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2

If you use unbounded operators, then $$ H = \int_{0}^{\infty}\lambda dE(\lambda). $$ The spectrum theorem for unbounded selfadjoint operators has the excellent provision that $$ \mathcal{D}(H) = \left\{ x \in X : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty\right\}. $$ You have $H=A^{2}$, where ...


1

The domain of $A^{\star}$ is identical to the domain of $A$ in this case. However, when you compose the two, then you get $\mathcal{D}(A^{\star}A)$ as $$ \mathcal{D}(A^{\star}A)= \{ f \in L^{2} : f\in \mathcal{D}(A) \mbox{ and } Af \in \mathcal{D}(A^{\star}) \}. $$ In particular, $f$ is twice absolutely continuous with $f(0)=f(1)$ and $f'(0)=f'(1)$ . ...


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If $H:X \to X$ is a bounded operator, there is indeed always such a decomposition available. We may construct one such decomposition as follows: By the spectral theorem, we have $H = UTU^*$ Where the operator $T$ is given by $$ [T(\phi)](x) = f(x)\phi(x) $$ For some $f:\Bbb R \to \sigma(H) \subset [0,\infty)$. We can simply define $\sqrt T$ by $$ [\sqrt ...


2

This is a continuation of what I posted earlier. What follows are two examples of how the theory is applied to the trigonometric functions where $V=0$. The equation is in the limit point case on $[0,\infty)$ because $e^{i\sqrt{\lambda}x}\in L^{2}[0,\infty)$ while $e^{-i\sqrt{\lambda}x}$ is not, where $\sqrt{\lambda}$ is the branch whose branch cut is along ...


3

The classical operator $$ L = -\frac{d^{2}}{dx^{2}}+V,\;\;\; a \le x \le b, $$ is different if $V$ is very singular. If $V \in L^{1}[a,b]$, then things are nice because there are 2 linearly-independet classical solutions of $Lf = \lambda f$ for every $\lambda$. That is, such solutions are continuous on $[a,b]$, their first derivatives are ...


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For help with the 'rigorous mathematics', there is a proof of the affine property here. Just substitute your $0$ for $\mu_x$, $I$ for $\Sigma_x$ and $\mu$ for $b$ and you should get the same result :) For your 2nd question about the intuition for the spectral decomposition, we begin by observing that $C = UDU^T$, and that $U$ provides an orthogonal basis ...


1

Classical Solutions: First, assume $V\in L^{1}[a,b]$, and show the existence of classical solutions of $$ -f''+Vf = \lambda f,\;\;\; f(a)=A,\;f'(a)=B. $$ This can be done by considering the equivalent integral equation $$ f(x)=A+B(x-a)+\int_{a}^{x}\int_{a}^{t}(V(u)-\lambda)f(u)\,du\,dv. $$ This is a fixed point problem for $C[a,b]$ ...


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The answer to the first question is simply the following: take $f(z)=z^4$ in the spectral mapping theorem and use $\mathfrak{F}^4=1$.


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The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius $r$ only, one has ...


1

Let $N=\{ (Ax,x) : \|x\|=1\}$. Suppose $\lambda \notin N^{c}$ so that there exists $\delta$ such that the following holds whenever $\|x\|=1$: $$ 0 < \delta \le |(Ax,x)-\lambda|=|((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\| = \|(A-\lambda I)x\| $$ Then $\|(A-\lambda I)x\| \ge \delta \|x\|$ for all $x$. (The same holds for ...


3

A bounded linear operator $A$ on a separable Hilbert space $X$ with orthonormal basis $\{ e_{j} \}_{j=1}^{\infty}$ is a Hilbert-Schmidt operator if $$ \sum_{j=1}^{\infty}\|Ae_{j}\|^{2} < \infty. $$ This condition is true for one orthonormal basis iff it is true for every other orthonormal basis. If you're studying $X=L^{2}[a,b]$, then an ...


1

Using $P_i^2 = P_i$ we get $$e^{P_i\log\rho_i} = \sum_{k=0}^\infty \frac{(P_i\log\rho_i)^k}{k!} = \mathbb{1}-P_i +P_i\sum\frac{\log^k\rho_i}{k!} = \mathbb{1}-P_i+P_i\rho_i$$ For the whole expression we find (using that $P_iP_k=\delta_{i,k}P_k$, and $\sum_iP_i=\mathbb{1}$) $$\exp \sum_iP_i\log\rho_i = \prod_i\mathbb{1}-P_i+P_i\rho_i = \mathbb{1} -\sum_iP_i + ...


2

Consider $L^{2}(\mathbb{R})$. The Fourier transform and its inverse implement the Spectral Theorem for the selfadjoint operator $Af = \frac{1}{i}\frac{d}{dx}f$ on the domain $\mathcal{D}(A)$ consisting of absolutely continuous $f \in L^{2}(\mathbb{R})$ for which $f'\in L^{2}(\mathbb{R})$. The spectral measure $E$ is $$ E[a,b]f = ...


4

Since the spectrum does not depend on the (C$^*$) algebra, we can assume that $\mathcal A=C^*(x)$. Using the Gelfand transform, we can identify $C^*(x)$ with $C(\sigma(x))$, with $x$ mapped to the function $z\mapsto z$. The point evaluation states $f\mapsto f(t)$ are precisely the pure states The pure states are the extremal points of the set of states of ...


0

Singular values of the SVD decomposition of the matrix A is the square root of the eigenvalues of the matrix (A multiplied by A transpose) or(A transpose multplied by A), the two ar identical with positive eigenvalues.


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? As you've suggested already yourself, is the answer not simply this: $T = S + D$ , with $$ S = \frac{1}{2} \left( T + T^* \right) \qquad ; \qquad D = \frac{1}{2} \left(T - T^* \right) $$ Where $S$ is self-adjoint and $D$ is a so-called anti self-adjoint operator: $D^* = -D$ . In three dimensional space, the anti self-adjoint operator ($3 \times 3$ matrix) ...


2

For a matrix $A \in \mathbb{R}^{n \times n}$ with strictly positive entries (actually you just need it to be irreducible), you can apply the Perron-Frobenius Theorem which asserts that the eigenvalue $\lambda$ with largest magnitude is positive, simple and the associated eigenvector has strictly positive entries. In particular, with these special settings ...


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Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of $A$ is $$ (\lambda I -A)\left(\begin{array}{c}f \\ g\end{array}\right)= \left(\begin{array}{cc}\lambda & -I \\ I-\Delta & \lambda\end{array}\right) \left(\begin{array}{c}f \\ ...



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