- #1
PFfan01
- 88
- 2
Suppose that operators A and B have complete sets of eigenfunctions, [A, B] = 0, and Ψ is an eigenfunction of A with the eigenvalue a, namely AΨ=aΨ. Then we have A(BΨ) = BAΨ = a (BΨ). They say (BΨ) is also the eigenfunction of A. Why? How do we know (BΨ) is not equal to zero?