# 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.

\(A|\varphi_n\rangle = a_n|\varphi_n\rangle. 0 = \langle \varphi_m|[A,B]|\varphi_n\rangle = \langle|AB-BA\varphi_n\rangle = a_m\langle \varphi_m|B|\varphi_n\rangle - a_n\langle\varphi_m|B|\varphi_n\rangle = (a_m-a_n)\langle \varphi_m|B|\varphi_n\rangle\) So, \(\langle \varphi_m|B|\varphi_n\rangle = 0\) 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 \(A\).

- \(AB|\psi\rangle = b_nA|b_n\rangle = b_na_n|a_n\rangle\)
- \(BA|\psi\rangle = a_nB|a_n\rangle = a_nb_n|b_n\rangle\)

These are not necessarily the same.