= Solution
{c}
Use the resonant <interaction picture> and define $g>0$ by the one-atom <Jaynes-Cummings model> in part (a). Equal coupling to both atoms gives the <Tavis-Cummings model>
$$
H_I=\hbar g\left[a(\sigma_+^{(1)}+\sigma_+^{(2)})
+a^\dagger(\sigma_-^{(1)}+\sigma_-^{(2)})\right].
$$
The initial state has two excitations and is symmetric under atom exchange. Excitation conservation and this symmetry confine the evolution to
$$
|A\rangle=|ee;0\rangle,\quad
|B\rangle=\frac{|eg\rangle+|ge\rangle}{\sqrt2}\otimes|1\rangle,\quad
|C\rangle=|gg;2\rangle.
$$
The antisymmetric atomic <spin singlet state> is uncoupled and unpopulated. Applying the <annihilation operator> and <creation operator> gives
$$
\frac{H_I}{\hbar}=g
\begin{pmatrix}0&\sqrt2&0\\\sqrt2&0&2\\0&2&0\end{pmatrix}.
$$
The first off-diagonal entry combines two indistinguishable atomic emission paths; the second also includes the $\sqrt2$ enhancement for creating the second <photon>. Its <eigenvalues> are $0,\pm\sqrt6g$. Put $\Omega=\sqrt6g$ and $K=H_I/\hbar$, so $K^3=\Omega^2K$. Its exponential is
$$
e^{-iKt}=I-\frac{i\sin\Omega t}{\Omega}K
+\frac{\cos\Omega t-1}{\Omega^2}K^2.
$$
Applying this to $|A\rangle$ yields
$$
\boxed{\begin{aligned}
|\psi(t)\rangle={}&\frac{2+\cos\Omega t}{3}|ee;0\rangle\\
&-\frac{i}{\sqrt3}\sin\Omega t\,
\frac{|eg\rangle+|ge\rangle}{\sqrt2}\otimes|1\rangle\\
&+\frac{\sqrt2}{3}(\cos\Omega t-1)|gg;2\rangle.
\end{aligned}}
$$
All three states have the same bare <energy> on <resonance>, so restoring the free Schrödinger-picture evolution adds only a common <global phase>. The expression starts at $|ee;0\rangle$, has initial derivative $-i\sqrt2g|B\rangle$, and its three squared <probability amplitudes> sum to one.
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