= Solution
Use units $\hbar=1$. A Hermitian quadrature coupling of a cavity mode to a continuum of external modes can be written
$$
H_{\mathrm{couple}}=i\int d\omega\,\kappa(\omega)(a+a^\dagger)\otimes(b_\omega^\dagger-b_\omega),
$$
with real coupling amplitude $\kappa$. The free <Quantum Hamiltonian> is $H_0=\omega_ca^\dagger a+\int\omega b_\omega^\dagger b_\omega\,d\omega$. In the <interaction picture>, the two photon-exchange products $ab_\omega^\dagger$ and $a^\dagger b_\omega$ acquire phases with frequency $\omega-\omega_c$, whereas $a^\dagger b_\omega^\dagger$ and $ab_\omega$ oscillate at $\omega+\omega_c$.
For weak coupling near resonance, the <rotating-wave approximation> drops those rapidly oscillating pair-creation and pair-annihilation terms and retains
$$
H_I(t)=i\int d\omega\,\kappa(\omega)\{a\otimes b_\omega^\dagger e^{i(\omega-\omega_c)t}-a^\dagger\otimes b_\omega e^{-i(\omega-\omega_c)t}\}.
$$
For a broadband flat coupling $\kappa=\sqrt{\gamma/(2\pi)}$, define the envelope <annihilation operator> $b(t)=(2\pi)^{-1/2}\int b_\omega e^{-i(\omega-\omega_c)t}d\omega$. The <Markov approximation> extends the detuning integral over the full real line and gives $[b(t),b^\dagger(s)]=\delta(t-s)$. Then the <cavity-reservoir interaction> is
$$
\boxed{H_I(t)=i\sqrt\gamma\{a(t)\otimes b^\dagger(t)-a^\dagger(t)\otimes b(t)\}.}
$$
The phases and normalization of the external field fix the sign convention. The <creation operators> and <annihilation operators> exchange one excitation between cavity and field, rather than creating or destroying a pair.
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