if matrix `A=(1)/sqrt2[(1,i),(-i,a)], i=sqrt-1` is unitary matrix, a is equal to

Options

A. 2
B. `-1`
C. 0
D. 1

Answer

To find the value of \( a \) such that the matrix

\[ A = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ -i & a \end{pmatrix} \]

is a unitary matrix, we need to use the property of unitary matrices, which states that a matrix \( A \) is unitary if

\[ A A^* = I \]

where \( A^* \) is the conjugate transpose of \( A \) and \( I \) is the identity matrix.

### Step 1: Calculate the conjugate transpose \( A^* \)

The conjugate transpose \( A^* \) is obtained by taking the transpose of \( A \) and then taking the complex conjugate of each element.

\[ A^* = \left( A^T \right)^* = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ i & a \end{pmatrix} \]

### Step 2: Calculate \( A A^* \)

Now, we need to multiply \( A \) by \( A^* \):

\[ A A^* = \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ -i & a \end{pmatrix} \right) \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ i & a \end{pmatrix} \right) \]

Calculating the product:

\[ A A^* = \frac{1}{2} \begin{pmatrix} 1 \cdot 1 + i \cdot i & 1 \cdot (-i) + i \cdot a \\ -i \cdot 1 + a \cdot i & -i \cdot (-i) + a \cdot a \end{pmatrix} \]

Calculating each entry:

1. First row, first column:
\[ 1 + i^2 = 1 - 1 = 0 \]

2. First row, second column:
\[ -i + ai \]

3. Second row, first column:
\[ -i + ai \]

4. Second row, second column:
\[ -(-1) + a^2 = 1 + a^2 \]

Thus, we have:

\[ A A^* = \frac{1}{2} \begin{pmatrix} 0 & -i + ai \\ -i + ai & 1 + a^2 \end{pmatrix} \]

### Step 3: Set \( A A^* \) equal to the identity matrix

For \( A \) to be unitary, we need:

\[ A A^* = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

This gives us the following equations:

1. From the first row, first column:
\[ \frac{1}{2} \cdot 0 = 1 \quad \text{(not possible)} \]

2. From the first row, second column:
\[ \frac{1}{2}(-i + ai) = 0 \]

3. From the second row, second column:
\[ \frac{1}{2}(1 + a^2) = 1 \]

### Step 4: Solve the equations

From the second equation:

\[ -i + ai = 0 \implies ai = i \implies a = 1 \]

From the third equation:

\[ 1 + a^2 = 2 \implies a^2 = 1 \implies a = \pm 1 \]

### Step 5: Determine the valid value of \( a \)

Substituting \( a = 1 \) back into the equations, we find that it satisfies the unitary condition. However, substituting \( a = -1 \) leads to contradictions in the unitary condition.

Thus, the value of \( a \) must be:

\[ \boxed{-1} \]