In the field of quantum optics when solving master equations it is well know that the populations1 are constants to linear order in the driving field. I.e. weak driving fields will only affect the coherences off the system.
I am interested in how this statement can be made more precise and what the most general system is that it applies too. So far I could only find derivations for example systems (e.g. two level system) and imprecise statements of this notion.
To be a bit more formal: Let there be a general multi-level system with unspecified couplings between the levels and a loss channel from each level. The system is driven by a harmonic field of given frequency and intensity and couples to (some of the) levels via a dipole approximation Hamiltonian. For such a system, under what conditions do the populations1 only change in second order of the driving field and how can this be proven in general?
As an example a typical Master equation equation for a two level system is
$$\dot{\rho} = \frac{1}{i \hbar} \left[H,\rho\right] - \frac{\gamma}{2} \left(\sigma^+ \sigma^- \rho + \rho \sigma^+ \sigma^- - 2 \sigma^- \rho \sigma^+ \right)$$
where $\sigma^\pm$ are the raising and lowering operators for the two level system, $\rho$ is the system density matrix in the interaction picture and $H=\hbar \Omega (\sigma^+ + \sigma^-)$ is the laser driving Hamiltonian. $\Omega$ is proportional to the driving field and the dipole matrix element.
If one solves this system (e.g. most simply finds the steady state) one will see that the populations1 are not affected by the driving in linear order of $\Omega$. Note that the question is not about an example, but about formulating this notion in its most general form and how to prove it.
1 The diagonal elements of the density matrix.
Answer
I will consider a general multi-level system. Like in your example, let us write $\rho$ for the density matrix of the system in the interaction picture. That means that the populations of the system are* the diagonal entries of $\rho$, $$ p_n(t) = \langle n | \rho(t) | n \rangle . $$
Generically, the system Hamiltonian (in the interaction picture) can be split into two parts: $$ H = H^0 + K . $$ $H^0$ is the diagonal part. For example, for the two-level system it is the Lamb/Stark shift proportional to $\sigma^z$, which was neglected in your example. On the other hand, $K$ is the off-diagonal part, i.e. the driving (in your example, the $\sigma^+$ and $\sigma^-$ terms).
Remark: $H$ is often time-dependent in practice. Time-dependence would not change the following much, and I want to keep the notation simple.
The time evolution of the density matrix is then $$ \dot \rho(t) = -\frac{\mathrm i}{\hbar} [H^0 + K, \rho(t)] + \hat D \rho(t) , $$ where $\hat D$ is the dissipative part which is not terribly important for this question. Using this, we can calculate how the populations evolve in time: $$ \dot p_n(t) = \langle n | [K, \rho(t)] | n \rangle + \langle n | \hat D \rho(t) | n \rangle . \tag{1} $$ Note that the first term is zero, because $\langle n | [H^0, \rho(t)] | n \rangle = E^0_n\, p_n(t) - p_n(t)\, E^0_n = 0$, where $E^0_n$ are the eigenvalues of $H^0$.
Assume now that the density matrix is diagonal initially: $$ \rho(0) = \sum_n p_n(0)\, |n \rangle\!\langle n| .$$ In that case, $$ \dot p_n(0) = \langle n | \hat D \rho(0) | n \rangle $$ does not depend on $K$. The first term is zero, because $\rho(0) | n \rangle = p_n(0) | n \rangle$ and we can use the same trick that we used earlier for $\langle n | [H^0, \rho(t)] | n \rangle$.
Dissipation does its work immediately, but the driving will only change the populations after a short while, when $\rho(\Delta t)$ is not diagonal any more: $$ \rho(\Delta t) = \rho(0) + \left( -\frac{\mathrm i}{\hbar} [K, \rho(0)] + \hat D\rho(0) \right) \Delta t $$ so that $$ \dot p_n(\Delta t) = -\frac{\mathrm i}{\hbar} \langle n | [K, [K, \rho(0)]] | n \rangle \Delta t + \cdots . $$
You see that the effect is quite general: The driving field only changes the populations at second order. And its easy to understand, why: (1) tells us that the change of the populations due to $K$ is proportional to the coherences. If the system is initially diagonal, the driving has to create these coherences first.
If the system is not initially diagonal, the whole statement is not true at all. Consider your example without the dissipative part, that can be solved easily. For $\rho(0) = \scriptstyle\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$, you get $$ \rho(t) = \frac 1 2 \begin{pmatrix} 1 - \cos(2\Omega t) & -\mathrm i \sin(2\Omega t) \\ \mathrm i \sin(2\Omega t) & 1 + \cos(2\Omega t) \end{pmatrix} , $$ the populations change in second order.
After a $\pi/2$-pulse, the system is in the state $\scriptstyle \frac 1 2 \begin{pmatrix} 1 & -\mathrm i \\ \mathrm i & 1 \end{pmatrix}$. Let us for example see what happens if we start in that state, $\rho(0) = \scriptstyle \frac 1 2 \begin{pmatrix} 1 & -\mathrm i \\ \mathrm i & 1 \end{pmatrix}$: $$ \rho(t) = \frac 1 2 \begin{pmatrix} 1+\sin(2\Omega t) & -\mathrm i \cos(2\Omega t) \\ \mathrm i \cos(2\Omega t) & 1-\sin(2\Omega t) \end{pmatrix} . $$ The populations change in the first order.
*Side note: This is only true in the interaction picture. Without the interaction picture, the populations are $p_n(t) = \langle E_n(t) | \rho(t) | E_n(t) \rangle$, where $|E_n(t)\rangle$ are the - in general time-dependent - eigenvectors of the system Hamiltonian.)
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