Collapse operators of two-level atom Currently I am learning about the Lindbladian. I want to derive the optical bloch equations for a two-level atom interacting with monochromatic light from the Lindbladian. However I am having troubles with that.
In general the Lindbladian is
$$
\dot{\rho}=-\frac{i}{\hbar}[H,\rho]+\sum^{N^2-1}_{i=1}\gamma_i\bigg(L_i\rho L_i^\dagger-\frac{1}{2}\Big\{L_i^\dagger L_i,\rho\Big\}\bigg).
$$
But here I am only interested in the Lindbladian (second term). In the optical bloch equations this term is represented by all the terms multiplied with a $\gamma$ ($\gamma_1=\gamma_2=\gamma$). As $N$ is the number of dimensions ($N=2$) there should be three different collapse operators in the second term. But I only came up with 2. Namely
$$
L_1=\sigma^-=\begin{pmatrix} 0&1\\0&0 \end{pmatrix}, \qquad L_2=\sigma^+=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
$$
The operators that correspond to switching states. Note that I can write them in this way because $|1\rangle=\begin{pmatrix}1\\0\end{pmatrix}, |2\rangle=\begin{pmatrix}0\\1\end{pmatrix}$ and with that $\rho=\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}$. But I have no idea what the third collapse operator could be. I also calculated the sum only with the two operators and assumed that the third collapse operator would be equal to 0. But I did not come to the right result. The right result should be
$$
\sum^{N^2-1}_{i=1}\gamma_i\bigg(L_i\rho L_i^\dagger-\frac{1}{2}\Big\{L_i^\dagger L_i,\rho\Big\}\bigg)=\gamma\begin{pmatrix}-\rho_{11}&\frac{1}{2}\rho_{21}\\\frac{1}{2}\rho_{12}&\rho_{11}\end{pmatrix}
$$
but I came up with
$$
\sum^{N^2-1}_{i=1}\gamma_i\bigg(L_i\rho L_i^\dagger-\frac{1}{2}\Big\{L_i^\dagger L_i,\rho\Big\}\bigg)=\gamma\begin{pmatrix}\rho_{22}&0\\0&\rho_{11}\end{pmatrix}
$$
does anyone know what went wrong here ? Are the operators wrong ? Or would the missing third operator correct my attempt ?
 A: Note that the presence of $N^2-1$ jump operators is a mathematical generality: the point is that one needs to write at most this number of distinct operators, since any other operator must be linearly dependent on some combination of $\{L_i\}$, i.e. $A = c_0 \mathbb{1} + \sum_{i=1}^{N^2-1} c_i L_i$. (The contribution from the identity operator, $c_0\mathbb{1}$, can be absorbed into the Hamiltonian.) This number of jump operators also represents the general case, since one can just set $\gamma_i = 0$ for any operators that don't contribute. In physical problems one may have far less than $N^2-1$ operators, in which case one usually does not bother writing them down.
For an optical-frequency transition in a radiation bath at room temperature (e.g. think of an atom floating in free space in a dark room), the probability of an atom being spontaneously excited by a stray photon is essentially zero. So the standard optical Bloch equations consider only energy loss by spontaneous emission, represented by your jump operator $L_1 = \sigma^-$. Therefore, you should set $\gamma_1 = \gamma$ and $\gamma_2 = \gamma_3 = 0$. It is irrelevant but you could choose, for example, $L_3 = \sigma_z$, which would represent additional dephasing noise.
Suppose that you have in addition a coherent driving field that is close to resonance with the atomic transition frequency. This is described by the Hamiltonian (working in a suitable interaction picture within the rotating wave approximation)
$$ H = \delta \sigma_z + \Omega\sigma_x,$$
where $\delta$ is the detuning of the drive from resonance, and $\Omega$ is the Rabi frequency (proportional to the driving electric field).
