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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 ...

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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, $$is different if V is very singular. If V \in L^{1}[a,b], then things are nice because there are 2 linearly-independet classical solutions of Lf = \lambda f for every \lambda. That is, such solutions are continuous on [a,b], their first derivatives are ... 0 For help with the 'rigorous mathematics', there is a proof of the affine property here. Just substitute your 0 for \mu_x, I for \Sigma_x and \mu for b and you should get the same result :) For your 2nd question about the intuition for the spectral decomposition, we begin by observing that C = UDU^T, and that U provides an orthogonal basis ... 1 Classical Solutions: First, assume V\in L^{1}[a,b], and show the existence of classical solutions of$$ -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. $$This is a fixed point problem for C[a,b] ... 0 The answer to the first question is simply the following: take f(z)=z^4 in the spectral mapping theorem and use \mathfrak{F}^4=1. 0 The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius r only, one has ... 1 Let N=\{ (Ax,x) : \|x\|=1\}. Suppose \lambda \notin N^{c} so that there exists \delta such that the following holds whenever \|x\|=1:$$ 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 + ...

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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 $$E[a,b]f = ... 4 Since the spectrum does not depend on the (C^*) algebra, we can assume that \mathcal A=C^*(x). Using the Gelfand transform, we can identify C^*(x) with C(\sigma(x)), with x mapped to the function z\mapsto z. The point evaluation states f\mapsto f(t) are precisely the pure states The pure states are the extremal points of the set of states of ... 0 Singular values of the SVD decomposition of the matrix A is the square root of the eigenvalues of the matrix (A multiplied by A transpose) or(A transpose multplied by A), the two ar identical with positive eigenvalues. 0 ? As you've suggested already yourself, is the answer not simply this: T = S + D , with$$ S = \frac{1}{2} \left( T + T^* \right) \qquad ; \qquad D = \frac{1}{2} \left(T - T^* \right) $$Where S is self-adjoint and D is a so-called anti self-adjoint operator: D^* = -D . In three dimensional space, the anti self-adjoint operator (3 \times 3 matrix) ... 2 For a matrix A \in \mathbb{R}^{n \times n} with strictly positive entries (actually you just need it to be irreducible), you can apply the Perron-Frobenius Theorem which asserts that the eigenvalue \lambda with largest magnitude is positive, simple and the associated eigenvector has strictly positive entries. In particular, with these special settings ... 1 Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of A is$$ (\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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