### 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.

- 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’’.
- 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).

- 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).
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).

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