Hamiltonian diagonalization

[aaaa202]

Homework Statement

The exercise: https://www.physicsforums.com/attachment.php?attachmentid=64229&d=1385257430

Homework Equations

Are in my attempt at a solution. I am sure it would be easier to use the transformation equation for the operator and plug it into the diagonalized Hamiltonian at the bottom. The problem is I don't really understand this procedure so I have started from the basic equation for diagonalizing a matrix. The problem is now that I will need the matrix elements of H in the old basis and I know really see what basis the Hamiltonian is even written in. Which is that?

The Attempt at a Solution

How I tried to solve it: https://www.physicsforums.com/attachment.php?attachmentid=64283&d=1385405657

 

[kurt.physics]

I started out by trying to diagonalize the Hamiltonian by finding the eigenvalues and eigenvectors. I found that the eigenvalues are $E_1 = 2, E_2 = -2$ and the corresponding eigenvectors are $|\psi_1 \rangle=\frac{1}{\sqrt2}(|0 \rangle + |1 \rangle)$ and $|\psi_2 \rangle=\frac{1}{\sqrt2}(|0 \rangle - |1 \rangle)$. I then tried to use these eigenvectors to find the transformation matrix U between the two bases. I assumed that the new basis would be $\{|\psi_1 \rangle, |\psi_2 \rangle\}$ and the old basis would be $\{|0 \rangle, |1 \rangle\}$. This gives me the transformation matrixU = $\begin{pmatrix} \frac{1}{\sqrt2} & \frac{1}{\sqrt2} \\ \frac{1}{\sqrt2} & -\frac{1}{\sqrt2} \end{pmatrix}$I then used this transformation matrix to find the matrix elements of the Hamiltonian in the new basis. H' = UHU$^{-1}$ = $\begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix}$This is the diagonalized form of the Hamiltonian.