Time-dependent perturbation theory derivation with 2-level system (Griffiths)

Question

I'm reviewing time-dependent perturbation theory (TDPT) via Griffiths QM book. I'm looking at section 9.12: he reviews the results for a two-level system where the system starts in one eigenstate, ie. $c_a(t=0) = 1$, $c_b(t=0)=0$ [9.15]. To zeroth order, then, $c_a^{(0)}(t) = 1$, $c_b^{(0)}(t)=0$. Plugging in the value for $c_b$ into the RHS of $\frac{d}{dt}c_a^{(1)} = -i/\hbar H'_{ab} e^{-i\omega_0 t} c_b^{(0)}$ (where $H'_{ab}$ is the matrix element connecting states a and b) one gets $\frac{d}{dt} c_a^{(1)} = 0$ [9.17], which Griffiths then says leads to $c_a^{(1)}(t)=1$. My question is about the very last part. Why do we get $c_a^{(1)}(t)=1$ instead of some other constant $C<=1$? I can't see a constraint in the problem which makes the result 1 rather than some arbitrary constant.

Answer 1

If possible, it's a good idea to type up the relevant text so that people without immediate access to the resource can answer (though an earlier version of Griffiths' book can be found in its entirety via a Google search).

In any case - he himself answers this question after the 2nd order calculation:

Notice that in my notation, $c_a^{(2)}(t)$ includes the zeroth order term $[\ldots]$

Because $c_a(0)=1$ and $c_b(0)=0$, the correction term of every order is zero when $t=0$. They will generically become non-zero as $t$ grows, but to first order, $\dot c_a^{(1)}=0$, which means that $c_a^{(1)}(t) = c_a^{(0)}(t) = 1 $.