Eigenstate measurements are deterministic

$\hat{f}\psi =\lambda \psi ⟹\mathbb{E}[f^m]={⟨(\hat{f}{)}^m⟩}_{\psi }={\lambda }^m$, where

• $\psi$ is an Eigenstate of a Measurement $f$
• $\psi \in H$ is a Quantum state
• $f$ is a Classical observable
• $\hat{f}$ is the corresponding Quantum observable
• $\lambda \in \mathbb{R},m\in {\mathbb{Z}}^{+}$

Proof

$$\begin{array}{rll}\mathbb{E}[f^m]& =⟨\psi ,(\hat{f}{)}^m\psi ⟩& \text{Due to the 3rd postulate of QM}\\ & =⟨\psi ,{\lambda }^m\psi ⟩& \\ & ={\lambda }^m⟨\psi ,\psi ⟩& \\ & ={\lambda }^m\Vert \psi \Vert & \\ & ={\lambda }^m& \end{array}$$

We used Postulate 3