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 ci |
vi >
A |vi > = α |vi >
B |vi > = β |vi >
BA |vi > = Bα |vi > = α B |vi > = αβ |vi >
AB |vi > = Aβ |vi >= β A |vi > = βα |vi >
(BA − AB) |vi > = (αβ − βα) |vi > = 0
