Why Is My Matrix Not Diagonal After Transformation?

[LagrangeEuler]

Homework Statement

Form unitary matrix from eigen vectors of $\sigma_y$ and using that unitary matrix diagonalize $\sigma_y$.
$$\sigma_y= \begin{bmatrix} 0 & -i & \\ i & 0 & \\ \end{bmatrix}$$[/B]

Homework Equations

Eigen vectors of $\sigma_y$ are
$$\vec{X}_1=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ i \\ \end{bmatrix}$$
and
$$\vec{X}_2=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \\ \end{bmatrix}$$

The Attempt at a Solution

I am not sure, where I am making a mistake. Unitary matrix $U$ is defined by
$$U=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 & \\ i & -i & \\ \end{bmatrix}$$
whereas
$$U^{\dagger}=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & -i & \\ 1 & i & \\ \end{bmatrix}$$.
And somehow

$U \sigma_y U^{\dagger}$
is not diagonal. Is there any explanation? Where I am making a mistake?


[/B]

 

[stevendaryl]

$U \sigma_y U^{\dagger}$
is not diagonal. Is there any explanation? Where I am making a mistake?

If $U$ is formed from the eigenvectors of $\sigma_y$, then $U^\dagger \sigma_y U$ will be diagonal, not $U \sigma_y U^\dagger$

In general, let $A$ be a square matrix, and let $u_1$, $u_2$ etc. be eigenvectors with eigenvalues $\lambda_1$, $\lambda_2$, etc. Then let $U$ be the matrix formed by sticking the eigenvectors together: $U = \left( \begin{array}\\ u_1 & u_2 & ...\end{array} \right)$. Then $A U$ will be the matrix

$A U = \left( \begin{array}\\ \lambda_1 u_1 & \lambda_2 u_2 & ...\end{array} \right)$.

Then $U^\dagger$ will be the matrix $U^\dagger = \left( \begin{array}\\ u_1^\dagger \\ u_2^\dagger \\ . \\ . \\. \end{array} \right)$

So since $u_n^\dagger u_m = 0$ or $1$, depending on whether $n=m$, you will have:

$U^\dagger A U$ will be the matrix $U^\dagger = \left( \begin{array}\\ \lambda_1 & 0 & ...\\ 0 & \lambda_2 & 0 & ... \\ 0 & 0 & \lambda_3 & ... \\ . \\. \end{array} \right)$