Finite completely positive trace-preserving maps — Examples

Fix $\gamma >0$ in ${\mathrm{s}}^{-1}$ and $t>0$ in seconds, so $0<e^{-\gamma t}<1$ by [F6]–[F8]. On ${\mathbb{C}}^2$, define $K_0=\mathrm{diag}(e^{-\gamma t\mathrm{/}2},1)$, $K_1=\sqrt{1-e^{-\gamma t}}\mathrm{\mid }1⟩⟨0\mathrm{\mid }$, and ${\mathrm{\Phi }}_t(X)=K_{0}X{K}_0^{\ast }+K_{1}X{K}_1^{\ast }$. Since $K_0^{\ast }{K}_0+K_1^{\ast }{K}_1=I_2$, [F2] and [F1] show that ${\mathrm{\Phi }}_t$ is a finite channel.

The matrix ${\rho }_0=I_2\mathrm{/}2$ is positive semidefinite with trace one. Direct multiplication gives ${\mathrm{\Phi }}_t({\rho }_0)=\mathrm{diag}(p,1-p)$, where $p=e^{-\gamma t}\mathrm{/}2\in (0,1\mathrm{/}2)$ by [F6]–[F8]. The entropy difference is $S({\rho }_0)-S({\mathrm{\Phi }}_t({\rho }_0))=k_B(\ln 2-h(p))$, where $h(p)=-p\ln p-(1-p)\ln (1-p)$. For the binary laws $P=(p,1-p)$ and $Q=(1\mathrm{/}2,1\mathrm{/}2)$, [F4] gives $D(P\mathrm{\parallel }Q)=\ln 2-h(p)$; since $P\mathrm{\ne }Q$, [F5] gives $D(P\mathrm{\parallel }Q)>0$. With $k_B>0$, this yields $S({\mathrm{\Phi }}_t({\rho }_0))<S({\rho }_0)$ and refutes the universal claim.

The transpose is complex-linear and preserves trace because it fixes the diagonal entries. For any $X\ge 0$, [F2] gives $X=Y{Y}^{\ast }$ for some matrix $Y$; hence $T(X)=X^{\mathsf{T}}=\bar{Y}{Y}^{\mathsf{T}}=\bar{Y}(\bar{Y}{)}^{\ast }\ge 0$. Thus $T$ is positive and trace-preserving.

Its Choi matrix is $J(T)={\sum }_{i,j=0}^1{E}_{ij}\otimes E_{ji}$, the swap operator $F(\mathrm{\mid }a⟩\otimes \mathrm{\mid }b⟩)=\mathrm{\mid }b⟩\otimes \mathrm{\mid }a⟩$. For $\eta =(\mathrm{\mid }0⟩\otimes \mathrm{\mid }1⟩-\mathrm{\mid }1⟩\otimes \mathrm{\mid }0⟩)\mathrm{/}\sqrt{2}$, one has $F\eta =-\eta$; therefore $J(T)\ge ̸0$. By [F3], $T$ is not completely positive.

Let $\mathrm{\mid }{\psi }_{\pm }⟩=(\mathrm{\mid }00⟩\pm \mathrm{\mid }11⟩)\mathrm{/}\sqrt{2}$ and ${\omega }_{\pm }=\mathrm{\mid }{\psi }_{\pm }⟩⟨{\psi }_{\pm }\mathrm{\mid }$; each ${\omega }_{\pm }$ is positive semidefinite with trace one. Using [F1], the cross terms $\mathrm{\mid }00⟩⟨11\mathrm{\mid }$ and $\mathrm{\mid }11⟩⟨00\mathrm{\mid }$ vanish under ${\mathrm{Tr}}_E$ because their environment indices differ, while the diagonal terms give ${\mathrm{Tr}}_E{\omega }_{+}={\mathrm{Tr}}_E{\omega }_{-}=I_2\mathrm{/}2$.

Define $U\mathrm{\mid }a,b⟩=\mathrm{\mid }a,a\oplus b⟩$ for $a,b\in \{0,1\}$, where $\oplus$ is addition modulo two. This CNOT map permutes the orthonormal product basis and is therefore unitary. It sends $\mathrm{\mid }{\psi }_{\pm }⟩$ to $\mathrm{\mid }\pm ⟩\otimes \mathrm{\mid }0⟩$, where $\mathrm{\mid }\pm ⟩=(\mathrm{\mid }0⟩\pm \mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Thus ${\mathrm{Tr}}_E(U{\omega }_{\pm }{U}^{\ast })=\mathrm{\mid }\pm ⟩⟨\pm \mathrm{\mid }$, and the outputs are distinct because their $(0,1)$ matrix entries are $1\mathrm{/}2$ and $-1\mathrm{/}2$. Equal system marginals therefore do not determine the final system state for these correlated preparations, refuting the claim. The product-preparation theorem [F2] is unaffected.

Let $Z=\mathrm{diag}(1,-1)$, take $g>0$ in ${\mathrm{s}}^{-1}$, and set $H=\mathrm{\hslash }gZ\otimes Z$. With ${\sigma }_E=I_2\mathrm{/}2=\frac{1}{2}{\sum }_{b=\pm 1}\mathrm{\mid }b⟩⟨b\mathrm{\mid }$ and $U_t=e^{-itH\mathrm{/}\mathrm{\hslash }}$, the power-series definition of the finite operator exponential in [F3] and the $Z$-eigenbasis give $U_t={\sum }_{b=\pm 1}\mathrm{diag}(\exp (-ibgt),\exp (ibgt))\otimes \mathrm{\mid }b⟩⟨b\mathrm{\mid }$. Applying [F1] to the two orthogonal environment sectors yields ${\mathrm{\Phi }}_t(\rho )=\frac{1}{2}(V_t\rho V_t^{\ast }+V_t^{\ast }\rho V_t)$, where $V_t=\mathrm{diag}(\exp (-igt),\exp (igt))$. In matrix coordinates, ${\mathrm{\Phi }}_t\text{\hspace{0.17em}⁣}\left(\begin{array}{cc}a& x\\ y& d\end{array}\right)=\left(\begin{array}{cc}a& \kappa (t)x\\ \kappa (t)y& d\end{array}\right)$ with $\kappa (t):=\frac{1}{2}(\exp (-2igt)+\exp (2igt))$. By [F3], each ${\mathrm{\Phi }}_t$ is a channel.

By [F7], choose a dimensionless $\theta \in [0,2\pi )$ with $(\cos \theta ,\sin \theta )=(0,1)$; hence $\theta >0$. Equation [F6] gives $\exp (i\theta )=i$, and [F5] with [F4] gives $\exp (-i\theta )=-i$ and $\exp (2i\theta )=\exp (-2i\theta )=-1$. Set $\tau =\theta \mathrm{/}(2g)>0$ and ${\rho }_{+}=\mathrm{\mid }+⟩⟨+\mathrm{\mid }$, with $\mathrm{\mid }+⟩=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. By step 1.1, $\kappa (\tau )=\frac{1}{2}(-i+i)=0$ and $\kappa (2\tau )=\frac{1}{2}(-1-1)=-1$. Therefore ${\mathrm{\Phi }}_{\tau }({\mathrm{\Phi }}_{\tau }({\rho }_{+}))=I_2\mathrm{/}2$ whereas ${\mathrm{\Phi }}_{2\tau }({\rho }_{+})=\mathrm{\mid }-⟩⟨-\mathrm{\mid }$, where $\mathrm{\mid }-⟩=(\mathrm{\mid }0⟩-\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. The outputs differ, so ${\mathrm{\Phi }}_{2\tau }\mathrm{\ne }{\mathrm{\Phi }}_{\tau }\circ {\mathrm{\Phi }}_{\tau }$; the family is not a semigroup.