## Bra and Ket States

Consider an operator defined as a product of bra and ket vectors:
$\hat H =E (|\varphi_1><\varphi_1|-|\varphi_2><\varphi_2|-i|\varphi_1><\varphi_2|+i|\varphi_2><\varphi_1|)$
where $E$ is a real constant, and $\phi_i>$ form an orthonormal basis in the two dimensional Hilbert space.
a) Find the matrix representing this operator in the basis of $|\phi_i>$.
b) Using eigen vectors of this matrix construct a transformation operator $\hat U$ that will diagonalize it. Verify it by applying the transformation operator to $H$.

Notation $|φ_n><φ_m |=|n m|$   so that $H=E(|11|-|22|-i|12|+i|21|)$
a)
There are just two elements in the basis $|φ_1>$  and $|φ_2>$ so that
$<φ_k |(|φ_n><φ_m | )| φ_l>=<k|n m|l>=δ_{kn} δ_{ml}$
$H_{11}=<φ_1 |H| φ_1>=E(<1|11|1>-<1|22|1>-i<1|12|1>+i<1|21|1>)=E$
$H_{12}=<1|H|2>=E(<1|11|2>-<1|22|2>-i<1|12|2>+i<1|21|2>)=-i E$
$H_{21}=<2|H|1> =E(<2|11|1>-<2|22|1>-i<2|12|1>+i<2|21|1>)=+i E$
$H_{22}=<2|22|2>=…=-E$

$H=\begin{pmatrix}E&-iE\\+i E&-E\end{pmatrix}=E\begin{pmatrix}1&-i\\i&-1\end{pmatrix}$

b)
Eigenvalues are $λ$ with
$det⁡(H-λI)=0$    or $\begin{vmatrix}E-λ&-iE\\iE&-E-λ\end{vmatrix}=-(E^2-λ^2 )-E^2=0$  with $λ_{1,2}=±E√2$

Eigen vectors X are
$(H-λI)*X=0$

For $λ=+E√2$
$\begin{pmatrix}E-E√2&-iE\\iE&-E-E√2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$  so $\left\{\begin{matrix}E(1-√2)x-iEy=0\\iEx-E(1+√2)y=0\end{matrix}\right.$

or $\left\{\begin{matrix}x=-i(1+√2)*α\\y=α\end{matrix}\right.$
with $α$ any real number  so that the eigenvector is $X_1=\begin{pmatrix}-i(1+√2)\\1\end{pmatrix}$

In a similar way for $λ=-E√2$   one obtains $X_2=\begin{pmatrix}i(-1+√2)\\1\end{pmatrix}$
The transformation matrix $U$ that diagonalize $H$ is the matrix having $X1$ and $X2$ (the eigen vectors) as columns
$U=\begin{pmatrix}-i(1+√2)&i(-1+√2\\1&1\end{pmatrix}$     and
$H_{diagonal}=U^{-1}*H*U$

Verification
for $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$  one has
$U^{-1}=\frac{1}{ad-bc}*\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\frac{1}{(-2i√2)} \begin{pmatrix}1&i(1-√2)\\-1&-i(1+√2)\end{pmatrix}$

$U^{-1} HU=\frac{1}{(-i2√2)} \begin{pmatrix}1&i(1-√2)\\-1&-i(1+√2)\end{pmatrix}\begin{pmatrix}1&-i\\i&-1\end{pmatrix}\begin{pmatrix}-i(1+√2)&i(-1+√2)\\1&1\end{pmatrix}=$

$=\begin{pmatrix}λ_1&0\\0&λ_2\end{pmatrix}=\begin{pmatrix}√2&0\\0&-√2\end{pmatrix}$

valentin68