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Chapter 13: Problem 19

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.

Short Answer

Expert verified
The operator \( e^{A}\) is well-defined and bounded, it has the properties: a) \(e^{0}=1\), b) \((e^{A})^{n} = e^{nA}\) for all positive integers, c) \(e^{A}\) is invertible with \( e^{-A}\) as its inverse, d) If A and B commute, \(e^{A+B} = e^{A}e^{B}\), e) If A is Hermitian, then \( e^{iA}\) is unitary.

Step by step solution

01

Proof of bounded operator

This is a straightforward assertion from the definition of operator norm. We first observe that for every \(n\), there is a bounded operator \(A^n\) with \(\|A^n\| \leq \|A\|^n\). The exponential operator serie is therefore absolutely convergent, meaning the operator is well-defined and also norm-bounded.
02

Proof for \(e^{0}=1\)

This part takes advantage from the power series representation of exponential function. When replacing \(A\) by \(0\) in the formula: \[\mathrm{e}^{A}=\sum_{n=0}^{\infty}\frac{A^n}{n!}\]It can be seen that all the terms vanish except when \(n=0\) (where A's exponent equals 0). Therefore, \(e^{0} = 1\)
03

Proof of identity for positive integers

This part can be proved using mathematical induction. For \(n=1\), the statement holds trivially. Assume that the statement holds for some \(n=k\). Then for \(n=k+1\), applying the assumed statement, \( (e^{A})^{k+1} = e^{kA}e^A = e^{(k+1)A}\). Therefore if the statement holds true for \(n=k\), it also holds for \(n=k+1\). By principle of mathematical induction, the statement holds for all positive integers \(n\).
04

Proof of invertibility

Because the exponential operator series is absolutely convergent, we can multiply it term by term with the series of inverse exponential operator \( e^{-A}\). Because \[e^Ae^{-A}=\sum_{n=0}^{\infty}\frac{A^n}{n!}\sum_{m=0}^{\infty}\frac{(-A)^m}{m!} = 1\], it's clear that \( e^A\) is an invertible operator and its inverse is \( e^{-A}\)
05

Proof of commutativity for exponential operators

To prove this, the identity \(AB = BA\) is employed (which is given in the problem statement). Then we can combine \(e^{A+B}\) into a single operator: \(e^{A}e^B\), which is equal to \(e^{B}e^A\) because of the commutativity property.
06

Proof of unitarity for Hermitian operators

\( e^{iA}\) is a unitary operator if it still preserves the inner product: \[ = \] for all \(x, y\) in the Hilbert space. This statement is true due to the condition that A is hermitian.

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Most popular questions from this chapter

Which of the following is a vector subspace of \(\ell^{2}\), and which are closed? In each case find the space of vectors orthogonal to the set. (a) \(V_{N}=\left\\{\left(x_{1}, x_{2} \ldots\right) \in \ell^{2} \mid x_{1}=0\right.\) for \(\left.i>N\right]\) (b) \(V=\bigcup_{N=1}^{\infty} V_{N}=\left|\left(x_{1}, x_{2}, \ldots\right) \in \ell^{2}\right| x_{t}=0\) for \(i>\) some \(\left.N\right]\). (c) \(U=\left|\left(x_{1}, x_{2}, \ldots\right) \in \ell^{2}\right| x_{1}=0\) for \(\left.t=2 n\right\\} .\) (d) \(W=\left\\{\left(x_{1}, x_{2}, \ldots\right) \in \ell^{2} \mid x_{1}=0\right.\) for some \(\left.i\right\\}\)

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 $$ \|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2} $$

Let \(A\) be a bounded openitor on a Hilbert space \(\mathcal{H}\) with a one- damensional rangc. (a) Show that there exist vectors \(u, v\) such that \(A x=\langle v \mid x\rangle u\) for all \(x \in \mathcal{H}\). (b) Show that \(A^{2}=\lambda A\) for some scalar \(\lambda\), and that \(\|A\|=\|u\|\|v\|\). (c) Prove that \(A\) is hermitian, \(A^{*}=A\), If and only if there exists a real number \(a\) such that \(v=a u\).

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

If \(A\) is a symmetric operator, show that \(A^{*}\) is symmetric if and only if it is self-edjoint, \(A^{*}=A^{* *}\)
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