Subjects Linear Algebra

Unitary Matrix 9Eb83D

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Unitary Matrix 9Eb83D


1. **Problem:** Show that the matrix $$A=\begin{pmatrix} \alpha + i\gamma & -\beta + i\delta \\ \beta + i\delta & \alpha - i\gamma \end{pmatrix}$$ is unitary if $$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 1$$. 2. **Recall:** A matrix $$A$$ is unitary if $$A^\dagger A = I$$, where $$A^\dagger$$ is the conjugate transpose of $$A$$ and $$I$$ is the identity matrix. 3. **Calculate the conjugate transpose:** $$A^\dagger = \begin{pmatrix} \overline{\alpha + i\gamma} & \overline{\beta + i\delta} \\ \overline{-\beta + i\delta} & \overline{\alpha - i\gamma} \end{pmatrix} = \begin{pmatrix} \alpha - i\gamma & \beta - i\delta \\ -\beta - i\delta & \alpha + i\gamma \end{pmatrix}$$ 4. **Multiply $$A^\dagger A$$:** $$\begin{aligned} A^\dagger A &= \begin{pmatrix} \alpha - i\gamma & \beta - i\delta \\ -\beta - i\delta & \alpha + i\gamma \end{pmatrix} \begin{pmatrix} \alpha + i\gamma & -\beta + i\delta \\ \beta + i\delta & \alpha - i\gamma \end{pmatrix} \\ &= \begin{pmatrix} (alpha - i\gamma)(\alpha + i\gamma) + (\beta - i\delta)(\beta + i\delta) & (\alpha - i\gamma)(-\beta + i\delta) + (\beta - i\delta)(\alpha - i\gamma) \\ (-\beta - i\delta)(\alpha + i\gamma) + (\alpha + i\gamma)(\beta + i\delta) & (-\beta - i\delta)(-\beta + i\delta) + (\alpha + i\gamma)(\alpha - i\gamma) \end{pmatrix} \end{aligned}$$ 5. **Simplify diagonal elements:** - For top-left: $$ (\alpha - i\gamma)(\alpha + i\gamma) + (\beta - i\delta)(\beta + i\delta) = \alpha^2 + \gamma^2 + \beta^2 + \delta^2 $$ - For bottom-right: $$ (-\beta - i\delta)(-\beta + i\delta) + (\alpha + i\gamma)(\alpha - i\gamma) = \beta^2 + \delta^2 + \alpha^2 + \gamma^2 $$ 6. **Simplify off-diagonal elements:** - Top-right: $$ (\alpha - i\gamma)(-\beta + i\delta) + (\beta - i\delta)(\alpha - i\gamma) = 0 $$ - Bottom-left: $$ (-\beta - i\delta)(\alpha + i\gamma) + (\alpha + i\gamma)(\beta + i\delta) = 0 $$ 7. **Since $$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 1$$, the diagonal elements equal 1 and off-diagonal elements equal 0, so:** $$ A^\dagger A = I $$ **Therefore, $$A$$ is unitary under the given condition.**