Quantum Mechanics - Wave Function

The state of a quantum mechanical system is described by a wave function. Consider two physical observables that have discrete eigenvalues: observable A with eigenvalues {α}, and observable B with eigenvalues {β}. Under what circumstances can all wave functions be expanded in a set of basis states, each of which is a simultaneous eigenfunction of both A and B?

A. Only if the values {α} and {β} are nondegenerate
B. Only if A and B commute
C. Only if A commutes with the Hamiltonian of the system
D. Only if B commutes with the Hamiltonian of the system
E. Under all circumstances

(GR9277 #50)

Solution:

For two physical quantities to be simultaneously observable, their operator representations must commute, [A, B] = 0.

Answer: B


Proof:

Wave function, ψ = ∑_i c_i |v_i >

A |v_i > = α |v_i >

B |v_i > = β |v_i >

BA |v_i > = Bα |v_i > = α B |v_i > = αβ |v_i >

AB |v_i > = Aβ |v_i >= β A |v_i > = βα |v_i >

(BA − AB) |v_i > = (αβ − βα) |v_i > = 0

[A, B] = 0