# How do we know (BΨ) is not equal to zero?

1. Feb 9, 2014

### PFfan01

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?

2. Feb 9, 2014

### dextercioby

Well Psi is the domain of B, so that the only way that B psi =0 is that psi =0, which is trivial, the null vector is excluded in any analysis.

3. Feb 9, 2014

### PFfan01

Thanks, but I am thinking in a different way.
Ψ and (BΨ) are both eigenfunctions of A with the same non-degenerate eigenvalue a, and they must be linearly dependent, namely BΨ=b*Ψ. Ψ is a common eigenfunction of A and B. But the eigenvalue b could be zero (b=0), so that BΨ=0. Right?

4. Feb 9, 2014

### dextercioby

If Bpsi =0, then it's no problem, the 0 vector is a trivial eigenvector for any linear operator.