[proofplan] The proof is a direct unpacking of the spectral theorem and the functional calculus for [self-adjoint operators](/page/Self-Adjoint%20Operators). The spectral theorem identifies $A$ with multiplication by the identity function $\lambda \mapsto \lambda$ against its projection-valued measure, which gives the first moment formula. The norm identity in the functional calculus gives the second moment formula. Finally, expanding the centered square and using $\mu_u^A(\mathbb{R}) = 1$ identifies the spectral second central moment with the Hilbert-space variance. [/proofplan] [step:Define the spectral measure associated to the normalized vector] Let $E_A: \mathcal{B}(\mathbb{R}) \to \mathcal{L}(H)$ denote the projection-valued spectral measure of $A$. Let $\mu_u^A: \mathcal{B}(\mathbb{R}) \to [0,1]$ be the scalar spectral measure defined by \begin{align*} \mu_u^A(B) := (E_A(B)u,u)_H \end{align*} for every $B \in \mathcal{B}(\mathbb{R})$. Since $E_A(\mathbb{R}) = I$ and $\|u\|_H = 1$, this is a probability measure: \begin{align*} \mu_u^A(\mathbb{R}) = (E_A(\mathbb{R})u,u)_H = (u,u)_H = 1. \end{align*} [/step] [step:Recover the expectation from the spectral integral] Define the identity function $m: \mathbb{R} \to \mathbb{R}$ by \begin{align*} m(\lambda) := \lambda \quad \text{for every } \lambda \in \mathbb{R}. \end{align*} For each $v \in H$, let $\mu_v^A: \mathcal{B}(\mathbb{R}) \to [0,\infty)$ denote the finite scalar spectral measure defined by \begin{align*} \mu_v^A(B) := (E_A(B)v,v)_H \quad \text{for every } B \in \mathcal{B}(\mathbb{R}). \end{align*} By the [spectral theorem for self-adjoint operators](/theorems/6911) and its [Borel functional calculus](/theorems/2696) formulation, applied to the self-adjoint operator $A$ with spectral measure $E_A$, the operator $A$ is the spectral integral of $m$, with domain \begin{align*} \mathcal{D}(A) = \left\{v \in H : \int_{\mathbb{R}} \lambda^2 \, d\mu_v^A(\lambda) < \infty \right\}. \end{align*} Since $u \in \mathcal{D}(A)$, the weak spectral integral identity gives \begin{align*} (Au,u)_H = \int_{\mathbb{R}} \lambda \, d\mu_u^A(\lambda). \end{align*} [guided] The first identity is the weak form of the statement that $A$ is obtained by integrating the identity function against its spectral measure. We define $m: \mathbb{R} \to \mathbb{R}$ by \begin{align*} m(\lambda) := \lambda \quad \text{for every } \lambda \in \mathbb{R}. \end{align*} For each $v \in H$, let $\mu_v^A: \mathcal{B}(\mathbb{R}) \to [0,\infty)$ be the finite scalar spectral measure defined by \begin{align*} \mu_v^A(B) := (E_A(B)v,v)_H \quad \text{for every } B \in \mathcal{B}(\mathbb{R}). \end{align*} The spectral theorem for self-adjoint operators, in the Borel functional calculus formulation, says that $A = m(A)$ and that a vector $v \in H$ belongs to $\mathcal{D}(A)$ exactly when the second scalar spectral moment is finite: \begin{align*} v \in \mathcal{D}(A) \iff \int_{\mathbb{R}} \lambda^2 \, d\mu_v^A(\lambda) < \infty. \end{align*} We may apply this theorem because $A$ is self-adjoint by hypothesis and $E_A$ is its projection-valued spectral measure. Since $u \in \mathcal{D}(A)$, the spectral integral defining $Au$ is valid. Pairing the spectral integral for $Au$ with $u$ gives \begin{align*} (Au,u)_H = \int_{\mathbb{R}} \lambda \, d\mu_u^A(\lambda). \end{align*} This is the expectation of the measurement distribution $\mu_u^A$ because $\mu_u^A$ is the scalar measure obtained from $E_A$ in the state $u$. [/guided] [/step] [step:Recover the second moment from the functional calculus norm identity] The same functional calculus gives the norm identity for $m(A)u = Au$: \begin{align*} \|Au\|_H^2 = \|m(A)u\|_H^2 = \int_{\mathbb{R}} |m(\lambda)|^2 \, d\mu_u^A(\lambda). \end{align*} Since $m(\lambda) = \lambda$ is real-valued, $|m(\lambda)|^2 = \lambda^2$. Hence \begin{align*} \|Au\|_H^2 = \int_{\mathbb{R}} \lambda^2 \, d\mu_u^A(\lambda). \end{align*} [/step] [step:Identify the Hilbert-space variance with the centered spectral second moment] Define the scalar \begin{align*} a := (Au,u)_H. \end{align*} Since $A$ is self-adjoint, $a \in \mathbb{R}$. Define the centered function $c: \mathbb{R} \to \mathbb{R}$ by \begin{align*} c(\lambda) := \lambda - a \quad \text{for every } \lambda \in \mathbb{R}. \end{align*} By the functional calculus, $c(A) = A - aI$ on $\mathcal{D}(A)$, so the norm identity gives \begin{align*} \|(A-aI)u\|_H^2 = \int_{\mathbb{R}} |\lambda-a|^2 \, d\mu_u^A(\lambda). \end{align*} Because $a$ and $\lambda$ are real, $|\lambda-a|^2 = (\lambda-a)^2$. Therefore \begin{align*} \operatorname{Var}_u(A) = \|(A-(Au,u)_H I)u\|_H^2 = \int_{\mathbb{R}}(\lambda-(Au,u)_H)^2 \, d\mu_u^A(\lambda). \end{align*} [/step]