# Unitary evolutions as CPTP maps

An example of a Complete Positive Trace Preserving Map (CPTP), in the context of Open Quantum Systems/ Information Theory, is provided by unitary evolutions:

$$\mathfrak{U}[\rho] = U \rho U^\dagger$$

for some unitary operator U, which corresponds to a Kraus decomposition with one single Kraus operator. I have to show that unitary evolutions are $\textbf{reversible}$.

That is, I am asked to show that a CPTP map $\Lambda$ has an inverse $\Lambda^{-1}$ if and only if $\Lambda$ is given by the equation shown above, for some unitary U.

I know the inverse unitary evolution should be fixed by $U^\dagger$. Also, I have thought that my starting point should be considering that I can only write $\sum_{\alpha} M_{\alpha} \sigma M_{\alpha}^{\dagger}$, for any operator of the NxN complex matrix space, if every $M_{\alpha}$ is proportional to the identity. It also seems to be interesting to use Polar decompositin of the Kraus operators, but even with all of this, I am not able to get the recipe to prove that. I would be very thankful to receive any help.

• The form you are working with I think is hard to prove theorems with. It might be better to consider the choi matrix of the CPTP. I suspect that if it has an inverse then the CPTP must have an inverse – Amara Nov 28 '16 at 17:41
• Left-inverse? (Natural for "reversible".) Right-inverse? Both? Should the input and output space have the same dimension, or can they be different? – Norbert Schuch Nov 28 '16 at 22:43

Suppose a linear CPTP $\Lambda$ gives a reversible but not unitary evolution. Then $\Lambda$ must map some pure state $|\psi\rangle$ into a mixed state $\rho = \Lambda(|\psi\rangle\langle \psi|)$. But if $\Lambda$ is reversible, its inverse $\Lambda^{-1}$ must map the mixed state $\rho$ back into $|\psi\rangle$, $\Lambda^{-1}(\rho) = |\psi\rangle\langle \psi |$. On the other hand, Kadison's inequality as applied to $\Lambda^{-1}$ requires $$\Lambda^{-1}(\rho^2) \ge \left(\Lambda^{-1}(\rho)\right)^2 = |\psi\rangle\langle \psi |$$ But from this follows $Tr\left[\Lambda^{-1}(\rho^2)\right] = 1$, and this can only happen if $\rho$ is itself a pure state, etc.