called -qubit Pauli matrices.
Thus, for any , we get the decomposition , where the inner product represents the one of .
多次元正規分布の条件付き分布
を 次元正規分布 に従う確率変数とし、 を -値確率変数とする。また、確率変数 は 次元正規分布 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、 に関する 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、, である。
証明. 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。, 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。□
マルコフ性のためのおべんきょう
In what follows, represents a probability space and the expectation under is denoted by .
Proposition. Let and be measurable spaces, be a bounded measurable function, and be a sub -field of . If an -valued r.v. is -measurable and an -valued r.v. is independent of , we have the following:
(1) , a.s.
(2) , -a.s.
Proof. Recalling the construction of integration in the general measure theory, we may assume that is the form for . Furthermore, the Dynkin's theorem allows to be the form for , .
(1) Since is -measurable and is independent of , 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 , it is followed by the same argument since is -measurable and is independent of .
(2) Suppose . Then, by the independence between and , 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 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 -system in ; that is, and it holds for any .
Denote by the collection of satisfying
\begin{align*}
E[\boldsymbol{1}_A(X,Y)|\mathcal{G}]
&= E[\boldsymbol{1}_A(x,Y)]\big|_{x=X}.
\end{align*}
Then, is a -system; that is, all of the followings are fulfilled:
- .
- for any , .
- for any , .
Then one can apply the Dynkin's theorem and get . The remainings are also shown in the same way as above.
Remark 2. The proposition above holds even if the assumption of is relaxed to that is integrable.
Remark 3. It can be seen directly that is a -field.
We now show the Markov property of a geometric Browinan motion.
Example. Let , where is a Brownian motion starting at the origin, and let be a bounded function. And denotes the Brownian filtration. By (1) of the proposition above, we get, for ,
\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 holds in distribution.