Commutation Problem: Eigenvectors, Basis & Hamiltonian

[Diracobama2181]

Homework Statement

Consider a finite set of operators $B_i$
Let H be a Hamiltonian which commutes with each Bi;
i.e., [H, Bi] = 0 for all i. Suppose the |a_n> 's form a complete set of eigenstates of H satisfying
H|a_n>= a_n|a_n>.
(a) Let us choose one particular value of i and one particular value of n. Under what
circumstances can it be deduced that Bi|a_n> is proportional to |a_n>?
(b) Show that if the above is true for all i and for all n, then [B_i, B_j] = 0 for all i, j.
(c) How can you reconcile the rule stated in part (b) with the fact that for angular momentum
operators L_i, we can have a situation where [Li, H] = 0 but [Li, Lj] \= 0 when i \= j.

Relevant Equations

H|a_n>=a_n|a_n>

a) This would be true whenever |a_n> is an eigenvector of B_i.
b) If this holds true for each eigenvector, then B_i and B_j must share the same basis. Therefore, they must commute. Is this reasoning correct?
C) Despite commuting with the hamiltonian. the energy states can be degenerate, which I believe would imply that Li and Lj does not commute. Not sure how to mathematically formalize this though.

 

[vela]

a) This would be true whenever |a_n> is an eigenvector of B_i.

The point of this part of the question is to figure out what has to be true so you can conclude that $\lvert a_n \rangle$ is an eigenvector of $B_i$.

 

[PeroK]

Homework Statement:: Consider a finite set of operators $B_i$
Let H be a Hamiltonian which commutes with each Bi;
i.e., [H, Bi] = 0 for all i. Suppose the |a_n> 's form a complete set of eigenstates of H satisfying
H|a_n>= a_n|a_n>.
(a) Let us choose one particular value of i and one particular value of n. Under what
circumstances can it be deduced that Bi|a_n> is proportional to |a_n>?
(b) Show that if the above is true for all i and for all n, then [B_i, B_j] = 0 for all i, j.
(c) How can you reconcile the rule stated in part (b) with the fact that for angular momentum
operators L_i, we can have a situation where [Li, H] = 0 but [Li, Lj] \= 0 when i \= j.
Homework Equations:: H|a_n>=a_n|a_n>

a) This would be true whenever |a_n> is an eigenvector of B_i.
b) If this holds true for each eigenvector, then B_i and B_j must share the same basis. Therefore, they must commute. Is this reasoning correct?
C) Despite commuting with the hamiltonian. the energy states can be degenerate, which I believe would imply that Li and Lj does not commute. Not sure how to mathematically formalize this though.

You need to do more maths for these answers. For example, for part b) you need to show that for any state (vector) $\psi$ we have $B_iB_j (\psi) = B_jB_i(\psi)$.