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.