Unitary operator as an exponent of self-adjoint

Question

I'm trying to proof that set of unitary operators on infinite Hilbert space is path-connected. That's why i need to show that for each unitary $U$ there is a self-adjoint $B$ such that $U=e^{iB}$. Any ideas on how to show that? Maybe by using the fact that every unitary operator is unitary equivalent to operator of multiplication by $f$ in some $L_{2}(\mu)$ ?

Answer 1

Fix a Borel function $\varphi:\mathbb T=\{z\in\mathbb C:|z|=1\}\to [0,2\pi)$ such that $e^{i\varphi(z)}=z$. Put $B=\varphi(U)$ (here we're using Borel functional calculus). Then $B$ is self-adjoint, and $U=\exp(iB)$.

Answer 2

Proposition. Let $\widehat{A}$ be a Hermitian operator. The bounded operator $\widehat{U} : L^2([a,b],\mu) \to L^2([a,b],\mu)$ on the Hilbert space $L^2([a,b],\mu)$ that assigns to any $\psi$ in the domain the vector $$ \psi'(r) = \left[\widehat{U}\psi\right]\!(r) = \left[\exp\left(i\widehat{A}\right)\psi\right]\!(r) $$ in the domain is, in fact, a unitary operator.

My answer relies on two definitions: the definition of a bounded operator and the definition of a unitary operator. These definitions are given below. With the first definition, I first prove Lemma 1, which is annexed below. With the aid of Lemma 1, I complete the proof of the proposition.

Definition [Bounded operator]. In functional analysis and operator theory, a bounded linear operator is a linear transformation $\widehat{U} : X \to Y$ between Hilbert space $X$ and $Y$ that maps bounded subsets of $X$ to bounded subsets of $Y$. If $X$ and $Y$ are normed vector spaces, then $\widehat{U}$ is bounded if, and only if, there exists some $M > 0$ such that for all $x \in X$, $$ \left\| \widehat{U}x \right\|_{Y} \leq M \left\|x\right\|_{X}. $$

Definition. [Unitary operator]
A unitary operator is a bounded linear operator $\widehat{U} : H \to H$ on a Hilbert space $H$ that satisfies $\widehat{U}^\dagger \widehat{U} = \widehat{U}\widehat{U}^\dagger = \widehat{ I }$, where $U^\dagger$ is the adjoint of $\widehat{U}$, and $\widehat{I} : H \to H$ is the identity operator.

Proof. In Lemma 1, I show that that $\widehat{U}$ is a bounded linear operator $\widehat{U} : L^2([a,b],\mu) \to L^2([a,b],\mu) $ on a Hilbert space $L^2([a,b],\mu) $. Now, note that \begin{align} \widehat{U}^\dagger \widehat{U} &= \left[\exp(i\widehat{A})\right]^\dagger \exp(i\widehat{A}) &&\text{given} \\ &= \exp(-i\widehat{A}^\dagger) \exp(i\widehat{A}) \\ &= \exp \left(-i\left[\widehat{A}^\dagger-\widehat{A}\right]\right) &&\text{$\widehat{A}$ is normal} \\ &= \exp \left(-i\left[\widehat{A} -\widehat{A}\right]\right) &&\text{$\widehat{A}$ is Hermitian} \\ &= \exp(-i \widehat{0} ) \\ &= \widehat{ I } &&\text{$\widehat{A}$ is normal}; \end{align} similarly, note that \begin{align} \widehat{U} \widehat{U}^\dagger &= \exp(i\widehat{A}) \left[\exp(i\widehat{A})\right]^\dagger &&\text{given} \\ &= \exp(i\widehat{A}) \exp \left(-i\widehat{A}^\dagger \right) \\ &= \exp \left(-i\left[\widehat{A}^\dagger-\widehat{A}\right]\right) &&\text{$A$ is normal} \\ &= \exp \left(-i\left[\widehat{A} -\widehat{A}\right]\right) &&\text{$A$ is Hermitian} \\ &= \exp(-i \widehat{0} ) \\ &= \widehat{ I } . \end{align} With this I show that the exponential operator satisfies the necessary conditions to be a unitary operator.

Q.E.D.

Lemma 1. $\widehat{U}$ is a bounded operator.

Proof. So in this Hilbert space, I have that $$ \|\psi(r)\|_{L^2([a,b],\mu)} = \int_a^b \psi^\dagger(r)\,\psi(r) \,dr. $$ Further, I have that \begin{align} \|\left[\widehat{U}\psi\right](r)\|_{L^2([a,b],\mu)} &= \int_a^b \left[\widehat{U}\psi\right]^\dagger\!(r) \left[\widehat{U}\psi\right]\!(r)~dr \\ &= \int_a^b \left[ \psi\right]^\dagger\!(r) \left[\exp(i\widehat{A}) \right]^\dagger\! \, \left[\exp(i\widehat{A})\psi\right]\!(r) ~dr \\ &= \int_a^b \psi^*\!(r) \exp\!\left(\left[i\widehat{A} \right]^\dagger\right) \, \left[\exp(i\widehat{A})\psi\right]\!(r) ~dr \\ &= \int_a^b \psi^*\!(r) \exp\!\left(-i \widehat{A}^\dagger\right) \, \left[\exp(+i\widehat{A})\psi\right]\!(r) ~dr . \end{align} Since $\widehat{A}$ is a normal operator, therefore $\widehat{A}$ commutes with $\widehat{A}^\dagger$ and \begin{align} \left\|\left[\widehat{U}\psi\right](r)\right\|_{L^2([a,b],\mu)} &= \int_a^b \psi^*\!(r) \left[\exp\!\left(-i \left[\widehat{A}^\dagger- \widehat{A}\right]\right)\psi\right]\!(r) ~dr . \end{align} In addition, since $\widehat{A}$ is a Hermitian operator, therefore $\widehat{A}^\dagger=\widehat{A}$ and \begin{align} \left\|\left[\widehat{U}\psi\right](r)\right\|_{L^2([a,b],\mu)} &= \int_a^b \psi^*\!(r) \left[\exp\!\left(-i \left[\widehat{A} - \widehat{A}\right]\right)\psi\right]\!(r) ~dr \\ &= \int_a^b \psi^*\!(r) \left[\widehat{I}\psi\right]\!(r) ~dr \\ &= \int_a^b \psi^*\!(r) \psi \!(r) ~dr . \end{align} Since $L^2([a,b],\mu)$ is a normed vector space, then $\widehat{U}$ is bounded since there exists some $M>0$, for example $M=1$ such that for all $\psi \in L^2([a,b],\mu)$, $$ \left\|\left[\widehat{U}\psi\right]\!(r)\left\|_{L^2([a,b],\mu)} \leq M \right\|\psi(r)\right\|_{L^2([a,b],\mu)}. $$ Q.E.D.

With this lemma proven, I proceed to prove the major proposition.

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