Quantum Mechanics – Can Quantum Zeno Dynamics Be Written as a Lindblad Master Equation?

Question

The interaction of the system with the environment is considered weak and Markovian in the Lindblad master equation. Whereas the measurements involved in the Zeno effect seem quite drastic, I think. So if the environment is performing the Zeno measurements on the system, would it mean the interaction with the environment has to be strong? Can we write a master equation for the Zeno effect/dynamics in that case? I had the following equation in mind for the system density operator:
$\dot{\rho}=-i[H,\rho]+\gamma\left( \Pi \rho \Pi – \frac{1}{2} \left\{ \Pi,\rho\right\} \right)$, where $\Pi$ is a projector onto the Hilbert space which is favored by the Zeno effect. My thought is that $\gamma$ here must be big for this to be the Zeno effect. Also, even if this form of the equation gives a proper description of the Zeno effect, am I missing terms corresponding to $(1-\Pi)$ as another jump operator. I'm confused.

Answer 1

It's true that we make the 'weak coupling approximation' when deriving the master equation, but this approximation is more about taming the bath dynamics and doesn't care too much about what the system is doing. In order to trace out the bath and write down equations of motion for the system alone, we assume that any system information that has leaked into the bath is rapidly 'washed away' by the fast bath dynamics. This is how we justify having a memoryless i.e. Markovian bath, and lets us approximate $\rho_\text{tot}(t) = \rho_\text{sys}(t) \otimes \rho_\text{bath}$. As long as the bath dynamics are faster than the system-bath interaction, this approximation will still hold.

Also regarding the master equation, the third term $\{\Pi,\rho\}/2$ corresponds to the no-jump evolution. In general, the jump operators are $c$ and $I-dt c^\dagger c/2$. Applying these to a density matrix with the proper weights gives

$$\rho \rightarrow c\rho c^\dagger dt + (I-dt c^\dagger c/2)\rho (I-dt c^\dagger c/2)\\ = \rho + \left(c\rho c^\dagger - \frac{1}{2}(c^\dagger c \rho+\rho c^\dagger c)\right)dt$$

($\Pi^\dagger \Pi = \Pi$ for projection operators).