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## Homework Statement

In the absence of degeneracy, prove that a sufficient condition for the equation below (1), where [itex]\left|a'\right>[/itex] is an eigenket of [itex]A[/itex], et al., is (2) or (3).

## Homework Equations

[tex]\sum_{b'} \left<c'|b'\right>\left<b'|a'\right>\left<a'|b'\right>\left<b'|c'\right> = \sum_{b',b''} \left<c'|b'\right>\left<b'|a'\right>\left<a'|b''\right>\left<b''|c'\right>\qquad (1)[/tex]

[tex]\left[A,B\right] = 0\qquad (2)[/tex]

[tex]\left[B,C\right] = 0\qquad (3)[/tex]

## The Attempt at a Solution

I know degeneracy is about one eigenvalue being associated with more than one eigevector (or eigenket). The summation looks intuitive to me. On the right side, a second completeness relation is used, where the completeness relation is

[tex]\mathbb{1} = \sum_{b''} \left|b''\right>\left<b''\right|[/tex]

I know the commutation relations of (2) and (3) can be written as

[tex]\left[X,Y\right] = XY - YX[/tex]

for any two operators [itex]X[/itex] and [itex]Y[/itex]. I know that since [itex]\left|a'\right>[/itex] is an eigenket of [itex]A[/itex], [itex]\left|b'\right>[/itex] is an eigenket of [itex]B[/itex] and [tex]\left|c'\right>[/itex] is an eigenket of [itex]C[/itex].

Half the solution is understanding the problem, and the problem seems to be saying that no eigenvalue has the same eigenvector (eigenket). Knowing that prove that the sum is valid if either pair of operators commute. I would greatly appreciate some help moving forward. Thanks!