Calculating eigenstates of an operator

[barefeet]

Homework Statement

Consider a two-dimensional space spanned by two orthonormal state vectors \mid \alpha \rangle and \mid \beta \rangle. An operator is expressed in terms of these vectors as
A = \mid \alpha \rangle \langle \alpha \mid + \lambda \mid \beta \rangle \langle \alpha \mid + \lambda^* \mid \alpha \rangle \langle \beta \mid + \mu \mid \beta \rangle \langle \beta \mid

Determine the eigenstates of A for the case where (i) \lambda = 1, \mu = \pm 1, (ii) \lambda = i, \mu = \pm 1. Do this problem also by expressing A as a 2 X 2 matrix with eigenstates as the column vectors.

Homework Equations

Just linear algebra rules.

The Attempt at a Solution

I started with \lambda = 1, \mu = 1. Then A is:
A = \mid \alpha \rangle \langle \alpha \mid + \mid \beta \rangle \langle \alpha \mid + \mid \alpha \rangle \langle \beta \mid + \mid \beta \rangle \langle \beta \mid

A \mid \alpha \rangle = \mid \alpha \rangle \langle \alpha \mid \alpha \rangle + \mid \beta \rangle \langle \alpha \mid \alpha \rangle + \mid \alpha \rangle \langle \beta \mid \alpha \rangle + \mid \beta \rangle \langle \beta \mid \alpha \rangle = \mid \alpha \rangle + \mid \beta \rangle

A \mid \beta \rangle = \mid \alpha \rangle \langle \alpha \mid \beta \rangle + \mid \beta \rangle \langle \alpha \mid \beta \rangle + \mid \alpha \rangle \langle \beta \mid \beta \rangle + \mid \beta \rangle \langle \beta \mid \beta \rangle = \mid \alpha \rangle + \mid \beta \rangle

The eigenstate is \mid a_n \rangle with eigenvalue a_n. Then the following holds:
A \mid a_n \rangle = a_n \mid a_n \rangle

The eigenstate \mid a_n \rangle can be expressed in the basis vectors \mid \alpha \rangle and \mid \beta \rangle:

\mid a_n \rangle = c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle

Then the earlier equation becomes:
A \mid a_n \rangle = A( c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle ) = a_n (c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle

But this is also:
A( c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle ) = c_1 A \mid \alpha \rangle + c_2 A \mid \beta \rangle = c_1 (\mid \alpha \rangle + \mid \beta \rangle) + c_2 (\mid \alpha \rangle + \mid \beta \rangle) \\ = (c_1 + c_2) \mid \alpha \rangle + (c_1 + c_2) \mid \beta \rangle

This gives the equations :
a_n c_1 = c_1 + c_2
a_n c_2 = c_1 + c_2

The only solution is if c_1 = c_2 = 0. Obviously I am doing something wrong but I can't see it.

 

[TSny]

This gives the equations :
a_n c_1 = c_1 + c_2
a_n c_2 = c_1 + c_2
The only solution is if c_1 = c_2 = 0. Obviously I am doing something wrong but I can't see it.

Are you sure that's the only solution?

 

[barefeet]

I see, but I still get one non trivial solution out of it.
If one is 0, then both are 0. So they are either both nonzero or both 0. If they are nonzero, I can write:
a_n = \frac{c_1 + c_2}{c_1} = \frac{c_1 + c_2}{c_2}
If c_1 + c_2 =0 then a_n =0 or the c's are 0. If c_1 + c_2 is nonzero, then I can divide by it and gives me c_1 = c_2 with a_n = 2 Unnormalized I can take as eigenstate \mid \alpha \rangle + \mid \beta \rangle.
Now I can't find the other eigenstate. \mid \alpha \rangle - \mid \beta \rangle is orthogonal to this eigenstate but letting A operate on it gives me 0. Or does this just mean that it is an eigenstate with eigenvalue 0?

 

[TSny]

Or does this just mean that it is an eigenstate with eigenvalue 0?

That's it.

Another approach is to subtract the two equations:
$a_nc_1 = c_1+c_2$
$a_nc_2 = c_1+c_2$

 

[barefeet]

Thanks again