Recent content by blakegriffin1

Right but isn't any wavefunction \psi an eigenfunction of \hat{x} as \hat{x} \psi = x \psi This should work for basically everything right since the operator \hat{x} doesn't really do anything besides apply "x" ?

I'm just revising some Quantum Mech and I have two questions. I know that if two operators commute say for instance [\hat{A},\hat{B}] = [\hat{B},\hat{A}] = 0 Then the observables that the operators extract from the wavefunction can be measured exactly (without losing information about the...