# 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.

Created: 2023-06-25 Sun 02:30

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