Question

If \( \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1+i}{\sqrt{3}} \\ \frac{1-i}{\sqrt{3}} & \frac{K}{\sqrt{3}} \end{bmatrix} \) is a unitary matrix, then the sum of all possible values of K is

• A. 0
• B. 1
• C. --1
• D. 2

Hint:

A matrix A is unitary if \(AA^\dagger = I\), where \(A^\dagger = (\bar{A})^T\).
For the product \(AA^\dagger\) to be the identity matrix, diagonal elements must be 1 and off-diagonal elements must be 0.

Answer 1

The Correct Option is C

Let the given matrix be \(A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1+i}{\sqrt{3}} \\ \frac{1-i}{\sqrt{3}} & \frac{K}{\sqrt{3}} \end{bmatrix}\). For A to be unitary, \(AA^\dagger = I\), where \(A^\dagger\) is the conjugate transpose of A. \(A^\dagger = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1+i}{\sqrt{3}} \\ \frac{1-i}{\sqrt{3}} & \frac{\bar{K}}{\sqrt{3}} \end{bmatrix}\). \(AA^\dagger = \frac{1}{3} \begin{bmatrix} 1 & 1+i \\ 1-i & K \end{bmatrix} \begin{bmatrix} 1 & 1+i \\ 1-i & \bar{K} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1+(1+i)(1-i) & (1+i) + (1+i)\bar{K} \\ (1-i) + K(1-i) & (1-i)(1+i) + K\bar{K} \end{bmatrix}\) \(AA^\dagger = \frac{1}{3} \begin{bmatrix} 1+2 & (1+i)(1+\bar{K}) \\ (1-i)(1+K) & 2+|K|^2 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 & (1+i)(1+\bar{K})
(1-i)(1+K) & 2+|K|^2 \end{bmatrix}\). For \(AA^\dagger = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\): From element (1,1): \(\frac{1}{3}(3) = 1\), which is correct. From element (1,2): \(\frac{1}{3}(1+i)(1+\bar{K}) = 0 \Rightarrow (1+i)(1+\bar{K}) = 0\). Since \(1+i \neq 0\), then \(1+\bar{K}=0 \Rightarrow \bar{K}=-1 \Rightarrow K=-1\). From element (2,1): \(\frac{1}{3}(1-i)(1+K) = 0 \Rightarrow (1-i)(1+K) = 0\). Since \(1-i \neq 0\), then \(1+K=0 \Rightarrow K=-1\). From element (2,2): \(\frac{1}{3}(2+|K|^2) = 1 \Rightarrow 2+|K|^2=3 \Rightarrow |K|^2=1\). If \(K=-1\), then \(|K|^2 = |-1|^2 = 1\), which is satisfied. The only possible value for K is -1. The sum of all possible values of K is -1. \[ \boxed{\text{--1}} \]