QuantumInfo.Finite.CPTPMap.Dual

Let $R$ be a semiring. For a matrix map $M: \text{Mat}_{d_{In} \times d_{In}}(R) \to \text{Mat}_{d_{Out} \times d_{Out}}(R)$, its dual map $M^* : \text{Mat}_{d_{Out} \times d_{Out}}(R) \to \text{Mat}_{d_{In} \times d_{In}}(R)$ is the unique linear map defined such that for all matrices $A$ and $B$, the trace identity $\text{tr}(M(A) B) = \text{tr}(A M^*(B))$ holds. Formally, this is constructed by identifying the space of matrices with its dual space using the trace form and taking the algebraic dual of the linear map $M$.

Let $R$ be a semiring. For any matrix map $M: \text{Mat}_{d_{In} \times d_{In}}(R) \to \text{Mat}_{d_{Out} \times d_{Out}}(R)$ and any matrices $A \in \text{Mat}_{d_{In} \times d_{In}}(R)$ and $B \in \text{Mat}_{d_{Out} \times d_{Out}}(R)$, the trace of the product of $M(A)$ and $B$ is equal to the trace of the product of $A$ and the dual map $M^*$ applied to $B$. That is, $\text{tr}(M(A)B) = \text{tr}(A M^*(B))$.

Let $M$ be a matrix map. If $M$ is Hermitian-preserving, then its dual map $M^*$ is also Hermitian-preserving.

Let $n$ be a finite type with decidable equality, and let $A$ and $B$ be $n \times n$ matrices over a field $\mathbb{k}$ (typically $\mathbb{R}$ or $\mathbb{C}$). If both $A$ and $B$ are positive semidefinite, then the trace of their product is non-negative, i.e., $0 \le \text{tr}(AB)$.

Let $M: \text{Mat}_{d_{In} \times d_{In}}(R) \to \text{Mat}_{d_{Out} \times d_{Out}}(R)$ be a matrix map between matrices over a semiring $R$. If $M$ is a positive map (meaning it maps positive semidefinite matrices to positive semidefinite matrices), then its dual map $M^*$ is also a positive map.

Let $M$ be a matrix map. If $M$ is trace-preserving, then its dual map $M^*$ is unital, meaning $M^*(I) = I$ where $I$ is the identity matrix.

Let $\mathbb{k}$ be a field. For any two matrix maps $M: \text{Mat}_{d_{in} \times d_{in}}(\mathbb{k}) \to \text{Mat}_{d_{out} \times d_{out}}(\mathbb{k})$ and $M': \text{Mat}_{d_{out} \times d_{out}}(\mathbb{k}) \to \text{Mat}_{d_{in} \times d_{in}}(\mathbb{k})$, if for all matrices $A$ and $B$, the trace equality $\text{tr}(M(A)B) = \text{tr}(A M'(B))$ holds, then $M'$ is equal to the dual map $M^*$.

Let $M$ be a matrix map from matrices over $d_{in}$ to matrices over $d_{out}$ with coefficients in a field $\mathbb{k}$. The Choi matrix of the dual map $M^*$, denoted $\mathcal{C}(M^*)$, is equal to the transpose of the Choi matrix of the original map $M$, $\mathcal{C}(M)^T$, after reindexing the row and column indices using the product commutation equivalence $(d_{out} \times d_{in}) \cong (d_{in} \times d_{out})$. Specifically, $$\mathcal{C}(M^*) = \text{reindex}(\mathcal{C}(M)^T)$$ where the reindexing swaps the input and output components of the index pairs.

Let $M$ be a matrix map between matrix spaces over $d_{in}$ and $d_{out}$ with coefficients in a field $\mathbb{k}$. If the Choi matrix of $M$, denoted $\mathcal{C}(M)$, is positive semidefinite, then the Choi matrix of its dual map $M^*$, denoted $\mathcal{C}(M^*)$, is also positive semidefinite.

For any finite type $d_{in}$ and any field $\mathbb{k}$, the dual of the identity matrix map $\text{id} : \text{Mat}_{d_{in}}(\mathbb{k}) \to \text{Mat}_{d_{in}}(\mathbb{k})$ is the identity matrix map itself. That is, $(\text{id})^* = \text{id}$.

Let $\mathbb{k}$ be a field, and let $A, B, C, D$ be finite types with decidable equality. For any matrix maps $M: \text{Mat}_{A \times A}(\mathbb{k}) \to \text{Mat}_{B \times B}(\mathbb{k})$ and $N: \text{Mat}_{C \times C}(\mathbb{k}) \to \text{Mat}_{D \times D}(\mathbb{k})$, the dual of their Kronecker product $M \otimes_{km} N$ is equal to the Kronecker product of their dual maps, $M^* \otimes_{km} N^*$. That is, $(M \otimes_{km} N)^* = M^* \otimes_{km} N^*$.

Let $M$ be a matrix map between finite-dimensional spaces. If $M$ is completely positive, then its dual map $M^\dagger$ (or $M^*$ in some contexts) is also completely positive.

Let $R$ be a commutative ring and $M$ a reflexive $R$-module with a basis $b$ indexed by a finite set $\iota$. Let $\phi_b : M \cong M^*$ be the dual basis isomorphism induced by $b$. Then the composition of the dual of the inverse isomorphism, $(\phi_b^{-1})^* : M^{**} \to M^*$, with the isomorphism $\phi_b : M \to M^*$ satisfies $(\phi_b^{-1})^* \circ \phi_b = \text{ev}$, where $\text{ev} : M \to M^{**}$ is the canonical evaluation map into the double dual.

Let $R$ be a commutative ring and $M$ be a reflexive module over $R$ with a finite basis $b$ indexed by $\iota$. Let $\Phi_b: M \to M^\vee$ denote the isomorphisms to the dual space induced by the basis $b$, and $\Phi_b^{-1}: M^\vee \to M$ its inverse. Let $\text{eval}: M \xrightarrow{\cong} M^{\vee\vee}$ be the natural evaluation isomorphism to the double dual. Then the composition of the inverse dual basis isomorphism with the dual of the dual basis isomorphism is the inverse of the evaluation map, i.e., $$\Phi_b^{-1} \circ \Phi_b^* = \text{eval}^{-1}$$ where $\Phi_b^*: M^{\vee\vee} \to M^\vee$ is the dual map of $\Phi_b$.

For a matrix map $M$ acting between types $d_{In}$ and $d_{Out}$ over a semiring $R$, the dual of its dual is equal to $M$ itself, i.e., $(M^*)^* = M$.

Let $M$ be a completely positive and trace-preserving (CPTP) map from the space of matrices of dimension $d_{in}$ to the space of matrices of dimension $d_{out}$ over $\mathbb{C}$. The dual $\text{dual}(M)$ is defined as the adjoint (dual) linear map $M^\dagger$. This map is a completely positive and unital (CPU) map from the space of matrices of dimension $d_{out}$ to the space of matrices of dimension $d_{in}$.

Let $M$ be a completely positive trace-preserving (CPTP) map from the space of $d_{in} \times d_{in}$ complex Hermitian matrices to the space of $d_{out} \times d_{out}$ complex Hermitian matrices. Let $M^\dagger$ (denoted as `M.dual`) be the dual map of $M$, which is a completely positive unital (CPU) map. For any $d_{out} \times d_{out}$ Hermitian matrix $T$, if $T$ is positive semi-definite ($0 \le T$), then its image under the dual map $M^\dagger(T)$ is also positive semi-definite ($0 \le M^\dagger(T)$).

Let $M$ be a completely positive and trace-preserving (CPTP) map from the space of complex Hermitian matrices of dimension $d_{in}$ to those of dimension $d_{out}$. Let $M^\dagger$ (denoted as `M.dual`) be its dual map. For any complex Hermitian matrix $T$ of dimension $d_{out}$ that represents a two-element POVM effect (i.e., $0 \le T \le \mathbb{I}$), its image under the dual map also satisfies $0 \le M^\dagger(T) \le \mathbb{I}$, where $\mathbb{I}$ denotes the identity matrix.

Let $\mathcal{E}$ be a completely positive trace-preserving (CPTP) map from the space of complex matrices of dimension $d_{in}$ to those of dimension $d_{out}$. For any quantum mixed state $\rho$ with dimension $d_{in}$ and any Hermitian operator $T$ of dimension $d_{out}$, the expectation value of $T$ with respect to the evolved state $\mathcal{E}(\rho)$ is equal to the expectation value of the evolved operator $\mathcal{E}^\dagger(T)$ with respect to the initial state $\rho$. That is, \[ \text{Tr}(\mathcal{E}(\rho) T) = \text{Tr}(\rho \mathcal{E}^\dagger(T)) \] where $\mathcal{E}^\dagger$ is the dual (completely positive unital) map of $\mathcal{E}$.

Let $f: \text{HermitianMat}_{d_{in}}(\mathbb{C}) \to_{\mathbb{R}} \text{HermitianMat}_{d_{out}}(\mathbb{C})$ be an $\mathbb{R}$-linear map between spaces of Hermitian matrices. The definition constructs a Hermiticity-Preserving Map (HPMap) $\mathcal{E}$ from $d_{in}$ to $d_{out}$ by extending $f$ to all complex matrices via the decomposition into Hermitian and anti-Hermitian parts. Specifically, for any matrix $x$, the map is defined as: $$\mathcal{E}(x) = f(\text{Re}(x)) + i \cdot f(\text{Im}(x))$$ where $\text{Re}(x) = \frac{x + x^*}{2}$ and $\text{Im}(x) = \frac{x - x^*}{2i}$ are the Hermitian components of $x$.

Let $f: \text{HermitianMat}_{d_{in}}(\mathbb{C}) \to_{\mathbb{R}} \text{HermitianMat}_{d_{out}}(\mathbb{C})$ be an $\mathbb{R}$-linear map between spaces of Hermitian matrices. Let $\text{HPMap.ofHermitianMat}(f)$ be the hermiticity-preserving map $\mathcal{E}$ obtained by extending $f$ to all complex matrices. Then, the $\mathbb{R}$-linear map on Hermitian matrices associated with $\mathcal{E}$ is equal to the original map $f$.

Let $f \in \text{HPMap}(d_{In}, d_{Out}, \mathbb{C})$ be a Hermitian-preserving map. If we take its restriction to the space of Hermitian matrices as an $\mathbb{R}$-linear map, denoted by $\text{LinearMapClass.linearMap}(f)$, and then reconstruct a Hermitian-preserving map from this linear map using the `ofHermitianMat` construction, the resulting map is equal to the original map $f$.

Given a Hermitian-preserving map $f : \text{HermitianMat}_{d_{In}}(\mathbb{C}) \to \text{HermitianMat}_{d_{Out}}(\mathbb{C})$, its Hermitian dual $f^\dagger$ (or `hermDual`) is the Hermitian-preserving map from $\text{HermitianMat}_{d_{Out}}(\mathbb{C})$ to $\text{HermitianMat}_{d_{In}}(\mathbb{C})$ defined as the adjoint of the linear map $f$ with respect to the Hilbert-Schmidt inner product. It satisfies the adjoint property $\langle f(A), B \rangle = \langle A, f^\dagger(B) \rangle$ for all Hermitian matrices $A$ and $B$.

For any Hermitian-preserving map $f : \text{HermitianMat}_{d_{In}}(\mathbb{C}) \to \text{HermitianMat}_{d_{Out}}(\mathbb{C})$, the Hermitian dual of its Hermitian dual is equal to the original map, i.e., $(f^\dagger)^\dagger = f$. Here, the Hermitian dual $f^\dagger$ is defined as the adjoint of $f$ with respect to the Hilbert-Schmidt inner product.

Let $f$ be a Hermitian-preserving map (an element of `HPMap`) from the space of Hermitian matrices of dimension $d_{In}$ to $d_{Out}$ over $\mathbb{C}$. For any Hermitian matrices $A \in \text{HermitianMat}_{d_{In}}(\mathbb{C})$ and $B \in \text{HermitianMat}_{d_{Out}}(\mathbb{C})$, the Hilbert-Schmidt inner product satisfies: $$\langle f(A), B \rangle = \langle A, f^\dagger(B) \rangle$$ where $f^\dagger$ (denoted as `f.hermDual`) is the Hermitian dual map of $f$.

Let $f: \text{HermitianMat}_{d_{In}}(\mathbb{C}) \to \text{HermitianMat}_{d_{Out}}(\mathbb{C})$ be a Hermitian-preserving map and $f^\dagger$ denote its Hermitian dual. For any Hermitian matrices $A \in \text{HermitianMat}_{d_{In}}(\mathbb{C})$ and $B \in \text{HermitianMat}_{d_{Out}}(\mathbb{C})$, the Hilbert-Schmidt inner product satisfies $\langle f(A), B \rangle = \langle A, f^\dagger(B) \rangle$.

Let $f$ be a Hermitian-preserving matrix map (of type `HPMap`) from complex matrices of dimension $d_{In}$ to $d_{Out}$. If the underlying linear map $f.\text{map}$ is positive (i.e., it maps positive semi-definite matrices to positive semi-definite matrices), then the underlying linear map of its Hermitian dual, $f.\text{hermDual}.\text{map}$, is also positive.

Let $f$ be a Hermitian-preserving map between complex Hermitian matrices of dimensions $d_{In}$ and $d_{Out}$. If the underlying linear map $f.\text{map}$ is trace-preserving, then the underlying linear map of its Hermitian dual $f^\dagger.\text{map}$ is unital, meaning it maps the identity matrix to the identity matrix ($f^\dagger(I) = I$).

Given a positive trace-preserving map (PTPMap) $M$ from $d_{In} \times d_{In}$ complex matrices to $d_{Out} \times d_{Out}$ complex matrices, its Hermitian dual $M^\dagger$ is a positive unital map (PUMap) from $d_{Out} \times d_{Out}$ matrices to $d_{In} \times d_{In}$ matrices. This dual map is constructed by taking the adjoint of the underlying Hermitian-preserving map and inherits the property of being unital from the trace-preserving property of $M$.

Let $M$ be a positive trace-preserving map ($\text{PTPMap}$) from the space of $d_{In} \times d_{In}$ complex Hermitian matrices to $d_{Out} \times d_{Out}$ complex Hermitian matrices. For any Hermitian matrix $T$ of size $d_{Out}$ such that $T$ is positive semidefinite ($0 \le T$), its image under the Hermitian dual map $M^\dagger T$ (or $M.hermDual T$) is also positive semidefinite ($0 \le M^\dagger T$).

Let $M$ be a point-trace-preserving (PTP) map from the space of complex matrices of dimension $d_{In}$ to $d_{Out}$. Let $T$ be a Hermitian matrix of dimension $d_{Out}$ such that $0 \le T \le I$, representing a two-element Positive Operator-Valued Measure (POVM) effect. Then its image under the Hermitian dual map, $M^\dagger(T)$, is a Hermitian matrix of dimension $d_{In}$ that also satisfies $0 \le M^\dagger(T) \le I$.

Let $\mathcal{E}$ be a positive trace-preserving map (PTPMap) from $d_{In}$ to $d_{Out}$. For any quantum mixed state $\rho$ of dimension $d_{In}$ and any Hermitian operator $T$ of dimension $d_{Out}$, the expectation value of $T$ in the state $\mathcal{E}(\rho)$ is equal to the expectation value of the dual map $\mathcal{E}^*$ applied to $T$ in the original state $\rho$: \[ \text{Tr}(\mathcal{E}(\rho) T) = \text{Tr}(\rho \mathcal{E}^*(T)) \] where $\mathcal{E}^*$ (denoted as `hermDual`) is the adjoint positive unital map (PUMap) of $\mathcal{E}$ with respect to the Frobenius inner product.