QuantumInfo.Finite.Capacity

A quantum channel (CPTP map) $\Lambda_1: \text{Mat}_{d_1}(\mathbb{C}) \to \text{Mat}_{d_2}(\mathbb{C})$ emulates another quantum channel $\Lambda_2: \text{Mat}_{d_3}(\mathbb{C}) \to \text{Mat}_{d_4}(\mathbb{C})$ if there exist an encoding CPTP map $E: \text{Mat}_{d_3}(\mathbb{C}) \to \text{Mat}_{d_1}(\mathbb{C})$ and a decoding CPTP map $D: \text{Mat}_{d_2}(\mathbb{C}) \to \text{Mat}_{d_4}(\mathbb{C})$ such that their composition satisfies $D \circ \Lambda_1 \circ E = \Lambda_2$.

Let $\mathcal{A}$ and $\mathcal{B}$ be two completely positive trace-preserving (CPTP) maps (quantum channels) from a system of dimension $d_1$ to a system of dimension $d_2$. We say that $\mathcal{A}$ $\epsilon$-approximates $\mathcal{B}$ if for every input quantum state (density matrix) $\rho \in \text{MState}(d_1)$, the fidelity between the output states $F(\mathcal{A}(\rho), \mathcal{B}(\rho))$ is at least $1 - \epsilon$, where $\epsilon \in \mathbb{R}$.

Let $\mathcal{A}: \mathcal{B}(\mathcal{H}_{in}) \to \mathcal{B}(\mathcal{H}_{out})$ be a completely positive trace-preserving (CPTP) map. A rate $R \in \mathbb{R}$ is said to be achievable by $\mathcal{A}$ if for every $\epsilon > 0$, there exists an integer $n > 0$ and a CPTP map $\mathcal{B}: \mathcal{B}(\mathbb{C}^d) \to \mathcal{B}(\mathbb{C}^d)$ such that: 1. The $n$-fold tensor product channel $\mathcal{A}^{\otimes n}$ emulates $\mathcal{B}$. That is, there exist encoding and decoding CPTP maps $\mathcal{E}$ and $\mathcal{D}$ such that $\mathcal{D} \circ \mathcal{A}^{\otimes n} \circ \mathcal{E} = \mathcal{B}$. 2. The transmission rate satisfies $\frac{\log_2 d}{n} \geq R$ (or equivalently, $\log_2 d \geq R \cdot n$). 3. The channel $\mathcal{B}$ is an $\epsilon$-approximation of the identity channel $\text{id}_d$, meaning that for every state $\rho$, the fidelity $F(\mathcal{B}(\rho), \rho) \geq 1 - \epsilon$.

For a quantum channel $\mathcal{A}$ mapping states between dimensions $d_1$ and $d_2$ (represented as a completely positive trace-preserving map), its quantum capacity $Q(\mathcal{A})$ is defined as the supremum of all achievable rates $R \in \mathbb{R}$. A rate $R$ is achievable if for every $\varepsilon > 0$, there exists an integer $n$ such that $n$ copies of the channel $\mathcal{A}^{\otimes n}$ can emulate a channel $\mathcal{B}$ with $\frac{1}{n}\log_2(\text{dim}(\mathcal{B})) \geq R$, where $\mathcal{B}$ is $\varepsilon$-approximately the identity channel.

For any completely positive trace-preserving (CPTP) map $\Lambda$, the map $\Lambda$ emulates itself. According to the definition of emulation, this implies there exist CPTP maps $\mathcal{D}$ and $\mathcal{E}$ such that $\mathcal{D} \circ \Lambda \circ \mathcal{E} = \Lambda$.

Let $\Lambda_1, \Lambda_2,$ and $\Lambda_3$ be completely positive trace-preserving (CPTP) maps (quantum channels). If $\Lambda_1$ emulates $\Lambda_2$ and $\Lambda_2$ emulates $\Lambda_3$, then $\Lambda_1$ emulates $\Lambda_3$. Here, a channel $A$ is said to emulate a channel $B$ if there exist CPTP maps $\mathcal{D}$ and $\mathcal{E}$ such that $\mathcal{D} \circ A \circ \mathcal{E} = B$.

Let $\Lambda$ be a completely positive trace-preserving (CPTP) map from the space of matrix states $\mathcal{M}_{d_1}$ to $\mathcal{M}_{d_2}$. Then $\Lambda$ $0$-approximates itself. Specifically, for any state $\rho$, the fidelity between $\Lambda(\rho)$ and $\Lambda(\rho)$ satisfies $F(\Lambda(\rho), \Lambda(\rho)) \geq 1 - 0$.

Let $\mathcal{A}$ and $\mathcal{B}$ be quantum channels (completely positive trace-preserving maps) from dimensions $d_1$ to $d_2$. If $\mathcal{A}$ is an $\epsilon_0$-approximation of $\mathcal{B}$ (meaning for every state $\rho$, the fidelity $F(\mathcal{A}(\rho), \mathcal{B}(\rho)) \ge 1 - \epsilon_0$), then for any $\epsilon_1 \in \mathbb{R}$ such that $\epsilon_0 \le \epsilon_1$, $\mathcal{A}$ is also an $\epsilon_1$-approximation of $\mathcal{B}$.

For any completely positive trace-preserving (CPTP) map $\Lambda$ between finite-dimensional Hilbert spaces, the channel $\Lambda$ achieves a transmission rate of $R = 0$ in the sense of quantum capacity.

Let $d_1$ be a finite type representing the dimension of a Hilbert space, and let $I$ be the identity completely positive trace-preserving (CPTP) map on that space. Then the map $I$ achieves a quantum communication rate $R = \log_2(|d_1|)$, where $|d_1|$ denotes the cardinality of the type $d_1$.

Let $\Lambda: \mathcal{L}(\mathbb{C}^{d_1}) \to \mathcal{L}(\mathbb{C}^{d_2})$ be a Completely Positive Trace-Preserving (CPTP) map (a quantum channel) where $d_1$ and $d_2$ are the dimensions of the input and output Hilbert spaces respectively. If a rate $R \in \mathbb{R}$ satisfies $R > \log_2(d_1)$, then the channel $\Lambda$ cannot achieve rate $R$.

Let $\Lambda$ be a completely positive trace-preserving (CPTP) map from a system with dimension $d_1$ to a system with dimension $d_2$. For any rate $R \in \mathbb{R}$, if $R > \log_2(d_2)$, then $\Lambda$ does not achieve rate $R$ (i.e., $\neg \text{AchievesRate}(\Lambda, R)$), where $d_2$ is the cardinality of the output type.

Let $\Lambda$ be a completely positive trace-preserving (CPTP) map from a system of dimension $d_1$ to a system of dimension $d_2$. The set of all achievable rates $R \in \mathbb{R}$ for the channel $\Lambda$ is bounded above.

For any quantum channel (completely positive trace-preserving map) $\Lambda$ with input dimension $d_1$ and output dimension $d_2$, its quantum capacity $Q(\Lambda)$ is non-negative, i.e., $0 \le Q(\Lambda)$.

Let $\Lambda$ be a Completely Positive Trace-Preserving (CPTP) map with input dimension $|d_1|$. The quantum capacity $Q(\Lambda)$ of the channel satisfies: $$Q(\Lambda) \le \log_2(|d_1|)$$ where $|d_1|$ denotes the cardinality of the finite type $d_1$ representing the input Hilbert space basis.

For any quantum channel $\Lambda$ (represented as a completely positive trace-preserving map from dimension $d_1$ to $d_2$) and any input state $\rho$ in the space of states of dimension $d_1$, the single-copy coherent information $I_c(\rho, \Lambda)$ is a lower bound for the quantum capacity $Q(\Lambda)$. That is, $I_c(\rho, \Lambda) \leq Q(\Lambda)$.

Let $\Lambda: \mathcal{B}(\mathcal{H}_{d_1}) \to \mathcal{B}(\mathcal{H}_{d_2})$ be a completely positive trace-preserving (CPTP) map. The quantum capacity of $\Lambda$, denoted $Q(\Lambda)$, is equal to the supremum of the set of rates $r$ such that there exists a natural number $n \in \mathbb{N}$ and an input state $\rho$ on the $n$-fold tensor product space $\mathcal{H}_{d_1}^{\otimes n}$ for which $r = \frac{1}{n} I_c(\rho, \Lambda^{\otimes n})$, where $I_c$ denotes the coherent information. Specifically, $$Q(\Lambda) = \sup \left\{ r \in \mathbb{R} \mid \exists n \in \mathbb{N}, \exists \rho \in \mathcal{S}(\mathcal{H}_{d_1}^{\otimes n}), r = \frac{1}{n} I_c(\rho, \Lambda^{\otimes n}) \right\},$$ where $\Lambda^{\otimes n}$ is the $n$-copy tensor product channel $\bigotimes_{i=1}^n \Lambda$.