Unitary Matrices
A unitary matrix is a complex matrix $ U $ whose product with its conjugate transpose $ U^{\dagger} $ yields the identity matrix $ I $: $$ U^{\dagger} U = U U^{\dagger} = I $$ In this expression, $ U^{\dagger} $ denotes the conjugate transpose of $ U $, and $ I $ represents the identity matrix.
Put simply, a matrix is unitary if its inverse $ U^{-1} $ is equal to its conjugate transpose $ U^{\dagger} $.
$$ U^{-1} = U^{\dagger} $$
In either case, the matrix product results in the identity matrix: $ U^{-1} U = I $ and $ U^{\dagger} U = I $.
Note. Unitary matrices are defined and studied over the complex field. In the real case, the corresponding concept is the orthogonal matrix, for which $ O^{-1} = O^T $.
A unitary matrix preserves both lengths and angles, just as a rotation preserves distances in real space.
A practical example
Consider the complex matrix
$$ U = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} $$
Its conjugate transpose is
$$ U^{\dagger} = \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} $$
How do we compute the conjugate transpose? To find the conjugate transpose of a matrix $$
U = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} $$ first exchange rows and columns to obtain the transpose. Since this matrix is diagonal (non-zero entries only along the main diagonal), the transpose remains unchanged: $$
U^T = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} $$ Then take the complex conjugate by changing the sign of each element’s imaginary part. For example, $ \overline{i} = -i $, $ \overline{1} = 1 $, and $ \overline{0} = 0 $. $$ U^{\dagger} =
\begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} $$
The row-by-column multiplication of the two matrices gives the identity matrix:
$$ U^{\dagger}U = \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (-i)(i) & 0 \cdot 0 \\ 0 \cdot 0 & 1 \cdot 1 \end{pmatrix} $$
$$ U^{\dagger}U = \begin{pmatrix} -i^2 & 0 \\ 0 & 1 \end{pmatrix} = I $$
In the complex field, the square of the imaginary unit is $ i^2 = -1 $, so $ -i^2 = -(-1) = 1 $.
$$ U^{\dagger}U = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I $$
Hence, the matrix $ U $ is unitary.
The unitary groups U and SU
The set of all $ n \times n $ unitary matrices forms a group under matrix multiplication, known as the unitary group, denoted by $ U(n) $.
$$ U(n) = \{ U \in \mathbb{C}^{n\times n} \mid U^{\dagger}U = I \} $$
Each element of $ U(n) $ is a complex $ n \times n $ matrix that satisfies the unitarity condition.
Note. The group $ U(n) $ satisfies the four fundamental group axioms:
- Closure: if $ U_1 $ and $ U_2 $ are unitary, their product $ U_1 U_2 $ is also unitary, since $ (U_1U_2)^{\dagger}(U_1U_2) = U_2^{\dagger}U_1^{\dagger}U_1U_2 = U_2^{\dagger}I U_2 = I $.
- Identity element: the identity matrix $ I_n $ is unitary because $ I^{\dagger}I = I $.
- Inverse element: every unitary matrix is invertible, and its inverse is given by $ U^{-1} = U^{\dagger} $. Thus, the inverse of any element in $ U(n) $ also belongs to the group.
- Associativity: matrix multiplication is always associative, so this property automatically holds.
The special unitary group $ SU(n) $ is the subgroup of $ U(n) $ consisting of matrices with determinant equal to 1:
$$ SU(n) = \{ U \in U(n) \mid \det(U) = 1 \} $$
It is called “special” because the determinant constraint removes any overall phase factor, ensuring a unit determinant.
And so the structure continues.
