# Kraus operator of dephasing channel

I'm trying to deduce the Kraus representation of the dephasing channel using the Choi operator (I know the Kraus operators can be guessed in this case, I want to understand the general case).

The dephasing channel maps a density operator $$\rho$$ as $$\rho\rightarrow D(\rho)=(1-p)\rho+ p\textrm{diag}(\rho_{00},\rho_{11})$$

The Choi operator acts on a channel as

$$C(D)=(I \otimes D)\sum_{k,j=0}^1 \vert k\rangle \langle j \vert \otimes \vert k\rangle \langle j \vert=\sum_{k,j=0}^1\vert k\rangle \langle j \vert \otimes D(\vert k\rangle \langle j \vert)=\\=|0\rangle\langle 0|\otimes|0\rangle\langle 0|+p|0\rangle\langle 1|\otimes|0\rangle\langle 1|+p|1\rangle\langle 0|\otimes|1\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|=\\=|00\rangle\langle00|+p|01\rangle\langle01|+p|10\rangle\langle10|+|11\rangle\langle11|= \sum_{j=0}^3 |\psi_j\rangle\langle\psi_j|$$

Now, to find the Kraus operators, I should just find some $$K_j$$ such that $$|\psi_j\rangle =(I\otimes K_j) \sum_{k=0}^1 \vert k\rangle \otimes \vert k\rangle$$. These operators are simply

$$K_0=|0\rangle\langle 0|\quad K_1=\sqrt{p}|1\rangle\langle 0| \quad K_2=\sqrt{p}|0\rangle\langle 1|\quad K_3=|1\rangle\langle 1|$$

And I should have $$D(\rho)=\sum_{j=1}^3 K_j\rho K_j^\dagger$$

But $$\sum_{j=1}^3 K_j\rho K_j^\dagger=(\rho_{00}+p\rho_{11})|0\rangle\langle0| + (\rho_{11}+p\rho_{00})|1\rangle\langle1|$$

Which is most certainly not what I should get. I'm sure I'm either doing a massive calculation error, or I have massively misunderstand everything. Moreover, doing this I should only be able to find 4 Kraus operator, while I know that the representation is not unique and in particular this channel can be represented by only two Kraus operators. Any help is appreciated.

## 1 Answer

Acting with the dephasing channel on the possible states of a single qubit:

\begin{align}D\left(\left|0\rangle\langle0\right|\right) &= \left|0\rangle\langle0\right| \\ D\left(\left|0\rangle\langle1\right|\right) &= \left(1-p\right)\left|0\rangle\langle1\right|\\ D\left(\left|1\rangle\langle0\right|\right) &= \left(1-p\right)\left|1\rangle\langle0\right|\\ D\left(\left|1\rangle\langle1\right|\right) &= \left|1\rangle\langle1\right|.\end{align}

This gives that \begin{align}C\left(D\right) &= \sum_{k,j=0}^1\vert k\rangle \langle j \vert \otimes D(\vert k\rangle \langle j \vert) \\ &= |0\rangle\langle 0|\otimes|0\rangle\langle 0|+\left(1-p\right)|0\rangle\langle 1|\otimes|0\rangle\langle 1|+\left(1-p\right)|1\rangle\langle 0|\otimes|1\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|\\ &= |00\rangle\langle00|+|00\rangle\langle11|+|11\rangle\langle00|+|11\rangle\langle11|- p|00\rangle\langle 11|-p|11\rangle\langle 00|\\ &=\sum_{k, j=0}^1\left(1-p\right)\vert k\rangle \langle j \vert \otimes \vert k\rangle \langle j \vert + p\left(|0\rangle\langle 0|\otimes|0\rangle\langle 0|+|1\rangle\langle 1|\otimes|1\rangle\langle 1|\right) \\ &= \sum_{j=0}^N |\psi_j\rangle\langle\psi_j|.\end{align}

Now using $$|\psi_j\rangle =(I\otimes K_j) \sum_{k=0}^1 \vert k\rangle \otimes \vert k\rangle$$ to get $$\sum_{j=0}^N |\psi_j\rangle\langle\psi_j| = \sum_{j=0}^N\sum_{k,l=0}^1\vert k\rangle\langle l\vert \otimes K_j\vert k\rangle \langle l\vert K_j^{\dagger},$$ which equals $$C\left(D\right)$$ when the Kraus operators $$K_0 = \sqrt{1-p}I,\, K_1 = \sqrt p |0\rangle\langle 0|$$ and $$K_2 = \sqrt p |1\rangle\langle 1|$$.

Taking an arbitrary (single qubit) density matrix $$\rho = \rho_{00}|0\rangle\langle 0| + \rho_{01}|0\rangle\langle 1| + \rho_{10}|1\rangle\langle 0| + \rho_{11}|1\rangle\langle 1|$$ and acting on this using the above Kraus operators gives \begin{align}D(\rho)&=\sum_{j=1}^3 K_j\rho K_j^\dagger \\ &=\left(1-p\right)\rho + p\rho_{00}|0\rangle\langle 0| + p\rho_{11}|1\rangle\langle 1|,\end{align} as expected for the dephasing channel.