The dephasing channel is often represented as

$$\mathcal{E}\left(\rho\right)=p_0\rho+(1-p_0)\sigma_z\rho\sigma_z$$ that acts on a single qubit.

Another way of writing it down is by using Kraus-Operators $E_0=\sqrt{1-p}\mathbb{1},E_1=\sqrt{p}\sigma_z$ and $$ \mathcal{E}\left(\rho\right)=E_0\rho E_0^{\dagger}+E_1\rho E_1^{\dagger}.$$

How does one write the dephasing channel with a unitary operation in the following way?:

$$ \mathcal{E}\left(\rho\right)=U \rho U^{\dagger}$$ with $U=e^{- i\omega \sigma_z}$?


2 Answers 2


The dephasing channel cannot be represented in the form $\mathcal{E}(\rho) = U\rho U^\dagger$. One way to see this is that the transformation $\rho \to U\rho U^\dagger$ preserves the eigenvalues of $\rho$, whereas $\mathcal{E}(\rho)$ does not.

Actually, the Kraus operators describe the evolution of open quantum system and include dissipation, whereas unitary transformations $\rho \to U\rho U^\dagger$ describe isolated quantum systems. So, in general, the channel defined with Kraus operators cannot be represented in unitary form.


You might be looking for the so-called Stinespring dilation. While not every quantum channel can be written as a unitary, it can always be written as an isometry (think of it as adding an additional system and applying a unitary transformation to the whole), plus a partial trace. This corresponds to some information leaking to the 'environment' that we cannot retrieve anymore.

This post here has some examples, including dephasing: https://quantumcomputing.stackexchange.com/questions/24511/what-are-examples-of-the-correspondence-between-channels-and-their-stinespring-d .


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