# Homework Help: QM eigenket

1. Oct 18, 2007

### zhaiyujia

1. The problem statement, all variables and given/known data
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?

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

3. 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

Last edited: Oct 18, 2007
2. Oct 18, 2007

### Hurkyl

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

3. Oct 18, 2007

### zhaiyujia

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

4. Oct 18, 2007

### Hurkyl

Staff Emeritus
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.