How to find the unitary operation of a depolarizing channel?

3

Suppose we have a depolarizing channel operation $$E(\rho)=\frac{p}{2}\textbf{1}+(1-p)\rho$$ acting on a Spin$\frac{1}{2}$ density matrix of the form $\rho=\frac{1}{2}(\textbf{1}+\textbf{s}\cdot\textbf{$\sigma$})$. I have found the Kraus operators to be: $$E_1=\sqrt{\left(1-\frac{3}{4}p\right)}\textbf{1}, E_2=\frac{\sqrt{p}}{2}\sigma_x,E_3=\frac{\sqrt{p}}{2}\sigma_y \text{ and } E_4=\frac{\sqrt{p}}{2}\sigma_z$$ I am now supposed to find the unitary matrix U such that the Operation can be expressed in a bigger system i.e. after adding a System S. As far as I understand it, the new operation can be written as: $$E(\rho)=\sum_kE_k\rho E_k^\dagger=\text{Tr}_S(U\rho\otimes\rho_EU^\dagger)$$ Supposing the new system S is prepared on a state $|e_0\rangle$, How do I find the correct unitary matrix?

I appreciate your cooperation.


Crossposted to physics: https://physics.stackexchange.com/questions/576952/how-to-find-the-unitary-operation-of-a-depolarizing-channel

Polarized photon

Posted 2020-09-01T16:18:02.667

Reputation: 33

1If you have access to a copy of Nielsen and Chuang this is explained in Box 8.1 at the end of section 8.2.3. – Condo – 2020-09-01T17:39:39.030

Answers

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From N&C: Assuming the environment is in some pure state we recall that Kraus representations comes from the unitary evolution $$\sum_{k}E_k\rho E_k^*=\sum_k \langle e_k |U\left(\rho\otimes|e_0\rangle \langle e_0|\right)U^*|e_k\rangle$$ for some unitary $U$. So we need a unitary such that $E_k=\langle e_k|U|e_0\rangle$. We can pick $$U=\begin{pmatrix} E_1 & \cdots & \cdots & \cdots \\ E_2 & \cdots & \cdots & \cdots \\ \vdots & \cdots & \cdots & \cdots \\ E_n & \cdots & \cdots & \cdots\end{pmatrix}$$ in the basis for $e_k$. To find your unitary you just need to find the remaining entries that make the resulting matrix unitary (this should be possible).

More uniquely, each Kraus representation has a Stinespring dilation, which is merely an isometry $V$ such that $$\sum_{k}E_k\rho E_k^*=tr_S(V\rho V^*)$$ and now $V$ can be written as the matrix $$V=\begin{pmatrix} E_1 \\ E_2 \\ \vdots \end{pmatrix}$$ in the basis for $e_k$, and you don't need to worry about the degrees of freedom on the environment. This freedom in the unitary determining the Kraus representations comes from the fact that there are many equivalent kraus representations of the same unitary representation (there is more about this in N&C).

Condo

Posted 2020-09-01T16:18:02.667

Reputation: 664