QuantumInfo.ForMathlib.HermitianMat.Unitary

The notation $\mathbf{U}[n]$ denotes the unitary group of degree $n$, which is the group of $n \times n$ matrices $U$ with entries in the complex numbers $\mathbb{C}$ such that $U^\dagger U = I$, where $U^\dagger$ is the conjugate transpose and $I$ is the identity matrix.

For any unitary matrix $u \in \mathbf{U}[\alpha]$, the value of its negation as a unitary matrix is equal to the negation of its underlying matrix, i.e., $(-u).val = -u$.

Given two complex matrices $a \in \mathbb{C}^{\alpha \times \alpha}$ and $b \in \mathbb{C}^{\beta \times \beta}$, the conjugate transpose (adjoint) of their Kronecker product is the Kronecker product of their respective conjugate transposes: $$(a \otimes_k b)^* = a^* \otimes_k b^*$$ where $\otimes_k$ denotes the Kronecker product and $\text{star}$ ($^*$) denotes the conjugate transpose.

Let $a$ and $b$ be unitary matrices of type $\mathbf{U}[\alpha]$ and $\mathbf{U}[\beta]$ respectively. Let $a.\text{val}$ and $b.\text{val}$ denote their underlying matrices in $M_{\alpha}(\mathbb{C})$ and $M_{\beta}(\mathbb{C})$. Then the Kronecker product of these matrices, $a.\text{val} \otimes_k b.\text{val}$, is a unitary matrix in $\mathbf{U}[\alpha \times \beta]$.

Given two unitary matrices $a \in \mathbf{U}(\alpha)$ and $b \in \mathbf{U}(\beta)$, their Kronecker product $a \otimes_k b$ is a unitary matrix in $\mathbf{U}(\alpha \times \beta)$. Here, $\mathbf{U}(n)$ denotes the unitary group of degree $n$ over the complex numbers $\mathbb{C}$.

For any two unitary matrices $a \in \mathbf{U}[\alpha]$ and $b \in \mathbf{U}[\beta]$, the notation $a \otimes_u b$ denotes their Kronecker product (tensor product) as an element of the unitary group $\mathbf{U}[\alpha \times \beta]$. This operation ensures that the resulting matrix remains unitary.

Let $a$ be a unitary matrix of type $\mathbf{U}[\alpha]$ and $b$ be a unitary matrix of type $\mathbf{U}[\beta]$. The entry of the unitary Kronecker product $a \otimes_u b$ at the row index $(i_1, j_1)$ and column index $(i_2, j_2)$ is equal to the product of the entries $a(i_1, i_2)$ and $b(j_1, j_2)$, where $i_1, i_2 \in \alpha$ and $j_1, j_2 \in \beta$.

Let $\mathbf{U}[\alpha]$ and $\mathbf{U}[\beta]$ denote the groups of unitary matrices indexed by types $\alpha$ and $\beta$ respectively. Let $1$ denote the identity unitary matrix of the appropriate dimension. The Kronecker product of the identity matrix in $\mathbf{U}[\alpha]$ and the identity matrix in $\mathbf{U}[\beta]$ is equal to the identity matrix in $\mathbf{U}[\alpha \times \beta]$. That is, $$1_{\mathbf{U}[\alpha]} \otimes_u 1_{\mathbf{U}[\beta]} = 1_{\mathbf{U}[\alpha \times \beta]}$$ where $\otimes_u$ denotes the Kronecker product specialized for unitary matrices.

Let $C$ be a $d \times d$ complex matrix such that $C C^\dagger = I$, where $C^\dagger$ denotes the conjugate transpose and $I$ is the identity matrix. Then for any row index $i$, the sum of the squared norms of the elements in that row is equal to 1, i.e., $\sum_j \|C_{ij}\|^2 = 1$.

For any $d \times d$ complex matrix $C$ such that $C^\dagger C = I$ (where $C^\dagger$ is the conjugate transpose and $I$ is the identity matrix), and for any column index $j \in d$, the sum of the squared norms of the entries in that column is equal to 1, i.e., $\sum_{i} \|C_{ij}\|^2 = 1$.