Question

Let the operator \(\mathbf{U}: \mathbb{C}^{2} \rightarrow \mathbb{C}^{2}\) be given by $$ \mathbf{U}\left(\begin{array}{l} \alpha_{1} \\ \alpha_{2} \end{array}\right)=\left(\begin{array}{l} i \frac{\alpha_{1}}{\sqrt{2}}-i \frac{\alpha_{2}}{\sqrt{2}} \\ \frac{\alpha_{1}}{\sqrt{2}}+\frac{\alpha_{2}}{\sqrt{2}} \end{array}\right) $$ Find \(\mathbf{U}^{\dagger}\) and test if \(\mathbf{U}\) is unitary.

Short answer

The conjugate transpose of matrix \(\mathbf{U}\) is \[ \mathbf{U}^{\dagger} = \left( \begin{array}{cc} -i/\sqrt{2} & 1/\sqrt{2} \\ i/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \]. By using matrix multiplication, we found out that \(\mathbf{U}\mathbf{U}^{\dagger}=\mathbf{I}\), the identity matrix. Hence, the matrix \( \mathbf{U} \) is unitary.

Step-by-step solution

Identify the Matrix U

The first step is to identify the matrix \(\mathbf{U}\). This can be done by writing out the transformation as a matrix. In this case, the operator \(\mathbf{U}\) is given to us as: \[ \mathbf{U}= \left( \begin{array}{cc} i/\sqrt{2} & -i/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \]

Find the Conjugate Transpose (\(\mathbf{U}^{\dagger}\)) of the Matrix

The conjugate transpose of a matrix, also known as the adjoint matrix, is the transpose of the matrix with every element replaced by its complex conjugate. This process involves two steps: first taking the transpose of the matrix and then finding the complex conjugate. In this case, the transpose of \(\mathbf{U}\) is \[ \mathbf{U}^T = \left( \begin{array}{cc} i/\sqrt{2} & 1/\sqrt{2} \\ -i/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \] And the complex conjugate of \(\mathbf{U}^T\) is \[ \mathbf{U}^{\dagger} = \left( \begin{array}{cc} -i/\sqrt{2} & 1/\sqrt{2} \\ i/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \]

Test if the matrix U is unitary

A matrix is unitary if the matrix times its conjugate transpose equals the identity matrix. In other words, if \(\mathbf{U}\mathbf{U}^{\dagger}=\mathbf{I}\), then the matrix \(\mathbf{U}\) is unitary. We have \[ \mathbf{U}\mathbf{U}^{\dagger} = \left( \begin{array}{cc} i/\sqrt{2} & -i/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{array} \right) \left( \begin{array}{cc} -i/\sqrt{2} & 1/\sqrt{2} \\ i/\sqrt{2} & 1/\sqrt{2} \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) = \mathbf{I} \] This gives us the identity matrix, so \(\mathbf{U}\) is indeed unitary.

Key concepts

Conjugate Transpose

Understanding the conjugate transpose is crucial in the world of linear algebra, especially when dealing with complex vector spaces. Imagine you have a matrix with complex numbers as its entries. To find its conjugate transpose, denoted by \( \mathbf{U}^\dagger \), you'll need to follow two steps: transpose the matrix and take the complex conjugate of each entry.

The transpose is like flipping the matrix over its diagonal, turning rows into columns and vice-versa. After transposing, you would then replace each element by its complex conjugate, which means flipping the sign of the imaginary part in complex numbers. For example, the complex conjugate of \( i \) is \( -i \), and for \( -i \) it is \( i \).

Applying Conjugate Transpose in the Solution

When you applied this to the given matrix \( \mathbf{U} \), the steps led to the matrix \( \mathbf{U}^\dagger \), which is essential for determining if \( \mathbf{U} \) is unitary. This process is pivotal in quantum mechanics and many areas of mathematics, where the behavior of complex systems is explored.

Complex Vector Spaces

Complex vector spaces extend the idea of vector spaces that you might know from real numbers to the field of complex numbers. This allows for a whole new landscape of mathematical and physical interpretations. In these spaces, vectors have components that are complex numbers, and they can be transformed by operators like our matrix \( \mathbf{U} \).

A vector space is, simply put, a collection of vectors that can be scaled and added together while still remaining in the same space. This means if you take two vectors in this space and add them or multiply by a scalar (which, in complex spaces, is a complex number), the resultant vector will still be a member of this space.

Connection to the Solution

In the given exercise, the transformation by \( \mathbf{U} \) operates within a complex vector space, mapping one two-dimensional complex vector to another. Each complex number in these vectors corresponds to a point in a two-dimensional plane, reflecting both magnitude and direction, crucial for understanding how \( \mathbf{U} \) alters these points.

Identity Matrix

The identity matrix, often notated as \( \mathbf{I} \), is a simple yet fundamental concept in linear algebra. It is the equivalent of the number '1' in matrix form. This special matrix has 1's on the diagonal and 0's everywhere else; it acts as a multiplicative identity for matrices. When you multiply any matrix by the identity matrix, the original matrix remains unchanged.

For simplicity, think of the identity matrix as a 'do-nothing' operation. It's like a mirror reflecting whatever vector or matrix you present to it, unchanged and intact.

Role in the Solution

The resolution of our matrix being unitary or not hinged on the product of \( \mathbf{U} \) and its conjugate transpose \( \mathbf{U}^\dagger \) resulting in the identity matrix \( \mathbf{I} \). This outcome validates that \( \mathbf{U} \) preserves the length of the vectors it operates on—a key feature of unitary operators.