I am following the simple tutorial below:
In this they look at single qubit with Hamiltonian $H = \frac{\Delta}{2}\sigma_{x}$ and 1 jump operator $c = \sqrt{g}\sigma_{z}$.
The Lidbladian ME for such a system will be:
$$ \dot \rho = -i[\frac{\Delta}{2}\sigma_{x}, \rho] + g(\sigma_{z}\rho\sigma_{z} - \rho). $$
I try to find the equation of motion for $\sigma_{z}$ by using the following
$$ \frac{d}{dt}\langle \sigma_z \rangle = \mathrm{Tr}[\sigma_{z}\dot\rho] $$
and then substituting the above equation. This leads to the two coupled differential equations
$$ \frac{d}{dt}\langle \sigma_{z} \rangle = \Delta \langle \sigma_y \rangle $$ $$ \frac{d}{dt}\langle \sigma_{y} \rangle = - (\Delta \langle \sigma_z \rangle + 2g\langle \sigma_y \rangle). $$
I try to solve these in the standard way by converting it to a matrix-vector equation and finding the eigenvalues and eigenvectors etc. However, I do not get remotely close to the analytical solution which they write as
$$ \langle \sigma_{z} \rangle = cos(2\pi t)e^{-gt}. $$
Is it possible to get this same analytical solution this way? And if so how?
