I've been reading the Time-Dependent Perturbation Theory in Griffiths' book. There are many places that puzzle me a lot.
The system is two-level($\psi_a,\psi_b,E_a<E_b$). Suppose the particle starts out in the lower state: $c_a(0)=1, \quad c_b(0)=0$. There is a time-dependent perturbation $\hat{H}^{\prime}$.
I can understand the zeroth $c_a^{(0)}(t)=1, \quad c_b^{(0)}(t)=0$, and the first order is $$\frac{d c_a^{(1)}}{d t}=0 \Rightarrow c_a^{(1)}(t)=1$$ $$\frac{d c_b^{(1)}}{d t}=-\frac{i}{\hbar} H_{b a}^{\prime} e^{i \omega_0 t} \Rightarrow c_b^{(1)}=-\frac{i}{\hbar} \int_0^t H_{b a}^{\prime}\left(t^{\prime}\right) e^{i \omega_0 t^{\prime}} d t^{\prime} \, .$$ I also understand that the error in the first-order approximation is evident in the fact that $$\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2 \neq 1 \, .$$
I don't understand when Griffiths mentions:
"the error in the first-order approximation is evident in the fact that $\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2 \neq 1$, However, $\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2$ is equal to 1 to first order in $\hat{H}^{\prime}$, which is all we can expect from a first-order approximation. And the same goes for the higher orders."
Why is $\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2$ equal to 1 again? What does the "first order in $\hat{H}^{\prime}$" specifically indicate?