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Consider the 4-by-4 matrix $\boldsymbol M = \boldsymbol M_0 + \boldsymbol M_1$, where

$\boldsymbol M_0 = \alpha \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{matrix} \right)$ and $\boldsymbol M_1 = \beta \left( \begin{matrix} 0 & \gamma & 0 & -\gamma^{*} \\ \gamma^{*} & 0 & -\gamma^{*} & 0 \\ 0 & \gamma & 0 & -\gamma^{*} \\ \gamma & 0 & -\gamma & 0 \\ \end{matrix} \right)$

where $\alpha$ and $\beta$ are constants and $\gamma = \gamma_x + i \gamma_y$ is complex.

Is it possible to unitary transform $\boldsymbol M$ into block off-diagonal form $\boldsymbol M_B$?

Namely, I want to find a unitary transform $\boldsymbol U$ so that I can write down $\boldsymbol M_B = \boldsymbol U \boldsymbol M \boldsymbol U^{*}$ (here $\boldsymbol U^{*}$ is the conjugate transpose).

Explicitly, the required block off-diagonal matrix is (in general form)

$\boldsymbol M_B = \left( \begin{matrix} 0 & \boldsymbol Q \\ \boldsymbol Q^{*} & 0 \end{matrix} \right)$ where $\boldsymbol Q = \left( \begin{matrix} Q_z & Q_x - i Q_y \\ Q_x + i Q_y & -Q_z \end{matrix} \right)$

Is there a general recipe to find such a unitary transformation matrix $\boldsymbol U$ which leads to the block off-diagonal form, $\boldsymbol M \to \boldsymbol M_B$?

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    $\begingroup$ What do you mean by "transform"? Are you looking for a similar matrix? If not, what are you trying to do? Are you trying to compute a determinant? $\endgroup$ – Ben Grossmann Feb 9 '16 at 14:42
  • $\begingroup$ @Omnomnomnom I try to find a unitary transform matrix to allow me to pass from my original matrix to its block-off diagonal form. I am not interested in determinants (unless it helps to acheive this goal!). I tried to edit the question for clarity. $\endgroup$ – Nigel1 Feb 10 '16 at 16:59
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I think this $4 \times 4$ matrix is the answer.

$$\frac{1}{\sqrt2} \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ -I & 0 & I & 0 \\ 0 & -I & 0 & I \end{pmatrix}$$

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