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

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.

• It looks as though the two maps are very similar. Have you considered describing them both in terms such as $\Lambda_p(\rho) := (1-p)\rho + p[ \sigma _z \rho \sigma_z]$, and considered what the effect of this map is on the coefficients of the density matrix? Commented Dec 4, 2018 at 17:12
• $\tilde{\Lambda}$ follows similar pattern? Commented Dec 4, 2018 at 19:00

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 $$\tilde\Lambda$$.

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