[proofplan] Use the Cayley transform to convert the unbounded self-adjoint operator into a bounded unitary operator whose spectral point $1$ is absent from the range of the transformed domain. Apply the spectral theorem for unitary operators to obtain a projection-valued measure on the unit circle, then transport it back to the real line through the inverse Cayley map. The [Borel functional calculus](/theorems/2696) gives the operator identity and the stated domain, and uniqueness follows because the Cayley transform determines the transported spectral projections. [/proofplan] [step:Form the Cayley transform and record its unitary properties] Let $i\in\mathbb C$ denote the imaginary unit, and let $I:H\to H$ denote the identity operator on $H$. Define the Cayley transform $U:\!H\to H$ of $A$ by \begin{align*} U=(A-iI)(A+iI)^{-1}. \end{align*} Since $A$ is self-adjoint, the Cayley transform theorem for [self-adjoint operators](/page/Self-Adjoint%20Operators) applies: the operators $A+iI:\mathcal D(A)\to H$ and $A-iI:\mathcal D(A)\to H$ are bijective with bounded inverses, the operator $U$ is unitary, and $1$ is not an eigenvalue of $U$. Hence $U$ is a bounded operator on $H$. [guided] The purpose of the Cayley transform is to replace the unbounded operator $A$ by a bounded normal operator. Let $i\in\mathbb C$ be the imaginary unit, and let $I:H\to H$ be the identity operator on $H$. Define \begin{align*} U=(A-iI)(A+iI)^{-1}:H\to H. \end{align*} We now verify why this formula is meaningful. The Cayley transform theorem for self-adjoint operators says that if $A:\mathcal D(A)\subset H\to H$ is self-adjoint on a complex [Hilbert space](/page/Hilbert%20Space), then $A+iI:\mathcal D(A)\to H$ and $A-iI:\mathcal D(A)\to H$ are bijective with bounded inverses, and the operator $(A-iI)(A+iI)^{-1}$ is unitary. Its inverse transform recovers $A$ on the appropriate domain, and $1$ is not an eigenvalue of the transformed unitary. The hypotheses are exactly those in the theorem statement: $H$ is a complex Hilbert space and $A$ is self-adjoint. Therefore $(A+iI)^{-1}:H\to\mathcal D(A)$ is bounded as an operator into $H$, the composition $(A-iI)(A+iI)^{-1}:H\to H$ is bounded and everywhere defined, and $U$ is unitary. The exclusion of the eigenvalue $1$ is the point that allows the inverse Cayley function to recover $A$ without losing domain information. [/guided] [/step] [step:Apply the bounded spectral theorem to the unitary Cayley transform] By the bounded spectral theorem for unitary operators, there is a unique projection-valued measure $F$ on the Borel subsets of the unit circle $\mathbb T\subset\mathbb C$ such that \begin{align*} U=\int_{\mathbb T}z\,dF(z). \end{align*} The theorem applies because $U:H\to H$ is bounded and unitary; in particular, $U$ is normal and its spectrum is contained in $\mathbb T$. [/step] [step:Transport the spectral measure from the unit circle to the real line] Define the inverse Cayley map $c:\mathbb R\to\mathbb T\setminus\{1\}$ by \begin{align*} c(\lambda)=\frac{\lambda-i}{\lambda+i}. \end{align*} For each Borel set $B\subset\mathbb R$, define \begin{align*} E_A(B)=F(c(B)). \end{align*} Because $c$ is a Borel isomorphism from $\mathbb R$ onto $\mathbb T\setminus\{1\}$ and $F(\{1\})=0$, the assignment $B\mapsto E_A(B)$ is a projection-valued measure on the Borel subsets of $\mathbb R$. [/step] [step:Recover the operator and its domain from the transported functional calculus] Let $m:\mathbb T\setminus\{1\}\to\mathbb R$ be the Borel function \begin{align*} m(z)=i\frac{1+z}{1-z}. \end{align*} Then $m(c(\lambda))=\lambda$ for every $\lambda\in\mathbb R$. Since $F(\{1\})=0$, the function $m$ is finite $F$-almost everywhere and is real-valued on $\mathbb T\setminus\{1\}$. The unbounded Borel functional calculus for the projection-valued measure $F$ therefore defines the self-adjoint operator $m(U)$ with domain \begin{align*} \left\{\psi\in H:\int_{\mathbb T}|m(z)|^2\,d(F(z)\psi,\psi)_H<\infty\right\}. \end{align*} Transporting this spectral integral through the Borel isomorphism $c:\mathbb R\to\mathbb T\setminus\{1\}$ gives \begin{align*} A=m(U)=\int_{\mathbb R}\lambda\,dE_A(\lambda). \end{align*} The same change of spectral variable gives the domain \begin{align*} \left\{\psi\in H:\int_{\mathbb R}\lambda^2\,d(E_A(\lambda)\psi,\psi)_H<\infty\right\}. \end{align*} This is the stated domain $\mathcal D(A)$. [/step] [step:Prove uniqueness of the projection-valued measure] Suppose $G$ is another projection-valued measure on $\mathbb R$ satisfying the same operator identity and domain formula. Pushing $G$ forward by $c$ gives a projection-valued measure $c_*G$ on $\mathbb T$ with no mass at $1$ and with \begin{align*} U=\int_{\mathbb T}z\,d(c_*G)(z). \end{align*} By uniqueness in the bounded spectral theorem for the unitary operator $U$, one has $c_*G=F$. Since $c:\mathbb R\to\mathbb T\setminus\{1\}$ is a Borel isomorphism and $F(\{1\})=0$, it follows that $G(B)=F(c(B))=E_A(B)$ for every Borel set $B\subset\mathbb R$. [/step] [step:Interpret spectral projections as measurement probabilities] For a Borel set $B\subset\mathbb R$, $E_A(B)$ is an [orthogonal projection](/theorems/437), so $0\le (E_A(B)\psi,\psi)_H\le \|\psi\|_H^2$. If $\|\psi\|_H=1$, then the map $B\mapsto (E_A(B)\psi,\psi)_H$ is a probability measure on $\mathbb R$, because $E_A(\mathbb R)=I$ and countable additivity of $E_A$ holds in the strong operator topology. Thus $(E_A(B)\psi,\psi)_H$ is the probability assigned by the spectral measure of $A$ to outcomes lying in $B$, which is the asserted quantum-mechanical interpretation. [/step]