QuantumInfo.Finite.Entropy.VonNeumann

Given a mixed quantum state $\rho$ of dimension $d$, the von Neumann entropy $S_{vn}(\rho)$ is defined as the Shannon entropy $H_s$ of its eigenvalue spectrum. Specifically, if the eigenvalues of the density matrix $\rho$ are $\{\lambda_1, \dots, \lambda_d\}$, then $S_{vn}(\rho) = -\sum_{i=1}^d \lambda_i \log \lambda_i$.

Given a bipartite quantum mixed state $\rho$ on a composite system with state space dimensions $d_A \times d_B$, the quantum conditional entropy $S(A|B)_\rho$ is defined as $$S(A|B)_\rho = S(\rho^{AB}) - S(\rho^B)$$ where $S$ denotes the von Neumann entropy $S_{vn}$, and $\rho^B = \text{Tr}_A(\rho)$ is the reduced density matrix of the second subsystem obtained by taking the partial trace over the first subsystem $A$ (the left system).

Given a bipartite quantum mixed state $\rho$ on a composite system with dimensions $d_A \times d_B$, the quantum mutual information $I(A:B)$ is defined as $$I(A:B) = S(\rho^A) + S(\rho^B) - S(\rho^{AB})$$ where $S$ denotes the von Neumann entropy $S_{vn}$. Here, $\rho^B = \text{Tr}_A(\rho)$ is the reduced density matrix obtained by the partial trace over the left subsystem (system $A$), and $\rho^A = \text{Tr}_B(\rho)$ is the reduced density matrix obtained by the partial trace over the right subsystem (system $B$).

Let $\rho$ be a quantum state (density matrix) in the Hilbert space $\mathcal{H}_{d_1}$ and $\Lambda: \mathcal{M}_{d_1} \to \mathcal{M}_{d_2}$ be a completely positive trace-preserving (CPTP) map (a quantum channel). The coherent information $I(\rho, \Lambda)$ is defined as the negative of the quantum conditional entropy $-S(A|B)_{\sigma}$ of the state $\sigma = (\Lambda \otimes \text{id})(\psi_\rho)$, where $\psi_\rho = |\text{purify}(\rho)\rangle\langle\text{purify}(\rho)|$ is the pure state density matrix on $\mathcal{H}_{d_1} \otimes \mathcal{H}_{d_1}$ representing the purification of $\rho$.

The quantum conditional mutual information $I(A:C|B)_\rho$ of a tripartite mixed state $\rho$ on the system $d_A \times d_B \times d_C$ is defined as the difference between the quantum conditional entropy of the first two subsystems and the quantum conditional entropy of the total system: $$I(A:C|B)_\rho = S(A|B)_{\rho_{AB}} - S(A|BC)_\rho$$ where $\rho_{AB} = \text{Tr}_C(\rho)$ is the reduced density matrix obtained by taking the partial trace over the third subsystem $d_C$, and $S(X|Y)$ denotes the von Neumann conditional entropy.

For any mixed quantum state $\rho$ (density matrix) in a $d$-dimensional Hilbert space, the von Neumann entropy $S(\rho)$ is non-negative, i.e., $S(\rho) \ge 0$.

For any mixed quantum state $\rho$ in a Hilbert space of dimension $d$, the von Neumann entropy $S_{vn}(\rho)$ is less than or equal to the natural logarithm of the dimension $d$, i.e., $S_{vn}(\rho) \le \log d$.

For any normalized quantum state vector $|\psi\rangle$ (represented as a `Ket d`), the von Neumann entropy $S_{vN}$ of the corresponding pure state $\rho = |\psi\rangle\langle\psi|$ (represented as `MState.pure ψ`) is equal to zero.

For a quantum mixed state $\rho$ of dimension $d$, the von Neumann entropy $S_{vn}(\rho)$ is equal to the negative of the Hilbert-Schmidt inner product of the logarithm of its associated Hermitian operator $\rho.M$ and the operator itself, i.e., $S_{vn}(\rho) = -\langle \log(\rho.M), \rho.M \rangle_{HS}$.

For a quantum mixed state $\rho$ of dimension $d$, the von Neumann entropy $S_{vn}(\rho)$ is equal to the trace of the operator obtained by applying the function $f(x) = -x \log x$ to the operator $\rho.M$ via the continuous functional calculus, i.e., $S_{vn}(\rho) = \operatorname{Tr}(f(\rho.M))$.

For a quantum mixed state $\rho$ in a Hilbert space of dimension $d$, where $d$ is a type with exactly one element (representing a 1-dimensional system), the von Neumann entropy $S_{vn}(\rho)$ is equal to $0$.

Let $\rho$ be a quantum mixed state on a system with basis $d_1$, and let $e : d_2 \simeq d_1$ be a bijection between two basis sets. If $\rho'$ is the mixed state obtained by relabeling $\rho$ according to $e$, then the von Neumann entropy of the relabeled state is equal to the von Neumann entropy of the original state, i.e., $S_{vn}(\rho') = S_{vn}(\rho)$.

Let $\rho$ be a quantum mixed state on a bipartite system indexed by $d_1 \times d_2$. Let $\text{SWAP}(\rho)$ be the mixed state on the system $d_2 \times d_1$ obtained by exchanging the two subsystems. Then the von Neumann entropy $S_{VN}$ of the swapped state is equal to the von Neumann entropy of the original state, i.e., $S_{VN}(\text{SWAP}(\rho)) = S_{VN}(\rho)$.

For any quantum mixed state $\rho$ defined on a tripartite system with dimensions $((d_1 \times d_2) \times d_3)$, the von Neumann entropy $S_{vn}$ is invariant under the associativity transformation. That is, $S_{vn}(\rho.\text{assoc}) = S_{vn}(\rho)$, where $\rho.\text{assoc}$ is the state relabeled to the system $d_1 \times (d_2 \times d_3)$.

Let $\rho$ be a quantum mixed state on a tripartite system with dimensions $d_1 \times (d_2 \times d_3)$. Let $\text{assoc}'(\rho)$ denote the state re-clustered into the system $(d_1 \times d_2) \times d_3$ via the standard associativity isomorphism. Then the von Neumann entropy $S_{vn}$ of the re-clustered state is equal to the von Neumann entropy of the original state: $$S_{vn}(\text{assoc}'(\rho)) = S_{vn}(\rho)$$

Let $\mathbf{k}$ be a field that is either $\mathbb{R}$ or $\mathbb{C}$ (an `RCLike` field). Let $\text{re}: \mathbf{k} \to \mathbb{R}$ be the map that sends an element to its real part (which, in this context, coincides with the map $\text{selfadjMap}$ from $\mathbf{k}$ to its maximal self-adjoint part $\mathbb{R}$). Then, the map $\text{re}$ is continuous.

Let $\mathbf{k}$ be a field that is either $\mathbb{R}$ or $\mathbb{C}$ (an `RCLike` field), and let $d$ be a finite index set. For any Hermitian matrix $A \in \text{HermitianMat}_d(\mathbf{k})$, let $\text{Tr}(A)$ denote the trace of the matrix. The function $\text{Tr}: \text{HermitianMat}_d(\mathbf{k}) \to \mathbb{R}$ is continuous.

The von Neumann entropy $S_{VN}(\rho)$ is a continuous function as a mapping from the space of quantum mixed states $\text{MState } d$ to the real numbers $\mathbb{R}$, where $\text{MState } d$ is equipped with the subspace topology derived from the space of $d \times d$ complex matrices.

Let $\psi$ be a quantum state vector (ket) in the tensor product space of dimensions $d_1$ and $d_2$, and let $\rho = |\psi\rangle\langle\psi|$ be the corresponding pure state density matrix. If we represent $\psi$ as a $d_1 \times d_2$ matrix $A$ (via `vecToMat`), then the underlying matrix of the reduced state obtained by taking the partial trace over the first subsystem, denoted $\text{Tr}_1(|\psi\rangle\langle\psi|)$, is equal to $A^T \overline{A}$, where $A^T$ is the transpose and $\overline{A}$ is the complex conjugate of $A$.

Let $|\psi\rangle$ be a ket in a bipartite Hilbert space with dimensions $d_1 \times d_2$. Let $\rho = |\psi\rangle\langle\psi|$ be the corresponding pure state density matrix. Then the von Neumann entropy of the reduced density matrix obtained by taking the partial trace over the left subsystem, $S_{vN}(\text{Tr}_1(\rho))$, is equal to the von Neumann entropy of the reduced density matrix obtained by taking the partial trace over the right subsystem, $S_{vN}(\text{Tr}_2(\rho))$.

For a bipartite pure state $|\psi\rangle$ in the composite Hilbert space $\mathcal{H}_{d_1} \otimes \mathcal{H}_{d_2}$, the quantum conditional entropy of its swap is equal to its own quantum conditional entropy. More precisely, let $\rho = |\psi\rangle\langle\psi|$ be the pure mixed state density matrix; then the quantum conditional entropy $S(A|B)$ is symmetric in the sense that $qConditionalEnt(\text{SWAP}(\rho)) = qConditionalEnt(\rho)$.

For any bipartite quantum mixed state $\rho$ defined on a composite system with dimensions $d_1 \times d_2$, the quantum mutual information of the swapped state $\rho^{\text{SWAP}}$ is equal to the quantum mutual information of the original state $\rho$. That is, $I(B:A) = I(A:B)$.

For any pure state $|\psi\rangle$ in a Hilbert space indexed by $d_1$, and any bijection $e : d_2 \times d_3 \simeq d_1$ that relabels the state into a bipartite system, the von Neumann entropy $S_{vN}$ of the reduced density matrix obtained by tracing over the first subsystem equals the von Neumann entropy of the reduced density matrix obtained by tracing over the second subsystem. Specifically, $S_{vN}(\text{Tr}_1((|\psi\rangle\langle\psi|)_{e})) = S_{vN}(\text{Tr}_2((|\psi\rangle\langle\psi|)_{e}))$, where $(|\psi\rangle\langle\psi|)_{e}$ denotes the state after relabeling.