What Condition Determines Eigenket of A?

[zhaiyujia]

Homework Statement

Suppose that |$$\alpha$$> and |$$\beta$$> are eigenkets(eigenfunctions) of a hermitian operator A. Under what condition can we conclude that |$$\alpha$$> + |$$\beta$$> is also an eigenket of A?

Homework Equations

It's quite basic, I don't think any addtional equations are needed except the definations.

The Attempt at a Solution

From the question we know that A| $$\alpha$$ > =a|$$\alpha$$> , A|$$\beta$$> =b|$$\beta$$>. And A is a hermitian operator:
<$$\alpha$$|A$$\beta$$>=<$$\alpha$$|b$$\beta$$>=b<$$\alpha$$|$$\beta$$>,
<$$\alpha$$|A$$\beta$$>=<A$$\alpha$$|$$\beta$$>=<a$$\alpha$$|$$\beta$$>=a<$$\alpha$$|$$\beta$$>,
Therefore a=b? or <$$\alpha$$|$$\beta$$>=0?
But it's nothing to do with |$$\alpha$$>+|$$\beta$$>+
It seems no addition is need to constrain on them

 

[Hurkyl]

It seems no addition is need to constrain on them

Of course not; you never applied the condition that $|\alpha\rangle + |\beta\rangle$ is to be an eigenket!

 

[zhaiyujia]

But what is the condition? if my first part is right:
A(|$$\alpha$$>+|$$\beta$$>)=a(|$$\alpha$$>+|$$\beta$$>)
is automatic right?

 

[Hurkyl]

Well, no, it's not automatic. Your first part said

"a = b" or "$\langle \alpha | \beta \rangle$ = 0".​

So, you have to consider both cases, not just the "a = b" case.