# Exercise 10

## from Example Sheet 4

### Exercise 10.

An interesting class of quantum channels are the entanglement-breaking (EB) channels. An EB channel $$\Lambda$$ is one for which $$(\operatorname{id}\otimes \Lambda)(\omega)$$ is separable, even for entangled $$\omega$$1. The Holevo capacity has been proved to be additive for EB channels.

1. Prove that any channel of the following form is EB: $\Lambda(\rho) = \sum_k \sigma_k \operatorname{tr}(E_k\rho),$(1) where $$\sigma_k$$ are density matrices and $$\{E_k\}$$ is a POVM. The above form has the following physical interpretation. Alice does a measurement (POVM) on the input state $$\rho$$ and communicates the outcomes $$k$$ to Bob via a classical channel; Bob then prepares an agreed upon state $$\sigma_k$$. Hence, EB channels are also called measure-and-prepare channels’’.
2. Prove that if the Choi state $$(I \otimes \Lambda) |\Omega\rangle\langle \Omega|$$ (where $$|\Omega\rangle$$ denotes the unnormalized maximally entangled state) is separable, then $$\Lambda$$ has the form (1).
1. First, let us consider a pure state $$| \psi \rangle_{AB}$$ with Schmidt decomposition $$| \psi \rangle_{AB} = \sum_i \sqrt{\lambda_i}| e_i \rangle| f_i \rangle$$. Then \begin{aligned} \operatorname{id}_A \otimes \Lambda(\psi_{AB}) &= \sum_{i,j} \sqrt{\lambda_i \lambda_j} \sum_k \operatorname{tr}(E_k | f_i \rangle\langle f_j |) | e_i \rangle\langle e_j | \otimes \sigma_k.\\ &= \sum_k \sum_{i,j} \sqrt{\lambda_i \lambda_j} \langle f_j | E_k | f_i \rangle | e_i \rangle\langle e_j | \otimes \sigma_k. \end{aligned} Now, let $$p_k = \operatorname{tr}[ \sum_{i,j} \sqrt{\lambda_i \lambda_j} \langle f_j | E_k | f_i \rangle | e_i \rangle\langle e_j | ] = \sum_i \lambda_i \langle f_i | E_k | f_i \rangle$$. Then since $$E_k\geq 0$$ and $$\lambda_i\geq 0$$ we have $$p_k\geq 0$$, and moreover \begin{aligned} \sum_k p_k &= \sum_i \sum_k \lambda_i \langle f_i | E_k | f_i \rangle \\ &= \sum_i \lambda_i \langle f_i |\sum_k E_k | f_i \rangle\\ &= \sum_i \lambda_i \langle f_i |I|f_i \rangle\\ &= \sum_i \lambda_i = 1. \end{aligned} Define $$\omega_k = \frac{1}{p_k} \sum_{i,j} \sqrt{\lambda_i \lambda_j} \langle f_j | E_k | f_i \rangle | e_i \rangle\langle e_j |$$. Let $$| \psi \rangle$$ be any vector. Then \begin{aligned} \langle \psi| \omega_k \psi \rangle &= \frac{1}{p_k} \sum_{i,j} \sqrt{\lambda_i \lambda_j} \langle f_j | E_k | f_i \rangle \langle \psi | e_i \rangle\langle e_j | \psi \rangle \\ &= \frac{1}{p_k} \sum_\ell \sum_{i,j} \sqrt{\lambda_i \lambda_j} \langle f_j | E_k^{1/2} | \ell \rangle \langle \ell | E_k^{1/2} | f_i \rangle \langle \psi | e_i \rangle\langle e_j | \psi \rangle\\ &= \frac{1}{p_k}\sum_\ell \left|\sum_i \sqrt{\lambda_i} \langle \ell | E_k^{1/2} | f_i \rangle \,\langle \psi | e_i \rangle \right|^2 \geq 0. \end{aligned} Thus, $$\omega_k\geq 0$$. Since we chose $$p_k$$ so that $$\operatorname{tr}\omega_k = 1$$, we have that $$\omega_k$$ is a density matrix. Then \begin{aligned} \operatorname{id}_A \otimes \Lambda(\psi_{AB}) &= \sum_k p_k \omega_k \otimes \sigma_k \end{aligned} is a separable state. To treat a mixed state $$\rho_{AB}$$ we simply consider a convex decomposition into pure states, $$\rho_{AB} = \sum_\alpha q_\alpha \psi_\alpha$$, and use linearity to see that $$\rho_{AB}$$ must be separable as well (i.e. using that the set of separable states is convex).
2. Assume $$(I \otimes \Lambda) |\Omega\rangle\langle \Omega|$$ is separable. Then \begin{aligned} (I \otimes \Lambda) |\Omega\rangle\langle \Omega| &= \sum_{i,j} | i \rangle\langle j | \otimes \Lambda(| i \rangle\langle j |) \\ &= \sum_k p_k \psi_k \otimes \phi_k \end{aligned} for pure states $$\psi_k,\phi_k$$ and a probability distribution $$\{p_k\}$$. We can multiply each side by $$| \ell \rangle\langle \ell' |\otimes I$$ and partial trace over the first system to find \begin{aligned} \Lambda(| \ell' \rangle\langle \ell |) &= \sum_k p_k \operatorname{tr}[| \ell \rangle\langle \ell' | \psi_k] \phi_k\\ &= \sum_k p_k \operatorname{tr}[(| \ell' \rangle\langle \ell |)^T \psi_k] \phi_k\\ &= \sum_k p_k \operatorname{tr}[(| \ell' \rangle\langle \ell |)^T (\bar \psi_k)^T] \phi_k\\ &= \sum_k p_k \operatorname{tr}[ (\bar \psi_k | \ell' \rangle\langle \ell |)^T ] \phi_k\\ &= \sum_k p_k \operatorname{tr}[ \bar \psi_k | \ell' \rangle\langle \ell | ] \phi_k\\ &= \sum_k p_k \operatorname{tr}[ | \ell' \rangle\langle \ell | \bar \psi_k ] \phi_k \end{aligned} using $$\psi_k = \psi_k^\dagger = (\bar \psi_k)^T$$, where $$A^T$$ is the transpose of a matrix $$A$$ in the basis $$\{| i \rangle\}$$ and $$\bar A$$ is the complex conjugate of $$A$$ in the same basis. In the second-to-last line we used that the trace is invariant under transpose, and in the last line, the cyclicity of the trace. Note the complex conjugate matrix $$\bar \psi_k$$ is still a pure state (one can check it is a self-adjoint, rank-one projector).

By linearity, we therefore have $\Lambda(\rho) =\sum_k p_k \operatorname{tr}[\rho \bar \psi_k] \phi_k$ Let $$E_k = p_k \bar \psi_k$$ so that $$\Lambda(\rho) = \sum_k \operatorname{tr}[\rho E_k] \phi_k$$. It remains to prove $$\{E_k\}$$ is a POVM. We have immediately that $$E_k \geq 0$$. To show $$\sum_k E_k= I$$, we note that we can write $\Lambda(\rho) = \sum_k p_k \langle \bar \psi_k | \rho | \bar \psi_k \rangle | \phi_k \rangle\langle \phi_k | = \sum_k p_k | \phi_k \rangle\langle \bar \psi_k | \rho | \bar \psi_k \rangle\langle \phi_k |.$ We can therefore define $$A_k := \sqrt{p_k} | \phi_k \rangle\langle \bar \psi_k |$$ yielding $$\{A_k\}$$ as a set of Kraus operators for $$\Lambda$$. Thus, since $$\Lambda$$ is TP, \begin{aligned} I &= \sum_k A_k^\dagger A_k \\ &= \sum_k p_k | \bar \psi_k \rangle\langle \phi_k | | \phi_k \rangle\langle \bar \psi_k | \\ &= \sum_k p_k | \bar \psi_k \rangle\langle \bar \psi_k | \\ &= \sum_k E_k \end{aligned} as desired. Thus, $$\{E_k\}$$ is a POVM, and indeed $$\Lambda$$ has the form (1).

1. It derives its name from the fact that the channel outputs a separable state whenever half of an entangled state is input to it.