Going from 1+1=2 to Quantum Mechanics Chaimongkol, 2026 Compatibility of Observables
If two Observable are compatible , then they can be measured simultaneously with one basic measurement.
This means that observing one doesn’t affect the other.
Two Operator A ^ \hat{A} A ^ B ^ \hat{B} B ^ commute , i.e. [ A ^ , B ^ ] = 0 [\hat{A},\hat{B}]=0 [ A ^ , B ^ ] = 0
A ^ , B ^ are compatible ⇔ [ A ^ , B ^ ] = 0 \hat{A},\hat{B}\quad\text{are compatible}\quad\Leftrightarrow\quad[\hat{A},\hat{B}]=0 A ^ , B ^ are compatible ⇔ [ A ^ , B ^ ] = 0 Proof
⇒ A ^ B ^ = ∑ n , m A n B m ∣ ϕ n ⟩ ⟨ ϕ n ∣ ϕ m ⟩ ⟨ ϕ m ∣ \Rightarrow\hat{A}\hat{B}=\sum_{n,m}A_nB_m\left\lvert \phi_n \right\rangle\left\langle \phi_n|\phi_m \right\rangle\left\langle \phi_m \right\rvert\\ ⇒ A ^ B ^ = n , m ∑ A n B m ∣ ϕ n ⟩ ⟨ ϕ n ∣ ϕ m ⟩ ⟨ ϕ m ∣ = ∑ n A n B n ∣ ϕ n ⟩ ⟨ ϕ n ∣ = B ^ A ^ =\sum_nA_nB_n\left\lvert \phi_n \right\rangle\left\langle \phi_n \right\rvert=\hat{B}\hat{A}\\ = n ∑ A n B n ∣ ϕ n ⟩ ⟨ ϕ n ∣ = B ^ A ^ ⇐ A ^ ∣ ϕ n ⟩ = A n ∣ ϕ n ⟩ \Leftarrow\hat{A}\left\lvert \phi_n \right\rangle=A_n\left\lvert \phi_n \right\rangle\\ ⇐ A ^ ∣ ϕ n ⟩ = A n ∣ ϕ n ⟩ A ^ B ^ ∣ ϕ n ⟩ = B ^ A ^ ∣ ϕ n ⟩ = A n B ^ ∣ ϕ n ⟩ \hat{A}\hat{B}\left\lvert \phi_n \right\rangle=\hat{B}\hat{A}\left\lvert \phi_n \right\rangle=A_n\hat{B}\left\lvert \phi_n \right\rangle A ^ B ^ ∣ ϕ n ⟩ = B ^ A ^ ∣ ϕ n ⟩ = A n B ^ ∣ ϕ n ⟩ Example
Ehrenfest Theorem