Let $p$ be an arbitrary real-valued parameter such that $0<p<1$. We will explore some nice properties of the one-qubit operation, commonly referred to as the amplitude damping channel, defined by the two Krauss operators
\[
A_0=\begin{pmatrix}1&0\\0&\sqrt{1-p} \end{pmatrix} \ \ \ \text{and} \ \ \ A_1=\begin{pmatrix} 0 &\sqrt{p} \\ 0 & 0 \end{pmatrix} .
\]
It is easy to verify that $A^\dagger_0A_0+A^\dagger_1A_1 = I$, so the operation $D_\rho$ that maps each $2 \times 2$ density matrix $\rho$ to
\[
D_p(\rho)=A_0\rho A^\dagger_0 +A_1\rho A^\dagger_1
\]
is indeed a valid quantum operation.
Lets find a state that is a fixed point of $D_p$, which is a one-qubit density matrix $\rho_0$ satisfying $D_p(\rho_0)=\rho_0$.
Let $\rho_0=\begin{pmatrix} a&b\\c&d\end{pmatrix}$ such that $Tr(\rho_0)=a+b=1$ so that $\rho_0$ be such a density matrix. Then
\[ \begin{align*}
D_p(\rho_0)&= A_0\rho_0A^\dagger_0+A_1\rho_0A^\dagger_1 \\
&=\begin{pmatrix} 1&0\\0&\sqrt{1-p} \end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix}\begin{pmatrix} 1&0\\0&\sqrt{1-p}\end{pmatrix}+\begin{pmatrix} 0 &\sqrt{p} \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix}\begin{pmatrix} 0 &0 \\ \sqrt{p} & 0 \end{pmatrix} \\
&=\begin{pmatrix} a & b\sqrt{1-p} \\ c\sqrt{1-p} & d(1-p) \end{pmatrix} + \begin{pmatrix} dp &0 \\0 & 0 \end{pmatrix} \\
&= \begin{pmatrix}a+dp & b\sqrt{1-p} \\ c\sqrt{1-p}& d(1-p) \end{pmatrix} \\
&= \begin{pmatrix} a&b\\c&d\end{pmatrix}.
\end{align*}\]
Equating the coefficients of the matrices in the last two lines gives
\[
\left\{\begin{array}{rl}
a=&a+dp, \\
b=& b\sqrt{1-p}, \\
c=& c\sqrt{1-p}, \\
d=&d(1-p) \\
\end{array}
\right.
\]
Since $0<p<1$, $1-p\neq 0,1$, and thus $b=c=d=0$. Furthermore, since $\rho_0$ is a density matrix, $Tr(\rho_0)=a+d=a=1$. Thus,
\[
\rho_0=\begin{pmatrix} 1&0\\0&0\end{pmatrix}=\ket{0}\bra{0}
\]
is the fixed point satisfying $D_p(\rho_0)=\rho_0$.
Now, lets show that the operation $D^{(2)}_p$, corresponding to applying $D_p$ twice in succession, that is $D^{(2)}_p(\rho)=D_p(D_p(\rho))$, is equivalent to applying $D_q$ once for some suitably chosen value $q$.
Before proceeding consider the following identities involving the Krauss operators $A_0$ and $A_1$:
\[ \begin{align*}
A_0=A_0^\dagger, \ \ \ \ \
& A_0^2=A_0^{\dagger2}=\begin{pmatrix}1&0\\0&1-p \end{pmatrix}, \ \ \ \ \ A_1^2=A_1^{\dagger2}=\begin{pmatrix}0&0\\0&0
\end{pmatrix} \\
& A_0A_1=A_1, \ \ \ \ \ A_1A_0=\begin{pmatrix}0&\sqrt{p^2-p}\\0&0\end{pmatrix}
\end{align*}\]
Then,
\[ \begin{align*}
D_p(D_p(\rho))&=A_0(A_0\rho A^\dagger_0 +A_1\rho A^\dagger_1) A^\dagger_0 +A_1(A_0\rho A^\dagger_0 +A_1\rho A^\dagger_1) A^\dagger_1 \\
&=A_0A_0\rho A^\dagger_0A^\dagger_0 +A_0A_1\rho A^\dagger_1A^\dagger_0 +A_1A_0\rho A^\dagger_0A^\dagger_1 +A_1A_1\rho A^\dagger_1A^\dagger_1 \\
&=A_0^2\rho A_0^{\dagger2}+(A_0A_1)\rho (A_0A_1)^\dagger +(A_1A_0)\rho (A_1A_0)^\dagger \\
&=A_0^2\rho A_0^{\dagger2}+A_1\rho A_1^\dagger +(A_1A_0)\rho (A_1A_0)^\dagger
\end{align*}\]
Now, observing that
\[\begin{align*}
A_0^2\rho A_0^{\dagger2}&=\begin{pmatrix}1&0\\0&1-p \end{pmatrix}\rho\begin{pmatrix}1&0\\0&1-p \end{pmatrix} \\
A_1\rho A_1A_1^\dagger +(A_1A_0)\rho (A_1A_0)^\dagger&=\begin{pmatrix} 0 &\sqrt{2p-p^2} \\ 0 & 0 \end{pmatrix}\rho\begin{pmatrix} 0 &\sqrt{2p-p^2} \\ 0 & 0 \end{pmatrix},
\end{align*}\]
and by letting $q:=2p-p^2$ allows $D_p(D_p(\rho))$ to be expressed as
\[
D_p(D_p(\rho))=\begin{pmatrix}1&0\\0&\sqrt{1-q} \end{pmatrix}\rho\begin{pmatrix}1&0\\0&\sqrt{1-q} \end{pmatrix}+\begin{pmatrix} 0 &\sqrt{q} \\ 0 & 0 \end{pmatrix}\rho\begin{pmatrix} 0 &0 \\ \sqrt{q} & 0 \end{pmatrix}=D_q(\rho).
\]
Hence $ D_p(D_p(\rho))=D_q(\rho)$, with $q=\sqrt{2p-p^2}$.
Generalizing the case just analyzed, we can also define the operation $D^{(k)}_p$, corresponding to applying $D_p$ $k$ times in succession. What are the Krauss operators of $D^{(k)}_p$ with their matrix entries written as closed-form expressions in terms of $p$ and $k$?
Calculating the effect of applying $D_p(\rho)$ $k-$times in succession results in the expression
\[
D_p^{(k)}(\rho)=A_0^kA_0^{k\dagger}+\SUM{n=1}{k}(A_1A_0^{n-1})\rho\SUM{n=1}{k}(A_1A_0^{n-1})^\dagger.
\]
Let
\[
B_0^{(k)}:=A_0^kA_0^{k\dagger}=\begin{pmatrix}1&0\\0&(1-p)^{k/2} \end{pmatrix} \ \ \text{and} \ \ B_1^{(k)}=\SUM{n=1}{k}(A_1A_0^{n-1})=\begin{pmatrix}a&b\\c&d\end{pmatrix}
\]
be the two Krauss operators of $D_p^{(k)}$ (here, $B_1^{(k)}$ has been expressed as some matrix with unknown coefficients which are to be determined) so that
\[
D_p^{(k)}(\rho)= B_0^{(k)}\rho B_0^{(k)\dagger}+ B_1^{(k)} \rho B_1^{(k)\dagger}.
\]
Since $D_p$ itself is a valid quantum operation, compositions of $D_{p}$ will also always be another valid quantum operation. Thus, the two Krauss operators for $D_p^{(k)}$ must satisfy the relation $I=B_0^{(k)\dagger} B_0^{(k)}+ B_1^{(k)^\dagger} B_1^{(k)}$. Thus. $B_1^{(k)}$ can be determined more precisely by the identity $ B_1^{(k)\dagger}B_1^{(k)}=I-B_0^{(k)^\dagger}B_0^{(k)}$. Since
\[
B_0^{(k)^\dagger}B_0^{(k)}=\begin{pmatrix}1&0\\0&(1-p)^k\end{pmatrix} \ \ \text{and} \ \ B_1^{(k)\dagger}B_1^{(k)}= \begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}
\]
then using the previous relation implies
\[
\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}
=\begin{pmatrix}0&0\\0&1-(1-p)^{k} \end{pmatrix}.
\]
From this, it is seen that
\[ \begin{align*}
a^2+c^2=0 &\implies a=c=0 \\
b^2+d^2&=1-(1-p)^{k},
\end{align*}\]
but in the case when $k=1$, $B_1^{(1)}=A_1$ so that $b=\sqrt{p}$ and $b^2+d^2=1-(1-p)=p$ implying that $d=0$. Hence, in general
\[
B_1^{(k)}=\begin{pmatrix}0&\sqrt{1-(1-p)^k}\\0&0\end{pmatrix} \ \ \text{and} \ \ B_0^{(k)}=\begin{pmatrix}1&0\\0&(1-p)^{k/2} \end{pmatrix}
\]
are the Krauss operators of $D_p^{(k)}$.
Now, lets answer the following question: Is $lim_{k\to \infty}D^(k)_p(\rho)=\rho_0$ for any initial state $\rho$, where $\rho_0$ is the fixed point of $D_p$ that was calculated in part above?
For $0<p<1$, it is always the case that $lim_{k\to\infty}(1-p)^k=0$. Therefore, in the limit,
\[
lim_{k\to\infty}B_0^{(k)}=\begin{pmatrix}1&0\\0&0 \end{pmatrix} \ \ \text{and} \ \ lim_{k\to\infty}B_1^{(k)}=\begin{pmatrix}0&1\\0&0 \end{pmatrix}.
\]
Let $\rho$ be an arbitrary density matrix expressed in the general form
\[
\rho=\begin{pmatrix}a&b\\c&d \end{pmatrix},
\]
where the constraint $a+d=1$ is imposed so that $Tr(\rho)=1$ as required for density operators by definition.
Then it follows that
\[\begin{align*}
lim_{k\to \infty}D^(k)_p(\rho)&=\begin{pmatrix}1&0\\0&0 \end{pmatrix}\rho\begin{pmatrix}1&0\\0&0 \end{pmatrix}+\begin{pmatrix}0&1\\0&0 \end{pmatrix}\rho\begin{pmatrix}0&0\\1&0 \end{pmatrix} =\begin{pmatrix}a+d&0\\0&0 \end{pmatrix}=\begin{pmatrix}1&0\\0&0 \end{pmatrix}=\rho_0,
\end{align*}\]
which is precisely equal to the fixed point calculated in above.
\[
A_0=\begin{pmatrix}1&0\\0&\sqrt{1-p} \end{pmatrix} \ \ \ \text{and} \ \ \ A_1=\begin{pmatrix} 0 &\sqrt{p} \\ 0 & 0 \end{pmatrix} .
\]
It is easy to verify that $A^\dagger_0A_0+A^\dagger_1A_1 = I$, so the operation $D_\rho$ that maps each $2 \times 2$ density matrix $\rho$ to
\[
D_p(\rho)=A_0\rho A^\dagger_0 +A_1\rho A^\dagger_1
\]
is indeed a valid quantum operation.
Lets find a state that is a fixed point of $D_p$, which is a one-qubit density matrix $\rho_0$ satisfying $D_p(\rho_0)=\rho_0$.
Let $\rho_0=\begin{pmatrix} a&b\\c&d\end{pmatrix}$ such that $Tr(\rho_0)=a+b=1$ so that $\rho_0$ be such a density matrix. Then
\[ \begin{align*}
D_p(\rho_0)&= A_0\rho_0A^\dagger_0+A_1\rho_0A^\dagger_1 \\
&=\begin{pmatrix} 1&0\\0&\sqrt{1-p} \end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix}\begin{pmatrix} 1&0\\0&\sqrt{1-p}\end{pmatrix}+\begin{pmatrix} 0 &\sqrt{p} \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix}\begin{pmatrix} 0 &0 \\ \sqrt{p} & 0 \end{pmatrix} \\
&=\begin{pmatrix} a & b\sqrt{1-p} \\ c\sqrt{1-p} & d(1-p) \end{pmatrix} + \begin{pmatrix} dp &0 \\0 & 0 \end{pmatrix} \\
&= \begin{pmatrix}a+dp & b\sqrt{1-p} \\ c\sqrt{1-p}& d(1-p) \end{pmatrix} \\
&= \begin{pmatrix} a&b\\c&d\end{pmatrix}.
\end{align*}\]
Equating the coefficients of the matrices in the last two lines gives
\[
\left\{\begin{array}{rl}
a=&a+dp, \\
b=& b\sqrt{1-p}, \\
c=& c\sqrt{1-p}, \\
d=&d(1-p) \\
\end{array}
\right.
\]
Since $0<p<1$, $1-p\neq 0,1$, and thus $b=c=d=0$. Furthermore, since $\rho_0$ is a density matrix, $Tr(\rho_0)=a+d=a=1$. Thus,
\[
\rho_0=\begin{pmatrix} 1&0\\0&0\end{pmatrix}=\ket{0}\bra{0}
\]
is the fixed point satisfying $D_p(\rho_0)=\rho_0$.
Now, lets show that the operation $D^{(2)}_p$, corresponding to applying $D_p$ twice in succession, that is $D^{(2)}_p(\rho)=D_p(D_p(\rho))$, is equivalent to applying $D_q$ once for some suitably chosen value $q$.
Before proceeding consider the following identities involving the Krauss operators $A_0$ and $A_1$:
\[ \begin{align*}
A_0=A_0^\dagger, \ \ \ \ \
& A_0^2=A_0^{\dagger2}=\begin{pmatrix}1&0\\0&1-p \end{pmatrix}, \ \ \ \ \ A_1^2=A_1^{\dagger2}=\begin{pmatrix}0&0\\0&0
\end{pmatrix} \\
& A_0A_1=A_1, \ \ \ \ \ A_1A_0=\begin{pmatrix}0&\sqrt{p^2-p}\\0&0\end{pmatrix}
\end{align*}\]
Then,
\[ \begin{align*}
D_p(D_p(\rho))&=A_0(A_0\rho A^\dagger_0 +A_1\rho A^\dagger_1) A^\dagger_0 +A_1(A_0\rho A^\dagger_0 +A_1\rho A^\dagger_1) A^\dagger_1 \\
&=A_0A_0\rho A^\dagger_0A^\dagger_0 +A_0A_1\rho A^\dagger_1A^\dagger_0 +A_1A_0\rho A^\dagger_0A^\dagger_1 +A_1A_1\rho A^\dagger_1A^\dagger_1 \\
&=A_0^2\rho A_0^{\dagger2}+(A_0A_1)\rho (A_0A_1)^\dagger +(A_1A_0)\rho (A_1A_0)^\dagger \\
&=A_0^2\rho A_0^{\dagger2}+A_1\rho A_1^\dagger +(A_1A_0)\rho (A_1A_0)^\dagger
\end{align*}\]
Now, observing that
\[\begin{align*}
A_0^2\rho A_0^{\dagger2}&=\begin{pmatrix}1&0\\0&1-p \end{pmatrix}\rho\begin{pmatrix}1&0\\0&1-p \end{pmatrix} \\
A_1\rho A_1A_1^\dagger +(A_1A_0)\rho (A_1A_0)^\dagger&=\begin{pmatrix} 0 &\sqrt{2p-p^2} \\ 0 & 0 \end{pmatrix}\rho\begin{pmatrix} 0 &\sqrt{2p-p^2} \\ 0 & 0 \end{pmatrix},
\end{align*}\]
and by letting $q:=2p-p^2$ allows $D_p(D_p(\rho))$ to be expressed as
\[
D_p(D_p(\rho))=\begin{pmatrix}1&0\\0&\sqrt{1-q} \end{pmatrix}\rho\begin{pmatrix}1&0\\0&\sqrt{1-q} \end{pmatrix}+\begin{pmatrix} 0 &\sqrt{q} \\ 0 & 0 \end{pmatrix}\rho\begin{pmatrix} 0 &0 \\ \sqrt{q} & 0 \end{pmatrix}=D_q(\rho).
\]
Hence $ D_p(D_p(\rho))=D_q(\rho)$, with $q=\sqrt{2p-p^2}$.
Generalizing the case just analyzed, we can also define the operation $D^{(k)}_p$, corresponding to applying $D_p$ $k$ times in succession. What are the Krauss operators of $D^{(k)}_p$ with their matrix entries written as closed-form expressions in terms of $p$ and $k$?
Calculating the effect of applying $D_p(\rho)$ $k-$times in succession results in the expression
\[
D_p^{(k)}(\rho)=A_0^kA_0^{k\dagger}+\SUM{n=1}{k}(A_1A_0^{n-1})\rho\SUM{n=1}{k}(A_1A_0^{n-1})^\dagger.
\]
Let
\[
B_0^{(k)}:=A_0^kA_0^{k\dagger}=\begin{pmatrix}1&0\\0&(1-p)^{k/2} \end{pmatrix} \ \ \text{and} \ \ B_1^{(k)}=\SUM{n=1}{k}(A_1A_0^{n-1})=\begin{pmatrix}a&b\\c&d\end{pmatrix}
\]
be the two Krauss operators of $D_p^{(k)}$ (here, $B_1^{(k)}$ has been expressed as some matrix with unknown coefficients which are to be determined) so that
\[
D_p^{(k)}(\rho)= B_0^{(k)}\rho B_0^{(k)\dagger}+ B_1^{(k)} \rho B_1^{(k)\dagger}.
\]
Since $D_p$ itself is a valid quantum operation, compositions of $D_{p}$ will also always be another valid quantum operation. Thus, the two Krauss operators for $D_p^{(k)}$ must satisfy the relation $I=B_0^{(k)\dagger} B_0^{(k)}+ B_1^{(k)^\dagger} B_1^{(k)}$. Thus. $B_1^{(k)}$ can be determined more precisely by the identity $ B_1^{(k)\dagger}B_1^{(k)}=I-B_0^{(k)^\dagger}B_0^{(k)}$. Since
\[
B_0^{(k)^\dagger}B_0^{(k)}=\begin{pmatrix}1&0\\0&(1-p)^k\end{pmatrix} \ \ \text{and} \ \ B_1^{(k)\dagger}B_1^{(k)}= \begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}
\]
then using the previous relation implies
\[
\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}
=\begin{pmatrix}0&0\\0&1-(1-p)^{k} \end{pmatrix}.
\]
From this, it is seen that
\[ \begin{align*}
a^2+c^2=0 &\implies a=c=0 \\
b^2+d^2&=1-(1-p)^{k},
\end{align*}\]
but in the case when $k=1$, $B_1^{(1)}=A_1$ so that $b=\sqrt{p}$ and $b^2+d^2=1-(1-p)=p$ implying that $d=0$. Hence, in general
\[
B_1^{(k)}=\begin{pmatrix}0&\sqrt{1-(1-p)^k}\\0&0\end{pmatrix} \ \ \text{and} \ \ B_0^{(k)}=\begin{pmatrix}1&0\\0&(1-p)^{k/2} \end{pmatrix}
\]
are the Krauss operators of $D_p^{(k)}$.
Now, lets answer the following question: Is $lim_{k\to \infty}D^(k)_p(\rho)=\rho_0$ for any initial state $\rho$, where $\rho_0$ is the fixed point of $D_p$ that was calculated in part above?
For $0<p<1$, it is always the case that $lim_{k\to\infty}(1-p)^k=0$. Therefore, in the limit,
\[
lim_{k\to\infty}B_0^{(k)}=\begin{pmatrix}1&0\\0&0 \end{pmatrix} \ \ \text{and} \ \ lim_{k\to\infty}B_1^{(k)}=\begin{pmatrix}0&1\\0&0 \end{pmatrix}.
\]
Let $\rho$ be an arbitrary density matrix expressed in the general form
\[
\rho=\begin{pmatrix}a&b\\c&d \end{pmatrix},
\]
where the constraint $a+d=1$ is imposed so that $Tr(\rho)=1$ as required for density operators by definition.
Then it follows that
\[\begin{align*}
lim_{k\to \infty}D^(k)_p(\rho)&=\begin{pmatrix}1&0\\0&0 \end{pmatrix}\rho\begin{pmatrix}1&0\\0&0 \end{pmatrix}+\begin{pmatrix}0&1\\0&0 \end{pmatrix}\rho\begin{pmatrix}0&0\\1&0 \end{pmatrix} =\begin{pmatrix}a+d&0\\0&0 \end{pmatrix}=\begin{pmatrix}1&0\\0&0 \end{pmatrix}=\rho_0,
\end{align*}\]
which is precisely equal to the fixed point calculated in above.