Compatible and Incompatible Observables

Definition

Two observables are compatible if their commutator is zero.

Theorem

If two observables are compatible, then their corresponding operators possess a set of common eigenstates.

Proof. Assume non-degenerate eigenvalues.

$A | φ_{n} ⟩ = a_{n} | φ_{n} ⟩ . 0 = ⟨ φ_{m} | [A, B] | φ_{n} ⟩ = ⟨ | A B - B A φ_{n} ⟩ = a_{m} ⟨ φ_{m} | B | φ_{n} ⟩ - a_{n} ⟨ φ_{m} | B | φ_{n} ⟩ = (a_{m} - a_{n}) ⟨ φ_{m} | B | φ_{n} ⟩$ So, $⟨ φ_{m} | B | φ_{n} ⟩ = 0$ for unequal eigenvalues. Thus, off-diagonal matrix elements of $B$ for $A$ ’s eigenbasis representation is zero for non-degenerate eigenvalues. Therefore, $B$ ’s representation relative to $A$ ’s basis is diagonal.

Consecutively Measuring

Compatible

Since they are compatible, measuring $A$ first then $B$ is the same as measuring $B$ then $A$ . The result of either measurement puts the result in the respective shared eigenstate.

Measuring $A$ gives $a_{n}$ and puts the state in the $n-$th eigenstate. Measuring $B$ gives $b_{n}$ and remains the eigenstate.
Measuring $B$ gives $b_{n}$ and puts the state in the $n-$th eigenstate. Measuring $A$ gives $a_{n}$ and remains the eigenstate.

Incompatible

Since they are incompatible, measuring $A$ first then $B$ has the state transform to an $A$-eigenstate then project it onto a $B$-eigenstate. Vice-versa for $B$ then $A$ .

$A B | ψ ⟩ = b_{n} A | b_{n} ⟩ = b_{n} a_{n} | a_{n} ⟩$
$B A | ψ ⟩ = a_{n} B | a_{n} ⟩ = a_{n} b_{n} | b_{n} ⟩$

These are not necessarily the same.