I know that Lindblad operator for $1$-qubit dephasing quantum channel is $L = \sqrt{\gamma} \sigma_z$ so that the corresponding master equation is $\dot{\rho} = \sigma_z \rho \sigma_z - \rho$ (e.g., on page 11-12 in this document). I'm trying to extend this to $n$-qubits, where each qubit is affected by $\sigma_z$ noise (with the same strength for simplicity). This is different from collective dephasing where $L = \sqrt{\gamma} \sigma_z \otimes \cdots \otimes \sigma_z$. What would be the Lindblad operators be in this case? My intuitive, straightforward guess for $n=2$ qubits case would be $L_1 = \sqrt{\gamma} \sigma_z \otimes I$ and $L_2 = \sqrt{\gamma} I \otimes \sigma_z$ where $0 \leq \gamma \leq 1$. But in this case, the master equation becomes $\dot{\rho} = \gamma((\sigma_z \otimes I)\rho(\sigma_z \otimes I) + (I \otimes \sigma_z)\rho(I \otimes \sigma_z) - 2\rho)$, and I'm not sure if this a valid Lindbladian in terms of the scaling/normalization. Is this correct? Is there any restriction in terms of normalization for the Lindbladian? Thanks in advance!


1 Answer 1


You are right in your guess. The normalization of the corresponding Kraus operators is already guaranteed by the form in which the Lindblad master equation is written:

\begin{equation} \dot{\rho} = -i [H, \rho] + \sum_k L_k \rho L^{\dagger}_k - \frac{1}{2}\{L_kL^{\dagger}_k, \rho \} \end{equation}

If you pick any set of $L_k$, this is equivalent (for small times) to Kraus evolution $\rho \mapsto E_0 \rho E^{\dagger}_0 + \sum_k E_k \rho E_k^{\dagger}$, where the Kraus operators are

\begin{align} E_0 & = 1 - \delta t( \frac{1}{2}\sum_k L_k L_k^{\dagger} + i H) \\ E_k & = \sqrt{\delta t} L_k \end{align}

Since $E_0 E_0^{\dagger} + \sum_k E_k E_k^{\dagger} = 1 + O(t^2)$, this is a valid set of Kraus operators for small times. Note that the set of $L_k$ is completely arbitrary and the relevant restrictions are already contained in the form in which the master equation is written.

You are also right that your choice correctly describes independent dephasing in the n-qubits. Consider that in a small time step $\delta t$ the state has a probability $\gamma \delta t$ of being dephased in each of the qubits. Then the open evolution of the state is

\begin{equation} \mathcal{E}(\rho) = (1 - n \gamma \delta t) \rho + \sum_j \gamma \delta t \sigma_z^j \rho \sigma_z^j \end{equation}

where $\sigma_z^j = 1 \otimes \dots \otimes 1 \otimes \sigma_z \otimes 1 \otimes \dots \otimes 1$. Taking the derivative, the master equation is indeed

\begin{equation} \dot{\rho} = - n\gamma \rho + \gamma \sum_j (\sigma_z^j \rho \sigma_z^j) = \gamma \sum_j (\sigma_z^j \rho \sigma_z^j - \frac{1}{2}\{ \sigma_z^j \sigma_z^j, \rho\}) \end{equation}

which correspond to $L_j = \sqrt{\gamma} \sigma_z^j$, as you said.

Following @Norbert Schuch comment, it is possible to consider the finite time map that corresponds to our dynamic, given by $\mathcal{E}_t = \mathcal{E}^1_t \otimes \dots \otimes \mathcal{E}^n_t$, where $\mathcal{E}^j_t$ is the dephasing channel for the j-th qubit. We already know that the derivative of the dephasing channel for a single qubit, evaluated at $t = 0$, is the Lindbladian $\mathcal{L}^j$ with operator $\sqrt{\gamma} \sigma_z^j$. The Lindbladian for the whole $\mathcal{E}$ is then

\begin{equation} \frac{d}{dt} \mathcal{E}_t \rvert_{t = 0} = \sum_j \frac{d}{dt} \mathcal{E}^j_t \rvert_{t = 0} = \sum_j \mathcal{L}^j \end{equation}

and we obtain the same result again. This result generalizes: for independent evolutions of each of the qubits, the finite time map is the tensor product and the Lindbladian (which is a derivative) is the sum.

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    $\begingroup$ Isn't it both easier and more accurate (in fact, exact?) to take the derivative of the $n$-fold tensor product of the single-qubit dephasing channel for finite times? $\endgroup$ Feb 3 at 19:29
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    $\begingroup$ Yes, I think so. Please check if my addition to my answer is correct and thank you for the comment! $\endgroup$ Feb 3 at 19:41

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