Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 32 / 3 / b / Solution

The transpose map $T(X)=X^T$ in a fixed basis is linear, trace preserving and positive: if $X\ge 0$, then for every $v$, $v^{†}{X}^{T}v=\overline{{\bar{v}}^{†}X\bar{v}}\ge 0$. It is nevertheless not a completely positive map. Apply $\mathrm{id}\otimes T$ to the two-qubit Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator iswith eigenvalues $1/2,1/2,1/2,-1/2$. The negative eigenvector is $(|01⟩-|10⟩)/\sqrt{2}$. Thus a positive input on system plus reference acquires a negative expectation value. Transposition is positive but not completely positive, so it cannot be a deterministic local quantum process on arbitrary entangled states.