I am working on a project for my degree to model a CNOT gate and I'm trying to do this with the Lindblad equation.

$$ \dot{\rho} = -\dfrac{i}{\hbar} [H, \rho] + \sum\limits_i \gamma _i (L_i \rho L _i ^{\dagger} - \dfrac{1}{2} \{ L _i ^{\dagger} L_i , \rho \}) $$

In the case of modelling a single qubit that might undergo a decay, a quantum jump, finding the jump operator is quite simple. However if we have a combined system of two qubits in a basis $|00>, |01>, |10>, |11> $ then it seems to me that the decay is more complicated. For example if it's in state $|01>$ then it might decay to $|00>$, but if it's in state $|11>$ then maybe it will decay to $|01>$ or $|10>$? And what if it is in a superposition of these states? It seems much more complicated than the single-qubit case where a quantum jump always causes the state to go to $|0>$ and nothing else.

I have no idea if this line of thinking is correct, or even makes sense. But I am reaching a bit above my level here in this so forgive me. I suppose what I am asking is what the jump operator(s?) would be in this situation?


1 Answer 1


Speaking mathematically, the Lindblad operators could be anything, so asking "what are the Lindblad operators for a 2-qubit system" is sort-of like asking "what is the Hamiltonian for a 2-qubit system": there are many options, and it depends on what your actual setup is.

One can narrow down the possible options for the Lindblad operators with information about the hierarchy of energy- / time-scales in the system. A typical scenario is that the system-environment interaction energy is much smaller than all other relevant energies. In this case, the Lindblad operators are ladder operators with respect to the Hamiltonian, i.e., $$ [H, L_i] = \hbar\omega_i\, L_i $$ for some frequencies $\omega_i$. For one qubit with $H \propto \sigma_z$, this actually leaves three options: $L_i \propto \sigma_\pm$ (spontaneous emission or excitation) or $L_i \propto \sigma_z$ (pure dephasing). For two qubits in this scenario, you would need to diagonalize your Hamiltonian $H$ to find the eigenstates $E_k$, and the Lindblad operators would have the form $L_i \propto | E_k \rangle\langle E_{k'} |$.

Another typical scenario is one where the qubit-qubit interaction is even weaker than the qubit-environment interaction. In this case you get what is called a "local Lindblad" equation, where the Lindblad operators are ladder operators for one of the qubits. That is, we get Lindblad operators of the form $L_i \propto \sigma_{\pm,z}^{(n)}$ acting only on qubit $n=1$ or $n=2$. Yet another interesting scenario is one where the qubits do not interact directly, but couple collectively to the same environment. That gives Lindblad operators like $L = \sum_n \sigma_-^{(n)}$, leading to the effect of superradiance.

I am not an expert on quantum circuits, but it seems to me that people in this field usually use local decay $\sigma_-^{(n)}$ or local dephasing $\sigma_z^{(n)}$ Lindblad operators, acting only one one qubit. For example, let's have a look at the qutip-qip software package which allows the simulation of noisy quantum circuits. When creating a "Processor" object, one can directly specify $T_1$ and $T_2$ times for each qubit, which correspond to local decay and dephasing channels. However, with a bit more effort, one can also specify an arbitrary Lindblad operator by adding "DecoherenceNoise".


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