2

I have been studying serial concatenations of quantum channels, i.e. $\mathcal{N}_{A\rightarrow B}=\mathcal{N}_1\circ\mathcal{N}_2=\mathcal{N}_{B'\rightarrow B}\circ\mathcal{N}_{A\rightarrow B'}$. Here $A$,$B'$ and $B$ refer to the initial system, the intermediate system and the output system.

I was wondering if acomplementary channel to the environment can be obtained by using the complementary channels of the individual channels. My reasoning would be that if we want to obtain complementary channel $\mathcal{N}^c_{A\rightarrow E}$, we can obtain such by the following concatenation: \begin{equation} \mathcal{N}^c_{A\rightarrow E}=\mathcal{N}^c_{B'\rightarrow E}\circ\mathcal{N}_{A\rightarrow B'}. \end{equation}

Is this correct? If not, where is the problem in my reasoning and is there any way to get such a complementary channel?

glS
  • 27,510
  • 7
  • 37
  • 125
Josu Etxezarreta Martinez
  • 4,136
  • 15
  • 42

1 Answers1

2

The formula has multiple problems: first, complementary channels are of course not unique and, even worse, the output dimension is not fixed; so to make things rigorous we would have to impose some quantifiers. Having said this consider the following counterexample which shows that no matter how we interpret the formula it cannot possibly be true:

If the proposed formula were to hold for all channels, then it would be true that $$ ({\rm id}\circ\mathcal D)^c={\rm id}^c\circ\mathcal D\,,\tag1 $$ that is, $\mathcal D^c={\rm id}^c\circ\mathcal D$ where ${\rm id}$ is the identity channel and $\mathcal D$ is qubit dephasing $\mathcal D(X):=|0\rangle\langle 0|X|0\rangle\langle 0|+|1\rangle\langle 1|X|1\rangle\langle 1|$. As we will see below this is equivalent to $$ \langle 0|\cdot|0\rangle|\psi\rangle\langle\psi|+\langle1|\cdot|1\rangle|\psi^\perp\rangle\langle\psi^\perp|={\rm tr}(\cdot)|\phi\rangle\langle\phi|\tag2 $$ for some vectors $\psi,\psi^\perp,\phi$ of some finite size such that $\langle\psi|\psi^\perp\rangle=0$. And the additional condition is precisely why this can never be true, no matter how the vectors $\psi,\psi^\perp,\phi$ are chosen. We can verify (2) via the following steps:

  1. Any complementary channel of the identity is of the form ${\rm id}^c={\rm tr}(\cdot)|\psi\rangle\langle\psi|$ for some state vector $|\psi\rangle$ of some arbitrary (finite) dimension: Start from any isometry $V$ such that ${\rm id}={\rm tr}_E(V(\cdot)V^\dagger )$. Because $V'x=x\otimes|0\rangle$ also represents ${\rm id}$ in this way there has to exist a unitary $U$ on the environment such that $V=({\bf1}\otimes U)V'={\bf1}\otimes|\phi\rangle$. This yields $${\rm id}^c={\rm tr}_A(V(\cdot)V^\dagger )={\rm tr}_A((\cdot)\otimes|\phi\rangle\langle\phi|)={\rm tr}(\cdot)|\phi\rangle\langle\phi| $$ where the unit vector $\phi$ (and even its dimension) is arbitrary.
  2. Now for the complementary channel of $\mathcal D$: start again from some Stinespring isometry $V$ of $\mathcal D$. Because $V'=|0\rangle\langle 0|\otimes|0\rangle+|1\rangle\langle 1|\otimes|1\rangle$ is another Stinespring isometry of $\mathcal D$ there again exists $U$ such that $$ V=({\bf1}\otimes U)V'=|0\rangle\langle 0|\otimes|\psi\rangle+|1\rangle\langle 1|\otimes|\psi^\perp\rangle $$ where $\psi,\psi^\perp$ are now arbitrary unit vectors (of the same, arbitrary dimension) such that $\langle\psi|\psi^\perp\rangle=0$. Finally, $$ \mathcal D^c={\rm tr}_A(V(\cdot)V^\dagger )=\langle 0|X|0\rangle|\psi\rangle\langle\psi|+\langle1|X|1\rangle|\psi^\perp\rangle\langle\psi^\perp| $$
Frederik vom Ende
  • 4,163
  • 3
  • 12
  • 49