Answer to Question #83991 in Physical Chemistry for Eshita

Question

Two operators A and B are related as [A,B]=0 . How accurately can one measure the physical property corresponding to A, for a system in the eigenstate of B ? Justify your answer. Assume the eigenstate are non-degenerate.

Answer 1

Solution:

A and B have common set of eigenstates.

For these common eigenstates one has ∆A = ∆B = 0 and hence ∆A ∆B = 0. There is a complete common set of собственныхn eigenfunctions, for which, therefore,

[A, B] ψn = (AB - BA) ψn = (anbn - bnan) ψn = 0.

Proof (⇐). Suppose that ea is an eigenfunction A. Then A (Bψa) = ABψa = BAψa = B aψa =

Bψa. Thus, Bψa is also an eigenfunction of A. Then you can select

(i) If a is nondegenerate, then Bψa ψa, say, Bψa = bψa, which means that isa is also

Own function B.

(ii) If a is degenerate (degeneration of s), consider the part of the Hilbert space that covers

functions ψar (r = 1, ..., s). For a given ψap (eigenfunction A), Bψap can also be

written in terms of ψar. Thus, we have a Hermitian operator B in the subspace

functions ψar. In this subspace, B can be diagonalized, and we can use the eigenvalues b1 ,. bs

as a second label, which leads to a common set of eigenfunctions.