📂Quantum Mechanics

Two operators with simultaneous eigenfunctions are commutative.

Summary

If two different operators have the same eigenfunction, then the two operators are commutative. In other words, if the following equation holds, it means $[A, B] = 0$.

$$ \begin{cases} A\psi=a\psi \\ B\psi=b\psi \end{cases} $$

In this case, $\psi$ is the normalized eigenfunction.

Converse

The converse of the above theorem also holds. That is, the commutativity of two operators and having a common eigenfunction is a necessary and sufficient condition.

Explanation

An eigenfunction $\psi$ common to two operators is also referred to in textbooks as a simultaneous eigenfunction.¹

Proof

$$ \begin{align*} AB\psi &= Ab\psi \\ &= bA\psi \\ &= ba\psi \\ &= ab\psi \\ &= aB\psi \\ &= Ba\psi \\ &= BA\psi \end{align*} $$

$$ \implies AB \psi - BA \psi = (AB-BA) \psi = 0 $$

$$ \implies (AB - BA) = [A, B] = 0 $$

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Converse

Suppose two operators $A$ and $B$ are commutative. Suppose also that the eigenvalue equations for each operator are as follows.

$$ A\psi_{a} = a\psi_{a} \\ B\psi_{b} = b\psi_{b} $$