# Does a quantum channel always preserve the identity matrix?

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 $$\mathcal{E}(I)= I$$?

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

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 $$\sum_k E_k E^\dagger_k=I$$

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 $$\sigma$$.

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.

No.

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

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 $$E_1=\begin{bmatrix}0 & \sqrt{\gamma}\\ 0 & 0\end{bmatrix}$$