I need to proof that the partial trace is a completely positive map. I have seen such proof before, but I just don't remember now. When I google it, they usually state it as a fact without any proof or say that the proof is easy. Can somebody give me a reference to this proof?
There is a general proof of the statement that taking the partial trace over a density operator gives a correct description (and is also unique) of the observable quantities for subsystems, in the book by Nielsen and Chuang: "Quantum Computation and Quantum Information", p 107. Since populations are represented by Hermitian operators of which the eigenvalues must be larger than or equal to zero, I assume you could take this also as proof that the partial trace gives a positive operator.
A somewhat simpler proof for the statement you want could be along the following lines: the diagonal elements of the full density operator, being populations, must all be larger than or equal to zero. Decomposing the density operator of the composite system in the product basis, and summing over the system you want to get rid off, gives for the diagonal elements of the reduced density matrix something like
\rho^{red}_{mm} = \sum_n \rho_{mn,mn}
Since all elements \rho_{mn,mn} are larger than or equal to zero, so is the diagonal element of the reduced density matrix.
I hope you get what I want to say, I could give a somewhat more detailed proof in a separate document, but don't have the time until thursday to do so. If you need it, let me know. May RG implement mathjax soon!
I think there is a proof on the Quantum Mechanics book by Cohen. See the Complement on Density Operator
There is a general proof of the statement that taking the partial trace over a density operator gives a correct description (and is also unique) of the observable quantities for subsystems, in the book by Nielsen and Chuang: "Quantum Computation and Quantum Information", p 107. Since populations are represented by Hermitian operators of which the eigenvalues must be larger than or equal to zero, I assume you could take this also as proof that the partial trace gives a positive operator.
A somewhat simpler proof for the statement you want could be along the following lines: the diagonal elements of the full density operator, being populations, must all be larger than or equal to zero. Decomposing the density operator of the composite system in the product basis, and summing over the system you want to get rid off, gives for the diagonal elements of the reduced density matrix something like
\rho^{red}_{mm} = \sum_n \rho_{mn,mn}
Since all elements \rho_{mn,mn} are larger than or equal to zero, so is the diagonal element of the reduced density matrix.
I hope you get what I want to say, I could give a somewhat more detailed proof in a separate document, but don't have the time until thursday to do so. If you need it, let me know. May RG implement mathjax soon!
A map is CP if and only if its Choi operator is positive-semidefinite. The Choi operator of the (partial) trace is the identity operator, which is positive-semidefinite. QED
Dear Researchers! Thank you very much for your input. I have read p. 107 from "Quantum Computation and Quantum Information" and it helped me understand the problem better.
Artur,
Have you also consulted Box 8.2 on page 369 of "Quantum Computation and Quantum Information", in which the difference is illustrated of positivity and complete positivity of a map? Unfortunately, this example is not very illuminating in proving that this difference can best be seen as a difference of `mapping a single probability' from `mapping a probability distribution {pm }' (rather than just a single probability pm). p. 107 is about positivity of a density operator, which is not a probability distribution.
Willem de Muynck
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