Observables are definite-valued for eigenstates only

Let $ˆ 𝑄$ be a Hermitian operator corresponding to an Observable $𝑄$ , and consider a system with normalized state $| 𝜓 ⟩$ . Then the outcome of measuring $𝑄$ is definitely $𝑞$ iff $ˆ 𝑄 | 𝜓 ⟩ = 𝑞 | 𝜓 ⟩$ .

Proof

$𝑄$ is definitely $𝑞$ iff $⟨ 𝑄 ⟩ = 𝑞$ and

0 = 𝜎 2 𝑄 = ⟨ (ˆ 𝑄 - ⟨ ˆ 𝑄 ⟩ 𝐈) 2 ⟩ = ⟨ 𝜓 | (ˆ 𝑄 - 𝑞 𝐈) 2 | 𝜓 ⟩ = ⟨ 𝜓 | (ˆ 𝑄 - 𝑞 𝐈) † (ˆ 𝑄 - 𝑞 𝐈) | 𝜓 ⟩

which holds iff $(ˆ 𝑄 - 𝑞 𝐈) | 𝜓 ⟩ = 0$ , i.e. $| 𝜓 ⟩$ is an eigenstate with eigenvalue $𝑞$ , which necessitates $⟨ ˆ 𝑄 ⟩ = 𝑞 ⟨ 𝜓 | 𝜓 ⟩ = 𝑞$

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