How to show there is a channel $\tilde\Lambda$ such that $\tilde\Lambda\circ\Lambda=\Lambda'$ with $\Lambda,\Lambda'$ dephasing channels?

Question

I'm taking a quantum information course and one of my exercises says to find $p,p'$ for which there is a channel $\tilde\Lambda(\Lambda(\rho))=\Lambda'(\rho)$, where $\Lambda$ and $\Lambda'$ are dephasing channels with $\Lambda(\rho)=(1-p)\rho+p\sigma_z\rho\sigma_z, \Lambda'(\rho)=(1-p')\rho+p'\sigma_z\rho\sigma_z$. I'm rather confused how to do that, would appreciate any tips and hints.

Answer 1

TL;DR: Quantum channel $\tilde\Lambda$ satisfying

$$ \tilde\Lambda\circ\Lambda=\Lambda'\tag1 $$

exists if and only if $p'$ lies between $p$ and $1-p$.

In particular, if $p=\frac12$ then $\tilde\Lambda$ exists if and only if $p'=\frac12$. In this case, $\tilde\Lambda$ is not unique. In fact, any channel that preserves diagonal elements, such as any dephasing channel, satisfies the required equality. On the other hand, if $p\ne\frac12$ then $\tilde\Lambda$ is unique and necessarily a dephasing channel.

Composition law for dephasing channels

For any $p\in\mathbb{R}$, define $\Lambda_p:L(\mathbb{C}^2)\to L(\mathbb{C}^2)$ to be the linear map that sends any $X\in L(\mathbb{C}^2)$ to

$$ \Lambda_p(X)=(1-p)\,X+p\,\sigma_z X\sigma_z\tag2 $$

where $\sigma_z=\mathrm{diag}(1, -1)$. For any $p,q\in[0,1]$, we have

$$ \begin{align} \Lambda_p(\Lambda_q(X)) &=(1-p)\,\Lambda_q(X)+p\,\sigma_z\Lambda_p(X)\sigma_z \\ &= (1-p)(1-q)\,X+(1-p)q\,\sigma_z X\sigma_z +p(1-q)\,\sigma_z X\sigma_z+pq\,X\\ &= (1-p-q+2pq)X + (p+q-2pq)\sigma_z X\sigma_z \\ &= \Lambda_{p+q-2pq}(X) \end{align}\tag3 $$

which can also be written as

$$ \Lambda_p\circ\Lambda_q = \Lambda_{p+q-2pq}.\tag4 $$

Inverse map

From $(2)$ we see that $\Lambda_0$ is the identity channel, so if $p\ne\frac12$ then by $(4)$, we have

$$ \Lambda_p^{-1}=\Lambda_{p/(2p-1)}.\tag5 $$

On the other hand, $\Lambda_{1/2}$ maps

$$ \begin{bmatrix}a&b\\c&d\end{bmatrix}\mapsto\begin{bmatrix}a&0\\0&d\end{bmatrix}\tag6 $$

and therefore is non-injective and has no inverse.

Solving for $\tilde\Lambda$ when $p\ne\frac12$

Let $p\in[0,1]$ be such that $\Lambda=\Lambda_p$ and $p'\in[0,1]$ be such that $\Lambda'=\Lambda_{p'}$. If $p\ne\frac12$, then

$$ \begin{align} \tilde\Lambda\circ\Lambda_p&=\Lambda_{p'} \\ \tilde\Lambda&=\Lambda_{p'}\circ\Lambda_p^{-1} \\ \tilde\Lambda&=\Lambda_{p'}\circ\Lambda_{p/(2p-1)} \\ \tilde\Lambda&=\Lambda_{(p'-p)/(1-2p)} \end{align}\tag7 $$

is the unique solution to $(1)$. By Choi's theorem, $\Lambda_{(p'-p)/(1-2p)}$ is a quantum channel$^1$ if and only if

$$ \frac{p'-p}{1-2p}\in[0,1]\tag8 $$

which is the case when $p>\frac12$ and $p'\in[1-p,p]$ or when $p<\frac12$ and $p'\in[p,1-p]$.

Solving for $\tilde\Lambda$ when $p=\frac12$

If $p=\frac12$, then there are two possibilities. If $p'=\frac12$ then any $\tilde\Lambda$ which preserves diagonal elements, such as any dephasing channel, satisfies $(1)$. On the other hand, if $p'\ne\frac12$ then no linear map $\tilde\Lambda$ satisfies $(1)$, because the right hand side of $(1)$ is an injection while the left hand side is non-injective for all $\tilde\Lambda$.

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^{$^1$ Quantum channel is a linear map which is completely positive and trace-preserving.}