# Exercise 4

## from Revision class solutions

### Exercise 4.

1. 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.
2. Using a maximally entangled state and the properties of the swap operator, prove that the transposition operator $$T$$ is positive but not completely positive.

1. See Theorem 1 of Notes 9.

2. 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.