How to recover the original density matrix from the output of amplitude damping channel?

Question

For amplitude damping, we have the below expression

$$\xi(\rho)=E_0\rho{E_0}^\dagger + E_1\rho{E_1}^\dagger.$$

How can I perform a matrix inverse operation on $\xi(\rho)$ at the receiver to get back the original density matrix $\rho$ back from above transformation?

This is my approach;

$${E_0}^\dagger\xi(\rho)E_0=\rho + {E_0}^\dagger{E_1}\rho{E_1}^\dagger{E_0}. $$ Let ${E_0}^\dagger{E_1}$ be $C$. So the final expression becomes,

$${E_0}^\dagger\xi(\rho)E_0=\rho + C\rho{C}\dagger. $$

I am unable to get rid of factor C. What should I do to get back $\rho$?

Answer 1

Inverting a quantum channel

Kraus representation does not make it convenient to find the inverse. However, quantum channels are linear maps, so we can represent them as matrices which can be inverted relatively easily. However, we should not in general expect that the resulting linear map is a quantum channel$^1$.

Now, let $K(\xi)$ denote the matrix representation of $\xi$, i.e. $K(\xi)$ is a matrix such that $$ K(\xi)\mathrm{vec}(\rho)=\mathrm{vec}(\xi(\rho))\tag1 $$ where $\mathrm{vec}(\rho)$ denotes the vectorization of $\rho$. If $\xi$ is invertible then $$ K(\xi^{-1})=K(\xi)^{-1}.\tag2 $$ By proposition $2.20$ on page $80$ in John Watrous' The Theory of Quantum Information, we have $$ K(\xi)=E_0\otimes\overline{E_0}+E_1\otimes\overline{E_1}.\tag3 $$

Inverse of the amplitude damping channel

In particular, if $\xi$ is the amplitude damping channel on a qubit then $$ E_0=\begin{bmatrix}1&0\\0&\sqrt{1-\gamma}\end{bmatrix}\quad E_1=\begin{bmatrix}0&\sqrt\gamma\\0&0\end{bmatrix}\tag4 $$ for some $\gamma\in[0,1]$ and $$ K(\xi)=\begin{bmatrix} 1&0&0&\gamma\\ 0&\sqrt{1-\gamma}&0&0\\ 0&0&\sqrt{1-\gamma}&0\\ 0&0&0&1-\gamma \end{bmatrix}.\tag5 $$ By $(2)$, we have $$ K(\xi^{-1})=\begin{bmatrix} 1&0&0&-\frac{\gamma}{1-\gamma}\\ 0&\frac{1}{\sqrt{1-\gamma}}&0&0\\ 0&0&\frac{1}{\sqrt{1-\gamma}}&0\\ 0&0&0&\frac{1}{1-\gamma} \end{bmatrix}\tag6 $$ as long as $\gamma\ne 1$. From this we recover a Kraus representation $\xi^{-1}(\rho)=F_0\rho F_0^\dagger-F_1\rho F_1^\dagger$ where $$ F_0=\begin{bmatrix}1&0\\0&\frac{1}{\sqrt{1-\gamma}}\end{bmatrix}\quad F_1=\begin{bmatrix}0&\sqrt{\frac{\gamma}{1-\gamma}}\\0&0\end{bmatrix}.\tag7 $$ It is clear from $(6)$ that if $\xi^{-1}$ is a quantum channel then $\gamma=0$.

---

^{$^1$ A quantum channel contracts the set of density matrices $\mathcal{D}$. Therefore, its inverse, if it exists, expands $\mathcal{D}$. Thus, the only way both can be quantum channels is if they both preserve $\mathcal{D}$, but then they both send extreme points to extreme points, i.e. pure states to pure states, so their Choi rank is one which means they are both unitary.}