QuantumInfo.ForMathlib.HermitianMat.Jordan

Given two Hermitian matrices $A$ and $B$ in the space $\text{HermitianMat}_d(\mathbb{k})$, the symmetric product $A \cdot B$ is defined as the Jordan product: $$A \cdot B = \frac{1}{2} (AB + BA)$$ where $AB$ and $BA$ denote standard matrix multiplication. The resulting matrix is also Hermitian.

Let $A$ and $B$ be Hermitian matrices of dimension $d$ over the field $\mathbb{k}$. The symmetric product (or Jordan product) $\circ$, defined by $A \circ B = \frac{1}{2}(AB + BA)$, is commutative, such that $A \circ B = B \circ A$.

Let $A$ be a Hermitian matrix. The symmetric product of $A$ and the zero matrix, defined as $A \odot 0 = \frac{1}{2}(A0 + 0A)$, is equal to the zero matrix.

Let $A$ be a Hermitian matrix. The symmetric product (Jordan product) of the zero matrix and $A$, defined as $0 \circ A = 2^{-1}(0 \cdot A + A \cdot 0)$, results in the zero matrix.

Let $A$ and $B$ be Hermitian matrices of dimension $d$ over a field $\mathbb{k}$. The matrix representation of their symmetric product $A \cdot B$ (also known as the Jordan product) is given by: $$(A \cdot B)_{\text{mat}} = 2^{-1} \cdot (A_{\text{mat}} B_{\text{mat}} + B_{\text{mat}} A_{\text{mat}})$$ where $A_{\text{mat}}$ and $B_{\text{mat}}$ are the underlying matrices of $A$ and $B$ respectively.

Let $A$ and $B$ be Hermitian matrices of dimension $d$ over a field $\mathbf{k}$. If the underlying matrices of $A$ and $B$ commute (i.e., $A.mat \cdot B.mat = B.mat \cdot A.mat$), then their symmetric product $A \circ B$ (defined as $2^{-1} (A.mat \cdot B.mat + B.mat \cdot A.mat)$) has an underlying matrix equal to the standard matrix product $A.mat \cdot B.mat$.

Let $A$ be a Hermitian matrix of dimension $d$ over a field $\mathbf{k}$. Let $\text{symmMul}(A, B)$ denote the symmetric product $2^{-1} (AB + BA)$ defined for Hermitian matrices. Then, the matrix representation of $\text{symmMul}(A, A)$ is equal to the square of the matrix representation of $A$, i.e., $A_{\text{mat}} \cdot A_{\text{mat}}$.

Let $A$ be a Hermitian matrix of dimension $d$ over a field $\mathbf{k}$. Let the symmetric product of two Hermitian matrices $A$ and $B$ be defined as $A \circ B = \frac{1}{2}(AB + BA)$, where $AB$ denotes standard matrix multiplication. Then, the symmetric product of $A$ with the identity matrix $I$ is equal to $A$, i.e., $A \circ I = A$.

Let $A$ be a Hermitian matrix. The symmetric multiplication (Jordan product) of the identity matrix $I$ and $A$, defined by $I \circ A = \frac{1}{2}(IA + AI)$, is equal to $A$.

For any Hermitian matrix $A$, let $A \circ B = \frac{1}{2}(AB + BA)$ denote the symmetric product of Hermitian matrices. Then, the symmetric product of $A$ and the negative identity matrix $-I$ satisfies $A \circ (-I) = -A$.

For any Hermitian matrix $A$, let the symmetric product (Jordan product) be defined as $A * B = \frac{1}{2}(AB + BA)$. Then the symmetric product of the negative identity matrix $-1$ and the matrix $A$ satisfies $\text{symmMul}(-1, A) = -A$.

For the space of Hermitian matrices of dimension $d$ over a field $\mathbb{k}$, denoted as $\text{HermitianMat}(d, \mathbb{k})$, there exists a commutative magma structure where the binary operation is defined by the symmetric product $A * B = \frac{1}{2}(AB + BA)$. This structure ensures that the multiplication is commutative, satisfying $A * B = B * A$ for all Hermitian matrices $A$ and $B$.

In the context of the `HermMul` scope, the multiplication operation $A * B$ between two Hermitian matrices $A, B \in \text{HermitianMat}(d, \mathbb{k})$ is equal to the symmetric product $A \cdot_{\text{symm}} B$, defined as $\frac{1}{2}(AB + BA)$.

Let $H_d(\mathbb{k})$ be the space of $d \times d$ Hermitian matrices over a field $\mathbb{k}$. When equipped with the symmetric product (also known as the Jordan product) defined by $A * B = \frac{1}{2}(AB + BA)$, the space $H_d(\mathbb{k})$ forms a commutative Jordan algebra. This means the multiplication is commutative, $A * B = B * A$, and satisfies the Jordan identity $(A * B) * A^2 = A * (B * A^2)$.

For the space of Hermitian matrices $\text{HermitianMat}_d(\mathbb{k})$ equipped with the symmetric product $A * B = \frac{1}{2}(AB + BA)$, the zero matrix $\mathbf{0}$ satisfies the property that $\mathbf{0} * A = \mathbf{0}$ and $A * \mathbf{0} = \mathbf{0}$ for any Hermitian matrix $A$. This endows the Hermitian matrices with the structure of a `MulZeroClass`.

For the set of Hermitian matrices of dimension $d$ over a field $\mathbb{k}$, the Jordan product defined by $A \circ B = \frac{1}{2}(AB + BA)$ satisfies the properties of a `MulZeroOneClass`. This means the operation has a multiplicative identity element $I$ such that $I \circ A = A \circ I = A$, and a zero element $0$ such that $0 \circ A = A \circ 0 = 0$ for any Hermitian matrix $A$.

The set of Hermitian matrices of dimension $d$ over a field $\mathbb{k}$, denoted as $\text{HermitianMat}(d, \mathbb{k})$, forms a non-unital non-associative ring. The multiplication operation in this structure is defined by the symmetric product $A * B = \frac{1}{2}(AB + BA)$, where $AB$ and $BA$ are standard matrix multiplications. This structure satisfies the left and right distributive laws: $A * (B + C) = A * B + A * C$ and $(A + B) * C = A * C + B * C$.

For a given dimension $d$ and a field $\mathbb{k}$ (typically $\mathbb{R}$ or $\mathbb{C}$), the set of Hermitian matrices $\text{Herm}_d(\mathbb{k})$ equipped with the symmetric product $A \circ B = \frac{1}{2}(AB + BA)$ forms a non-associative ring. This structure satisfies the axioms of an additive abelian group and possesses a distributive, non-associative multiplication with a multiplicative identity (the identity matrix $I$).

Let $\text{HermitianMat}_d(\mathbb{k})$ be the space of $d \times d$ Hermitian matrices over a field $\mathbb{k}$ endowed with the symmetric Jordan product $A \ast B = \frac{1}{2}(AB + BA)$. There exists an instance of a scalar tower for $\mathbb{R}$, $\text{HermitianMat}_d(\mathbb{k})$, and $\text{HermitianMat}_d(\mathbb{k})$. Specifically, for any $r \in \mathbb{R}$ and $A, B \in \text{HermitianMat}_d(\mathbb{k})$, the scalar multiplication satisfies $(r \cdot A) \ast B = r \cdot (A \ast B)$.