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Does a quantum channel (a completely positive trace-preserving map) always map the identity to the identity?

In other words, suppose that $ \mathcal{E}: \mathbb{C}^{N \times N} \to \mathbb{C}^{N \times N} $ is a quantum channel and $ I $ is the $ N \times N $ identity matrix. Then must we have that $ \mathcal{E}(I)= I $?

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    $\begingroup$ Adding this long comment does not improve the question. $\endgroup$ Feb 26 at 22:05
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    $\begingroup$ Ok I took it out and just posted it as an answer since it is too long for a comment $\endgroup$ Feb 27 at 0:39

5 Answers 5

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No, no reason it should. For instance, the amplitude damping $$ \varepsilon_{AD}(\rho)=E_0\rho E_0^\dagger+E_1\rho E_1^\dagger $$ with $$ E_0=\begin{bmatrix}1 & 0\\ 0 & \sqrt{1-\gamma}\end{bmatrix} $$ $$ E_1=\begin{bmatrix}0 & \sqrt{\gamma}\\ 0 & 0\end{bmatrix} $$

We have $$ \begin{alignat*}{1} \varepsilon_{AD}(I)&=E_0 E_0^\dagger+E_1 E_1^\dagger \\ &=\begin{bmatrix}1+\gamma & 0\\ 0 & 1-\gamma\end{bmatrix}\\ &\neq I\qquad \text{if }\gamma\neq 0 \end{alignat*} $$

What must verify trace preserving operations is $$ \sum_k E^\dagger_kE_k=I $$ not $$ \sum_k E_k E^\dagger_k=I $$

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No, such maps are referred to as unital maps.

A counterexample is the replacement map $$ \mathcal{E}(X) = \mathrm{tr}(X)\sigma $$ defined for some density matrix $\sigma$.

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No. When this is the case the channel is said to be unital. A necessary and sufficient condition for a map $\mathcal E$ being unital is its adjoint $\mathcal E^\dagger$ being trace-preserving.

For example, a replacement channel like $\Phi(\rho)=\operatorname{tr}(\rho) \sigma$ is not unital unless $\sigma=I/d$.

You can have a look for example at chapter 4 of Watrous' book, which is about unital channels and majorization.

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No.

Just try some of the examples at Canonical examples of quantum channels, such as #5 or #7.

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    $\begingroup$ Your linked question and answer are great references! $\endgroup$ Feb 27 at 0:42
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Long comment related to something I really like about the accepted answer:

By Kraus' theorem any completely positive map can be written in the form $$ \mathcal{E}(\rho)=\sum_{k} E_k \rho E_k^{\dagger} $$ where $ \sum_k E_k^\dagger E_k \leq I $. And conversely given any set of operators $ \{ E_k \} $ such that $ \sum_k E_k^\dagger E_k \leq I $, then the linear map $$ \rho \mapsto \sum_{k} E_k \rho E_k^{\dagger} $$ will be completely positive.

Anyway for a quantum channel it is not just that the map $ \mathcal{E} $ is completely positive but also we ask that it be trace preserving. That is $$ Tr(\rho)=Tr(\mathcal{E}(\rho))=Tr(\sum_{k} E_k \rho E_k^{\dagger})= \sum_k Tr(E_k \rho E_k^{\dagger})=\sum_k Tr( \rho E_k^{\dagger}E_k)=Tr(\sum_k \rho E_k^{\dagger}E_k)=Tr(\rho \sum_k E_k^{\dagger}E_k) $$ So take the standard traceless basis + identity matrix we see that $ \rho $ and $ \sum_k E_k^{\dagger}E_k $ are trace orthogonal for all traceless choices of $ \rho $ and so $ \sum_k E_k^{\dagger}E_k $ must be proportional to $ I $ and so since $ Tr(\rho)= Tr(\rho \sum_k E_k^{\dagger}E_k) $ it must in fact be that case that $ \sum_k E_k^{\dagger}E_k $ is exactly equal to $ I $. $$ \sum_k E_k^{\dagger}E_k=I $$ In other words, a completely positive map is exactly a $$ \rho \mapsto \sum_{k} E_k \rho E_k^{\dagger} $$ for $ \sum_k E_k^{\dagger}E_k \leq I $ while a completely positive trace-preserving map is exactly a
$$ \rho \mapsto \sum_{k} E_k \rho E_k^{\dagger} $$ for $ \sum_k E_k^{\dagger}E_k = I $.

It is kind of interesting that a completely positive map is unital (maps identity to identity) if and only if $$ I=\mathcal{E}(I)=\sum_{k} E_k I E_k^{\dagger}=\sum_{k} E_k E_k^{\dagger} $$

In other words, a completely positive map is trace preserving if and only if $$ I=\sum_{k} E_k^\dagger E_k $$ while a completely positive map is unital if and only if $$ I=\sum_{k} E_k E_k^{\dagger} $$ These two facts are equivalent to the fact, stated in the answer by gIS, that a completely positive map is unital if and only if its adjoint is trace-preserving.

Although these conditions look superficially similar, and they coincide if the $ E_k $ are all hermitian or unitary or more generally if the $ E_k $ are all normal operators, they are in general completely different! And I love that the accepted answer brings up both these conditions and demonstrates this important distinction by using the non-normal operator $ E_1=\begin{bmatrix}0 & \sqrt{\gamma}\\ 0 & 0\end{bmatrix} $

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