### Exercise 4.

Prove that any completely positive trace-preserving map \(\Phi\), acting on states \(\rho\) in a Hilbert space \(\mathcal{H}_A\) can be written in the Kraus form: \[ \Phi(\rho) = \sum_k A_k \rho A_k^\dagger, \] where \(A_k\) are linear operators satisfying \(\sum_k A_k^\dagger A_k = I\) and \(I\) is the identity operator.

*Hint: consider a maximally entangled state.*Using a maximally entangled state and the properties of the swap operator, prove that the transposition operator \(T\) is positive but not completely positive.

See Theorem 1 of Notes 9.

First, the transpose operator on square matrices is positive because it is spectrum preserving. We can see this by noticing for a matrix \(A\), the eigenvalues of \(A\) are the roots of its characteristic polynomial \(\lambda\mapsto \det(\lambda I - A)\). But \[ \det(\lambda I - A^T) = \det( (\lambda I - A)^T ) = \det(\lambda I -A) \] since transpose is linear and determinant is invariant under transpose. Thus, both \(A\) and its transpose have the same characeristic polynomial and therefore the same eigenvalues. Hence the transpose maps positive semi-definite matrices to positive semi-definite matrices.

Next, the swap operator acts as \(\mathbb{F}| i \rangle\otimes | j \rangle = | j \rangle\otimes | i \rangle\). We can then write a matrix representation of the swap operator as \[ \mathbb{F}= \sum_{i,j} | j i \rangle \langle i j | = \sum_{i,j} | i j \rangle \langle j i | \] where we have swiched the roles of the indices \(i\) and \(j\) in the second formulation. By comparing to the MES in this basis, \(\Omega =\frac{1}{d} \sum_{i,j} | ij \rangle\langle ij |\), we see \(\frac{1}{d}\mathbb{F}= (\operatorname{id}\otimes T) \Omega\), where \(T\) is the transpose operator in the basis \(\{| i \rangle\}\). If \(T\) were completely positive, then since \(\Omega\geq 0\), we would have \((\operatorname{id}\otimes T)\Omega\geq 0\). But this cannot be since \((\operatorname{id}\otimes T) = \frac{1}{d}\mathbb{F}\) and the swap operator has negative eigenvalues. We can see this by noticing that \(\mathbb{F}\) is self-adjoint so it has real eiegnvalues and squares to the identity, so all its eigenvalues must square to 1. Therefore, all of its eigenvalues must be \(\pm 1\). But if all its eigenvalues were \(+1\), then \(\mathbb{F}=I\) which is not the case; thus, \(\mathbb{F}\) has negative eigenvalues, so \(T\) is not completely positive.