 Solution

Differentiate ${\dot{c}}_2=-i\gamma e^{2i\Omega t}{c}_1$ and substitute the equation for ${\dot{c}}_1$. The result isPutting $c_2=e^{i\Omega t}u$ removes the first derivative and gives $\ddot{u}+{\kappa }^{2}u=0$, where $\kappa =\sqrt{{\Omega }^2+{\gamma }^2}$. For $\kappa >0$ the general solution isThe initial spinor fixes $c_1(0)=1,c_2(0)=0$, and the first-order coupled equation fixes the second initial datum ${\dot{c}}_2(0)=-i\gamma$. Thus $A=0$, $B=-i\gamma /\kappa$, andTheir squared magnitudes sum to one. If both $\Omega$ and $\gamma$ vanish, the initial amplitudes simply remain $(1,0)$; the formulas have that continuous limit.

Source convention correction. For positive $\gamma$, the printed expression for $c_2$ lacks the factor $-i$ required by the stated basis and initial vector. A concrete test is its derivative at $t=0$: the printed expression gives $+\gamma$, whereas the Schrödinger equation gives $-i\gamma$. Multiplying only $c_2$ by a phase changes the relative quantum phase. One may change the overall phase of the entire spinor, but must then transform $c_1$ and its initial amplitude as well. The printed modulus nevertheless gives the correct transition probability.