Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 28

For any pair of hermitian operators \(A\) and \(B\) on a Hilbert space \(\mathcal{H}\), define \(A \leq B\) iff \((u \mid A u) \leq(u \mid B u)\) for all \(u \in \mathcal{H}\). Show that this is a partial order on the set of hermatian operators pay particular attention to the symmetry property, \(A \leq B\) and \(B \leq A\) umplies \(A=B\). (a) For multiplication operators on \(L^{2}(X)\) show that \(A_{0} \leq A_{B}\). Iff \(\alpha(x) \leq \beta(x)\) a.e. on \(X\). (b) For projection operators show that the definition given here reduces to that given in the text, \(P_{N} \leq P_{N}\) iff \(M \subseteq N\)

Short Answer

Expert verified
The binary relation defined between hermitian operators is a partial order, satisfying reflexivity, antisymmetry, and transitivity. The antisymmetry was proved by considering positive and negative differences between the operators. For multiplication operators, it was shown that \(A_{0}\) is less than or equal to \(A_{B}\) if function \(\alpha\) is everywhere less or equal to \(\beta\). For projection operators, if \(P_{M}\) is less or equal to \(P{N}\), then subspace \(M\) must be a subset of subspace \(N\).

Step by step solution

01

Proving antisymmetry

Let's assume that \(A \leq B\) and \(B \leq A\). That means \((u|Au) \leq (u|Bu)\) and \((u|Bu) \leq (u|Au)\) for all \(u \in \mathcal{H}\). Then, \((u|Au)-(u|Bu) \leq 0\) and \((u|Bu)-(u|Au) \leq 0\) for all \(u\). This implies \((u|(B-A)u) = 0\) for all \(u\), therefore, any eigenvalue of \((B-A)\) must be zero. If the only eigenvalue is zero, then \(B-A = 0\) or \(A = B\)
02

Multiplication Operators

Within the context of multiplication operators on \(L^{2}(X)\), let \(A_{0}\) and \(A_{B}\) be given by multiplication by functions \(\alpha, \beta: X → \mathbb{R}\). If \(A_{0} \leq A_{B}\), for any \(y ∈ L^2(X)\), we have \((y| A_0 y) \leq (y|A_B y)\). Integral over \(X\) of \(|\alpha(x)y(x)|^2 \leq \) integral over \(X\) of \(|\beta(x)y(x)|^2\). Dividing by \(\| y \|_2^2\) and using Fatou's Lemma, we pointwise obtain \(\alpha(x) \leq \beta(x)\), a.e on \(X\).
03

Projection Operators

Let's assume that \(P_{M} \leq P_{N}\), \((u|P_M u) \leq (u|P_N u)\) for all \(u \in \mathcal{H}\). Since \(P_M\) and \(P_N\) are projection operators, their outputs lie in the subspaces \(M\) and \(N\) respectively. We know that \(\|P_N u\|^2 \geq \|P_M u\|^2\) for all \(u\). This inequality holds because the norm of a projection of \(u\) onto a subspace is less than or equal to the norm of \(u\) itself. But since the inequality holds for all \(u\), this implies that the subspace \(M\) must be contained within \(N\) or \(M \subseteq N\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The norm \(\|\phi\|\) of a bounded linear operator \(\phi: \mathcal{H} \rightarrow \mathrm{C}\) is defined as the greatest lower bound of all \(M\) such that \(|\phi(u)| \leq M\|u\|\) for all \(u \in \mathcal{H}\). If \(\phi(u)=(v \mid u)\) show that \(\|\phi\|=\|v\|\). Hence show that ahe bounded lanear functional norm satisfies the parallelogram law $$ \|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2} $$

If \(\left(A, D_{A}\right)\) is a densely defined openator and \(D_{A}\) is dense in \(\mathcal{H}_{1}\) show that \(A \subseteq A^{* *}\).

On the vector space \(\mathcal{F}^{\prime}[a, b]\) of complex continuous differentiable functions on the, interval \([a, b]\), set $$ \langle f| g)=\int_{a}^{b} \overline{f^{\prime}(x)} g^{\prime}(x) \mathrm{dr} \text { where } f^{\prime}=\frac{\mathrm{d} f}{\mathrm{~d} x}, \quad g^{\prime}=\frac{\mathrm{d} g}{\mathrm{~d} x} $$ Show that this is not an inner product, but becomes one if restricted to the space of functions \(f \in\) \(F^{\prime}[a, b]\) having \(f(c)=0\) for seme fixed \(a \leq c \leq b\). Is it a Hilbert space? Give a similar analysis for the case \(a=-\infty, b=\infty\), and restricting functions to those of compact support.

For every bounded operator \(A\) on a Hilbert space \(\mathcal{H}\) show that the exponential operator $$ \mathrm{e}^{A}=\sum_{n=0}^{\infty} \frac{A^{n}}{n !} $$ is well-defined and bounded on \(\mathcal{H}\). Show that (a) \(e^{0}=1\) (b) For all posituve integers \(n,\left(c^{A}\right)^{n}=e^{n A}\). (c) \(\mathrm{e}^{A}\) is invertuble for all bounded operators \(A\) (even if \(A\) is not mivertible) and \(e^{-A}=\left(e^{4}\right)^{-1} .\) (d) If \(A\) and \(B\) are commuting operators then \(e^{A+B}=\mathrm{e}^{A} \mathrm{e}^{\theta}\) (c) If \(A\) is hermitian then \(e^{i A}\) is unitary.

Show that if \(\left(A, D_{A}\right)\) and \(\left(B, D_{B}\right)\) arc operators on dense domains in \(H\) then \(B^{*} A^{*} \subseteq\) \((A B)^{\circ}\)
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free