Properties of Hermitian operators in complex vector spaces

[MSUmath]

Homework Statement

Given a Hermitian operator A = $\sum \left|a\right\rangle a \left\langle a\right|$ and B any operator (in general, not Hermitian) such that $\left[A,B\right]$ = $\lambda$B show that B$\left|a\right\rangle$ = const. $\left|a\right\rangle$

Homework Equations

Basically those listed above plus possibly the Hermitian condition and eigenvalue definition which I will not list since they are well known.

The Attempt at a Solution

I have tried expanding it out in terms of the commutator, but this seems like the wrong approach. I am not sure that there is a way to calculate it directly. I do not think the proof is very involved but I am approaching it the wrong way and can't seem to get anywhere. This is listed as a fundamental property of a Hermitian operator in a complex vector space in my text, though it does not prove it and other references seem unconcerned. I am also not very comfortable using this kind of spectral decomposition, which is a bit of a problem.

 

[MSUmath]

I should point out that the end of the above problem is to show that B$\left|a\right\rangle$ = const. $\left|a +\lambda\right\rangle$

 

[vela]

Homework Statement

Given a Hermitian operator A = $\sum \left|a\right\rangle a \left\langle a\right|$ and B any operator (in general, not Hermitian) such that $\left[A,B\right]$ = $\lambda$B show that B$\left|a\right\rangle$ = const. $\left|a\right\rangle$

I should point out that the end of the above problem is to show that B$\left|a\right\rangle$ = const. $\left|a +\lambda\right\rangle$

These two statements are inconsistent. In both cases, you're applying B to |a>, but you're getting different answers.

You can prove the second statement using the commutation relation.

 

[MSUmath]

They are inconsistent because the statement in the first post is incorrect. The correct relation to prove is the second.