# Calculating eigenstates of an operator

1. Feb 3, 2015

### barefeet

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

2. Relevant equations
Just linear algebra rules.

3. 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.

2. Feb 3, 2015

### TSny

Are you sure that's the only solution?

3. Feb 3, 2015

### 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?

4. Feb 3, 2015

### TSny

That's it.

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

5. Feb 3, 2015

Thanks again