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2

The canonical commutation relations for position and momentum operators in quantum mechanics is $$[\hat x,\hat p]=i \hbar.$$ Here the vector space is the Hilbert space over $\mathbb C$.


7

Yes, there is even a typical example. If $V$ is the space of all polynomials on $\mathbb C$, and $D,x:V\to V$ are defined by: $$(Df)(x)=\frac{df}{dx}\\(xf)(x)=xf(x).$$ Then $Dx-xD=I$. You could also take $V$ to be all meromorphic functions over $\mathbb C$, or the space of power series over $\mathbb C$. Such pairs of operators actually are the underlying ...


2

Let $V = \mathbb{K}[x]$ be the space of polynomials in one variable with coefficients from your ground field $\mathbb{K}$, and consider the two operators $\frac{d}{dx}$ (formal differentiation of polynomials) and $x$ (multiplication of polynomials by $x$).


1

Let $X$ be a compact Hausdorff space with more than one point. Then $C(X)$ is a natural, normal uniform algebra on $X$. This means that the character space of $C(X)$ is exactly $X$ and for each closed set $E\subseteq X$ and each closed set $F\subseteq X\setminus E$, there exists a function $f\in C(X)$ such that $f(y)=0$ for all $y\in E$ and $f(y)=1$ for each ...


0

Let $A$ be a non-commutative C*-algebra that is also an integral domain. If $a$ be a self-adjoint element of $A$, then it is well-known that C*(a), C*-algebra generated by $a$, is a commutative C*-algebra. Since, $A$ is an integral domain, C*(a) is an integral domain, too. If we accept the above conjecture that any commutatve C*-algebra which is also an ...


0

Using the Gram-Schmidt process, construct orthogonal vectors $f_n$ from the $A^nf$. Note that $\Vert f_n \Vert\le \Vert A^nf \Vert$. From the Gram-Schmidt equations it is apparent that, on the orthonormal vectors $\frac{f_n}{\Vert f_n\Vert}$, $A$ is a Jacobi matrix operator $J$ with $n,n+1$ elements $b_n := \frac{\Vert f_{n+1}\Vert}{\Vert f_n\Vert}$ and ...


0

While I do not know whether it will help you much, you may view a dictionary between topological concepts and their equivalent algebraic concepts on pages 6 and 13 of "Very Basic Non-commutative Geometry" by Masoud Khalkhali.


0

For two vectors $x,y\in H$, the notation $x\otimes y$ is frequently used to denote the rank-one operator $z\longmapsto \langle z,y\rangle x$. It is related to tensor products, but not in the obvious way: it is seeing $x\otimes y$ as an element of $H\otimes H^*$ (most often, no one pays attention to this and just uses the notation). I think it is just a ...


4

The Boolean algebra of connected components is equivalent to the projections (the elements with $p^2=p$) in the algebra of functions, with multiplication of functions representing intersection and $(p_1,p_2) \to p_1 + p_2 - p_1p_2$ being the union of sets of components. I think there is a version of algebraic K-theory for topological algebras whose value ...


2

Let $L$ denote the matrix $$ L = \pmatrix{u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3} $$ My best guess is that whatever the author is getting at has something to do with the fact that $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_N $$ or, if $N$ is odd, $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + ...


1

Yes and you don't even have to take the closure. For instance if $A$ is a non-simple C*-algebra with trivial cetnre that has unique (faithful) trace (for instance $A=C^*(G)$ for a sufficiently non-commutative amenable group such as the group of permutations of integers that move at most finitely many entries), then $A\otimes \mathcal{Z}$, where $\mathcal{Z}$ ...


1

Since you have to relate the norm with the spectrum, I don't think this has an elementary algebraic proof. Facts needed (from the theory of Banach algebras): For any $A\in\mathcal S$, $\sigma(A)=\{\phi(A):\ \phi\in S(\mathcal A)\}$, where $S(\mathcal A)$ is the state space. For any $A\in\mathcal S$, $\|A\|=\text{spr}(A)=\max\{|\lambda|:\ ...


2

Let $E=\{1,1/2,1/3,\cdots\}$ and let $\mu_{a}$ be the finite atomic meausre on $[0,1]$ that is supported on $E$ with $\mu\{1/n\}=1/n^{2}$. Let $\mu = \mu_{a}+m$ where $m$ is Lebesgue measure on $[0,1]$. Let $X=L^{2}_{\mu}[0,1]$. Let $\chi$ be the characteristic function of $E$, and define operators $A, B \in \mathcal{L}(X)$ by $$ Af = ...


1

It is correct for positive operators, for more information see Gohberg, Krein, Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space page 27



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